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Sets, Relations and Functions — JEE Previous Year Questions

Every Sets, Relations and Functions question asked in JEE Main and JEE Advanced across the last 186 papers — 313 questions, each with its correct answer. Free to read, no account needed.

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313

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165/186

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89%

All 313 Sets, Relations and Functions questions

Most recent papers first.

Q1·MathematicsIntegerJEE Advanced 2026
Let S={1,2,3,…,10}S = \{1, 2, 3, \ldots, 10\}S={1,2,3,…,10}. Consider the set X={R:R is an equivalence relation on the set S such that R has exactly 42 elements}X = \{R : R \text{ is an equivalence relation on the set } S \text{ such that } R \text{ has exactly 42 elements}\}X={R:R is an equivalence relation on the set S such that R has exactly 42 elements}. Then the number of elements in XXX is ______.

Correct answer: 2520

Step-by-step solution →
Q2·MathematicsNumericalJEE Advanced 2026
Let N\mathbb{N}N denote the set of all positive integers. Consider the sets A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5} and B={1,2,3,4,5,6,7}B = \{1, 2, 3, 4, 5, 6, 7\}B={1,2,3,4,5,6,7}. Let SSS be the set of all functions f:A→Bf : A \to Bf:A→B such that f(2)≠2f(2) \neq 2f(2)=2 and f(4)≠4f(4) \neq 4f(4)=4. Consider the set T={f∈S:T = \{f \in S :T={f∈S: there exists a function g:B→Ng : B \to \mathbb{N}g:B→N such that g(f(x))=2xg(f(x)) = 2^{x}g(f(x))=2x for all x∈A}x \in A\}x∈A}. Then the number of elements in the set TTT is _____.

Correct answer: 1860

Step-by-step solution →
Q3·MathematicsNumericalJEE Main 2026
Let fff be a polynomial function such that log⁡2(f(x))=(log⁡2(2+23+29+…∞))⋅log⁡3(1+f(x)f(1/x))\log_2(f(x)) = \left(\log_2\left(2 + \dfrac{2}{3} + \dfrac{2}{9} + \ldots \infty\right)\right) \cdot \log_3\left(1 + \dfrac{f(x)}{f(1/x)}\right)log2​(f(x))=(log2​(2+32​+92​+…∞))⋅log3​(1+f(1/x)f(x)​), x>0x > 0x>0 and f(6)=37f(6) = 37f(6)=37. Then ∑n=110f(n)\displaystyle\sum_{n=1}^{10} f(n)n=1∑10​f(n) is equal to ________.

Correct answer: 395

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
Consider the relation R on the set {−2,−1,0,1,2}\{-2, -1, 0, 1, 2\}{−2,−1,0,1,2} defined by (a,b)∈R(a, b) \in R(a,b)∈R if and only if 1+ab>01 + ab > 01+ab>0. Then, among the statements: I. The number of elements in R is 17 II. R is an equivalence relation
  1. (A)Only I is true
  2. (B)Only II is true
  3. (C)Both I and II are true
  4. (D)Neither I nor II is true

Correct answer: (A)

Step-by-step solution →
Q5·MathematicsNumericalJEE Main 2026
Let R={(x,y)∈N×N:log⁡e(x+y)≤2}R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : \log_{e}(x + y) \le 2\}R={(x,y)∈N×N:loge​(x+y)≤2}. Then the minimum number of elements, required to be added in RRR to make it a transitive relation, is __________.

Correct answer: 15

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined as f(x)=2x2−3x+23x2+x+3f(x) = \frac{2x^{2} - 3x + 2}{3x^{2} + x + 3}f(x)=3x2+x+32x2−3x+2​. Then fff is :
  1. (A)both one-one and onto
  2. (B)one-one but not onto
  3. (C)onto but not one-one
  4. (D)neither one-one nor onto

Correct answer: (D)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Let [·] denote the greatest integer function. If the domain of the function f(x)=sin⁡−1(x+[x]3)f(x) = \sin^{-1}\left(\frac{x + [x]}{3}\right)f(x)=sin−1(3x+[x]​) is [α, β), then α2^22 + β2^22 is equal to:
  1. (A)2
  2. (B)5
  3. (C)10
  4. (D)13

Correct answer: (B)

Step-by-step solution →
Q8·MathematicsNumericalJEE Main 2026
Let A = {1, 4, 7} and B = {2, 3, 8}. Then the number of elements, in the relation R={((a1,b1),(a2,b2))∈((A×B)×(A×B)):a1+b2 divides a2+b1}R = \{((a_1, b_1), (a_2, b_2)) \in ((A \times B) \times (A \times B)) : a_1 + b_2 \text{ divides } a_2 + b_1\}R={((a1​,b1​),(a2​,b2​))∈((A×B)×(A×B)):a1​+b2​ divides a2​+b1​} is _______.

Correct answer: 18

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let for some α ∈ R\mathbb{R}R, f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function satisfying f(x+y)=f(x)+2y2+y+αxyf(x + y) = f(x) + 2y^{2} + y + \alpha xyf(x+y)=f(x)+2y2+y+αxy for all x,y∈Rx, y \in \mathbb{R}x,y∈R. If f(0)=−1f(0) = -1f(0)=−1 and f(1)=2f(1) = 2f(1)=2, then the value of ∑n=15(α+f(n))\sum_{n=1}^{5}(\alpha + f(n))∑n=15​(α+f(n)) is:
  1. (A)110
  2. (B)140
  3. (C)150
  4. (D)170

Correct answer: (B)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
For the function f:[1,∞)→[1,∞)f : [1, \infty) \to [1, \infty)f:[1,∞)→[1,∞) defined by f(x)=(x−1)4+1f(x) = (x-1)^{4} + 1f(x)=(x−1)4+1, among the two statements: (I) The set S={x∈[1,∞):f(x)=f−1(x)}S = \{x \in [1, \infty) : f(x) = f^{-1}(x)\}S={x∈[1,∞):f(x)=f−1(x)} contains exactly two elements, and (II) The set S={x∈[1,∞):f(x)=f−1(x+1)}S = \{x \in [1, \infty) : f(x) = f^{-1}(x+1)\}S={x∈[1,∞):f(x)=f−1(x+1)} is an empty set,
  1. (A)only (I) is TRUE
  2. (B)only (II) is TRUE
  3. (C)both (I) and (II) are TRUE
  4. (D)neither (I) nor (II) is TRUE

Correct answer: (A)

Step-by-step solution →
Q11·MathematicsNumericalJEE Main 2026
If the domain of the function f(x)=log⁡(0.6)(∣2x−5x2−4∣)f(x) = \sqrt{\log_{(0.6)}\left(\left|\frac{2x - 5}{x^{2} - 4}\right|\right)}f(x)=log(0.6)​(​x2−42x−5​​)​ is (−∞,a]∪{b}∪[c,d)∪(e,∞)(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)(−∞,a]∪{b}∪[c,d)∪(e,∞), then the value of a+b+c+d+ea + b + c + d + ea+b+c+d+e is ________.

Correct answer: 4

Step-by-step solution →
Q12·MathematicsNumericalJEE Main 2026
Let A={2,3,4,5,6}A = \{2, 3, 4, 5, 6\}A={2,3,4,5,6}. Let R be a relation on the set A×AA \times AA×A given by (x,y)R(z,w)(x, y)R(z, w)(x,y)R(z,w) if and only if xxx divides zzz and y≤wy \le wy≤w. Then the number of elements in R is ________.

Correct answer: 120

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
If g(x)=3x2+2x−3g(x) = 3x^2 + 2x - 3g(x)=3x2+2x−3, f(0)=−3f(0) = -3f(0)=−3 and 4g(f(x))=3x2−32x+724g(f(x)) = 3x^2 - 32x + 724g(f(x))=3x2−32x+72, then f(g(2))f(g(2))f(g(2)) is equal to:
  1. (A)256\dfrac{25}{6}625​
  2. (B)−256-\dfrac{25}{6}−625​
  3. (C)72\dfrac{7}{2}27​
  4. (D)−72-\dfrac{7}{2}−27​

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
The sum of all the elements in the range of f(x)=Sgn(sinx)+Sgn(cosx)+Sgn(tanx)+Sgn(cotx), x≠nπ2x \neq \frac{n\pi}{2}x=2nπ​, n ∈\in∈ Z\mathbf{Z}Z, where Sgn(t) = {1,ift>0−1ift<0\begin{cases} 1, & \text{if} \quad t > 0 \\ -1 & \text{if} \quad t < 0 \end{cases}{1,−1​ift>0ift<0​, is
  1. (A)4
  2. (B)2
  3. (C)–2
  4. (D)0

Correct answer: (B)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
Given below are two statements : Statement I : The function f : R → R defined by f(x)=x1+∣x∣f(x) = \dfrac{x}{1+|x|}f(x)=1+∣x∣x​ is one-one. Statement II : The function f : R → R defined by f(x)=x2+4x−30x2−8x+18f(x) = \dfrac{x^{2}+4x-30}{x^{2}-8x+18}f(x)=x2−8x+18x2+4x−30​ is many-one. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are false.
  2. (B)Both Statement I and Statement II are true.
  3. (C)Statement I is false but Statement II is true .
  4. (D)Statement I is true but Statement II is false.

Correct answer: (B)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
Let R be a relation defined on the set {1,2,3,4}×{1,2,3,4}\{1,2,3,4\}\times\{1,2,3,4\}{1,2,3,4}×{1,2,3,4} by R={((a,b),(c,d)):2a+3b=3c+4d}R=\{((a,b),(c,d)):2a+3b=3c+4d\}R={((a,b),(c,d)):2a+3b=3c+4d}. Then the number of elements in R is
  1. (A)666
  2. (B)181818
  3. (C)121212
  4. (D)151515

Correct answer: (C)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
If the domain of the function f(x)=sin⁡−1(1x2−2x−2)f(x) = \sin^{-1}\left(\frac{1}{x^{2} - 2x - 2}\right)f(x)=sin−1(x2−2x−21​), is (−∞,α]∪[β,γ]∪[δ,∞)(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)(−∞,α]∪[β,γ]∪[δ,∞), then α+β+γ+δ\alpha + \beta + \gamma + \deltaα+β+γ+δ is equal to
  1. (A)2
  2. (B)4
  3. (C)3
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
If the domain of the function f(x)=log⁡(10x2−17x+7)(18x2−11x+1)f(x)=\log_{(10x^{2}-17x+7)}\left(18x^{2}-11x+1\right)f(x)=log(10x2−17x+7)​(18x2−11x+1) is (−∞, a)∪(b, c)∪(d, ∞)−{e}(-\infty,\ a)\cup(b,\ c)\cup(d,\ \infty)-\{e\}(−∞, a)∪(b, c)∪(d, ∞)−{e}, then 90(a+b+c+d+e)90(a+b+c+d+e)90(a+b+c+d+e) equals:
  1. (A)170170170
  2. (B)177177177
  3. (C)307307307
  4. (D)316316316

Correct answer: (D)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
Let f be a function such that 3f(x)+2f(m19x)=5x3f(x) + 2f\left( \dfrac{m}{19x} \right) = 5x3f(x)+2f(19xm​)=5x, x≠0x \ne 0x=0, where m=∑i=19(i)2m = \sum\limits_{i=1}^{9} (i)^{2}m=i=1∑9​(i)2. Then f(5)−f(2)f(5) - f(2)f(5)−f(2) is equal to
  1. (A)−9-9−9
  2. (B)36
  3. (C)18
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2026
Consider two sets A = {x∈z:∣(∣x−3∣−3)∣≤1}\{x \in z : |(|x - 3| - 3)| \leq 1\}{x∈z:∣(∣x−3∣−3)∣≤1} and B= {x∈R−{1,2}:(x−2)(x−4)x−1log⁡e(∣x−2∣)=0}\left\{ x \in \mathbb{R} - \{1,2\} : \dfrac{(x-2)(x-4)}{x-1}\log_e(|x-2|) = 0 \right\}{x∈R−{1,2}:x−1(x−2)(x−4)​loge​(∣x−2∣)=0}. Then the number of onto functions fff : A→B is equal to :
  1. (A)62
  2. (B)79
  3. (C)32
  4. (D)81

Correct answer: (A)

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2026
Let A={-2, -1, 0,1, 2, 3, 4}. Let R be a relation on A defined by xRy if and only if 2x+y≤22x + y \le 22x+y≤2. Let lll be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations respectively. Then lll + m + n is equal to :
  1. (A)32
  2. (B)34
  3. (C)33
  4. (D)35

Correct answer: (C)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
Let the relation R on the set M = {1, 2, 3,.......16} be given by R={(x,y):4y=5x−3, x,y∈M}R = \{(x,y) : 4y = 5x - 3,\ x, y \in M\}R={(x,y):4y=5x−3, x,y∈M}. Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to
  1. (A)1
  2. (B)2
  3. (C)4
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2026
Let the domain of the function f(x)=log⁡3log⁡5(7−log⁡2(x2−10x+85))+sin⁡−1(∣3x−717−x∣)f(x) = \log_3 \log_5 \left(7 - \log_2 (x^2 - 10x + 85)\right) + \sin^{-1}\left(\left|\frac{3x - 7}{17 - x}\right|\right)f(x)=log3​log5​(7−log2​(x2−10x+85))+sin−1(​17−x3x−7​​) be (α, β]. Then α + β is equal to :
  1. (A)10
  2. (B)12
  3. (C)9
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2026
Let f(x)=[x]2−[x+3]−3f(x) = [x]^2 - [x + 3] - 3f(x)=[x]2−[x+3]−3, x∈Rx \in \mathbb{R}x∈R where [•] is the greatest integer function. Then
  1. (A)f(x)>0f(x) > 0f(x)>0 only for x∈[4,∞)x \in [4, \infty)x∈[4,∞)
  2. (B)f(x)<0f(x) < 0f(x)<0 only for x∈[−1,3)x \in [-1, 3)x∈[−1,3)
  3. (C)∫02f(x)dx=−6\int_0^2 f(x)dx = -6∫02​f(x)dx=−6
  4. (D)f(x)=0f(x) = 0f(x)=0 for finitely many values of x.

Correct answer: (B)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2026
Let fff and g be functions satisfying f(x+y)=f(x) f(y)f(x+y) = f(x)\, f(y)f(x+y)=f(x)f(y), f(1)=7f(1) = 7f(1)=7 and g(x+y)=g(xy)g(x+y) = g(xy)g(x+y)=g(xy), g(1)=1g(1) = 1g(1)=1, for all x,y∈Nx, y \in \mathbb{N}x,y∈N. ∑x=1n(f(x)g(x))=19607\sum_{x=1}^{n}\left(\frac{f(x)}{g(x)}\right) = 19607∑x=1n​(g(x)f(x)​)=19607, then n is equal to :
  1. (A)7
  2. (B)5
  3. (C)6
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2026
The number of elements in the relation R={(x,y):4x2+y2<52, x, y∈Z}R = \{(x,y): 4x^{2} + y^{2} < 52,\ x,\ y \in Z\}R={(x,y):4x2+y2<52, x, y∈Z} is
  1. (A)77
  2. (B)89
  3. (C)67
  4. (D)86

Correct answer: (A)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2026
The number of relations, defined on the set {a,b,c,d}\{a, b, c, d\}{a,b,c,d}, which are both reflexive and symmetric, is equal to:
  1. (A)256
  2. (B)16
  3. (C)1024
  4. (D)64

Correct answer: (D)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2026
Let A={2,3,5,7,9}A = \{2, 3, 5, 7, 9\}A={2,3,5,7,9}. Let R be the relation on A defined by xxx R yyy if and only if 2x≤3y2x \le 3y2x≤3y. Let ℓ\ellℓ be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then ℓ+m\ell + mℓ+m is equal to:
  1. (A)23
  2. (B)25
  3. (C)21
  4. (D)27

Correct answer: (B)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2026
Let A={x:∣x2−10∣≤6}A = \{x : |x^2 - 10| \le 6\}A={x:∣x2−10∣≤6} and B={x:∣x−2∣>1}B = \{x : |x - 2| > 1\}B={x:∣x−2∣>1}. Then
  1. (A)A∪B=(−∞,1]∪(2,∞)A \cup B = (-\infty, 1] \cup (2, \infty)A∪B=(−∞,1]∪(2,∞)
  2. (B)A−B=[2,3)A - B = [2, 3)A−B=[2,3)
  3. (C)A∩B=[−4,−2]∪[3,4]A \cap B = [-4, -2] \cup [3, 4]A∩B=[−4,−2]∪[3,4]
  4. (D)B−A=(−∞,−4)∪(−2,1)∪(4,∞)B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)B−A=(−∞,−4)∪(−2,1)∪(4,∞)

Correct answer: (D)

Step-by-step solution →
Q30·MathematicsMultiple correctJEE Advanced 2025
Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f:N→Zf: N \to Zf:N→Z and g:Z→Ng: Z \to Ng:Z→N defined by f(n)={(n+1)/2if n is odd,(4−n)/2if n is even,f(n) = \begin{cases} (n+1)/2 & \text{if } n \text{ is odd,} \\ (4-n)/2 & \text{if } n \text{ is even,} \end{cases}f(n)={(n+1)/2(4−n)/2​if n is odd,if n is even,​ and g(n)={3+2nif n≥0,−2nif n<0.g(n) = \begin{cases} 3 + 2n & \text{if } n \geq 0, \\ -2n & \text{if } n < 0. \end{cases}g(n)={3+2n−2n​if n≥0,if n<0.​ Define (g∘f)(b)=g(f(n))(g \circ f)(b) = g(f(n))(g∘f)(b)=g(f(n)) for all n∈Nn \in Nn∈N, and (f∘g(n))=f(g(n))(f \circ g(n)) = f(g(n))(f∘g(n))=f(g(n)) for all n∈Zn \in Zn∈Z. Then which of the following statements is (are) TRUE?
  1. (A)g∘fg \circ fg∘f is NOT one-one and g∘fg \circ fg∘f is NOT onto
  2. (B)f∘gf \circ gf∘g is NOT one-one but f∘gf \circ gf∘g is onto
  3. (C)g is one-one and g is onto
  4. (D)fff is NOT one-one but fff is onto

Correct answer: (A), (D)

Step-by-step solution →
Q31·MathematicsNumericalJEE Advanced 2025
Let the set of all relation R on the set {a,b,c,d,e,f}\{a, b, c, d, e, f\}{a,b,c,d,e,f}, such that R is reflexive and symmetric, and R contains exactly 10 elements be denoted by S. Then the number of elements is S is __________ .

Correct answer: 105

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
Let A={0,1,2,3,4,5}A=\{0,1,2,3,4,5\}A={0,1,2,3,4,5}. Let RRR be a relation on AAA defined by (x,y)∈R(x,y)\in R(x,y)∈R if and only if max⁡{x,y}∈{3,4}\max\{x,y\}\in\{3,4\}max{x,y}∈{3,4}. Then among the statements (S1)(S_1)(S1​): The number of elements in RRR is 18, and (S2)(S_2)(S2​): The relation RRR is symmetric but neither reflexive nor transitive:
  1. (A)both are true
  2. (B)both are false
  3. (C)only (S2)(S_2)(S2​) is true
  4. (D)only (S1)(S_1)(S1​) is true

Correct answer: (C)

Step-by-step solution →
Q33·MathematicsIntegerJEE Main 2025
Let the domain of the function f(x)=cos⁡−1(4x+53x−7)f(x)=\cos^{-1}\left(\dfrac{4x+5}{3x-7}\right)f(x)=cos−1(3x−74x+5​) be [α,β][\alpha,\beta][α,β] and the domain of g(x)=log⁡2(2−6log⁡27(2x+5))g(x)=\log_2\big(2-6\log_{27}(2x+5)\big)g(x)=log2​(2−6log27​(2x+5)) be (γ,δ)(\gamma,\delta)(γ,δ). Then ∣7(α+β)+4(γ+δ)∣\big|7(\alpha+\beta)+4(\gamma+\delta)\big|​7(α+β)+4(γ+δ)​ is equal to ___

Correct answer: 96

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
Let A={(α,β)∈R×R:∣α−1∣≤4 and ∣β−5∣≤6}A=\{(\alpha,\beta)\in\mathbb{R}\times\mathbb{R}:|\alpha-1|\le 4 \text{ and } |\beta-5|\le 6\}A={(α,β)∈R×R:∣α−1∣≤4 and ∣β−5∣≤6} and B={(α,β)∈R×R:16(α−2)2+9(β−6)2≤144}B=\{(\alpha,\beta)\in\mathbb{R}\times\mathbb{R}:16(\alpha-2)^2+9(\beta-6)^2\le 144\}B={(α,β)∈R×R:16(α−2)2+9(β−6)2≤144}. Then
  1. (A)B⊂AB\subset AB⊂A
  2. (B)A∪B={(x,y):−4≤x≤4, −1≤y≤11}A\cup B=\{(x,y):-4\le x\le 4,\ -1\le y\le 11\}A∪B={(x,y):−4≤x≤4, −1≤y≤11}
  3. (C)neither A⊂BA\subset BA⊂B nor B⊂AB\subset AB⊂A
  4. (D)A⊂BA\subset BA⊂B

Correct answer: (A)

Step-by-step solution →
Q35·MathematicsIntegerJEE Main 2025
The number of relations on the set A={1,2,3}A=\{1,2,3\}A={1,2,3} containing at most 6 elements including (1,2)(1,2)(1,2), which are reflexive and transitive but not symmetric, is ______.

Correct answer: 6

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
If the range of the function f(x)=5−xx2−3x+2f(x)=\dfrac{5-x}{x^2-3x+2}f(x)=x2−3x+25−x​, x≠1,2x\ne 1,2x=1,2, is (−∞,α]∪[β,∞)(-\infty,\alpha]\cup[\beta,\infty)(−∞,α]∪[β,∞), then α2+β2\alpha^2+\beta^2α2+β2 is equal to:
  1. (A)190
  2. (B)192
  3. (C)188
  4. (D)194

Correct answer: (D)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a continuous function satisfying f(0)=1f(0)=1f(0)=1 and f(2x)−f(x)=xf(2x)-f(x)=xf(2x)−f(x)=x for all x∈Rx\in\mathbb{R}x∈R. If lim⁡n→∞{f(x)−f(x2n)}=G(x)\displaystyle\lim_{n\to\infty}\left\{f(x)-f\left(\dfrac{x}{2^n}\right)\right\}=G(x)n→∞lim​{f(x)−f(2nx​)}=G(x), then ∑r=110G(r2)\displaystyle\sum_{r=1}^{10}G(r^2)r=1∑10​G(r2) is equal to:
  1. (A)540
  2. (B)385
  3. (C)420
  4. (D)215

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
Let the domains of the functions f(x)=log⁡4log⁡3log⁡7(8−log⁡2(x2+4x+5))f(x)=\log_4\log_3\log_7\left(8-\log_2(x^2+4x+5)\right)f(x)=log4​log3​log7​(8−log2​(x2+4x+5)) and g(x)=sin⁡−1(7x+10x−2)g(x)=\sin^{-1}\left(\dfrac{7x+10}{x-2}\right)g(x)=sin−1(x−27x+10​) be (α,β)(\alpha,\beta)(α,β) and [γ,δ][\gamma,\delta][γ,δ], respectively. Then α2+β2+γ2+δ2\alpha^2+\beta^2+\gamma^2+\delta^2α2+β2+γ2+δ2 is equal to:
  1. (A)15
  2. (B)13
  3. (C)16
  4. (D)14

Correct answer: (A)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
Let f,g:(1,∞)→Rf, g:(1,\infty)\to\mathbb{R}f,g:(1,∞)→R be defined as f(x)=2x+35x+2f(x)=\dfrac{2x+3}{5x+2}f(x)=5x+22x+3​ and g(x)=2−3x1−xg(x)=\dfrac{2-3x}{1-x}g(x)=1−x2−3x​. If the range of the function f∘g:[2,4]→Rf\circ g:[2,4]\to\mathbb{R}f∘g:[2,4]→R is [α,β][\alpha,\beta][α,β], then 1β−α\dfrac{1}{\beta-\alpha}β−α1​ is equal to
  1. (A)68
  2. (B)29
  3. (C)2
  4. (D)56

Correct answer: (D)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
Let A={−3,−2,−1,0,1,2,3}A=\{-3,-2,-1,0,1,2,3\}A={−3,−2,−1,0,1,2,3} and RRR be a relation on AAA defined by xRyxRyxRy if and only if 2x−y∈{0,1}2x-y\in\{0,1\}2x−y∈{0,1}. Let lll be the number of elements in RRR. Let mmm and nnn be the minimum number of elements required to be added in RRR to make it reflexive and symmetric relations, respectively. Then l+m+nl+m+nl+m+n is equal to:
  1. (A)18
  2. (B)17
  3. (C)15
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
Let A={−3,−2,−1,0,1,2,3}A=\{-3,-2,-1,0,1,2,3\}A={−3,−2,−1,0,1,2,3}. Let RRR be a relation on AAA defined by xRyxRyxRy if and only if 0≤x2+2y≤40\le x^2+2y\le 40≤x2+2y≤4. Let lll be the number of elements in RRR and mmm be the minimum number of elements required to be added in RRR to make it a reflexive relation. Then l+ml+ml+m is equal to:
  1. (A)19
  2. (B)20
  3. (C)17
  4. (D)18

Correct answer: (D)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
Let A={−2,−1,0,1,2,3}A=\{-2,-1,0,1,2,3\}A={−2,−1,0,1,2,3}. Let RRR be a relation on AAA defined by xRyxRyxRy if and only if y=max⁡{x,1}y=\max\{x,1\}y=max{x,1}. Let lll be the number of elements in RRR. Let mmm and nnn be the minimum number of elements required to be added in RRR to make it reflexive and symmetric relations, respectively. Then l+m+nl+m+nl+m+n is equal to:
  1. (A)12
  2. (B)11
  3. (C)13
  4. (D)14

Correct answer: (A)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2025
If the domain of the function f(x)=log⁡e ⁣(2x−35+4x)+sin⁡−1 ⁣(4+3x2−x)f(x)=\log_e\!\left(\dfrac{2x-3}{5+4x}\right)+\sin^{-1}\!\left(\dfrac{4+3x}{2-x}\right)f(x)=loge​(5+4x2x−3​)+sin−1(2−x4+3x​) is [α,β][\alpha,\beta][α,β], then α2+4β\alpha^2+4\betaα2+4β is equal to:
  1. (A)5
  2. (B)4
  3. (C)3
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
If the domain of the function f(x)=log⁡7 ⁣(1−log⁡4(x2−9x+18))f(x)=\log_7\!\big(1-\log_4(x^2-9x+18)\big)f(x)=log7​(1−log4​(x2−9x+18)) is (α,β)∪(γ,δ)(\alpha,\beta)\cup(\gamma,\delta)(α,β)∪(γ,δ), then α+β+γ+δ\alpha+\beta+\gamma+\deltaα+β+γ+δ is equal to:
  1. (A)18
  2. (B)16
  3. (C)15
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2025
Let fff be a function such that f(x)+3f ⁣(24x)=4xf(x)+3f\!\left(\dfrac{24}{x}\right)=4xf(x)+3f(x24​)=4x, x≠0x\ne 0x=0. Then f(3)+f(8)f(3)+f(8)f(3)+f(8) is equal to:
  1. (A)11
  2. (B)10
  3. (C)12
  4. (D)13

Correct answer: (A)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2025
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a function defined by f(x)=∣ ∣x+2∣−2∣x∣ ∣f(x)=\big|\,|x+2|-2|x|\,\big|f(x)=​∣x+2∣−2∣x∣​. If mmm is the number of points of local minima and nnn is the number of points of local maxima of fff, then m+nm+nm+n is equal to:
  1. (A)5
  2. (B)3
  3. (C)2
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2025
If the domain of the function f(x)=110+3x−x2+1x+∣x∣f(x)=\dfrac{1}{\sqrt{10+3x-x^2}}+\dfrac{1}{\sqrt{x+|x|}}f(x)=10+3x−x2​1​+x+∣x∣​1​ is (a,b)(a,b)(a,b), then (1+a)2+b2(1+a)^2+b^2(1+a)2+b2 is equal to:
  1. (A)26
  2. (B)29
  3. (C)25
  4. (D)30

Correct answer: (A)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2025
Let AAA be the set of all functions f:Z→Zf:Z\to Zf:Z→Z and RRR be a relation on AAA such that R={(f,g):f(0)=g(1) and f(1)=g(0)}R=\{(f,g):f(0)=g(1)\text{ and }f(1)=g(0)\}R={(f,g):f(0)=g(1) and f(1)=g(0)}. Then RRR is:
  1. (A)Symmetric and transitive but not reflexive
  2. (B)Symmetric but neither reflexive nor transitive
  3. (C)Reflexive but neither symmetric nor transitive
  4. (D)Transitive but neither reflexive nor symmetric

Correct answer: (B)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2025
Let f:R→Rf:R\to Rf:R→R be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\sin x\cos y)(f(2x+2y)-f(2x-2y))=(\cos x\sin y)(f(2x+2y)+f(2x-2y))(sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y)), for all x,y∈Rx,y\in Rx,y∈R. If f′(0)=12f'(0)=\dfrac{1}{2}f′(0)=21​, then the value of 24 f′′(5π3)24\,f''\left(\dfrac{5\pi}{3}\right)24f′′(35π​) is:
  1. (A)2
  2. (B)−3-3−3
  3. (C)3
  4. (D)−2-2−2

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2025
Let A={1,2,3,…,10}A=\{1,2,3,\dots,10\}A={1,2,3,…,10} and RRR be a relation on AAA such that R={(a,b):a=2b+1}R=\{(a,b):a=2b+1\}R={(a,b):a=2b+1}. Let (a1,a2),(a2,a3),(a3,a4),…,(ak,ak+1)(a_1,a_2),(a_2,a_3),(a_3,a_4),\dots,(a_k,a_{k+1})(a1​,a2​),(a2​,a3​),(a3​,a4​),…,(ak​,ak+1​) be a sequence of kkk elements of RRR such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer kkk, for which such a sequence exists, is equal to:
  1. (A)6
  2. (B)7
  3. (C)5
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2025
Let S=N∪{0}S=N\cup\{0\}S=N∪{0}. Define a relation R from S to R by: R={(x,y):log⁡ey=xlog⁡e(25), x∈S, y∈R}R=\left\{(x,y):\log_e y=x\log_e\left(\dfrac{2}{5}\right),\ x\in S,\ y\in R\right\}R={(x,y):loge​y=xloge​(52​), x∈S, y∈R}. Then, the sum of all the elements in the range of R is equal to
  1. (A)32\dfrac{3}{2}23​
  2. (B)53\dfrac{5}{3}35​
  3. (C)109\dfrac{10}{9}910​
  4. (D)52\dfrac{5}{2}25​

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2025
If the domain of the function log⁡5(18x−x2−77)\log_5(18x-x^2-77)log5​(18x−x2−77) is (α,β)(\alpha,\beta)(α,β) and the domain of the function log⁡(x−1)(2x2+3x−2x2−3x−4)\log_{(x-1)}\left(\dfrac{2x^2+3x-2}{x^2-3x-4}\right)log(x−1)​(x2−3x−42x2+3x−2​) is (γ,δ)(\gamma,\delta)(γ,δ), then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is equal to:
  1. (A)195
  2. (B)174
  3. (C)186
  4. (D)179

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2025
Define a relation R on the interval [0,π2)\left[0,\dfrac{\pi}{2}\right)[0,2π​) by x R y if and only if sec⁡2x−tan⁡2y=1\sec^2 x-\tan^2 y=1sec2x−tan2y=1. Then R is:
  1. (A)an equivalence relation
  2. (B)both reflexive and transitive but not symmetric
  3. (C)both reflexive and symmetric but not transitive
  4. (D)reflexive but neither symmetric nor transitive

Correct answer: (A)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2025
Let f:R−{0}→(−∞,1)f:R-\{0\}\to(-\infty,1)f:R−{0}→(−∞,1) be a polynomial of degree 2, satisfying f(x) f(1x)=f(x)+f(1x)f(x)\,f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)f(x)f(x1​)=f(x)+f(x1​). If f(K)=−2Kf(K)=-2Kf(K)=−2K, then the sum of squares of all possible values of K is:
  1. (A)1
  2. (B)6
  3. (C)7
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2025
Let f:[0,3]→Af:[0,3]\to Af:[0,3]→A be defined by f(x)=2x3−15x2+36x+7f(x)=2x^3-15x^2+36x+7f(x)=2x3−15x2+36x+7 and g:[0,∞)→Bg:[0,\infty)\to Bg:[0,∞)→B be defined by g(x)=x2025x2025+1g(x)=\frac{x^{2025}}{x^{2025}+1}g(x)=x2025+1x2025​. If both the functions are onto and S={x∈Z:x∈AS=\{x\in Z:x\in AS={x∈Z:x∈A or x∈B}x\in B\}x∈B}, then n(S)n(S)n(S) is equal to:
  1. (A)30
  2. (B)36
  3. (C)29
  4. (D)31

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2025
If f(x)=2x2x+2f(x)=\frac{2^x}{2^x+\sqrt{2}}f(x)=2x+2​2x​, x∈Rx\in Rx∈R, then ∑k=181f(k82)\sum_{k=1}^{81} f\left(\frac{k}{82}\right)∑k=181​f(82k​) is equal to:
  1. (A)41
  2. (B)812\frac{81}{2}281​
  3. (C)82
  4. (D)81281\sqrt{2}812​

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2025
The relation R={(x,y):x,y∈zR=\{(x, y): x, y\in zR={(x,y):x,y∈z and x+yx+yx+y is even}\}} is:
  1. (A)reflexive and transitive but not symmetric
  2. (B)reflexive and symmetric but not transitive
  3. (C)an equivalence relation
  4. (D)symmetric and transitive but not reflexive

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2025
Let f:R→Rf:R\to Rf:R→R be a function defined by f(x)=(2+3a)x2+a+2a−1x+bf(x)=(2+3a)x^2+\frac{a+2}{a-1}x+bf(x)=(2+3a)x2+a−1a+2​x+b, a≠1a\neq1a=1. If f(x+y)=f(x)+f(y)+1−27xyf(x+y)=f(x)+f(y)+1-\frac{2}{7}xyf(x+y)=f(x)+f(y)+1−72​xy, then the value of 28∑i=15∣f(i)∣28\sum_{i=1}^{5}|f(i)|28∑i=15​∣f(i)∣ is:
  1. (A)715
  2. (B)735
  3. (C)545
  4. (D)675

Correct answer: (D)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2025
Let f(x)=2x+2+1622x+1+2x+4+32f(x)=\dfrac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}f(x)=22x+1+2x+4+322x+2+16​. Then the value of 8(f(115)+f(215)+…+f(5915))8\left(f\left(\dfrac{1}{15}\right)+f\left(\dfrac{2}{15}\right)+\ldots+f\left(\dfrac{59}{15}\right)\right)8(f(151​)+f(152​)+…+f(1559​)) is equal to
  1. (A)118118118
  2. (B)929292
  3. (C)102102102
  4. (D)108108108

Correct answer: (A)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2025
The function f:(−∞,∞)→(−∞,1)f:(-\infty,\infty)\to(-\infty,1)f:(−∞,∞)→(−∞,1), defined by f(x)=2x−2−x2x+2−xf(x)=\dfrac{2^x-2^{-x}}{2^x+2^{-x}}f(x)=2x+2−x2x−2−x​ is :
  1. (A)One-one but not onto
  2. (B)Onto but not one-one
  3. (C)Both one-one and onto
  4. (D)Neither one-one nor onto

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2025
Let A={x∈(0,π)−{π2}:log⁡(2/π)∣sin⁡x∣+log⁡(2/π)∣cos⁡x∣=2}A=\left\{x\in(0,\pi)-\left\{\tfrac{\pi}{2}\right\}:\log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\right\}A={x∈(0,π)−{2π​}:log(2/π)​∣sinx∣+log(2/π)​∣cosx∣=2} and B={x≥0:x(x−4)−3∣x−2∣+6=0}B=\left\{x\ge 0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\right\}B={x≥0:x​(x​−4)−3∣x​−2∣+6=0}. Then n(A∪B)n(A\cup B)n(A∪B) is equal to:
  1. (A)4
  2. (B)2
  3. (C)8
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2025
Let X=R×RX=\mathbb{R}\times\mathbb{R}X=R×R. Define a relation RRR on XXX as: (a1,b1) R (a2,b2)⇔b1=b2(a_1,b_1)\,R\,(a_2,b_2)\Leftrightarrow b_1=b_2(a1​,b1​)R(a2​,b2​)⇔b1​=b2​. Statement-I: RRR is an equivalence relation. Statement-II: For some (a,b)∈X(a,b)\in X(a,b)∈X, the set S={(x,y)∈X:(x,y) R (a,b)}S=\{(x,y)\in X:(x,y)\,R\,(a,b)\}S={(x,y)∈X:(x,y)R(a,b)} represents a line parallel to y=xy=xy=x. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement-I and Statement-II are false.
  2. (B)Statement-I is true but Statement-II is false.
  3. (C)Both Statement-I and Statement-II are true.
  4. (D)Statement-I is false but Statement-II is true.

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2025
Let A={(x,y)∈R×R:∣x+y∣≥3}A=\{(x,y)\in\mathbb{R}\times\mathbb{R}:|x+y|\ge 3\}A={(x,y)∈R×R:∣x+y∣≥3} and B={(x,y)∈R×R:∣x∣+∣y∣≤3}B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:|x|+|y|\le 3\}B={(x,y)∈R×R:∣x∣+∣y∣≤3}. If C={(x,y)∈A∩B:x=0 or y=0}C=\{(x,y)\in A\cap B: x=0 \text{ or } y=0\}C={(x,y)∈A∩B:x=0 or y=0}, then ∑(x,y)∈C∣x+y∣\displaystyle\sum_{(x,y)\in C}|x+y|(x,y)∈C∑​∣x+y∣ is :
  1. (A)15
  2. (B)18
  3. (C)24
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2025
Let f(x)=log⁡exf(x)=\log_e xf(x)=loge​x and g(x)=x4−2x3+3x2−2x+22x2−2x+1g(x)=\dfrac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}g(x)=2x2−2x+1x4−2x3+3x2−2x+2​. Then the domain of f∘gf\circ gf∘g is
  1. (A)R\mathbb{R}R
  2. (B)(0,∞)(0,\infty)(0,∞)
  3. (C)[0,∞)[0,\infty)[0,∞)
  4. (D)[1,∞)[1,\infty)[1,∞)

Correct answer: (A)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2025
Let R={(1,2),(2,3),(3,3)}R=\{(1,2),(2,3),(3,3)\}R={(1,2),(2,3),(3,3)} be a relation defined on the set {1,2,3,4}\{1,2,3,4\}{1,2,3,4}. Then the minimum number of elements, needed to be added in RRR so that RRR becomes an equivalence relation, is
  1. (A)101010
  2. (B)888
  3. (C)999
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2025
The number of non-empty equivalence relations on the set {1,2,3}\{1,2,3\}{1,2,3} is:
  1. (A)6
  2. (B)7
  3. (C)5
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2025
Let A={1,2,…,10}A=\{1,2,\dots,10\}A={1,2,…,10} and B={mn:m,n∈A, m<n, gcd⁡(m,n)=1}B=\{\tfrac{m}{n}:m,n\in A,\ m<n,\ \gcd(m,n)=1\}B={nm​:m,n∈A, m<n, gcd(m,n)=1}. Then n(B)n(B)n(B) equals:
  1. (A)31
  2. (B)36
  3. (C)37
  4. (D)29

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsIntegerJEE Main 2025
Let A={1,2,3}A=\{1,2,3\}A={1,2,3}. The number of relations on AAA, containing (1,2)(1,2)(1,2) and (2,3)(2,3)(2,3), which are reflexive and transitive but not symmetric, is ______.

Correct answer: 3

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2025
Let A={1,2,3,4}A=\{1,2,3,4\}A={1,2,3,4} and B={1,4,9,16}B=\{1,4,9,16\}B={1,4,9,16}. Then the number of many-one functions f:A→Bf:A\to Bf:A→B such that 1∈f(A)1\in f(A)1∈f(A) is equal to:
  1. (A)127
  2. (B)151
  3. (C)163
  4. (D)139

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Advanced 2024
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)={x2sin⁡(πx2); if x≠00; if x=0f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{x^2}\right) & ; \text{ if } x \neq 0 \\ 0 & ; \text{ if } x = 0 \end{cases}f(x)={x2sin(x2π​)0​; if x=0; if x=0​ Then which of the following statements is TRUE?
  1. (A)f(x) = 0 has infinitely many solutions in the interval [11010,∞)\left[\frac{1}{10^{10}}, \infty\right)[10101​,∞)
  2. (B)f(x) = 0 has no solutions in the interval [1π,∞)\left[\frac{1}{\pi}, \infty\right)[π1​,∞)
  3. (C)The set of solutions of f(x) = 0 in the interval (0,11010)\left(0, \frac{1}{10^{10}}\right)(0,10101​) is finite
  4. (D)f(x) = 0 has more than 25 solutions in the interval (1π2,1π)\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)(π21​,π1​)

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsIntegerJEE Advanced 2024
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function such that f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y)f(x+y)=f(x)+f(y) for all x,y∈Rx, y \in \mathbb{R}x,y∈R, and g:R→(0,∞)g : \mathbb{R} \to (0, \infty)g:R→(0,∞) be a function such that g(x+y)=g(x)g(y)g(x + y) = g(x)g(y)g(x+y)=g(x)g(y) for all x,y∈Rx, y \in \mathbb{R}x,y∈R. If f(−35)=12f\left(\frac{-3}{5}\right) = 12f(5−3​)=12 and g(−13)=2g\left(\frac{-1}{3}\right) = 2g(3−1​)=2, then the value of (f(14)+g(−2)−8)g(0)\left(f\left(\frac{1}{4}\right) + g(-2) - 8\right) g(0)(f(41​)+g(−2)−8)g(0) is ______

Correct answer: 51

Step-by-step solution →
Q72·MathematicsIntegerJEE Advanced 2024
Let a=32a = 3\sqrt{2}a=32​ and b=151/66b = \frac{1}{5^{1/6}\sqrt{6}}b=51/66​1​. If x,y∈Rx, y \in Rx,y∈R are such that 3x+2y=log⁡a(18)5/43x + 2y = \log_{a}(18)^{5/4}3x+2y=loga​(18)5/4 and 2x−y=log⁡b(1080)2x - y = \log_{b}\left(\sqrt{1080}\right)2x−y=logb​(1080​), then 4x+5y4x + 5y4x+5y is equal to ______ .

Correct answer: 8

Step-by-step solution →
Q73·MathematicsMultiple correctJEE Advanced 2024
Let S={a+b2:a,b∈Z}S = \left\{a + b\sqrt{2} : a, b \in Z\right\}S={a+b2​:a,b∈Z}, T1={(−1+2)n:n∈Z}T_{1} = \left\{\left(-1 + \sqrt{2}\right)^{n} : n \in Z\right\}T1​={(−1+2​)n:n∈Z}, and T2={(1+2)n:n∈N}T_{2} = \left\{\left(1 + \sqrt{2}\right)^{n} : n \in N\right\}T2​={(1+2​)n:n∈N}. Then which of the following statements is(are) TRUE ?
  1. (A)Z∪T1∪T2⊂SZ \cup T_{1} \cup T_{2} \subset SZ∪T1​∪T2​⊂S
  2. (B)T1∩(0,12024)=ϕT_{1} \cap \left(0, \frac{1}{2024}\right) = \phiT1​∩(0,20241​)=ϕ, where ϕ\phiϕ denotes the empty set
  3. (C)T2∩(2024,∞)≠ϕT_{2} \cap (2024, \infty) \neq \phiT2​∩(2024,∞)=ϕ
  4. (D)For any given a,b∈Za, b \in Za,b∈Z, cos⁡(π(a+b2))+isin⁡(π(a+b2))∈Z\cos\left(\pi\left(a + b\sqrt{2}\right)\right) + i \sin\left(\pi\left(a + b\sqrt{2}\right)\right) \in Zcos(π(a+b2​))+isin(π(a+b2​))∈Z if and only if b=0b = 0b=0, where i=−1i = \sqrt{-1}i=−1​

Correct answer: (A), (C), (D)

Step-by-step solution →
Q74·MathematicsNumericalJEE Main 2024
Let A={2,3,6,7}A = \{2, 3, 6, 7\}A={2,3,6,7} and B={4,5,6,8}B = \{4, 5, 6, 8\}B={4,5,6,8}. Let RRR be a relation defined on A×BA \times BA×B by (a1,b1) R (a2,b2)(a_1, b_1)\, R\, (a_2, b_2)(a1​,b1​)R(a2​,b2​) is and only if a1+a2=b1+b2a_1 + a_2 = b_1 + b_2a1​+a2​=b1​+b2​. Then the number of elements in RRR is ________.

Correct answer: 25

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2024
Let f(x)=x2+9f(x) = x^2 + 9f(x)=x2+9, g(x)=xx−9g(x) = \frac{x}{x-9}g(x)=x−9x​ and a=f∘g(10)a = f\circ g(10)a=f∘g(10), b=g∘f(3)b = g\circ f(3)b=g∘f(3). If eee and lll denote the eccentricity and the length of the latus rectum of the ellipse x2a+y2b=1\frac{x^2}{a} + \frac{y^2}{b} = 1ax2​+by2​=1, then 8e2+l28e^2 + l^28e2+l2 is equal to.
  1. (A)16
  2. (B)8
  3. (C)6
  4. (D)12

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2024
If the domain of the function f(x)=sin⁡−1(x−12x+3)f(x) = \sin^{-1}\left(\frac{x - 1}{2x + 3}\right)f(x)=sin−1(2x+3x−1​) is R−(α,β)\mathbb{R} - (\alpha, \beta)R−(α,β), then 12αβ12\alpha\beta12αβ is equal to:
  1. (A)36
  2. (B)24
  3. (C)40
  4. (D)32

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2024
Let f(x)={−aif −a≤x≤0x+aif 0<x≤af(x)=\begin{cases}-a & \text{if } -a\le x\le 0\\ x+a & \text{if } 0<x\le a\end{cases}f(x)={−ax+a​if −a≤x≤0if 0<x≤a​ where a>0a>0a>0 and g(x)=f(∣x∣)−∣f(x)∣2g(x)=\dfrac{f(|x|)-|f(x)|}{2}g(x)=2f(∣x∣)−∣f(x)∣​. Then the function g:[−a,a]→[−a,a]g:[-a,a]\to[-a,a]g:[−a,a]→[−a,a] is
  1. (A)neither one-one nor onto.
  2. (B)both one-one and onto.
  3. (C)one-one.
  4. (D)onto

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2024
Let A={2,3,6,8,9,11}A=\{2,3,6,8,9,11\}A={2,3,6,8,9,11} and B={1,4,5,10,15}B=\{1,4,5,10,15\}B={1,4,5,10,15}. Let R be a relation on A×BA\times BA×B defined by (a,b) R (c,d)(a,b)\,R\,(c,d)(a,b)R(c,d) if and only if 3ad−7bc3ad-7bc3ad−7bc is an even integer. Then the relation R is
  1. (A)reflexive but not symmetric.
  2. (B)transitive but not symmetric.
  3. (C)reflexive and symmetric but not transitive.
  4. (D)an equivalence relation.

Correct answer: (C)

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2024
Let f(x)=17−sin⁡5xf(x)=\frac{1}{7-\sin 5x}f(x)=7−sin5x1​ be a function defined on RRR. Then the range of the function f(x)f(x)f(x) is equal to:
  1. (A)[18,15]\left[\frac{1}{8},\frac{1}{5}\right][81​,51​]
  2. (B)[17,16]\left[\frac{1}{7},\frac{1}{6}\right][71​,61​]
  3. (C)[17,15]\left[\frac{1}{7},\frac{1}{5}\right][71​,51​]
  4. (D)[18,16]\left[\frac{1}{8},\frac{1}{6}\right][81​,61​]

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2024
The function f(x)=x2+2x−15x2−4x+9f(x)=\dfrac{x^{2}+2x-15}{x^{2}-4x+9}f(x)=x2−4x+9x2+2x−15​, x∈Rx\in\mathbb{R}x∈R is
  1. (A)both one-one and onto
  2. (B)onto but not one-one
  3. (C)neither one-one nor onto
  4. (D)one-one but not onto

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
Let the relations R1R_{1}R1​ and R2R_{2}R2​ on the set X={1,2,3,…,20}X=\{1,2,3,\ldots,20\}X={1,2,3,…,20} be given by R1={(x,y):2x−3y=2}R_{1}=\{(x,y):2x-3y=2\}R1​={(x,y):2x−3y=2} and R2={(x,y):−5x+4y=0}R_{2}=\{(x,y):-5x+4y=0\}R2​={(x,y):−5x+4y=0}. If MMM and NNN are the minimum number of elements required to be added in R1R_{1}R1​ and R2R_{2}R2​, respectively, in order to make the relations symmetric, then M+NM+NM+N equals
  1. (A)888
  2. (B)161616
  3. (C)121212
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2024
Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5}. Let R be a relation on A defined by xRyxRyxRy if and only if 4x≤5y4x \le 5y4x≤5y. Let m be the number of elements in R and n be the minimum number of elements from A×AA \times AA×A that are required to be added to R to make it a symmetric relation. Then m+nm + nm+n is equal to:
  1. (A)24
  2. (B)23
  3. (C)25
  4. (D)26

Correct answer: (C)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2024
Let A={n∈[100,700]∩N:n is neither a multiple of 3 nor a multiple of 4}A=\{n\in[100,700]\cap\mathbb{N}:n\text{ is neither a multiple of 3 nor a multiple of 4}\}A={n∈[100,700]∩N:n is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in AAA is
  1. (A)300300300
  2. (B)280280280
  3. (C)310310310
  4. (D)290290290

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2024
Let f,g:R→Rf,g:\mathbb{R}\to\mathbb{R}f,g:R→R be defined as f(x)=∣x−1∣f(x)=|x-1|f(x)=∣x−1∣ and g(x)=exg(x)=e^xg(x)=ex for x≥0x\ge 0x≥0, g(x)=x+1g(x)=x+1g(x)=x+1 for x<0x<0x<0. Then the function f(g(x))f(g(x))f(g(x)) is:
  1. (A)neither one-one nor onto
  2. (B)one-one but not onto
  3. (C)both one-one and onto
  4. (D)onto but not one-one

Correct answer: (A)

Step-by-step solution →
Q85·MathematicsNumericalJEE Main 2024
If S={a∈R:∣2a−1∣=3[a]+2{a}}S = \{a \in \mathbb{R} : |2a - 1| = 3[a] + 2\{a\}\}S={a∈R:∣2a−1∣=3[a]+2{a}}, where [t][t][t] denotes the greatest integer less than or equal to ttt and {t}\{t\}{t} represents the fractional part of ttt, then 72∑a∈Sa72\displaystyle\sum_{a \in S} a72a∈S∑​a is equal to ___.

Correct answer: 18

Step-by-step solution →
Q86·MathematicsNumericalJEE Main 2024
Consider the function f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R defined by f(x)=2x1+9x2f(x)=\dfrac{2x}{\sqrt{1+9x^2}}f(x)=1+9x2​2x​. If the composition of fff, (f∘f∘f∘…∘f)⏟10 times(x)=210x1+9αx2\underbrace{(f\circ f\circ f\circ\ldots\circ f)}_{10\text{ times}}(x)=\dfrac{2^{10}x}{\sqrt{1+9\alpha x^2}}10 times(f∘f∘f∘…∘f)​​(x)=1+9αx2​210x​, then the value of 3α+1\sqrt{3\alpha+1}3α+1​ is equal to

Correct answer: 1024

Step-by-step solution →
Q87·MathematicsNumericalJEE Main 2024
In a survey of 220220220 students of a higher secondary school, it was found that at least 125125125 and at most 130130130 studied Mathematics; at least 858585 and at most 959595 studied Physics; at least 757575 and at most 909090 studied Chemistry; 303030 studied both Physics and Chemistry; 505050 studied both Chemistry and Mathematics; 404040 studied both Mathematics and Physics and 101010 studied none of these three subjects. Let mmm and nnn respectively be the least and the most number of students who studied all the three subjects. Then m+nm+nm+n is equal to ___

Correct answer: 45

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2024
Let a relation R on N×N\mathbb{N}\times\mathbb{N}N×N be defined as (x1,y1) R (x2,y2)(x_1,y_1)\,R\,(x_2,y_2)(x1​,y1​)R(x2​,y2​) if and only if x1≤x2x_1\le x_2x1​≤x2​ or y1≤y2y_1\le y_2y1​≤y2​. Consider the two statements: (I) R is reflexive but not symmetric. (II) R is transitive. Then which one of the following is true?
  1. (A)Only (II) is correct.
  2. (B)Only (I) is correct.
  3. (C)Both (I) and (II) are correct.
  4. (D)Neither (I) nor (II) is correct.

Correct answer: (B)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2024
If the domain of the function sin⁡−1(3x−222x−19)+log⁡e(3x2−8x+5x2−3x−10)\sin^{-1}\left(\dfrac{3x-22}{2x-19}\right)+\log_e\left(\dfrac{3x^2-8x+5}{x^2-3x-10}\right)sin−1(2x−193x−22​)+loge​(x2−3x−103x2−8x+5​) is (α,β](\alpha,\beta](α,β], then 3α+10β3\alpha+10\beta3α+10β is equal to:
  1. (A)979797
  2. (B)100100100
  3. (C)959595
  4. (D)989898

Correct answer: (A)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2024
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R and g:R→Rg:\mathbb{R}\to\mathbb{R}g:R→R be defined as f(x)={log⁡ex, x>0e−x, x≤0f(x)=\begin{cases}\log_e x & ,\ x>0\\ e^{-x} & ,\ x\le0\end{cases}f(x)={loge​xe−x​, x>0, x≤0​ and g(x)={x, x≥0ex, x<0g(x)=\begin{cases}x & ,\ x\ge0\\ e^x & ,\ x<0\end{cases}g(x)={xex​, x≥0, x<0​. Then g∘f:R→Rg\circ f:\mathbb{R}\to\mathbb{R}g∘f:R→R is:
  1. (A)one-one but not onto
  2. (B)neither one-one nor onto
  3. (C)onto but not one-one
  4. (D)both one-one and onto

Correct answer: (B)

Step-by-step solution →
Q91·MathematicsNumericalJEE Main 2024
Let A={1,2,3,…,20}A=\{1,2,3,\ldots,20\}A={1,2,3,…,20}. Let R1R_1R1​ and R2R_2R2​ be two relations on AAA such that R1={(a,b):bR_1=\{(a,b):bR1​={(a,b):b is divisible by a}a\}a}, R2={(a,b):aR_2=\{(a,b):aR2​={(a,b):a is an integral multiple of b}b\}b}. Then, number of elements in R1−R2R_1-R_2R1​−R2​ is equal to ___

Correct answer: 46

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2024
Consider the relations R1R_1R1​ and R2R_2R2​ defined as aR1b⇔a2+b2=1aR_1 b\Leftrightarrow a^2+b^2=1aR1​b⇔a2+b2=1 for all a,b∈Ra,b\in\mathbb Ra,b∈R and (a,b)R2(c,d)⇔a+d=b+c(a,b)R_2(c,d)\Leftrightarrow a+d=b+c(a,b)R2​(c,d)⇔a+d=b+c for all (a,b),(c,d)∈N×N(a,b),(c,d)\in\mathbb N\times\mathbb N(a,b),(c,d)∈N×N. Then
  1. (A)Only R1R_1R1​ is an equivalence relation
  2. (B)Only R2R_2R2​ is an equivalence relation
  3. (C)R1R_1R1​ and R2R_2R2​ both are equivalence relations
  4. (D)Neither R1R_1R1​ nor R2R_2R2​ is an equivalence relation

Correct answer: (B)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2024
If the domain of the function f(x)=x2−25(4−x2)+log⁡10(x2+2x−15)f(x)=\dfrac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15)f(x)=(4−x2)x2−25​​+log10​(x2+2x−15) is (−∞,α)∪[β,∞)(-\infty,\alpha)\cup[\beta,\infty)(−∞,α)∪[β,∞), then α2+β3\alpha^2+\beta^3α2+β3 is equal to:
  1. (A)140
  2. (B)175
  3. (C)150
  4. (D)125

Correct answer: (C)

Step-by-step solution →
Q94·MathematicsNumericalJEE Main 2024
Let A={1,2,3,4}A=\{1,2,3,4\}A={1,2,3,4} and R={(1,2),(2,3),(1,4)}R=\{(1,2),(2,3),(1,4)\}R={(1,2),(2,3),(1,4)} be a relation on A. Let S be the equivalence relation on A such that R⊂SR\subset SR⊂S and the number of elements in S is n. Then, the minimum value of n is ______.

Correct answer: 16

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2024
Let A={1,2,3,……100}A=\{1,2,3,\ldots\ldots100\}A={1,2,3,……100}. Let R be a relation on A defined by (x,y)∈R(x,y)\in R(x,y)∈R if and only if 2x=3y2x=3y2x=3y. Let R1R_1R1​ be a symmetric relation on A such that R⊂R1R\subset R_1R⊂R1​ and the number of elements in R1R_1R1​ is nnn. Then the minimum value of nnn is ______.

Correct answer: 66

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2024
If f(x)=4x+36x−4f(x)=\dfrac{4x+3}{6x-4}f(x)=6x−44x+3​, x≠23x\ne\dfrac{2}{3}x=32​ and (f∘f)(x)=g(x)(f\circ f)(x)=g(x)(f∘f)(x)=g(x), where g:R−{23}→R−{23}g:R-\left\{\dfrac{2}{3}\right\}\to R-\left\{\dfrac{2}{3}\right\}g:R−{32​}→R−{32​}, then (g∘g∘g)(4)(g\circ g\circ g)(4)(g∘g∘g)(4) is equal to
  1. (A)−1920-\dfrac{19}{20}−2019​
  2. (B)1920\dfrac{19}{20}2019​
  3. (C)−4-4−4
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2024
Let A={1,2,3,…,7}A=\{1,2,3,\ldots,7\}A={1,2,3,…,7} and let P(A)P(A)P(A) denote the power set of AAA. If the number of functions f:A→P(A)f:A\to P(A)f:A→P(A) such that a∈f(a)a\in f(a)a∈f(a), ∀a∈A\forall a\in A∀a∈A is mnm^nmn, m,n∈Nm,n\in\mathbb{N}m,n∈N and mmm is least, then m+nm+nm+n is equal to ___

Correct answer: 44

Step-by-step solution →
Q98·MathematicsNumericalJEE Main 2024
A group of 404040 students appeared in an examination of 333 subjects – Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 202020 students passed in Mathematics, 252525 students passed in Physics, 161616 students passed in Chemistry, at most 111111 students passed in both Mathematics and Physics, at most 151515 students passed in both Physics and Chemistry, at most 151515 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is ___

Correct answer: 10

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2024
If the domain of the function f(x)=log⁡e(2x+34x2+x−3)+cos⁡−1(2x−1x+2)f(x)=\log_e\left(\dfrac{2x+3}{4x^2+x-3}\right)+\cos^{-1}\left(\dfrac{2x-1}{x+2}\right)f(x)=loge​(4x2+x−32x+3​)+cos−1(x+22x−1​) is (α,β](\alpha,\beta](α,β], then the value of 5β−4α5\beta-4\alpha5β−4α is equal to:
  1. (A)101010
  2. (B)121212
  3. (C)111111
  4. (D)999

Correct answer: (B)

Step-by-step solution →
Q100·MathematicsNumericalJEE Main 2024
The number of symmetric relations defined on the set {1,2,3,4}\{1,2,3,4\}{1,2,3,4} which are not reflexive is ______.

Correct answer: 960

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2024
If the domain of the function f(x)=cos⁡−1(2−∣x∣4)+(log⁡e(3−x))−1f(x)=\cos^{-1}\left(\dfrac{2-|x|}{4}\right)+\left(\log_e(3-x)\right)^{-1}f(x)=cos−1(42−∣x∣​)+(loge​(3−x))−1 is [−α,β]−{γ}[-\alpha,\beta]-\{\gamma\}[−α,β]−{γ}, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to:
  1. (A)121212
  2. (B)999
  3. (C)111111
  4. (D)888

Correct answer: (C)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2024
If f(x)={2+2x,−1≤x<01−x3,0≤x≤3f(x)=\begin{cases}2+2x, & -1\le x<0\\ 1-\dfrac{x}{3}, & 0\le x\le 3\end{cases}f(x)={2+2x,1−3x​,​−1≤x<00≤x≤3​ ; g(x)={−x,−3≤x≤0x,0<x≤1g(x)=\begin{cases}-x, & -3\le x\le 0\\ x, & 0<x\le 1\end{cases}g(x)={−x,x,​−3≤x≤00<x≤1​, then range of (f∘g)(x)(f\circ g)(x)(f∘g)(x) is
  1. (A)(0,1](0,1](0,1]
  2. (B)[0,3)[0,3)[0,3)
  3. (C)[0,1][0,1][0,1]
  4. (D)[0,1)[0,1)[0,1)

Correct answer: (C)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2024
Let R be a relation on Z×ZZ\times ZZ×Z defined by (a,b)R(c,d)(a,b)R(c,d)(a,b)R(c,d) if and only if ad−bcad-bcad−bc is divisible by 5. Then R is
  1. (A)Reflexive and symmetric but not transitive
  2. (B)Reflexive but neither symmetric not transitive
  3. (C)Reflexive, symmetric and transitive
  4. (D)Reflexive and transitive but not symmetric

Correct answer: (A)

Step-by-step solution →
Q104·MathematicsNumericalJEE Main 2024
Let the set C={(x,y)∣x2−2y=2023, x,y∈N}C=\{(x,y)\mid x^2-2^y=2023,\ x,y\in\mathbb{N}\}C={(x,y)∣x2−2y=2023, x,y∈N}. Then ∑(x,y)∈C(x+y)\displaystyle\sum_{(x,y)\in C}(x+y)(x,y)∈C∑​(x+y) is equal to ___.

Correct answer: 46

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2024
If RRR is the smallest equivalence relation on the set {1,2,3,4}\{1,2,3,4\}{1,2,3,4} such that {(1,2),(1,3)}⊂R\{(1,2),(1,3)\}\subset R{(1,2),(1,3)}⊂R, then the number of elements in RRR is:
  1. (A)10
  2. (B)12
  3. (C)8
  4. (D)15

Correct answer: (A)

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2024
Let S={1,2,3,…,10}S=\{1,2,3,\ldots,10\}S={1,2,3,…,10}. Suppose MMM is the set of all the subsets of SSS, then the relation R={(A,B):A∩B≠φ; A,B∈M}R=\{(A,B):A\cap B\neq\varphi;\,A,B\in M\}R={(A,B):A∩B=φ;A,B∈M} is:
  1. (A)symmetric and reflexive only
  2. (B)reflexive only
  3. (C)symmetric and transitive only
  4. (D)symmetric only

Correct answer: (D)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2024
Let f:R−{−12}→Rf:\mathbb{R}-\left\{-\dfrac12\right\}\to\mathbb{R}f:R−{−21​}→R and g:R−{−52}→Rg:\mathbb{R}-\left\{-\dfrac52\right\}\to\mathbb{R}g:R−{−25​}→R be defined as f(x)=2x+32x+1f(x)=\dfrac{2x+3}{2x+1}f(x)=2x+12x+3​ and g(x)=∣x∣+12x+5g(x)=\dfrac{|x|+1}{2x+5}g(x)=2x+5∣x∣+1​. Then the domain of the function f∘gf\circ gf∘g is :
  1. (A)R−{−52}\mathbb{R}-\left\{-\dfrac52\right\}R−{−25​}
  2. (B)R\mathbb{R}R
  3. (C)R−{−74}\mathbb{R}-\left\{-\dfrac74\right\}R−{−47​}
  4. (D)R−{−52,−74}\mathbb{R}-\left\{-\dfrac52,-\dfrac74\right\}R−{−25​,−47​}

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2024
The function f:N−{1}→Nf:\mathbb N-\{1\}\to\mathbb Nf:N−{1}→N defined by f(n)=f(n)=f(n)= the highest prime factor of nnn, is:
  1. (A)both one-one and onto
  2. (B)one-one only
  3. (C)onto only
  4. (D)neither one-one nor onto

Correct answer: (D)

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2024
Let A and B be two finite sets with mmm and nnn elements respectively. The total number of subsets of the set A is 56 more than the total number of subsets of B. Then the distance of the point P(m,n)P(m,n)P(m,n) from the point Q(−2,−3)Q(-2,-3)Q(−2,−3) is
  1. (A)10
  2. (B)6
  3. (C)4
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q110·MathematicsMultiple correctJEE Advanced 2023
Let S=(0,1)∪(1,2)∪(3,4)S = (0, 1) \cup (1, 2) \cup (3, 4)S=(0,1)∪(1,2)∪(3,4) and T={0,1,2,3}T = \{0, 1, 2, 3\}T={0,1,2,3} . Then which of the following statements is(are) true?
  1. (A)There are infinitely many functions from SSS to TTT
  2. (B)There are infinitely many strictly increasing functions from SSS to TTT
  3. (C)The number of continuous functions from SSS to TTT is at most 120
  4. (D)Every continuous function from SSS to TTT is differentiable

Correct answer: (A), (C), (D)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2023
Negation of p∧(q∧∼(p∧q))p\wedge(q\wedge\sim(p\wedge q))p∧(q∧∼(p∧q)) is
  1. (A)∼(p∨q)\sim(p\vee q)∼(p∨q)
  2. (B)p∨qp\vee qp∨q
  3. (C)(∼(p∧q))∧q(\sim(p\wedge q))\wedge q(∼(p∧q))∧q
  4. (D)(∼(p∧q))∨p(\sim(p\wedge q))\vee p(∼(p∧q))∨p

Correct answer: (D)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2023
If the domain of the function f(x)=log⁡e(4x2+11x+6)+sin⁡−1(4x+3)+cos⁡−1(10x+63)f(x) = \log_e\left(4x^2 + 11x + 6\right) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right)f(x)=loge​(4x2+11x+6)+sin−1(4x+3)+cos−1(310x+6​) is (α,β](\alpha, \beta](α,β], then 36 ∣α+β∣36\,|\alpha + \beta|36∣α+β∣ is equal to:
  1. (A)636363
  2. (B)454545
  3. (C)727272
  4. (D)545454

Correct answer: (B)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2023
Let [x][x][x] denote the greatest integer function and f(x)=max⁡{1+x+[x], 2+x, x+2[x]}f(x)=\max\{1+x+[x],\,2+x,\,x+2[x]\}f(x)=max{1+x+[x],2+x,x+2[x]}, 0≤x≤20\le x\le 20≤x≤2. Let mmm be the number of points in [0,2][0,2][0,2], where fff is not continuous and nnn be the number of points in (0,2)(0,2)(0,2), where fff is not differentiable. Then (m+n)2+2(m+n)^2+2(m+n)2+2 is equal to:
  1. (A)11
  2. (B)2
  3. (C)6
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsNumericalJEE Main 2023
Let A={1,2,3,4}A = \{1, 2, 3, 4\}A={1,2,3,4} and RRR be a relation on the set A×AA \times AA×A defined by R={((a,b),(c,d)):2a+3b=4c+5d}R = \{((a, b), (c, d)) : 2a + 3b = 4c + 5d\}R={((a,b),(c,d)):2a+3b=4c+5d}. Then the number of elements in RRR is:

Correct answer: 6

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2023
The negation of the statement ((A∧(B∨C))⇒(A∨B))⇒A((A\wedge(B\vee C))\Rightarrow(A\vee B))\Rightarrow A((A∧(B∨C))⇒(A∨B))⇒A is:
  1. (A)equivalent to ∼A\sim A∼A
  2. (B)equivalent to ∼C\sim C∼C
  3. (C)equivalent to B∨∼CB\vee\sim CB∨∼C
  4. (D)a fallacy

Correct answer: (A)

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2023
For x∈Rx\in\mathbb{R}x∈R, two real valued functions f(x)f(x)f(x) and g(x)g(x)g(x) are such that g(x)=x+1g(x)=\sqrt x+1g(x)=x​+1 and f(g(x))=x+3−xf(g(x))=x+3-\sqrt xf(g(x))=x+3−x​. Then f(0)f(0)f(0) is equal to:
  1. (A)111
  2. (B)−3-3−3
  3. (C)555
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q117·MathematicsNumericalJEE Main 2023
Let A={−4,−3,−2,0,1,3,4}A = \{-4, -3, -2, 0, 1, 3, 4\}A={−4,−3,−2,0,1,3,4} and R={(a,b)∈A×A:b=∣a∣ or b2=a+1}R = \{(a, b) \in A \times A : b = |a| \text{ or } b^2 = a + 1\}R={(a,b)∈A×A:b=∣a∣ or b2=a+1} be a relation on A. Then the minimum number of elements, that must be added to the relation R so that it becomes reflexive and symmetric, is

Correct answer: 7

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2023
The statement (∼(p⇔∼q))∧q(\sim(p \Leftrightarrow \sim q)) \wedge q(∼(p⇔∼q))∧q is equivalent to
  1. (A)(∼p)∧(∼q)(\sim p) \wedge (\sim q)(∼p)∧(∼q)
  2. (B)p∧(∼q)p \wedge (\sim q)p∧(∼q)
  3. (C)(∼p)∨q(\sim p) \vee q(∼p)∨q
  4. (D)p∨qp \vee qp∨q

Correct answer: (A)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2023
Among the two statements (S1): (p⇒q)∧(q∧(∼q))(p\Rightarrow q)\wedge(q\wedge(\sim q))(p⇒q)∧(q∧(∼q)) is a contradiction and (S2): (p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q))(p\wedge q)\vee((\sim p)\wedge q)\vee(p\wedge(\sim q))\vee((\sim p)\wedge(\sim q))(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q)) is a tautology
  1. (A)only (S2) is true
  2. (B)only (S1) is true
  3. (C)both are false
  4. (D)both are true

Correct answer: (D)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2023
Let D be the domain of the function f(x)=sin⁡−1(log⁡3x(6+2log⁡3x−5x))f(x)=\sin^{-1}\left(\log_{3x}\left(\dfrac{6+2\log_3 x}{-5x}\right)\right)f(x)=sin−1(log3x​(−5x6+2log3​x​)). If the range of the function g:D→Rg:D\to\mathbb{R}g:D→R defined by g(x)=x−[x]g(x)=x-[x]g(x)=x−[x] ([x][x][x] is the greatest integer function) is (α,β)(\alpha,\beta)(α,β), then α2+5β\alpha^2+\dfrac{5}{\beta}α2+β5​ is equal to
  1. (A)46
  2. (B)135
  3. (C)136
  4. (D)45

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsNumericalJEE Main 2023
The number of relations, on the set {1,2,3}\{1,2,3\}{1,2,3} containing (1,2)(1,2)(1,2) and (2,3)(2,3)(2,3), which are reflexive and transitive but not symmetric, is _________.

Correct answer: 4

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2023
Let A={1,3,4,6,9}A=\{1,3,4,6,9\}A={1,3,4,6,9} and B={2,4,5,8,10}B=\{2,4,5,8,10\}B={2,4,5,8,10}. Let RRR be a relation defined on A×BA\times BA×B such that R={((a1,b1),(a2,b2)):a1≤b2 and b1≤a2}R=\{((a_1,b_1),(a_2,b_2)):a_1\le b_2 \text{ and } b_1\le a_2\}R={((a1​,b1​),(a2​,b2​)):a1​≤b2​ and b1​≤a2​}. Then the number of elements in the set RRR is
  1. (A)26
  2. (B)160
  3. (C)180
  4. (D)52

Correct answer: (B)

Step-by-step solution →
Q123·MathematicsNumericalJEE Main 2023
The number of points, where the curve f(x)=e8x−e6x−3e4x−e2x+1f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1f(x)=e8x−e6x−3e4x−e2x+1, x∈Rx\in\mathbb{R}x∈R cuts the x-axis, is equal to

Correct answer: 2

Step-by-step solution →
Q124·MathematicsNumericalJEE Main 2023
The number of ordered triplets of the truth values of p,qp, qp,q and rrr such that the truth value of the statement (p∨q)∧(p∨r)⇒(q∨r)(p \vee q) \wedge (p \vee r) \Rightarrow (q \vee r)(p∨q)∧(p∨r)⇒(q∨r) is True, is equal to _______ .

Correct answer: 7

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2023
The domain of the function f(x)=1[x]2−3[x]−10f(x)=\dfrac{1}{\sqrt{[x]^2-3[x]-10}}f(x)=[x]2−3[x]−10​1​ is (where [x][x][x] denotes the greatest integer less than or equal to xxx)
  1. (A)(−∞,−3)∪(5,∞)(-\infty,-3)\cup(5,\infty)(−∞,−3)∪(5,∞)
  2. (B)(−∞,−2)∪(6,∞)(-\infty,-2)\cup(6,\infty)(−∞,−2)∪(6,∞)
  3. (C)(−∞,−3]∪(6,∞)(-\infty,-3]\cup(6,\infty)(−∞,−3]∪(6,∞)
  4. (D)(−∞,−3]∪[6,∞)(-\infty,-3]\cup[6,\infty)(−∞,−3]∪[6,∞)

Correct answer: (C)

Step-by-step solution →
Q126·MathematicsNumericalJEE Main 2023
Let A={1,2,3,4,5}A=\{1,2,3,4,5\}A={1,2,3,4,5} and B={1,2,3,4,5,6}B=\{1,2,3,4,5,6\}B={1,2,3,4,5,6}. Then the number of functions f:A→Bf:A\to Bf:A→B satisfying f(1)+f(2)=f(4)−1f(1)+f(2)=f(4)-1f(1)+f(2)=f(4)−1 is equal to

Correct answer: 360

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2023
An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
  1. (A)10
  2. (B)9
  3. (C)21
  4. (D)15

Correct answer: (C)

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2023
The converse of the statement ((∼p)∧q)⇒r((\sim p)\wedge q)\Rightarrow r((∼p)∧q)⇒r is
  1. (A)(∼r)⇒p∧q(\sim r)\Rightarrow p\wedge q(∼r)⇒p∧q
  2. (B)(∼r)⇒((∼p)∧q)(\sim r)\Rightarrow((\sim p)\wedge q)(∼r)⇒((∼p)∧q)
  3. (C)r⇒((∼p)∧q)r\Rightarrow((\sim p)\wedge q)r⇒((∼p)∧q)
  4. (D)(p∨(∼q))⇒(∼r)(p\vee(\sim q))\Rightarrow(\sim r)(p∨(∼q))⇒(∼r)

Correct answer: (D)

Step-by-step solution →
Q129·MathematicsSingle correctJEE Main 2023
The negation of the statement (p∨q)∧(q∨(∼r))(p\lor q)\land(q\lor(\sim r))(p∨q)∧(q∨(∼r)) is
  1. (A)((∼p)∨r)∧(∼q)((\sim p)\lor r)\land(\sim q)((∼p)∨r)∧(∼q)
  2. (B)((∼p)∨(∼q))∧(∼r)((\sim p)\lor(\sim q))\land(\sim r)((∼p)∨(∼q))∧(∼r)
  3. (C)((∼p)∨(∼q))∨(∼r)((\sim p)\lor(\sim q))\lor(\sim r)((∼p)∨(∼q))∨(∼r)
  4. (D)(p∨r)∧(∼q)(p\lor r)\land(\sim q)(p∨r)∧(∼q)

Correct answer: (A)

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2023
The statement ∼[p∨(∼(p∧q))]\sim[p\vee(\sim(p\wedge q))]∼[p∨(∼(p∧q))] is equivalent to:
  1. (A)∼(p∧q)∧q\sim(p\wedge q)\wedge q∼(p∧q)∧q
  2. (B)∼(p∧q)\sim(p\wedge q)∼(p∧q)
  3. (C)∼p∨q\sim p\vee q∼p∨q
  4. (D)(p∧q)∧(∼p)(p\wedge q)\wedge(\sim p)(p∧q)∧(∼p)

Correct answer: (D)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2023
If f(x)=(tan⁡1°)x+log⁡e(123)xlog⁡e(1234)−(tan⁡1°), x>0f(x)=\dfrac{(\tan 1°)x+\log_e(123)}{x\log_e(1234)-(\tan 1°)},\ x>0f(x)=xloge​(1234)−(tan1°)(tan1°)x+loge​(123)​, x>0, then the least value of f(f(x))+f(f(4x))f(f(x))+f\left(f\left(\dfrac4x\right)\right)f(f(x))+f(f(x4​)) is
  1. (A)888
  2. (B)444
  3. (C)222
  4. (D)000

Correct answer: (B)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2023
Let A={2,3,4}A=\{2,3,4\}A={2,3,4} and B={8,9,12}B=\{8,9,12\}B={8,9,12}. Then the number of elements in the relation R={((a1,b1),(a2,b2))∈(A×B, A×B):a1 divides b2 and a2 divides b1}R=\{((a_1,b_1),(a_2,b_2))\in(A\times B,\ A\times B):a_1\text{ divides }b_2\text{ and }a_2\text{ divides }b_1\}R={((a1​,b1​),(a2​,b2​))∈(A×B, A×B):a1​ divides b2​ and a2​ divides b1​} is:
  1. (A)363636
  2. (B)121212
  3. (C)181818
  4. (D)242424

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2023
Negation of p⇒q⇒q⇒pp\Rightarrow q\Rightarrow q\Rightarrow pp⇒q⇒q⇒p is
  1. (A)(∼p)∨q(\sim p)\vee q(∼p)∨q
  2. (B)(∼q)∧p(\sim q)\wedge p(∼q)∧p
  3. (C)q∧(∼p)q\wedge(\sim p)q∧(∼p)
  4. (D)p∨(∼q)p\vee(\sim q)p∨(∼q)

Correct answer: (C)

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2023
The negation of p∧(∼q)∨(∼p)p\wedge(\sim q)\vee(\sim p)p∧(∼q)∨(∼p) is equivalent to
  1. (A)p∧qp\wedge qp∧q
  2. (B)p∧(∼q)p\wedge(\sim q)p∧(∼q)
  3. (C)p∧(q∧(∼p))p\wedge(q\wedge(\sim p))p∧(q∧(∼p))
  4. (D)p∨(q∨(∼p))p\vee(q\vee(\sim p))p∨(q∨(∼p))

Correct answer: (A)

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2023
Let A={1,2,3,4,5,6,7}A=\{1,2,3,4,5,6,7\}A={1,2,3,4,5,6,7}. Then the relation R={(x,y)∈A×A:x+y=7}R=\{(x,y)\in A\times A:x+y=7\}R={(x,y)∈A×A:x+y=7} is
  1. (A)transitive but neither symmetric nor reflexive
  2. (B)reflexive but neither symmetric nor transitive
  3. (C)an equivalence relation
  4. (D)symmetric but neither reflexive nor transitive

Correct answer: (D)

Step-by-step solution →
Q136·MathematicsNumericalJEE Main 2023
Let A={0,3,4,6,7,8,9,10}A=\{0,3,4,6,7,8,9,10\}A={0,3,4,6,7,8,9,10} and RRR be the relation defined on AAA such that R={(x,y)∈A×A:x−y is odd positive integer or x−y=2}R=\{(x,y)\in A\times A: x-y \text{ is odd positive integer or } x-y=2\}R={(x,y)∈A×A:x−y is odd positive integer or x−y=2}. The minimum number of elements that must be added to the relation RRR, so that it is a symmetric relation, is equal to

Correct answer: 19

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2023
Let the number of elements in sets AAA and BBB be five and two respectively. Then the number of subsets of A×BA\times BA×B each having at least 3 and at most 6 element is :
  1. (A)792
  2. (B)752
  3. (C)782
  4. (D)772

Correct answer: (A)

Step-by-step solution →
Q138·MathematicsNumericalJEE Main 2023
Let A={1,2,3,4,…,10}A=\{1,2,3,4,\dots,10\}A={1,2,3,4,…,10} and B={0,1,2,3,4,5}B=\{0,1,2,3,4,5\}B={0,1,2,3,4,5}. The number of elements in the relation R={(a,b)∈A×A:2(a−b)2+3(a−b)∈B}R=\{(a,b)\in A\times A:2(a-b)^{2}+3(a-b)\in B\}R={(a,b)∈A×A:2(a−b)2+3(a−b)∈B} is _____.

Correct answer: 18

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2023
The statement (P⇒Q)∧(R⇒Q)(P\Rightarrow Q)\wedge(R\Rightarrow Q)(P⇒Q)∧(R⇒Q) is logically equivalent to:
  1. (A)(P∨R)⇒Q(P\vee R)\Rightarrow Q(P∨R)⇒Q
  2. (B)(P⇒R)∧(Q⇒R)(P\Rightarrow R)\wedge(Q\Rightarrow R)(P⇒R)∧(Q⇒R)
  3. (C)(P⇒R)∨(Q⇒R)(P\Rightarrow R)\vee(Q\Rightarrow R)(P⇒R)∨(Q⇒R)
  4. (D)(P∧R)⇒Q(P\wedge R)\Rightarrow Q(P∧R)⇒Q

Correct answer: (A)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2023
Let A={x∈R:[x+3]+[x+4]≤3}A=\{x\in\mathbb{R}:[x+3]+[x+4]\le 3\}A={x∈R:[x+3]+[x+4]≤3}, B={x∈R:3x(∑r=1∞310r)x<3−x}B=\left\{x\in\mathbb{R}:3^{x}\left(\displaystyle\sum_{r=1}^{\infty}\dfrac{3}{10^{r}}\right)^{x}<3^{-x}\right\}B={x∈R:3x(r=1∑∞​10r3​)x<3−x}, where [t][t][t] denotes greatest integer function. Then:
  1. (A)A=∅, B≠∅A=\varnothing,\,B\ne\varnothingA=∅,B=∅
  2. (B)A=BA=BA=B
  3. (C)B⊂A, A≠BB\subset A,\,A\ne BB⊂A,A=B
  4. (D)A⊂B, A≠BA\subset B,\,A\ne BA⊂B,A=B

Correct answer: (B)

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2023
Let the sets AAA and BBB denote the domain and range respectively of the function f(x)=1⌈x⌉−xf(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}f(x)=⌈x⌉−x​1​, where ⌈x⌉\lceil x\rceil⌈x⌉ denotes the smallest integer greater than or equal to xxx. Then among the statements (S1): A∩B=(1,∞)−NA\cap B=(1,\infty)-\mathbb{N}A∩B=(1,∞)−N and (S2): A∪B=(1,∞)A\cup B=(1,\infty)A∪B=(1,∞)
  1. (A)only (S1) is true
  2. (B)both (S1) and (S2) are true
  3. (C)neither (S1) nor (S2) is true
  4. (D)only (S2) is true

Correct answer: (A)

Step-by-step solution →
Q142·MathematicsNumericalJEE Main 2023
Let the point (p,p+1)(p,p+1)(p,p+1) lie inside the region E={(x,y):3−x≤y≤9−x2, 0≤x≤3}E=\{(x,y):3-x\le y\le\sqrt{9-x^{2}},\,0\le x\le 3\}E={(x,y):3−x≤y≤9−x2​,0≤x≤3}. If the set of all values of ppp is the interval (a,b)(a,b)(a,b), then b2+b−a2b^{2}+b-a^{2}b2+b−a2 is equal to _____.

Correct answer: 3

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2023
Among the statements: (S1): (p⇒q)∨(∼p∧q)(p\Rightarrow q)\lor(\sim p\land q)(p⇒q)∨(∼p∧q) is a tautology (S2): (q⇒p)⇒(∼p∧q)(q\Rightarrow p)\Rightarrow(\sim p\land q)(q⇒p)⇒(∼p∧q) is a contradiction
  1. (A)neither (S1) nor (S2) is True
  2. (B)only (S1) is True
  3. (C)only (S2) is True
  4. (D)both (S1) and (S2) are True

Correct answer: (A)

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2023
Let P(S)P(S)P(S) denote the power set of S={1,2,3,…,10}S=\{1,2,3,\ldots,10\}S={1,2,3,…,10}. Define the relations R1R_1R1​ and R2R_2R2​ on P(S)P(S)P(S) as AR1BAR_1BAR1​B if (A∩Bc)∪(B∩Ac)=∅(A\cap B^c)\cup(B\cap A^c)=\varnothing(A∩Bc)∪(B∩Ac)=∅ and AR2BAR_2BAR2​B if A∪Bc=B∪AcA\cup B^c=B\cup A^cA∪Bc=B∪Ac, ∀A,B∈P(S)\forall A,B\in P(S)∀A,B∈P(S). Then:
  1. (A)only R1R_1R1​ is an equivalence relation
  2. (B)only R2R_2R2​ is an equivalence relation
  3. (C)both R1R_1R1​ and R2R_2R2​ are equivalence relations
  4. (D)both R1R_1R1​ and R2R_2R2​ are not equivalence relations

Correct answer: (C)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2023
Which of the following statements is a tautology?
  1. (A)p∨(p∧q)p\vee(p\wedge q)p∨(p∧q)
  2. (B)(p∧(p→q))→∼q(p\wedge(p\to q))\to\sim q(p∧(p→q))→∼q
  3. (C)(p∧q)→(∼(p)→q)(p\wedge q)\to(\sim(p)\to q)(p∧q)→(∼(p)→q)
  4. (D)p→(p∧(p→q))p\to(p\wedge(p\to q))p→(p∧(p→q))

Correct answer: (C)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2023
Let RRR be a relation on R\mathbb{R}R, given by R={(a,b):3a−3b+7 is an irrational number}R=\{(a,b):3a-3b+\sqrt{7}\text{ is an irrational number}\}R={(a,b):3a−3b+7​ is an irrational number}. Then RRR is
  1. (A)an equivalence relation
  2. (B)reflexive and symmetric but not transitive
  3. (C)reflexive but neither symmetric nor transitive
  4. (D)reflexive and transitive but not symmetric

Correct answer: (C)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2023
Let f:R−{0,1}→Rf:\mathbb{R}-\{0,1\}\to\mathbb{R}f:R−{0,1}→R be a function such that f(x)+f(11−x)=1+xf(x)+f\left(\dfrac{1}{1-x}\right)=1+xf(x)+f(1−x1​)=1+x. Then f(2)f(2)f(2) is equal to
  1. (A)92\dfrac9229​
  2. (B)74\dfrac7447​
  3. (C)94\dfrac9449​
  4. (D)73\dfrac7337​

Correct answer: (C)

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2023
The negation of the expression q∨((∼q)∧p)q\vee((\sim q)\wedge p)q∨((∼q)∧p) is equivalent to
  1. (A)(∼p)∨(∼q)(\sim p)\vee(\sim q)(∼p)∨(∼q)
  2. (B)p∧(∼q)p\wedge(\sim q)p∧(∼q)
  3. (C)(∼p)∨q(\sim p)\vee q(∼p)∨q
  4. (D)(∼p)∧(∼q)(\sim p)\wedge(\sim q)(∼p)∧(∼q)

Correct answer: (D)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2023
Among the relations S={(a,b):a,b∈R−{0}, 2+ab>0}S=\left\{(a,b):a,b\in\mathbb{R}-\{0\},\,2+\dfrac{a}{b}>0\right\}S={(a,b):a,b∈R−{0},2+ba​>0} and T={(a,b):a,b∈R, a2−b2∈Z}T=\{(a,b):a,b\in\mathbb{R},\,a^2-b^2\in\mathbb{Z}\}T={(a,b):a,b∈R,a2−b2∈Z},
  1. (A)neither SSS nor TTT is transitive
  2. (B)SSS is transitive but TTT is not
  3. (C)TTT is symmetric but SSS is not
  4. (D)both SSS and TTT are symmetric

Correct answer: (C)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2023
(S1) (p⇒q)∨(p∧(∼q))(p\Rightarrow q)\vee(p\wedge(\sim q))(p⇒q)∨(p∧(∼q)) is a tautology. (S2) ((∼p)⇒(∼q))∧((∼p)∨q)((\sim p)\Rightarrow(\sim q))\wedge((\sim p)\vee q)((∼p)⇒(∼q))∧((∼p)∨q) is a contradiction. Then
  1. (A)both (S1) and (S2) are correct
  2. (B)only (S1) is correct
  3. (C)only (S2) is correct
  4. (D)both (S1) and (S2) are wrong

Correct answer: (B)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2023
Let f:R−{2,6}→Rf:\mathbb{R}-\{2,6\}\to\mathbb{R}f:R−{2,6}→R be real valued function defined as f(x)=x2+2x+1x2−8x+12f(x)=\dfrac{x^2+2x+1}{x^2-8x+12}f(x)=x2−8x+12x2+2x+1​. Then range of fff is
  1. (A)(−∞,−214]∪(0,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup(0,\infty)(−∞,−421​]∪(0,∞)
  2. (B)(−∞,−214]∪[1,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup[1,\infty)(−∞,−421​]∪[1,∞)
  3. (C)(−∞,−214]∪[214,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup\left[\dfrac{21}{4},\infty\right)(−∞,−421​]∪[421​,∞)
  4. (D)(−∞,−214]∪[0,∞)\left(-\infty,-\dfrac{21}{4}\right]\cup[0,\infty)(−∞,−421​]∪[0,∞)

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2023
Let R be a relation on N×NN\times NN×N defined by (a,b)R(c,d)(a,b)R(c,d)(a,b)R(c,d) if and only if ad(b−c)=bc(a−d)ad(b-c)=bc(a-d)ad(b−c)=bc(a−d). Then R is
  1. (A)transitive but neither reflexive nor symmetric
  2. (B)symmetric but neither reflexive nor transitive
  3. (C)symmetric and transitive but not reflexive
  4. (D)reflexive and symmetric but not transitive

Correct answer: (B)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2023
If the domain of the function f(x)=[x]1+x2f(x)=\frac{[x]}{1+x^{2}}f(x)=1+x2[x]​, where [x][x][x] is greatest integer ≤x\le x≤x, is [2,6)[2,6)[2,6), then its range is
  1. (A)(526,25]\left(\frac{5}{26},\frac{2}{5}\right](265​,52​]
  2. (B)(537,25]−{929,27109,1889,953}\left(\frac{5}{37},\frac{2}{5}\right]-\left\{\frac{9}{29},\frac{27}{109},\frac{18}{89},\frac{9}{53}\right\}(375​,52​]−{299​,10927​,8918​,539​}
  3. (C)(537,25]\left(\frac{5}{37},\frac{2}{5}\right](375​,52​]
  4. (D)(526,25]−{929,27109,1889,953}\left(\frac{5}{26},\frac{2}{5}\right]-\left\{\frac{9}{29},\frac{27}{109},\frac{18}{89},\frac{9}{53}\right\}(265​,52​]−{299​,10927​,8918​,539​}

Correct answer: (C)

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2023
y=f(x)=sin⁡3(π3(cos⁡(π32(−4x3+5x2+1)32)))y=f(x)=\sin^{3}\left(\frac{\pi}{3}\left(\cos\left(\frac{\pi}{3\sqrt{2}}(-4x^{3}+5x^{2}+1)^{\frac{3}{2}}\right)\right)\right)y=f(x)=sin3(3π​(cos(32​π​(−4x3+5x2+1)23​))). Then, at x=1x=1x=1,
  1. (A)2 y′−3π2y=0\sqrt{2}\,y'-3\pi^{2}y=02​y′−3π2y=0
  2. (B)y′+3π2y=0y'+3\pi^{2}y=0y′+3π2y=0
  3. (C)2y′+3π2y=02y'+3\pi^{2}y=02y′+3π2y=0
  4. (D)2y′+3π2y=02y'+\sqrt{3}\pi^{2}y=02y′+3​π2y=0

Correct answer: (C)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2023
The number of values of r∈{p,q,∼p,∼q}r\in\{p,q,\sim p,\sim q\}r∈{p,q,∼p,∼q} for which ((p∧q)⇒(r∨q))∧((p∧r)⇒q)((p\wedge q)\Rightarrow(r\vee q))\wedge((p\wedge r)\Rightarrow q)((p∧q)⇒(r∨q))∧((p∧r)⇒q) is a tautology, is:
  1. (A)333
  2. (B)444
  3. (C)111
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2023
Let A={1,2,3,5,8,9}A=\{1, 2, 3, 5, 8, 9\}A={1,2,3,5,8,9}. Then the number of possible functions f:A→Af: A \to Af:A→A such that f(m⋅n)=f(m)⋅f(n)f(m \cdot n)=f(m) \cdot f(n)f(m⋅n)=f(m)⋅f(n) for every m,n∈Am, n \in Am,n∈A with m⋅n∈Am \cdot n \in Am⋅n∈A is equal to _______.

Correct answer: 432

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2023
The minimum number of elements that must be added to the relation R={(a,b),(b,c)}R=\{(a,b),(b,c)\}R={(a,b),(b,c)} on the set {a,b,c}\{a,b,c\}{a,b,c} so that it becomes symmetric and transitive is:
  1. (A)333
  2. (B)444
  3. (C)555
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2023
The range of the function f(x)=3−x+2+xf(x)=\sqrt{3 - x} + \sqrt{2 + x}f(x)=3−x​+2+x​ is:
  1. (A)[22,11][2\sqrt{2}, \sqrt{11}][22​,11​]
  2. (B)[5,13][\sqrt{5}, \sqrt{13}][5​,13​]
  3. (C)[2,7][\sqrt{2}, \sqrt{7}][2​,7​]
  4. (D)[5,10][\sqrt{5}, \sqrt{10}][5​,10​]

Correct answer: (D)

Step-by-step solution →
Q159·MathematicsNumericalJEE Main 2023
Let f1(x)=3x+22x+3f^1(x)=\dfrac{3x+2}{2x+3}f1(x)=2x+33x+2​, x∈R−{−32}x\in\mathbb{R}-\left\{\dfrac{-3}{2}\right\}x∈R−{2−3​}. For n≥2n\geq2n≥2, define fn(x)=f1 of fn−1(x)f^n(x)=f^1\text{ of }f^{n-1}(x)fn(x)=f1 of fn−1(x). If f5(x)=ax+bbx+af^5(x)=\dfrac{ax+b}{bx+a}f5(x)=bx+aax+b​, gcd⁡(a,b)=1\gcd(a,b)=1gcd(a,b)=1, then a+ba+ba+b is equal to

Correct answer: 3125

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2023
Among the statements: (S1) ((p∨q)⇒r)⇔(p⇒r)((p\vee q)\Rightarrow r)\Leftrightarrow(p\Rightarrow r)((p∨q)⇒r)⇔(p⇒r) and (S2) ((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))((p\vee q)\Rightarrow r)\Leftrightarrow((p\Rightarrow r)\vee(q\Rightarrow r))((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))
  1. (A)only (S2) is a tautology
  2. (B)only (S1) is a tautology
  3. (C)neither (S1) nor (S2) is a tautology
  4. (D)both (S1) and (S2) are tautologies

Correct answer: (C)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2023
Consider the following statements: P: I have fever; Q: I will not take medicine; R: I will take rest. The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to:
  1. (A)((∼P)∨∼Q)∧((∼P)∨R)((\sim P) \vee \sim Q) \wedge ((\sim P) \vee R)((∼P)∨∼Q)∧((∼P)∨R)
  2. (B)(P∨Q)∧((∼P)∨R)(P \vee Q) \wedge ((\sim P) \vee R)(P∨Q)∧((∼P)∨R)
  3. (C)((∼P)∨∼Q)∧((∼P)∨∼R)((\sim P) \vee \sim Q) \wedge ((\sim P) \vee \sim R)((∼P)∨∼Q)∧((∼P)∨∼R)
  4. (D)(P∨∼Q)∧(P∨∼R)(P \vee \sim Q) \wedge (P \vee \sim R)(P∨∼Q)∧(P∨∼R)

Correct answer: (A)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2023
The domain of f(x)=log⁡(x+1)(x−2)e2log⁡ex−(2x+3)f(x) = \dfrac{\log_{(x+1)}(x - 2)}{e^{2\log_e x} - (2x + 3)}f(x)=e2loge​x−(2x+3)log(x+1)​(x−2)​, x∈Rx \in \mathbb{R}x∈R is
  1. (A)R−{3}\mathbb{R} - \{3\}R−{3}
  2. (B)(−1,∞)−{3}(-1, \infty) - \{3\}(−1,∞)−{3}
  3. (C)(2,∞)−{3}(2, \infty) - \{3\}(2,∞)−{3}
  4. (D)R−{−1,3}\mathbb{R} - \{-1, 3\}R−{−1,3}

Correct answer: (C)

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2023
If ppp, qqq and rrr three propositions, then which of the following combination of truth values of ppp, qqq and rrr makes the logical expression {(p∨q)∧((∼p)∨r)}→((∼q)∨r)\{(p \vee q) \wedge ((\sim p) \vee r)\} \to ((\sim q) \vee r){(p∨q)∧((∼p)∨r)}→((∼q)∨r) false?
  1. (A)p = T, q = T, r = F
  2. (B)p = T, q = F, r = T
  3. (C)p = F, q = T, r = F
  4. (D)p = T, q = F, r = F

Correct answer: (C)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2023
Let RRR be a relation defined on N\mathbb{N}N as a R ba\,R\,baRb if 2a+3b2a+3b2a+3b is a multiple of 5, a,b∈N5,\,a,b\in\mathbb{N}5,a,b∈N. Then RRR is:
  1. (A)an equivalence relation
  2. (B)transitive but not symmetric
  3. (C)not reflexive
  4. (D)symmetric but not transitive

Correct answer: (A)

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2023
Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be a function such that f(x)=x2+2x+1x2+1f(x) = \dfrac{x^2 + 2x + 1}{x^2 + 1}f(x)=x2+1x2+2x+1​. Then
  1. (A)f(x)f(x)f(x) is one-one in [1,∞)[1, \infty)[1,∞) but not in (−∞,∞)(-\infty, \infty)(−∞,∞)
  2. (B)f(x)f(x)f(x) is one-one in (−∞,∞)(-\infty, \infty)(−∞,∞)
  3. (C)f(x)f(x)f(x) is many-one in (−∞,−1)(-\infty, -1)(−∞,−1)
  4. (D)f(x)f(x)f(x) is many-one in (1,∞)(1, \infty)(1,∞)

Correct answer: (A)

Step-by-step solution →
Q166·MathematicsNumericalJEE Main 2023
Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)−1f(x + y) = f(x) + f(y) - 1f(x+y)=f(x)+f(y)−1, ∀x,y∈R\forall x, y \in \mathbb{R}∀x,y∈R. If f′(0)=2f'(0) = 2f′(0)=2, then ∣f(−2)∣|f(-2)|∣f(−2)∣ is equal to ________ .

Correct answer: 3

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2023
The statement B⇒((∼A)∨B)B\Rightarrow ((\sim A)\lor B)B⇒((∼A)∨B) is equivalent to:
  1. (A)A⇒(A⇔B)A\Rightarrow (A\Leftrightarrow B)A⇒(A⇔B)
  2. (B)A⇒((∼A)⇒B)A\Rightarrow ((\sim A)\Rightarrow B)A⇒((∼A)⇒B)
  3. (C)B⇒(A⇒B)B\Rightarrow (A\Rightarrow B)B⇒(A⇒B)
  4. (D)B⇒((∼A)⇒B)B\Rightarrow ((\sim A)\Rightarrow B)B⇒((∼A)⇒B)

Correct answer: (A)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2023
Let f(x)=2xn+λf(x) = 2x^n + \lambdaf(x)=2xn+λ, λ∈R\lambda \in \mathbb{R}λ∈R, n∈Nn \in \mathbb{N}n∈N, and f(4)=133f(4) = 133f(4)=133, f(5)=255f(5) = 255f(5)=255. Then the sum of all the positive integer divisors of (f(3)−f(2))(f(3) - f(2))(f(3)−f(2)) is
  1. (A)60
  2. (B)59
  3. (C)61
  4. (D)58

Correct answer: (A)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2023
Let Δ,∇∈{∧,∨}\Delta, \nabla \in \{\wedge, \vee\}Δ,∇∈{∧,∨} be such that (p→q) Δ (p ∇ q)(p \to q)\,\Delta\,(p\,\nabla\,q)(p→q)Δ(p∇q) is a tautology. Then
  1. (A)Δ=∨, ∇=∨\Delta = \vee,\ \nabla = \veeΔ=∨, ∇=∨
  2. (B)Δ=∨, ∇=∧\Delta = \vee,\ \nabla = \wedgeΔ=∨, ∇=∧
  3. (C)Δ=∧, ∇=∨\Delta = \wedge,\ \nabla = \veeΔ=∧, ∇=∨
  4. (D)Δ=∧, ∇=∧\Delta = \wedge,\ \nabla = \wedgeΔ=∧, ∇=∧

Correct answer: (A)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2023
The statement (p∧(∼q))⇒(p⇒(∼q))(p\wedge(\sim q))\Rightarrow(p\Rightarrow(\sim q))(p∧(∼q))⇒(p⇒(∼q)) is:
  1. (A)a tautology
  2. (B)a contradiction
  3. (C)equivalent to p∨qp\vee qp∨q
  4. (D)equivalent to (∼p)∨(∼q)(\sim p)\vee(\sim q)(∼p)∨(∼q)

Correct answer: (A)

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2023
The number of functions f:{1,2,3,4}→{a∈Z:∣a∣≤8}f : \{1,2,3,4\} \to \{a \in \mathbb{Z} : |a| \le 8\}f:{1,2,3,4}→{a∈Z:∣a∣≤8} satisfying f(n)+1nf(n+1)=1f(n) + \dfrac{1}{n}f(n+1) = 1f(n)+n1​f(n+1)=1, ∀n∈{1,2,3}\forall n \in \{1,2,3\}∀n∈{1,2,3} is
  1. (A)1
  2. (B)4
  3. (C)2
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2023
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a function defined by f(x)=log⁡m{2(sin⁡x−cos⁡x)+m−2}f(x)=\log_{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}f(x)=logm​​{2​(sinx−cosx)+m−2}, for some mmm, such that the range of fff is [0,2][0,2][0,2]. Then the value of mmm is
  1. (A)555
  2. (B)444
  3. (C)333
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q173·MathematicsNumericalJEE Main 2023
For some a,b,c∈Na,b,c\in\mathbb{N}a,b,c∈N, let f(x)=ax−3f(x)=ax-3f(x)=ax−3 and g(x)=xb+cg(x)=x^b+cg(x)=xb+c, x∈Rx\in\mathbb{R}x∈R. If (f∘g)−1(x)=(x−72)1/3(f\circ g)^{-1}(x)=\left(\dfrac{x-7}{2}\right)^{1/3}(f∘g)−1(x)=(2x−7​)1/3, then (f∘g)(ac)+(g∘f)(b)(f\circ g)(ac)+(g\circ f)(b)(f∘g)(ac)+(g∘f)(b) is equal to _______.

Correct answer: 2039

Step-by-step solution →
Q174·MathematicsNumericalJEE Main 2023
The minimum number of elements that must be added to the relation R={(a,b),(b,c),(b,d)}R=\{(a,b),(b,c),(b,d)\}R={(a,b),(b,c),(b,d)} on the set {a,b,c,d}\{a,b,c,d\}{a,b,c,d} so that it is an equivalence relation, is

Correct answer: 13

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2023
The equation x2−4x+[x]+3=x[x]x^2-4x+[x]+3=x[x]x2−4x+[x]+3=x[x], where [x][x][x] denotes the greatest integer function, has:
  1. (A)a unique solution in (−∞,1)(-\infty,1)(−∞,1)
  2. (B)no solution
  3. (C)exactly two solutions in (−∞,∞)(-\infty,\infty)(−∞,∞)
  4. (D)a unique solution in (−∞,∞)(-\infty,\infty)(−∞,∞)

Correct answer: (D)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2023
Let ppp and qqq be two statements. Then ∼(p∧(p⇒∼q))\sim(p\wedge(p\Rightarrow\sim q))∼(p∧(p⇒∼q)) is equivalent to
  1. (A)p∨(p∧q)p\vee(p\wedge q)p∨(p∧q)
  2. (B)p∨(p∧(∼q))p\vee(p\wedge(\sim q))p∨(p∧(∼q))
  3. (C)(∼p)∨q(\sim p)\vee q(∼p)∨q
  4. (D)p∨((∼p)∧q)p\vee((\sim p)\wedge q)p∨((∼p)∧q)

Correct answer: (C)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2023
If f(x)=22x22x+2, x∈Rf(x)=\dfrac{2^{2x}}{2^{2x}+2},\ x\in\mathbb{R}f(x)=22x+222x​, x∈R, then f(12023)+f(22023)+⋯+f(20222023)f\left(\dfrac{1}{2023}\right)+f\left(\dfrac{2}{2023}\right)+\cdots+f\left(\dfrac{2022}{2023}\right)f(20231​)+f(20232​)+⋯+f(20232022​) is equal to
  1. (A)101110111011
  2. (B)201020102010
  3. (C)101010101010
  4. (D)201120112011

Correct answer: (A)

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2023
The compound statement (∼(P∧Q))∨((∼P)∧Q)⇒((∼P)∧(∼Q))(\sim(P\wedge Q))\vee((\sim P)\wedge Q)\Rightarrow((\sim P)\wedge(\sim Q))(∼(P∧Q))∨((∼P)∧Q)⇒((∼P)∧(∼Q)) is equivalent to
  1. (A)(∼Q)∨P(\sim Q)\vee P(∼Q)∨P
  2. (B)((∼P)∨Q)∧(∼Q)((\sim P)\vee Q)\wedge(\sim Q)((∼P)∨Q)∧(∼Q)
  3. (C)(∼P)∨Q(\sim P)\vee Q(∼P)∨Q
  4. (D)((∼P)∨Q)∧((∼Q)∨P)((\sim P)\vee Q)\wedge((\sim Q)\vee P)((∼P)∨Q)∧((∼Q)∨P)

Correct answer: (D)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2023
The relation R={(a,b):gcd⁡(a,b)=1, 2a≠b, a,b∈Z}R=\{(a,b):\gcd(a,b)=1,\ 2a\ne b,\ a,b\in\mathbb{Z}\}R={(a,b):gcd(a,b)=1, 2a=b, a,b∈Z} is:
  1. (A)reflexive but not symmetric
  2. (B)transitive but not reflexive
  3. (C)symmetric but not transitive
  4. (D)neither symmetric nor transitive

Correct answer: (D)

Step-by-step solution →
Q180·MathematicsNumericalJEE Advanced 2022
In a study about a pandemic, data of 900 persons was collected. It was found that 190 persons had symptom of fever, 220 persons had symptom of cough, 220 persons had symptom of breathing problem, 330 persons had symptom of fever or cough or both, 350 persons had symptom of cough or breathing problem or both, 340 persons had symptom of fever or breathing problem or both, 30 persons had all three symptoms (fever, cough and breathing problem). If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is __________.

Correct answer: 0.80

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2022
Let R be a relation from the set {1,2,3………,60}\{1, 2, 3\ldots\ldots\ldots,60\}{1,2,3………,60} to itself such that R={(a,b):b=pq,R = \{(a, b) : b = pq,R={(a,b):b=pq, where p,q≥3p, q \geq 3p,q≥3 are prime numbers}\}}. Then, the number of elements in R is :
  1. (A)600
  2. (B)660
  3. (C)540
  4. (D)720

Correct answer: (B)

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2022
The statement (p⇒q)∨(p⇒r)\left(p \Rightarrow q\right) \vee \left(p \Rightarrow r\right)(p⇒q)∨(p⇒r) is NOT equivalent to:
  1. (A)(p∧(∼r))⇒q\left(p \wedge (\sim r)\right) \Rightarrow q(p∧(∼r))⇒q
  2. (B)(∼q)⇒((∼r)∨p)(\sim q) \Rightarrow \left((\sim r) \vee p\right)(∼q)⇒((∼r)∨p)
  3. (C)p⇒(q∨r)p \Rightarrow (q \vee r)p⇒(q∨r)
  4. (D)(p∧(∼q))⇒r\left(p \wedge (\sim q)\right) \Rightarrow r(p∧(∼q))⇒r

Correct answer: (B)

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2022
The statement (p∧q)⇒(p∧r)(p \wedge q) \Rightarrow (p \wedge r)(p∧q)⇒(p∧r) is equivalent to :
  1. (A)q⇒(p∧r)q \Rightarrow (p \wedge r)q⇒(p∧r)
  2. (B)p⇒(p∧r)p \Rightarrow (p \wedge r)p⇒(p∧r)
  3. (C)(p∧r)⇒(p∧q)(p \wedge r) \Rightarrow (p \wedge q)(p∧r)⇒(p∧q)
  4. (D)(p∧q)⇒r(p \wedge q) \Rightarrow r(p∧q)⇒r

Correct answer: (D)

Step-by-step solution →
Q184·MathematicsNumericalJEE Main 2022
Let S={4, 6, 9}S = \{4,\,6,\,9\}S={4,6,9} and T={9, 10, 11, …, 1000}T = \{9,\,10,\,11,\,\ldots,\,1000\}T={9,10,11,…,1000}. If A={a1+a2+…+ak:k∈N,  a1, a2, a3, …,ak∈S}A = \left\{a_{1}+a_{2}+\ldots+a_{k} : k \in N,\; a_{1},\,a_{2},\,a_{3},\,\ldots,a_{k} \in S\right\}A={a1​+a2​+…+ak​:k∈N,a1​,a2​,a3​,…,ak​∈S}, then the sum of all the elements in the set T−AT - AT−A is equal to _______.

Correct answer: 11

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2022
The domain of the function f(x)=sin⁡−1(x2−3x+2x2+2x+7)f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)f(x)=sin−1(x2+2x+7x2−3x+2​) is :
  1. (A)[1,∞)[1, \infty)[1,∞)
  2. (B)(−1,2](-1, 2](−1,2]
  3. (C)[−1,∞)[-1, \infty)[−1,∞)
  4. (D)(−∞,2](-\infty, 2](−∞,2]

Correct answer: (C)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2022
Let <b>p :</b> Ramesh listens to music. <b>q :</b> Ramesh is out of his village <b>r :</b> It is Sunday <b>s :</b> It is Saturday Then the statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday” can be expressed as
  1. (A)((∼q)∧(r∨s))⇒p\left((\sim q)\wedge(r \vee s)\right) \Rightarrow p((∼q)∧(r∨s))⇒p
  2. (B)(q∧(r∨s))⇒p\left(q \wedge (r \vee s)\right) \Rightarrow p(q∧(r∨s))⇒p
  3. (C)p⇒(q∧(r∨s))p \Rightarrow \left(q \wedge (r \vee s)\right)p⇒(q∧(r∨s))
  4. (D)p⇒((∼q)∧(r∨s))p \Rightarrow \left((\sim q)\wedge(r \vee s)\right)p⇒((∼q)∧(r∨s))

Correct answer: (D)

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2022
Let α\alphaα, β\betaβ and γ\gammaγ be three positive real numbers. Let f(x)=αx5+βx3+γxf(x) = \alpha x^{5} + \beta x^{3} + \gamma xf(x)=αx5+βx3+γx, x∈Rx \in Rx∈R and g:R→Rg : R \to Rg:R→R be such that g(f(x))=xg(f(x)) = xg(f(x))=x for all x∈Rx \in Rx∈R. If a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​,..., ana_{n}an​ be in arithmetic progression with mean zero, then the value of f(g(1n∑i=1nf(ai)))f\left( g\left( \frac{1}{n} \sum_{i=1}^{n} f\left(a_{i}\right) \right) \right)f(g(n1​∑i=1n​f(ai​))) is equal to :
  1. (A)0
  2. (B)3
  3. (C)9
  4. (D)27

Correct answer: (A)

Step-by-step solution →
Q188·MathematicsSingle correctJEE Main 2022
Let the operations ∗,⊙∈{∧,∨}*, \odot \in \{\wedge, \vee\}∗,⊙∈{∧,∨}. If (p∗q)⊙(p⊙∼q)(p * q) \odot (p \odot \sim q)(p∗q)⊙(p⊙∼q) is a tautology, then the ordered pair (∗,⊙)(*, \odot)(∗,⊙) is :
  1. (A)(∨,∧)(\vee, \wedge)(∨,∧)
  2. (B)(∨,∨)(\vee, \vee)(∨,∨)
  3. (C)(∧,∧)(\wedge, \wedge)(∧,∧)
  4. (D)(∧,∨)(\wedge, \vee)(∧,∨)

Correct answer: (B)

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2022
Let S={x∈[−6,3]−{−2,2}:∣x+3∣−1∣x∣−2≥0}S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\}S={x∈[−6,3]−{−2,2}:∣x∣−2∣x+3∣−1​≥0} and T={x∈Z:x2−7∣x∣+9≤0}T=\left\{x \in Z: x^{2}-7|x|+9 \leq 0\right\}T={x∈Z:x2−7∣x∣+9≤0}. Then the number of elements in S∩TS \cap TS∩T is
  1. (A)7
  2. (B)5
  3. (C)4
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2022
For α∈N\alpha \in Nα∈N, consider a relation R on N given by R={(x,y):3x+αy is a multiple of 7}R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}R={(x,y):3x+αy is a multiple of 7}. The relation R is an equivalence relation if and only if :
  1. (A)α=14\alpha = 14α=14
  2. (B)α\alphaα is a multiple of 4
  3. (C)4 is the remainder when α\alphaα is divided by 10
  4. (D)4 is the remainder when α\alphaα is divided by 7

Correct answer: (D)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2022
The domain of the function f(x)=sin⁡−1[2x2−3]+log⁡2(log⁡12(x2−5x+5))f(x) = \sin^{-1}[2x^{2}-3] + \log_{2}\left(\log_{\frac{1}{2}}\left(x^{2}-5x+5\right)\right)f(x)=sin−1[2x2−3]+log2​(log21​​(x2−5x+5)), where [t][t][t] is the greatest integer function, is :
  1. (A)(−52,5−52)\left(-\sqrt{\frac{5}{2}}, \frac{5-\sqrt{5}}{2}\right)(−25​​,25−5​​)
  2. (B)(5−52,5+52)\left(\frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)(25−5​​,25+5​​)
  3. (C)(1,5−52)\left(1, \frac{5-\sqrt{5}}{2}\right)(1,25−5​​)
  4. (D)[1,5+52)\left[1, \frac{5+\sqrt{5}}{2}\right)[1,25+5​​)

Correct answer: (C)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2022
(p∧r)⇔(p∧(∼q))\left(p\wedge r\right)\Leftrightarrow\left(p\wedge\left(\sim q\right)\right)(p∧r)⇔(p∧(∼q)) is equivalent to (∼p)\left(\sim p\right)(∼p) when r is
  1. (A)ppp
  2. (B)∼p\sim p∼p
  3. (C)qqq
  4. (D)∼q\sim q∼q

Correct answer: (C)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2022
Let f,g:N−{1}→Nf, g : \mathbb{N}-\{1\} \to \mathbb{N}f,g:N−{1}→N be functions defined by f(a)=αf(a) = \alphaf(a)=α, where α\alphaα is the maximum of the powers of those primes p such that pαp^{\alpha}pα divides aaa, and g(a)=a+1g(a) = a+1g(a)=a+1, for all a∈N−{1}a \in \mathbb{N}-\{1\}a∈N−{1}. Then, the function f+gf + gf+g is
  1. (A)one-one but not onto
  2. (B)onto but not one-one
  3. (C)both one-one and onto
  4. (D)neither one-one nor onto

Correct answer: (D)

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2022
Let R1R_1R1​ and R2R_2R2​ be two relations defined on R\mathbb{R}R by a R1 b⇔ab≥0a\,R_1\,b \Leftrightarrow ab \ge 0aR1​b⇔ab≥0 and a R2 b⇔a≥ba\,R_2\,b \Leftrightarrow a \ge baR2​b⇔a≥b, then
  1. (A)R1R_1R1​ is an equivalence relation but not R2R_2R2​
  2. (B)R2R_2R2​ is an equivalence relation but not R1R_1R1​
  3. (C)both R1R_1R1​ and R2R_2R2​ are equivalence relations
  4. (D)neither R1R_1R1​ nor R2R_2R2​ is an equivalence relation

Correct answer: (D)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2022
If the truth value of the statement (P∧(∼R))→((∼R)∧Q)\left(P \wedge (\sim R)\right) \rightarrow \left((\sim R) \wedge Q\right)(P∧(∼R))→((∼R)∧Q) is F, then the truth value of which of the following is F ?
  1. (A)P∨Q→∼RP \vee Q \rightarrow \sim RP∨Q→∼R
  2. (B)R∨Q→∼PR \vee Q \rightarrow \sim PR∨Q→∼P
  3. (C)∼(P∨Q)→∼R\sim\left(P \vee Q\right) \rightarrow \sim R∼(P∨Q)→∼R
  4. (D)∼(R∨Q)→∼P\sim\left(R \vee Q\right) \rightarrow \sim P∼(R∨Q)→∼P

Correct answer: (D)

Step-by-step solution →
Q196·MathematicsNumericalJEE Main 2022
Let A={1,2,3,4,5,6,7}A=\{1,2,3,4,5,6,7\}A={1,2,3,4,5,6,7} and B={3,6,7,9}B=\{3,6,7,9\}B={3,6,7,9}. Then the number of elements in the set {C⊆A:C∩B≠ϕ}\{C\subseteq A:C\cap B\neq\phi\}{C⊆A:C∩B=ϕ} is________

Correct answer: 112

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2022
Let f: R → R be a continuous function such that f (3x) – f (x) = x. If f (8) = 7, then f (14) is equal to :
  1. (A)4
  2. (B)10
  3. (C)11
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2022
The statement (~(p ⇔ ~q)) ∧ q is :
  1. (A)a tautology
  2. (B)a contradiction
  3. (C)equivalent to (p ⇒ q) ∧ q
  4. (D)equivalent to (p ⇒ q) ∧ p

Correct answer: (D)

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2022
Negation of the Boolean expression p⇔(q⇒p)p\Leftrightarrow(q\Rightarrow p)p⇔(q⇒p) is
  1. (A)(∼p)∧q(\sim p)\wedge q(∼p)∧q
  2. (B)p∧(∼q)p\wedge(\sim q)p∧(∼q)
  3. (C)(∼p)∨(∼q)(\sim p)\vee(\sim q)(∼p)∨(∼q)
  4. (D)(∼p)∧(∼q)(\sim p)\wedge(\sim q)(∼p)∧(∼q)

Correct answer: (D)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2022
Which of the following statements is a tautology ?
  1. (A)((∼p)∨q)⇒p((\sim p) \vee q) \Rightarrow p((∼p)∨q)⇒p
  2. (B)p⇒((∼p)∨q)p \Rightarrow ((\sim p) \vee q)p⇒((∼p)∨q)
  3. (C)((∼p)∨q)⇒q((\sim p) \vee q) \Rightarrow q((∼p)∨q)⇒q
  4. (D)q⇒((∼p)∨q)q \Rightarrow ((\sim p) \vee q)q⇒((∼p)∨q)

Correct answer: (D)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2022
Let a set A=A1∪A2∪…∪AkA = A_{1} \cup A_{2} \cup \ldots \cup A_{k}A=A1​∪A2​∪…∪Ak​, where Ai∩Aj=ϕA_{i} \cap A_{j} = \phiAi​∩Aj​=ϕ for i≠ji \ne ji=j 1≤i,j≤k1 \le i, j \le k1≤i,j≤k. Define the relation R from A to A by R={(x,y):y∈AiR = \{(x, y) : y \in A_{i}R={(x,y):y∈Ai​ if and only if x∈Ai,1≤i≤k}x \in A_{i}, 1 \le i \le k\}x∈Ai​,1≤i≤k}. Then, R is :
  1. (A)reflexive, symmetric but not transitive
  2. (B)reflexive, transitive but not symmetric
  3. (C)reflexive but not symmetric and transitive
  4. (D)an equivalence relation

Correct answer: (D)

Step-by-step solution →
Q202·MathematicsNumericalJEE Main 2022
Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If f(g(x))=8x2−2xf(g(x)) = 8x^2 - 2xf(g(x))=8x2−2x, and g(f(x))=4x2+6x+1g(f(x)) = 4x^2 + 6x + 1g(f(x))=4x2+6x+1, then the value of f(2)+g(2)f(2) + g(2)f(2)+g(2) is______.

Correct answer: 18

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2022
Negation of the Boolean statement (p∨q)⇒((∼r)∨p)(p \vee q) \Rightarrow \left((\sim r) \vee p\right)(p∨q)⇒((∼r)∨p) is equivalent to:
  1. (A)p∧(∼q)∧rp \wedge (\sim q) \wedge rp∧(∼q)∧r
  2. (B)(∼p)∧(∼q)∧r(\sim p) \wedge (\sim q) \wedge r(∼p)∧(∼q)∧r
  3. (C)(∼p)∧q∧r(\sim p) \wedge q \wedge r(∼p)∧q∧r
  4. (D)p∧q∧(∼r)p \wedge q \wedge (\sim r)p∧q∧(∼r)

Correct answer: (C)

Step-by-step solution →
Q204·MathematicsSingle correctJEE Main 2022
Let Δ∈{∧,∨,⇒,⇔}\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}Δ∈{∧,∨,⇒,⇔} be such that (p∧q)Δ((p∨q)⇒q)(p \wedge q)\Delta((p \vee q) \Rightarrow q)(p∧q)Δ((p∨q)⇒q) is a tautology. Then Δ is equal to :
  1. (A)∧\wedge∧
  2. (B)∨\vee∨
  3. (C)⇒\Rightarrow⇒
  4. (D)⇔\Leftrightarrow⇔

Correct answer: (C)

Step-by-step solution →
Q205·MathematicsNumericalJEE Main 2022
Let c, k ∈ R. If f(x)=(c+1)x2+(1−c2)x+2kf(x) = (c + 1)x^2 + (1 - c^2)x + 2kf(x)=(c+1)x2+(1−c2)x+2k and f(x+y)=f(x)+f(y)−xyf(x + y) = f(x) + f(y) - xyf(x+y)=f(x)+f(y)−xy, for all x, y ∈ R, then the value of ∣2(f(1)+f(2)+f(3)+…+f(20))∣|2(f(1) + f(2) + f(3) + \ldots + f(20))|∣2(f(1)+f(2)+f(3)+…+f(20))∣ is equal to ________.

Correct answer: 3395

Step-by-step solution →
Q206·MathematicsSingle correctJEE Main 2022
Let R1={(a,b)∈N×N:∣a−b∣≤13}R_1 = \{(a,b) \in N \times N : |a-b| \leq 13\}R1​={(a,b)∈N×N:∣a−b∣≤13} and R2={(a,b)∈N×N:∣a−b∣≠13}R_2 = \{(a,b) \in N \times N : |a-b| \neq 13\}R2​={(a,b)∈N×N:∣a−b∣=13}. Then on N:
  1. (A)Both R1R_1R1​ and R2R_2R2​ are equivalence relations
  2. (B)Neither R1R_1R1​ nor R2R_2R2​ is an equivalence relation
  3. (C)R1R_1R1​ is an equivalence relation but R2R_2R2​ is not
  4. (D)R2R_2R2​ is an equivalence relation but R1R_1R1​ is not

Correct answer: (B)

Step-by-step solution →
Q207·MathematicsNumericalJEE Main 2022
Let R1R_1R1​ and R2R_2R2​ be relations on the set {1, 2, …, 50} such that R1={(p, pn):p is a prime and n≥0 is an integer}R_1 = \{(p,\, p^n) : p \text{ is a prime and } n \geq 0 \text{ is an integer}\}R1​={(p,pn):p is a prime and n≥0 is an integer} and R2={(p, pn):p is a prime and n=0 or 1}R_2 = \{(p,\, p^n) : p \text{ is a prime and } n = 0 \text{ or } 1\}R2​={(p,pn):p is a prime and n=0 or 1}. Then, the number of elements in R1−R2R_1 - R_2R1​−R2​ is ______.

Correct answer: 8

Step-by-step solution →
Q208·MathematicsSingle correctJEE Main 2022
Let p, q, r be three logical statements. Consider the compound statements S1:((∼p)∨q)∨((∼p)∨r)S_1 : ((\sim p) \vee q) \vee ((\sim p) \vee r)S1​:((∼p)∨q)∨((∼p)∨r) and S2:p→(q∨r)S_2 : p \rightarrow (q \vee r)S2​:p→(q∨r) Then, which of the following is NOT true ?
  1. (A)If S2S_2S2​ is True, then S1S_1S1​ is True
  2. (B)If S2S_2S2​ is False, then S1S_1S1​ is False
  3. (C)If S2S_2S2​ is False, then S1S_1S1​ is True
  4. (D)If S1S_1S1​ is False, then S2S_2S2​ is False

Correct answer: (C)

Step-by-step solution →
Q209·MathematicsNumericalJEE Main 2022
The maximum number of compound propositions, out of p∨r∨sp \vee r \vee sp∨r∨s, p∨r∨∼sp \vee r \vee \sim sp∨r∨∼s, p∨∼q∨sp \vee \sim q \vee sp∨∼q∨s, ∼p∨∼r∨s\sim p \vee \sim r \vee s∼p∨∼r∨s, ∼p∨∼r∨∼s\sim p \vee \sim r \vee \sim s∼p∨∼r∨∼s, ∼p∨q∨∼s\sim p \vee q \vee \sim s∼p∨q∨∼s, q∨r∨∼sq \vee r \vee \sim sq∨r∨∼s, q∨∼r∨∼sq \vee \sim r \vee \sim sq∨∼r∨∼s, ∼p∨∼q∨∼s\sim p \vee \sim q \vee \sim s∼p∨∼q∨∼s that can be made simultaneously true by an assignment of the truth values to ppp, qqq, rrr and sss, is equal to

Correct answer: 9

Step-by-step solution →
Q210·MathematicsNumericalJEE Main 2022
Let S={1,2,3,4}S = \{1, 2, 3, 4\}S={1,2,3,4}. Then the number of elements in the set {f:S×S→S:f is onto and f(a,b)=f(b,a)≥a ∀ (a,b)∈S×S}\{f : S \times S \to S : f \text{ is onto and } f(a, b) = f(b, a) \geq a\ \forall\, (a, b) \in S \times S\}{f:S×S→S:f is onto and f(a,b)=f(b,a)≥a ∀(a,b)∈S×S} is

Correct answer: 37

Step-by-step solution →
Q211·MathematicsSingle correctJEE Main 2022
Let a function f:N→N\mathrm{f}:\mathbb{N} \to \mathbb{N}f:N→N be defined by f(n)={2n,n=2,4,6,8,.....n−1,n=3,7,11,15,.....n+12,n=1,5,9,13,.....\mathrm{f(n)}=\begin{cases} 2\mathrm{n}, & \mathrm{n}=2,4,6,8,..... \\ \mathrm{n}-1, & \mathrm{n}=3,7,11,15,..... \\ \dfrac{\mathrm{n}+1}{2}, & \mathrm{n}=1,5,9,13,..... \end{cases}f(n)=⎩⎨⎧​2n,n−1,2n+1​,​n=2,4,6,8,.....n=3,7,11,15,.....n=1,5,9,13,.....​ then, f is
  1. (A)one-one but not onto
  2. (B)onto but not one-one
  3. (C)neither one-one nor onto
  4. (D)one-one and onto

Correct answer: (D)

Step-by-step solution →
Q212·MathematicsNumericalJEE Main 2022
Let f:R→Rf : R \rightarrow Rf:R→R be a function defined f(x)=2e2xe2x+ef(x)=\frac{2e^{2x}}{e^{2x}+e}f(x)=e2x+e2e2x​. Then f(1100)+f(2100)+f(3100)+.....+f(99100)f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+.....+f\left(\frac{99}{100}\right)f(1001​)+f(1002​)+f(1003​)+.....+f(10099​) is equal to________.

Correct answer: 99

Step-by-step solution →
Q213·MathematicsSingle correctJEE Main 2022
The Boolean expression (∼(p∧q))∨q\left(\sim\left(p \wedge q\right)\right)\vee q(∼(p∧q))∨q is equivalent to :
  1. (A)q→(p∧q)q \rightarrow \left(p \wedge q\right)q→(p∧q)
  2. (B)p→qp \rightarrow qp→q
  3. (C)p→(p→q)p \rightarrow \left(p \rightarrow q\right)p→(p→q)
  4. (D)p→(p∨q)p \rightarrow \left(p \vee q\right)p→(p∨q)

Correct answer: (D)

Step-by-step solution →
Q214·MathematicsNumericalJEE Main 2022
Let S={1,2,3,4,5,6,7,8,9,10}S=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}S={1,2,3,4,5,6,7,8,9,10}. Define f:S→Sf:S\to Sf:S→S as f(n)={2n,if n=1,2,3,4,52n−11if n=6,7,8,9,10f(n)=\begin{cases}2n, & \text{if } n=1,2,3,4,5\\ 2n-11 & \text{if } n=6,7,8,9,10\end{cases}f(n)={2n,2n−11​if n=1,2,3,4,5if n=6,7,8,9,10​. Let g:S→Sg:S\to Sg:S→S be a function such that fog(n)={n+1, if n is oddn−1, if n is even\text{fog}(n)=\begin{cases}n+1 & \text{, if } n \text{ is odd}\\ n-1 & \text{, if } n \text{ is even}\end{cases}fog(n)={n+1n−1​, if n is odd, if n is even​, then g(10) ((g(1)+g(2)+g(3)+g(4)+g(5)))g(10)\ ((g(1)+g(2)+g(3)+g(4)+g(5)))g(10) ((g(1)+g(2)+g(3)+g(4)+g(5))) is equal to:

Correct answer: 190

Step-by-step solution →
Q215·MathematicsSingle correctJEE Main 2022
Which of the following statement is a tautology?
  1. (A)((∼q)∧p)∧q((\sim q)\wedge p)\wedge q((∼q)∧p)∧q
  2. (B)((∼q)∧p)∧(p∧(∼p))((\sim q)\wedge p)\wedge(p\wedge(\sim p))((∼q)∧p)∧(p∧(∼p))
  3. (C)((∼q)∧p)∨(p∨(∼p))((\sim q)\wedge p)\vee(p\vee(\sim p))((∼q)∧p)∨(p∨(∼p))
  4. (D)(p∧q)∧(∼(p∧q))(p\wedge q)\wedge(\sim(p\wedge q))(p∧q)∧(∼(p∧q))

Correct answer: (C)

Step-by-step solution →
Q216·MathematicsSingle correctJEE Main 2022
Let f(x)=x−1x+1f(x) = \frac{x-1}{x+1}f(x)=x+1x−1​, x∈R−{0,−1,1}x \in R - \{0, -1, 1\}x∈R−{0,−1,1}. If fn+1(x)=f(fn(x))f^{n+1}(x) = f(f^{n}(x))fn+1(x)=f(fn(x)) for all n∈Nn \in Nn∈N, then f6(6)+f7(7)f^{6}(6) + f^{7}(7)f6(6)+f7(7) is equal to :
  1. (A)76\frac{7}{6}67​
  2. (B)−32-\frac{3}{2}−23​
  3. (C)712\frac{7}{12}127​
  4. (D)−1112-\frac{11}{12}−1211​

Correct answer: (B)

Step-by-step solution →
Q217·MathematicsSingle correctJEE Main 2022
Let f : ℝ → ℝ be defined as f(x) = x-1 and g : ℝ − {1, −1} → ℝ be defined as g(x)=x2x2−1g(x)=\frac{x^2}{x^2-1}g(x)=x2−1x2​. Then the function fog is :
  1. (A)one-one but not onto function
  2. (B)onto but not one-one function
  3. (C)both one-one and onto function
  4. (D)neither one-one nor onto function

Correct answer: (D)

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2022
Let r ∈ {p, q, ~p, ~q} be such that the logical statement r∨(∼p)⇒(p∧q)∨rr\vee(\sim p)\Rightarrow(p\wedge q)\vee rr∨(∼p)⇒(p∧q)∨r is a tautology. Then 'r' is equal to :
  1. (A)p
  2. (B)q
  3. (C)~p
  4. (D)~q

Correct answer: (C)

Step-by-step solution →
Q219·MathematicsSingle correctJEE Main 2022
Let Δ,∇∈{∧,∨}\Delta, \nabla \in \{\wedge, \vee\}Δ,∇∈{∧,∨} be such that p∇q⇒((pΔq)∇r)p \nabla q \Rightarrow ((p \Delta q) \nabla r)p∇q⇒((pΔq)∇r) is a tautology. Then (p∇q)Δr(p \nabla q) \Delta r(p∇q)Δr is logically equivalent to :
  1. (A)(pΔr)∨q(p \Delta r) \vee q(pΔr)∨q
  2. (B)(pΔr)∧q(p \Delta r) \wedge q(pΔr)∧q
  3. (C)(p∧r)Δq(p \wedge r) \Delta q(p∧r)Δq
  4. (D)(p∇r)∧q(p \nabla r) \wedge q(p∇r)∧q

Correct answer: (A)

Step-by-step solution →
Q220·MathematicsSingle correctJEE Main 2022
The negation of the Boolean expression ((∼q)∧p)⇒((∼p)∨q)((\sim q) \wedge p) \Rightarrow ((\sim p) \vee q)((∼q)∧p)⇒((∼p)∨q) is logically equivalent to
  1. (A)p⇒qp \Rightarrow qp⇒q
  2. (B)q⇒pq \Rightarrow pq⇒p
  3. (C)∼(p⇒q)\sim\left(p \Rightarrow q\right)∼(p⇒q)
  4. (D)∼(q⇒p)\sim\left(q \Rightarrow p\right)∼(q⇒p)

Correct answer: (C)

Step-by-step solution →
Q221·MathematicsSingle correctJEE Main 2022
Consider the following two propositions: P1 : ~(p→ ~q) P2 : (p∧ ~ q)∧((~p)∨ q) If the proposition p → ((~p)∨ q) is evaluated as FALSE, then:
  1. (A)P1 is TRUE and P2 is FALSE
  2. (B)P1 is FALSE and P2 is TRUE
  3. (C)Both P1 and P2 are FALSE
  4. (D)Both P1 and P2 are TRUE

Correct answer: (C)

Step-by-step solution →
Q222·MathematicsNumericalJEE Main 2022
Let f : R → R be a function defined by f(x)=(2(1−x252)(2+x25))150f(x) = \left(2\left(1 - \frac{x^{25}}{2}\right)\left(2 + x^{25}\right)\right)^{\frac{1}{50}}f(x)=(2(1−2x25​)(2+x25))501​. If the function g(x)=f(f(f(x)))+f(f(x))g(x) = f\left(f\left(f(x)\right)\right) + f\left(f(x)\right)g(x)=f(f(f(x)))+f(f(x)), the the greatest integer less than or equal to g (1) is ______

Correct answer: 2

Step-by-step solution →
Q223·MathematicsSingle correctJEE Main 2022
Let A={x∈R:∣x+1∣<2}A = \{x \in R : |x+1| < 2\}A={x∈R:∣x+1∣<2} and B={x∈R:∣x−1∣≥2}B = \{x \in R : |x-1| \ge 2\}B={x∈R:∣x−1∣≥2}. Then which one of the following statements is NOT true ?
  1. (A)A−B=(−1,1)A - B = (-1,1)A−B=(−1,1)
  2. (B)B−A=R−(−3,1)B - A = R - (-3,1)B−A=R−(−3,1)
  3. (C)A∩B=(−3,−1]A \cap B = (-3,-1]A∩B=(−3,−1]
  4. (D)A∪B=R−[1,3)A \cup B = R - [1,3)A∪B=R−[1,3)

Correct answer: (B)

Step-by-step solution →
Q224·MathematicsSingle correctJEE Main 2022
The domain of the function f(x)=cos⁡−1(x2−5x+6x2−9)log⁡e(x2−3x+2)f(x)=\dfrac{\cos^{-1}\left(\dfrac{x^{2}-5x+6}{x^{2}-9}\right)}{\log_{e}(x^{2}-3x+2)}f(x)=loge​(x2−3x+2)cos−1(x2−9x2−5x+6​)​ is
  1. (A)(−∞,1)∪(2,∞)(-\infty,1)\cup(2,\infty)(−∞,1)∪(2,∞)
  2. (B)(2,∞)(2,\infty)(2,∞)
  3. (C)[−12,1)∪(2,∞)\left[-\dfrac{1}{2},1\right)\cup(2,\infty)[−21​,1)∪(2,∞)
  4. (D)[−12,1)∪(2,∞)−{3+52,3−52}\left[-\dfrac{1}{2},1\right)\cup(2,\infty)-\left\{\dfrac{3+\sqrt{5}}{2},\dfrac{3-\sqrt{5}}{2}\right\}[−21​,1)∪(2,∞)−{23+5​​,23−5​​}

Correct answer: (D)

Step-by-step solution →
Q225·MathematicsSingle correctJEE Main 2022
The number of choices of Δ∈{∧,∨,⇒,⇔}\Delta\in\{\wedge,\vee,\Rightarrow,\Leftrightarrow\}Δ∈{∧,∨,⇒,⇔}, such that (pΔq)⇒((pΔ∼q)∨((∼p)Δq))(p\Delta q)\Rightarrow((p\Delta\sim q)\vee((\sim p)\Delta q))(pΔq)⇒((pΔ∼q)∨((∼p)Δq)) is a tautology, is
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q226·MathematicsSingle correctJEE Main 2022
Consider the following statements : A : Rishi is a judge. B : Rishi is honest. C : Rishi is not arrogant. The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is
  1. (A)B→(A∨C)B \to (A \vee C)B→(A∨C)
  2. (B)(∼B)∧(A∧C)(\sim B) \wedge (A \wedge C)(∼B)∧(A∧C)
  3. (C)B→((∼A)∨(∼C))B \to ((\sim A) \vee (\sim C))B→((∼A)∨(∼C))
  4. (D)B→(A∧C)B \to (A \wedge C)B→(A∧C)

Correct answer: (B)

Step-by-step solution →
Q227·MathematicsSingle correctJEE Main 2021
Which of the following is equivalent to the Boolean expression p ∧ ~q ?
  1. (A)~ (q → p)
  2. (B)~ p → ~q
  3. (C)~ (p → ~q)
  4. (D)~ (p → q)

Correct answer: (D)

Step-by-step solution →
Q228·MathematicsSingle correctJEE Main 2021
Negation of the statement (p∨r)⇒(q∨r)(p \vee r) \Rightarrow (q \vee r)(p∨r)⇒(q∨r) is :
  1. (A)p∧∼q∧∼rp \wedge \sim q \wedge \sim rp∧∼q∧∼r
  2. (B)∼p∧q∧∼r\sim p \wedge q \wedge \sim r∼p∧q∧∼r
  3. (C)∼p∧q∧r\sim p \wedge q \wedge r∼p∧q∧r
  4. (D)p∧q∧rp \wedge q \wedge rp∧q∧r

Correct answer: (A)

Step-by-step solution →
Q229·MathematicsSingle correctJEE Main 2021
Which of the following is not correct for relation R on the set of real numbers ?
  1. (A)(x,y)∈R⇔0<∣x∣−∣y∣≤1(x, y) \in R \Leftrightarrow 0 < |x| - |y| \le 1(x,y)∈R⇔0<∣x∣−∣y∣≤1 is neither transitive nor symmetric.
  2. (B)(x,y)∈R⇔0<∣x−y∣≤1(x, y) \in R \Leftrightarrow 0 < |x - y| \le 1(x,y)∈R⇔0<∣x−y∣≤1 is symmetric and transitive.
  3. (C)(x,y)∈R⇔∣x∣−∣y∣≤1(x, y) \in R \Leftrightarrow |x| - |y| \le 1(x,y)∈R⇔∣x∣−∣y∣≤1 is reflexive but not symmetric.
  4. (D)(x,y)∈R⇔∣x−y∣≤1(x, y) \in R \Leftrightarrow |x - y| \le 1(x,y)∈R⇔∣x−y∣≤1 is reflexive and symmetric.

Correct answer: (B)

Step-by-step solution →
Q230·MathematicsSingle correctJEE Main 2021
Let ∗,□∈{∧,∨}*, \square \in \{\wedge, \vee\}∗,□∈{∧,∨} be such that the Boolean expression (p∗∼q)⇒(p□q)(p * \sim q) \Rightarrow (p \square q)(p∗∼q)⇒(p□q) is a tautology. Then :
  1. (A)∗=∨, □=∨* = \vee,\ \square = \vee∗=∨, □=∨
  2. (B)∗=∧, □=∧* = \wedge,\ \square = \wedge∗=∧, □=∧
  3. (C)∗=∧, □=∨* = \wedge,\ \square = \vee∗=∧, □=∨
  4. (D)∗=∨, □=∧* = \vee,\ \square = \wedge∗=∨, □=∧

Correct answer: (C)

Step-by-step solution →
Q231·MathematicsSingle correctJEE Main 2021
Let f:N→Nf : \mathbf{N} \to \mathbf{N}f:N→N be a function such that f(m+n)=f(m)+f(n)f(m + n) = f(m) + f(n)f(m+n)=f(m)+f(n) for every m,n∈Nm, n \in \mathbf{N}m,n∈N. If f(6)=18f(6) = 18f(6)=18, then f(2)⋅f(3)f(2) \cdot f(3)f(2)⋅f(3) is equal to :
  1. (A)6
  2. (B)54
  3. (C)18
  4. (D)36

Correct answer: (B)

Step-by-step solution →
Q232·MathematicsSingle correctJEE Main 2021
The statement (p∧(p→q)∧(q→r))→r(p \wedge (p \to q) \wedge (q \to r)) \to r(p∧(p→q)∧(q→r))→r is :
  1. (A)a tautology
  2. (B)equivalent to p→∼rp \to \sim rp→∼r
  3. (C)a fallacy
  4. (D)equivalent to q→∼rq \to \sim rq→∼r

Correct answer: (A)

Step-by-step solution →
Q233·MathematicsSingle correctJEE Main 2021
The Boolean expression (p∧q)⇒((r∧q)∧p)(p \wedge q) \Rightarrow ((r \wedge q) \wedge p)(p∧q)⇒((r∧q)∧p) is equivalent to :
  1. (A)(p∧q)⇒(r∧q)(p \wedge q) \Rightarrow (r \wedge q)(p∧q)⇒(r∧q)
  2. (B)(q∧r)⇒(p∧q)(q \wedge r) \Rightarrow (p \wedge q)(q∧r)⇒(p∧q)
  3. (C)(p∧q)⇒(r∨q)(p \wedge q) \Rightarrow (r \vee q)(p∧q)⇒(r∨q)
  4. (D)(p∧r)⇒(p∧q)(p \wedge r) \Rightarrow (p \wedge q)(p∧r)⇒(p∧q)

Correct answer: (A)

Step-by-step solution →
Q234·MathematicsNumericalJEE Main 2021
If A={x∈R:∣x−2∣>1}A = \{x \in \mathbf{R} : |x - 2| > 1\}A={x∈R:∣x−2∣>1}, B={x∈R:x2−3>1}B = \left\{x \in \mathbf{R} : \sqrt{x^{2} - 3} > 1\right\}B={x∈R:x2−3​>1}, C={x∈R:∣x−4∣≥2}C = \left\{x \in \mathbf{R} : |x - 4| \geq 2\right\}C={x∈R:∣x−4∣≥2} and Z\mathbf{Z}Z is the set of all integers, then the number of subsets of the set (A∩B∩C)c∩Z(A \cap B \cap C)^{c} \cap \mathbf{Z}(A∩B∩C)c∩Z is ___________.

Correct answer: 256

Step-by-step solution →
Q235·MathematicsSingle correctJEE Main 2021
Let ℤ be the set of all integers, A = {(x, y) ∈ ℤ × ℤ : (x−2)2+y2≤4(x - 2)^{2} + y^{2} \le 4(x−2)2+y2≤4}, B = {(x, y) ∈ ℤ × ℤ : x2+y2≤4x^{2} + y^{2} \le 4x2+y2≤4} and C = {(x, y) ∈ ℤ × ℤ : (x−2)2+(y−2)2≤4(x - 2)^{2} + (y - 2)^{2} \le 4(x−2)2+(y−2)2≤4} If the total number of relation from A ∩ B to A ∩ C is 2p2^{p}2p, then the value of p is :
  1. (A)16
  2. (B)25
  3. (C)49
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q236·MathematicsSingle correctJEE Main 2021
Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set :
  1. (A){80,83,86,89}\{80, 83, 86, 89\}{80,83,86,89}
  2. (B){84,86,88,90}\{84, 86, 88, 90\}{84,86,88,90}
  3. (C){79,81,83,85}\{79, 81, 83, 85\}{79,81,83,85}
  4. (D){84,87,90,93}\{84, 87, 90, 93\}{84,87,90,93}

Correct answer: (C)

Step-by-step solution →
Q237·MathematicsSingle correctJEE Main 2021
Consider the two statements : (S1):(p→q)∨(∼q→p)(S1) : (p \rightarrow q) \vee (\sim q \rightarrow p)(S1):(p→q)∨(∼q→p) is a tautology. (S2):(p∧∼q)∧(∼p∨q)(S2) : (p \wedge \sim q) \wedge (\sim p \vee q)(S2):(p∧∼q)∧(∼p∨q) is a fallacy. Then :
  1. (A)only (S1) is true.
  2. (B)both (S1) and (S2) are false.
  3. (C)both (S1) and (S2) are true.
  4. (D)only (S2) is true.

Correct answer: (C)

Step-by-step solution →
Q238·MathematicsSingle correctJEE Main 2021
If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(∼r))→(p∧q)\left((p \vee q) \wedge (q \to r) \wedge (\sim r)\right) \to (p \wedge q)((p∨q)∧(q→r)∧(∼r))→(p∧q) is false, then the truth values of the statements p, q, r respectively can be:
  1. (A)T F T
  2. (B)F F T
  3. (C)T F F
  4. (D)F T F

Correct answer: (C)

Step-by-step solution →
Q239·MathematicsSingle correctJEE Main 2021
Let N be the set of natural numbers and a relation R on N be defined by R = {(x,y)∈N×N:x3−3x2y−xy2+3y3=0}\left\{ (x, y) \in \mathbb{N} \times \mathbb{N} : x^3 - 3x^2 y - xy^2 + 3y^3 = 0 \right\}{(x,y)∈N×N:x3−3x2y−xy2+3y3=0}. Then the relation R is :
  1. (A)reflexive but neither symmetric nor transitive
  2. (B)an equivalence relation
  3. (C)symmetric but neither reflexive nor transitive
  4. (D)reflexive and symmetric, but not transitive

Correct answer: (A)

Step-by-step solution →
Q240·MathematicsSingle correctJEE Main 2021
The compound statement (P∨Q)∧(∼P)⇒Q\left(P \vee Q\right) \wedge \left(\sim P\right) \Rightarrow Q(P∨Q)∧(∼P)⇒Q is equivalent to :
  1. (A)P∨QP \vee QP∨Q
  2. (B)P∧∼QP \wedge \sim QP∧∼Q
  3. (C)∼(P⇒Q)⇔P∧∼Q\sim\left(P \Rightarrow Q\right) \Leftrightarrow P \wedge \sim Q∼(P⇒Q)⇔P∧∼Q
  4. (D)∼(P⇒Q)\sim\left(P \Rightarrow Q\right)∼(P⇒Q)

Correct answer: (C)

Step-by-step solution →
Q241·MathematicsSingle correctJEE Main 2021
Which of the following is the negation of the statement "for all M > 0, there exists x ∈ S such that x ≥ M"?
  1. (A)there exists M > 0, there exists x ∈ S such that x < M
  2. (B)there exists M > 0, such that x ≥ M for all x ∈ S
  3. (C)there exists M > 0, there exists x ∈ S such that x ≥ M
  4. (D)there exists M > 0, such that x < M for all x ∈ S

Correct answer: (D)

Step-by-step solution →
Q242·MathematicsNumericalJEE Main 2021
Let A = {n ∈ N|n2^22 ≤ n + 10,000}, B = {3k + 1|k ∈ N} and C = {2k|k ∈ N} , then the sum of all the elements of the set A ∩ (B − C) is equal to……

Correct answer: 832

Step-by-step solution →
Q243·MathematicsNumericalJEE Main 2021
Let S={1,2,3,4,5,6,7}S=\left\{1,2,3,4,5,6,7\right\}S={1,2,3,4,5,6,7}. Then the number of possible function f:S→Sf:S \to Sf:S→S such that f(m⋅n)=f(m)⋅f(n)f\left(m \cdot n\right)=f\left(m\right) \cdot f\left(n\right)f(m⋅n)=f(m)⋅f(n) for every m,n∈Sm,n \in Sm,n∈S and m⋅n∈Sm \cdot n \in Sm⋅n∈S is equal to ____.

Correct answer: 490

Step-by-step solution →
Q244·MathematicsSingle correctJEE Main 2021
Consider functions f:A→Bf : A \to Bf:A→B and g:B→C(A,B,C⊆R)g : B \to C\left(A,B,C \subseteq R\right)g:B→C(A,B,C⊆R) such that (gof)−1\left(gof\right)^{-1}(gof)−1 exists, then :
  1. (A)f and g both are onto
  2. (B)f is onto and g is one-one
  3. (C)f is one-one and g is onto
  4. (D)f and g both are one-one

Correct answer: (C)

Step-by-step solution →
Q245·MathematicsSingle correctJEE Main 2021
Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following :
  1. (A)The match will not be played or weather is good and ground is not wet.
  2. (B)The match will not be played or weather is not good and ground is wet.
  3. (C)If the match will not be played, then either weather is not good or ground is wet.
  4. (D)The match will be played and weather is not good or ground is wet.

Correct answer: (D)

Step-by-step solution →
Q246·MathematicsSingle correctJEE Main 2021
Let g:N→Ng:N \rightarrow Ng:N→N be defined as g(3n+1)=3n+2,g(3n+1)=3n+2,g(3n+1)=3n+2, g(3n+2)=3n+3,g(3n+2)=3n+3,g(3n+2)=3n+3, g(3n+3)=3n+1g(3n+3)=3n+1g(3n+3)=3n+1, for all n ≥ 0 Then which of the following statements is true?
  1. (A)There exists a function f:N→Nf:N \rightarrow Nf:N→N such that gof=fgof=fgof=f
  2. (B)gogog = g
  3. (C)There exists a one-one function f:N→Nf:N \rightarrow Nf:N→N such that fog=ffog=ffog=f
  4. (D)There exists on onto function f:N→Nf:N \rightarrow Nf:N→N such that fog=ffog=ffog=f

Correct answer: (D)

Step-by-step solution →
Q247·MathematicsSingle correctJEE Main 2021
The Boolean expression (p⇒q)∧(q⇒∼p)(p \Rightarrow q) \wedge (q \Rightarrow \sim p)(p⇒q)∧(q⇒∼p) is equivalent to:
  1. (A)q
  2. (B)p
  3. (C)~ p
  4. (D)~ q

Correct answer: (C)

Step-by-step solution →
Q248·MathematicsSingle correctJEE Main 2021
Let [x][x][x] denote the greatest integer less than or equal to x. Then, the values of x∈Rx \in Rx∈R satisfying the equation [ex]2+[ex+1]−3=0\left[e^x\right]^2 + \left[e^x + 1\right] - 3 = 0[ex]2+[ex+1]−3=0 lie in the interval :
  1. (A)[1,e)[1, e)[1,e)
  2. (B)[log⁡e2,log⁡e3)[\log_e 2, \log_e 3)[loge​2,loge​3)
  3. (C)[0,log⁡e2)[0, \log_e 2)[0,loge​2)
  4. (D)[0,1/ e)[0, 1/\,e)[0,1/e)

Correct answer: (C)

Step-by-step solution →
Q249·MathematicsNumericalJEE Main 2021
The sum of all the elements in the set {n∈{1,2,.......,100}|H.C.F. of n and 2040 is 1}\left\{n \in \{1,2,.......,100\} \middle| \text{H.C.F. of n and 2040 is 1}\right\}{n∈{1,2,.......,100}∣H.C.F. of n and 2040 is 1} is equal to...........

Correct answer: 1251

Step-by-step solution →
Q250·MathematicsSingle correctJEE Main 2021
Which of the following Boolean expressions is not a tautology?
  1. (A)(∼p⇒q)∨(∼q⇒p)(\sim p \Rightarrow q) \vee (\sim q \Rightarrow p)(∼p⇒q)∨(∼q⇒p)
  2. (B)(q⇒p)∨(∼q⇒p)(q \Rightarrow p) \vee (\sim q \Rightarrow p)(q⇒p)∨(∼q⇒p)
  3. (C)(p⇒∼q)∨(∼q⇒p)(p \Rightarrow \sim q) \vee (\sim q \Rightarrow p)(p⇒∼q)∨(∼q⇒p)
  4. (D)(p⇒q)∨(∼q⇒p)(p \Rightarrow q) \vee (\sim q \Rightarrow p)(p⇒q)∨(∼q⇒p)

Correct answer: (A)

Step-by-step solution →
Q251·MathematicsSingle correctJEE Main 2021
Let f:R−{α6}→Rf:R-\left\{\frac{\alpha}{6}\right\}\to Rf:R−{6α​}→R be defined by f(x)=5x+36x−αf(x)=\frac{5x+3}{6x-\alpha}f(x)=6x−α5x+3​. Then the value of α\alphaα for which (fof)(x)=x,(fof)(x)=x,(fof)(x)=x, for all x∈R−{α6}x\in R-\left\{\frac{\alpha}{6}\right\}x∈R−{6α​}, is :
  1. (A)666
  2. (B)888
  3. (C)No such α\alphaα exists
  4. (D)555

Correct answer: (D)

Step-by-step solution →
Q252·MathematicsSingle correctJEE Main 2021
Consider the following three statements : (A) if 3 + 3 = 7 then 4 + 3 = 8. (B) If 5 + 3 = 8 then earth is flat. (C) If both (A) and (B) are true then 5 + 6 = 17. Then, which of the following statements is correct?
  1. (A)(A) and (C) are true while (B) is false
  2. (B)(A) and (B) are false while (C) is true
  3. (C)(A) is true while (B) and (C) are false
  4. (D)(A) is false, but (B) and (C) are true

Correct answer: (A)

Step-by-step solution →
Q253·MathematicsSingle correctJEE Main 2021
Let [x] denote the greatest integer ≤x\leq x≤x, where x∈Rx \in Rx∈R. If the domain of the real valued function f(x)=∣[x]∣−2∣[x]∣−3f(x) = \sqrt{\frac{\big|[x]\big|-2}{\big|[x]\big|-3}}f(x)=​[x]​−3​[x]​−2​​ is (−∞,a)∪[b,c)∪[4,∞)(-\infty, a) \cup [b, c) \cup [4, \infty)(−∞,a)∪[b,c)∪[4,∞), a<b<ca < b < ca<b<c, then the value of a+b+ca+b+ca+b+c is:
  1. (A)-3
  2. (B)8
  3. (C)-2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q254·MathematicsSingle correctJEE Main 2021
The Boolean expression (p∧∼q)⇒(q∨∼p)(p \wedge \sim q) \Rightarrow (q \vee \sim p)(p∧∼q)⇒(q∨∼p) is equivalent to :
  1. (A)q⇒pq \Rightarrow pq⇒p
  2. (B)∼q⇒p\sim q \Rightarrow p∼q⇒p
  3. (C)p⇒qp \Rightarrow qp⇒q
  4. (D)p⇒∼qp \Rightarrow \sim qp⇒∼q

Correct answer: (C)

Step-by-step solution →
Q255·MathematicsSingle correctJEE Main 2021
The real valued function f(x)=cosec−1xx−[x]f\left(x\right) = \dfrac{\mathrm{cosec}^{-1}x}{\sqrt{x - \left[x\right]}}f(x)=x−[x]​cosec−1x​, where [x][x][x] denotes the greatest integer less than or equal to x, is defined for all x belonging to :
  1. (A)all reals except integers
  2. (B)all non-integers except the interval [−1,1][-1,1][−1,1]
  3. (C)all integers except 0,−1,10, -1, 10,−1,1
  4. (D)all reals except the Interval [−1,1][-1,1][−1,1]

Correct answer: (B)

Step-by-step solution →
Q256·MathematicsSingle correctJEE Main 2021
If P and Q are two statements, then which of the following compound statement is a tautology ?
  1. (A)((P ⇒ Q) ∧ ~ Q) ⇒ Q
  2. (B)((P ⇒ Q) ∧ ~ Q) ⇒ ~ P
  3. (C)((P ⇒ Q) ∧ ~ Q) ⇒ P
  4. (D)((P ⇒ Q) ∧ ~ Q) ⇒ (P ∧ Q)

Correct answer: (B)

Step-by-step solution →
Q257·MathematicsSingle correctJEE Main 2021
Define a relation R over a class of n × n real matrices A and B as "ARB iff there exists a non-singular matrix P such that PAP−1=BPAP^{-1} = BPAP−1=B". Then which of the following is true ?
  1. (A)R is symmetric, transitive but not reflexive,
  2. (B)R is reflexive, symmetric but not transitive
  3. (C)R is an equivalence relation
  4. (D)R is reflexive, transitive but not symmetric

Correct answer: (C)

Step-by-step solution →
Q258·MathematicsSingle correctJEE Main 2021
Let f : R − {3} → R − {1} be defined by f(x)=x−2x−3f(x) = \frac{x-2}{x-3}f(x)=x−3x−2​. Let g : R → R be given as g(x) = 2x − 3. Then, the sum of all the values of x for which f−1(x)+g−1(x)=132f^{-1}(x) + g^{-1}(x) = \frac{13}{2}f−1(x)+g−1(x)=213​ is equal to
  1. (A)7
  2. (B)2
  3. (C)5
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q259·MathematicsSingle correctJEE Main 2021
If the functions are defined as f(x)=xf\left(x\right) = \sqrt{x}f(x)=x​ and g(x)=1−xg\left(x\right) = \sqrt{1 - x}g(x)=1−x​, then what is the common domain of the following functions : f+gf + gf+g, f−gf - gf−g, f/gf/gf/g, g/fg/fg/f, g−fg - fg−f where (f±g)(x)=f(x)±g(x)(f \pm g)(x) = f(x) \pm g(x)(f±g)(x)=f(x)±g(x), (f/g)(x)=f(x)g(x)(f/g)(x) = \dfrac{f\left(x\right)}{g\left(x\right)}(f/g)(x)=g(x)f(x)​
  1. (A)0≤x≤10 \leq x \leq 10≤x≤1
  2. (B)0≤x<10 \leq x < 10≤x<1
  3. (C)0<x<10 < x < 10<x<1
  4. (D)0<x≤10 < x \leq 10<x≤1

Correct answer: (C)

Step-by-step solution →
Q260·MathematicsSingle correctJEE Main 2021
If the Boolean expression (p∧q)⊛(p⊗q)(p \wedge q) \circledast (p \otimes q)(p∧q)⊛(p⊗q) is a tautology, then ⊛\circledast⊛ and ⊗\otimes⊗ are respectively given by
  1. (A)→,→\rightarrow, \rightarrow→,→
  2. (B)∧,∨\wedge, \vee∧,∨
  3. (C)∨,→\vee, \rightarrow∨,→
  4. (D)∧,→\wedge, \rightarrow∧,→

Correct answer: (A)

Step-by-step solution →
Q261·MathematicsSingle correctJEE Main 2021
If the Boolean expression (p⇒q)⇔(q∗(∼p))(p \Rightarrow q) \Leftrightarrow (q * (\sim p))(p⇒q)⇔(q∗(∼p)) is a tautology, then the Boolean expression p∗(∼q)p * (\sim q)p∗(∼q) is equivalent to :
  1. (A)q⇒pq \Rightarrow pq⇒p
  2. (B)∼q⇒p\sim q \Rightarrow p∼q⇒p
  3. (C)p⇒∼qp \Rightarrow \sim qp⇒∼q
  4. (D)p⇒qp \Rightarrow qp⇒q

Correct answer: (A)

Step-by-step solution →
Q262·MathematicsSingle correctJEE Main 2021
In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?
  1. (A)P and Q
  2. (B)P and R
  3. (C)None of these
  4. (D)Q and R

Correct answer: (C)

Step-by-step solution →
Q263·MathematicsSingle correctJEE Main 2021
Which of the following Boolean expression is a tautology ?
  1. (A)(p ∧ q) ∨ (p ∨ q)
  2. (B)(p ∧ q) ∨ (p → q)
  3. (C)(p ∧ q) ∧ (p → q)
  4. (D)(p ∧ q) → (p → q)

Correct answer: (D)

Step-by-step solution →
Q264·MathematicsSingle correctJEE Main 2021
Let A={2,3,4,5,....,30}A = \{2, 3, 4, 5, ...., 30\}A={2,3,4,5,....,30} and '≃\simeq≃' be an equivalence relation on A×AA \times AA×A, defined by (a,b)≃(c,d)(a, b) \simeq (c, d)(a,b)≃(c,d), if and only if ad=bcad = bcad=bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :
  1. (A)555
  2. (B)666
  3. (C)888
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q265·MathematicsSingle correctJEE Main 2021
Let F1F_{1}F1​(A, B, C) = (A ∧ ~B) ∨ [~C ∧ (A ∨ B)] ∨ ~A and F2F_{2}F2​(A, B) = (A ∨ B) ∨ (B → ~A) be two logical expressions. Then :
  1. (A)F1F_{1}F1​ is not a tautology but F2F_{2}F2​ is a tautology
  2. (B)F1F_{1}F1​ is a tautology but F2F_{2}F2​ is not a tautology
  3. (C)F1F_{1}F1​ and F2F_{2}F2​ both area tautologies
  4. (D)Both F1F_{1}F1​ and F2F_{2}F2​ are not tautologies

Correct answer: (A)

Step-by-step solution →
Q266·MathematicsSingle correctJEE Main 2021
Let A = {1,2,3……,10} and f: A→ A be defined as f(k)={k+1if k is oddkif k is evenf(k) = \begin{cases} k + 1 & \text{if k is odd} \\ k & \text{if k is even} \end{cases}f(k)={k+1k​if k is oddif k is even​ Then the number of possible functions g : A→A such that gof = f is :
  1. (A)10510^{5}105
  2. (B)10C5^{10}C_{5}10C5​
  3. (C)555^{5}55
  4. (D)5!

Correct answer: (A)

Step-by-step solution →
Q267·MathematicsSingle correctJEE Main 2021
Let f(x)=sin⁡−1xf(x) = \sin^{-1} xf(x)=sin−1x and g(x)=x2−x−22x2−x−6g(x) = \frac{x^{2}-x-2}{2x^{2}-x-6}g(x)=2x2−x−6x2−x−2​. If g(2)=lim⁡x→2g(x)g(2) = \lim_{x \to 2} g(x)g(2)=limx→2​g(x), then the domain of the function fog is :
  1. (A)(−∞,−2]∪[−43,∞)(-\infty, -2] \cup \left[-\frac{4}{3}, \infty\right)(−∞,−2]∪[−34​,∞)
  2. (B)(−∞,−1]∪[2,∞)(-\infty, -1] \cup [2, \infty)(−∞,−1]∪[2,∞)
  3. (C)(−∞,−2]∪[−1,∞)(-\infty, -2] \cup [-1, \infty)(−∞,−2]∪[−1,∞)
  4. (D)(−∞,−2]∪[−32,∞)(-\infty, -2] \cup \left[-\frac{3}{2}, \infty\right)(−∞,−2]∪[−23​,∞)

Correct answer: (A)

Step-by-step solution →
Q268·MathematicsSingle correctJEE Main 2021
Let R = {P,Q)|P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1,−1) is the set:
  1. (A)S={(x,y) ∣ x2+y2=1}S=\{(x,y)\,|\,x^{2}+y^{2}=1\}S={(x,y)∣x2+y2=1}
  2. (B)S={(x,y) ∣ x2+y2=4}S=\{(x,y)\,|\,x^{2}+y^{2}=4\}S={(x,y)∣x2+y2=4}
  3. (C)S={(x,y) ∣ x2+y2=2}S=\{(x,y)\,|\,x^{2}+y^{2}=\sqrt{2}\}S={(x,y)∣x2+y2=2​}
  4. (D)S={(x,y) ∣ x2+y2=2}S=\{(x,y)\,|\,x^{2}+y^{2}=2\}S={(x,y)∣x2+y2=2}

Correct answer: (D)

Step-by-step solution →
Q269·MathematicsSingle correctJEE Main 2021
Let f, g: N→NN \rightarrow NN→N such that f(n+1)=f(n)+f(1)f(n + 1) = f(n) + f(1)f(n+1)=f(n)+f(1) ∀n∈N\forall n \in N∀n∈N and g be any arbitrary function. Which of the following statements is NOT true ?
  1. (A)f is one-one
  2. (B)If fog is one-one, then g is one-one
  3. (C)If g is onto, then fog is one-one
  4. (D)If f is onto, then f(n)=n ∀n∈Nf(n) = n\ \forall n \in Nf(n)=n ∀n∈N

Correct answer: (C)

Step-by-step solution →
Q270·MathematicsSingle correctJEE Main 2021
The contrapositive of the statement “If you will work, you will earn money” is:
  1. (A)If you will not earn money, you will not work
  2. (B)You will earn money, if you will not work
  3. (C)If you will earn money, you will work
  4. (D)To earn money, you need to work

Correct answer: (A)

Step-by-step solution →
Q271·MathematicsSingle correctJEE Main 2021
A function f(x) is given by f(x) = 5x_{x}x​5+x^{x}x5, then the sum of the series f (201 ) + f (202 ) + f (203 ) + ...... + f (3920 ) is equal to:
  1. (A)192
  2. (B)492
  3. (C)392
  4. (D)292

Correct answer: (C)

Step-by-step solution →
Q272·MathematicsSingle correctJEE Main 2021
The statement A→(B→A)A \rightarrow (B \rightarrow A)A→(B→A) is equivalent to:
  1. (A)A→(A∧B)A \rightarrow (A \wedge B)A→(A∧B)
  2. (B)A→(A∨B)A \rightarrow (A \vee B)A→(A∨B)
  3. (C)A→(A→B)A \rightarrow (A \rightarrow B)A→(A→B)
  4. (D)A→(A↔B)A \rightarrow (A \leftrightarrow B)A→(A↔B)

Correct answer: (B)

Step-by-step solution →
Q273·MathematicsSingle correctJEE Main 2021
Let f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)=2x−1f(x) = 2x-1f(x)=2x−1 and g:R−{1}→Rg : R - \{1\} \rightarrow Rg:R−{1}→R be defined as g(x)=x−12x−1g(x) = \frac{x - \frac{1}{2}}{x - 1}g(x)=x−1x−21​​. Then the composition function f(g(x))f(g(x))f(g(x)) is :
  1. (A)both one-one and onto
  2. (B)onto but not one-one
  3. (C)neither one-one nor onto
  4. (D)one-one but not onto

Correct answer: (D)

Step-by-step solution →
Q274·MathematicsSingle correctJEE Main 2021
The statement among the following that is a tautology is:
  1. (A)A∧(A∨B)A \wedge (A \vee B)A∧(A∨B)
  2. (B)B→[A∧(A→B)]B \rightarrow \left[A \wedge (A \rightarrow B)\right]B→[A∧(A→B)]
  3. (C)A∨(A∧B)A \vee (A \wedge B)A∨(A∧B)
  4. (D)[A∧(A→B)]→B\left[A \wedge (A \rightarrow B)\right] \rightarrow B[A∧(A→B)]→B

Correct answer: (D)

Step-by-step solution →
Q275·MathematicsSingle correctJEE Main 2021
The negation of the statement ∼p∧(p∨q)\sim p\wedge(p\vee q)∼p∧(p∨q) is :
  1. (A)∼p∧q\sim p\wedge q∼p∧q
  2. (B)p∧∼qp\wedge\sim qp∧∼q
  3. (C)∼p∨q\sim p\vee q∼p∨q
  4. (D)p∨∼qp\vee\sim qp∨∼q

Correct answer: (D)

Step-by-step solution →
Q276·MathematicsNumericalJEE Main 2021
If a+α=1,b+β=2a+\alpha=1, b+\beta=2a+α=1,b+β=2 and af(x)+αf(1x)=bx+βx,x≠0,af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x}, x\neq 0,af(x)+αf(x1​)=bx+xβ​,x=0, then the value of the expression f(x)+f(1x)x+1x\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}x+x1​f(x)+f(x1​)​ is__________.

Correct answer: 2

Step-by-step solution →
Q277·MathematicsNumericalJEE Advanced 2020
Let the function f:[0,1]→Rf : [0,1] \to \mathbb{R}f:[0,1]→R be defined by f(x)=4x4x+2f(x) = \frac{4^{x}}{4^{x} + 2}f(x)=4x+24x​ Then the value of f(140)+f(240)+f(340)+⋯+f(3940)−f(12)f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \cdots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right)f(401​)+f(402​)+f(403​)+⋯+f(4039​)−f(21​) is ________

Correct answer: 19.00

Step-by-step solution →
Q278·MathematicsSingle correctJEE Advanced 2020
If the function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is defined by f(x)=∣x∣(x−sin⁡x)f(x) = |x|(x - \sin x)f(x)=∣x∣(x−sinx), then which of the following statements is TRUE ?
  1. (A)fff is one-one, but NOT onto
  2. (B)fff is onto, but NOT one-one
  3. (C)fff is BOTH one-one and onto
  4. (D)fff is NEITHER one-one NOR onto

Correct answer: (C)

Step-by-step solution →
Q279·MathematicsSingle correctJEE Main 2020
For a suitably chosen real constant a, let a function, f:R−{−a}→Rf : R - \{-a\} \to Rf:R−{−a}→R be defined by f(x)=a−xa+xf(x) = \frac{a - x}{a + x}f(x)=a+xa−x​. Further suppose that for any real number x≠−ax \neq -ax=−a and f(x)≠−af(x) \neq -af(x)=−a, (fof)(x)=x(fof)(x) = x(fof)(x)=x. Then f(−12)f\left(-\frac{1}{2}\right)f(−21​) is equal to:
  1. (A)13\frac{1}{3}31​
  2. (B)−13-\frac{1}{3}−31​
  3. (C)−3-3−3
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q280·MathematicsNumericalJEE Main 2020
Set A has m elements and Set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m.n is ____.

Correct answer: 28

Step-by-step solution →
Q281·MathematicsNumericalJEE Main 2020
Suppose that a function f:R→Rf : R \to Rf:R→R satisfies f(x+y)=f(x)f(y)f(x + y) = f(x)f(y)f(x+y)=f(x)f(y) for all x,y∈Rx, y \in Rx,y∈R and f(1)=3f(1) = 3f(1)=3. If ∑i=1nf(i)=363\sum_{i=1}^{n} f(i) = 363∑i=1n​f(i)=363, then n is equal to __________.

Correct answer: 5

Step-by-step solution →
Q282·MathematicsSingle correctJEE Main 2020
A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be:
  1. (A)63
  2. (B)54
  3. (C)38
  4. (D)36

Correct answer: (D)

Step-by-step solution →
Q283·MathematicsNumericalJEE Main 2020
Let A=(a,b,c)A = (a, b, c)A=(a,b,c) and B=(1,2,3,4)B = (1, 2, 3, 4)B=(1,2,3,4). Then the number of elements in the set C={f:A→B∣2∈f(A) and f is not one-one}C = \{f : A \to B \mid 2 \in f(A) \text{ and } f \text{ is not one-one}\}C={f:A→B∣2∈f(A) and f is not one-one} is __________.

Correct answer: 19.00

Step-by-step solution →
Q284·MathematicsSingle correctJEE Main 2020
A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. if x% of the people read both the newspapers, then a possible value of x can be:
  1. (A)29
  2. (B)55
  3. (C)37
  4. (D)65

Correct answer: (B)

Step-by-step solution →
Q285·MathematicsSingle correctJEE Main 2020
Let [t] denote the greatest integer ≤t\le t≤t. Then the equation in x, [x]2+2[x+2]−7=0[x]^{2} + 2[x+2] - 7 = 0[x]2+2[x+2]−7=0 has:
  1. (A)infinitely many solutions
  2. (B)exactly two solutions
  3. (C)no integral solution
  4. (D)exactly four integral solutions

Correct answer: (A)

Step-by-step solution →
Q286·MathematicsSingle correctJEE Main 2020
Let ⋃i=150Xi=⋃i=1nYi=T\bigcup_{i=1}^{50} X_{i} = \bigcup_{i=1}^{n} Y_{i} = T⋃i=150​Xi​=⋃i=1n​Yi​=T, where each XiX_{i}Xi​ contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets XiX_{i}Xi​'s and exactly 6 of sets YiY_{i}Yi​'s then n is equal to:
  1. (A)15
  2. (B)30
  3. (C)45
  4. (D)50

Correct answer: (B)

Step-by-step solution →
Q287·MathematicsSingle correctJEE Main 2020
Consider the two sets: A = {m ∈ R : both the roots of x2−(m+1)x+m+4=0x^{2}-(m+1)x+m+4=0x2−(m+1)x+m+4=0 are real} and B = [−3, 5). Which of the following is not true?
  1. (A)A∩B={−3}A\cap B=\{-3\}A∩B={−3}
  2. (B)B−A=(−3, 5)B-A=(-3,\ 5)B−A=(−3, 5)
  3. (C)A−B=(−∞, −3)∪(5, ∞)A-B=(-\infty,\ -3)\cup(5,\ \infty)A−B=(−∞, −3)∪(5, ∞)
  4. (D)A∪B=RA\cup B=RA∪B=R

Correct answer: (C)

Step-by-step solution →
Q288·MathematicsSingle correctJEE Main 2020
Let R1R_1R1​ and R2R_2R2​ be two relations defined as follows: R1={(a,b)∈R2:a2+b2∈Q}R_1 = \{(a,b) \in R^2 : a^2 + b^2 \in Q\}R1​={(a,b)∈R2:a2+b2∈Q} and R2={(a,b)∈R2:a2+b2∉Q}R_2 = \{(a,b) \in R^2 : a^2 + b^2 \notin Q\}R2​={(a,b)∈R2:a2+b2∈/Q}, where QQQ is the set of all rational numbers. Then:
  1. (A)Neither R1R_1R1​ nor R2R_2R2​ is transitive.
  2. (B)R1R_1R1​ is transitive but R2R_2R2​ is not transitive.
  3. (C)R1R_1R1​ and R2R_2R2​ are both transitive
  4. (D)R2R_2R2​ is transitive but R1R_1R1​ is not transitive.

Correct answer: (A)

Step-by-step solution →
Q289·MathematicsSingle correctJEE Main 2020
If A={x∈R:∣x∣<2}A = \{x \in R : |x| < 2\}A={x∈R:∣x∣<2} and B={x∈R:∣x−2∣≥3}B = \{x \in R : |x - 2| \ge 3\}B={x∈R:∣x−2∣≥3}; then:
  1. (A)B−A=R−(−2,5)B - A = R - (-2, 5)B−A=R−(−2,5)
  2. (B)A∩B=(−2,−1)A \cap B = (-2, -1)A∩B=(−2,−1)
  3. (C)A−B=[−1,2)A - B = [-1, 2)A−B=[−1,2)
  4. (D)A∪B=R−(2,5)A \cup B = R - (2, 5)A∪B=R−(2,5)

Correct answer: (A)

Step-by-step solution →
Q290·MathematicsSingle correctJEE Main 2020
Let f:(1,3)→Rf:(1,3)\to Rf:(1,3)→R be a function defined by f(x)=x[x]1+x2f(x)=\dfrac{x[x]}{1+x^{2}}f(x)=1+x2x[x]​, where [x] denotes the greatest integer ≤x\le x≤x. Then the range of f is:
  1. (A)(25,45]\left(\dfrac{2}{5},\dfrac{4}{5}\right](52​,54​]
  2. (B)(25,12)∪(35,45]\left(\dfrac{2}{5},\dfrac{1}{2}\right)\cup\left(\dfrac{3}{5},\dfrac{4}{5}\right](52​,21​)∪(53​,54​]
  3. (C)(35,45)\left(\dfrac{3}{5},\dfrac{4}{5}\right)(53​,54​)
  4. (D)(25,35]∪(34,45)\left(\dfrac{2}{5},\dfrac{3}{5}\right]\cup\left(\dfrac{3}{4},\dfrac{4}{5}\right)(52​,53​]∪(43​,54​)

Correct answer: (B)

Step-by-step solution →
Q291·MathematicsSingle correctJEE Main 2020
The inverse function of f(x)=82x−8−2x82x+8−2x,x∈(−1,1)f(x)=\dfrac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}, x\in(-1,1)f(x)=82x+8−2x82x−8−2x​,x∈(−1,1), is
  1. (A)14(log⁡8e)log⁡e(1−x1+x)\dfrac{1}{4}(\log_{8}e)\log_{e}\left(\dfrac{1-x}{1+x}\right)41​(log8​e)loge​(1+x1−x​)
  2. (B)14log⁡e(1−x1+x)\dfrac{1}{4}\log_{e}\left(\dfrac{1-x}{1+x}\right)41​loge​(1+x1−x​)
  3. (C)14log⁡e(1+x1−x)\dfrac{1}{4}\log_{e}\left(\dfrac{1+x}{1-x}\right)41​loge​(1−x1+x​)
  4. (D)14(log⁡8e)log⁡e(1+x1−x)\dfrac{1}{4}(\log_{8}e)\log_{e}\left(\dfrac{1+x}{1-x}\right)41​(log8​e)loge​(1−x1+x​)

Correct answer: (D)

Step-by-step solution →
Q292·MathematicsNumericalJEE Main 2020
Let X={n∈N:1≤n≤50}X=\{n \in N : 1 \le n \le 50\}X={n∈N:1≤n≤50}. If A={n∈X:n is a multiple of 2}A=\{n \in X : n \text{ is a multiple of } 2\}A={n∈X:n is a multiple of 2} and B={n∈X,n is a multiple of 7}B=\{n \in X, n \text{ is a multiple of } 7\}B={n∈X,n is a multiple of 7}, then the number of elements in the smallest subset of X containing both A and B is __________.

Correct answer: 29

Step-by-step solution →
Q293·MathematicsSingle correctJEE Main 2020
If g(x)=x2+x−1g(x)=x^{2}+x-1g(x)=x2+x−1 and (gof)(x)=4x2−10x+5(gof)(x)=4x^{2}-10x+5(gof)(x)=4x2−10x+5, then f(54)f\left(\frac{5}{4}\right)f(45​) is equal to
  1. (A)32\frac{3}{2}23​
  2. (B)12\frac{1}{2}21​
  3. (C)−32-\frac{3}{2}−23​
  4. (D)−12-\frac{1}{2}−21​

Correct answer: (D)

Step-by-step solution →
Q294·MathematicsSingle correctJEE Main 2019
For x∈(0,32)x \in \left(0, \dfrac{3}{2}\right)x∈(0,23​), let f(x)=xf(x) = \sqrt{x}f(x)=x​, g(x)=tan⁡xg(x) = \tan xg(x)=tanx and h(x)=1−x21+x2h(x) = \dfrac{1-x^{2}}{1+x^{2}}h(x)=1+x21−x2​. If ϕ(x)=((h∘f)∘g)(x)\phi(x) = ((h \circ f) \circ g)(x)ϕ(x)=((h∘f)∘g)(x), then ϕ(π3)\phi\left(\dfrac{\pi}{3}\right)ϕ(3π​) is equal to :
  1. (A)tan⁡11π12\tan \dfrac{11\pi}{12}tan1211π​
  2. (B)tan⁡π12\tan \dfrac{\pi}{12}tan12π​
  3. (C)tan⁡5π12\tan \dfrac{5\pi}{12}tan125π​
  4. (D)tan⁡7π12\tan \dfrac{7\pi}{12}tan127π​

Correct answer: (A)

Step-by-step solution →
Q295·MathematicsSingle correctJEE Main 2019
Let A, B and C be sets such that ϕ≠A∩B⊆C\phi \neq A \cap B \subseteq Cϕ=A∩B⊆C. Then which of the following statements is not true?
  1. (A)If (A−C)⊆B(A-C) \subseteq B(A−C)⊆B, then A⊆BA \subseteq BA⊆B
  2. (B)If (A−B)⊆C(A-B) \subseteq C(A−B)⊆C, then A⊆CA \subseteq CA⊆C
  3. (C)(C∪A)∩(C∪B)=C(C \cup A) \cap (C \cup B) = C(C∪A)∩(C∪B)=C
  4. (D)B∩C≠ϕB \cap C \neq \phiB∩C=ϕ

Correct answer: (A)

Step-by-step solution →
Q296·MathematicsSingle correctJEE Main 2019
For x∈Rx \in Rx∈R, let [x][x][x] denote the greatest integer ≤x\le x≤x, then the sum of the series [−13]+[−13−1100]+[−13−2100]+⋯+[−13−99100]\left[-\dfrac{1}{3}\right] + \left[-\dfrac{1}{3} - \dfrac{1}{100}\right] + \left[-\dfrac{1}{3} - \dfrac{2}{100}\right] + \cdots + \left[-\dfrac{1}{3} - \dfrac{99}{100}\right][−31​]+[−31​−1001​]+[−31​−1002​]+⋯+[−31​−10099​]
  1. (A)−135-135−135
  2. (B)−153-153−153
  3. (C)−133-133−133
  4. (D)−131-131−131

Correct answer: (C)

Step-by-step solution →
Q297·MathematicsSingle correctJEE Main 2019
Let f(x) = x2^{2}2, x ∈ R. For any A ⊆ R, define g(A) = {x ∈ R : f(x) ∈ A}. If S = [0, 4], then which one of the following statements is not true?
  1. (A)f(g(S)) ≠ f(S)
  2. (B)f(g(S)) = S
  3. (C)g(f(S)) ≠ S
  4. (D)g(f(S)) = g(S)

Correct answer: (D)

Step-by-step solution →
Q298·MathematicsSingle correctJEE Main 2019
The domain of the definition of the function f(x)=14−x2+log⁡(x3−x)f\left(x\right)=\dfrac{1}{4-x^{2}}+\log\left(x^{3}-x\right)f(x)=4−x21​+log(x3−x) is
  1. (A)(1,2)∪(2,∞)\left(1,2\right)\cup\left(2,\infty\right)(1,2)∪(2,∞)
  2. (B)(−1,0)∪(1,2)∪(3,∞)\left(-1,0\right)\cup\left(1,2\right)\cup\left(3,\infty\right)(−1,0)∪(1,2)∪(3,∞)
  3. (C)(−1,0)∪(1,2)∪(2,∞)\left(-1,0\right)\cup\left(1,2\right)\cup\left(2,\infty\right)(−1,0)∪(1,2)∪(2,∞)
  4. (D)(−2,−1)∪(−1,0)∪(2,∞)\left(-2,-1\right)\cup\left(-1,0\right)\cup\left(2,\infty\right)(−2,−1)∪(−1,0)∪(2,∞)

Correct answer: (C)

Step-by-step solution →
Q299·MathematicsSingle correctJEE Main 2019
If the function f:R−{1,−1}→Af: R-\{1, -1\} \to Af:R−{1,−1}→A defined by f(x)=x21−x2f(x) = \dfrac{x^{2}}{1-x^{2}}f(x)=1−x2x2​, is surjective, then A is equal to:
  1. (A)R−[−1,0)R - [-1, 0)R−[−1,0)
  2. (B)R−(−1,0)R - (-1, 0)R−(−1,0)
  3. (C)R−{−1}R - \{-1\}R−{−1}
  4. (D)[0,∞][0, \infty][0,∞]

Correct answer: (A)

Step-by-step solution →
Q300·MathematicsSingle correctJEE Main 2019
If f(x)=log⁡e(1−x1+x)f(x)=\log_{e}\left(\frac{1-x}{1+x}\right)f(x)=loge​(1+x1−x​), ∣x∣<1|x|<1∣x∣<1, then f(2x1+x2)f\left(\frac{2x}{1+x^{2}}\right)f(1+x22x​) is equal to:
  1. (A)2f(x)2f(x)2f(x)
  2. (B)(f(x))2\left(f(x)\right)^{2}(f(x))2
  3. (C)2f(x2)2f(x^{2})2f(x2)
  4. (D)−2f(x)-2f(x)−2f(x)

Correct answer: (A)

Step-by-step solution →
Q301·MathematicsSingle correctJEE Main 2019
Let f(x)=axf(x)=a^{x}f(x)=ax (a>0)(a>0)(a>0) be written as f(x)=f1(x)+f2(x)f(x)=f_{1}(x)+f_{2}(x)f(x)=f1​(x)+f2​(x), where f1(x)f_{1}(x)f1​(x) is an even function and f2(x)f_{2}(x)f2​(x) is an odd function. Then f1(x+y)+f1(x−y)f_{1}(x+y)+f_{1}(x-y)f1​(x+y)+f1​(x−y) equals
  1. (A)2f1(x)f2(y)2f_{1}(x)f_{2}(y)2f1​(x)f2​(y)
  2. (B)2f1(x)f1(y)2f_{1}(x)f_{1}(y)2f1​(x)f1​(y)
  3. (C)2f1(x+y)f2(x−y)2f_{1}(x+y)f_{2}(x-y)2f1​(x+y)f2​(x−y)
  4. (D)2f1(x+y)f1(x−y)2f_{1}(x+y)f_{1}(x-y)2f1​(x+y)f1​(x−y)

Correct answer: (B)

Step-by-step solution →
Q302·MathematicsSingle correctJEE Main 2019
Let Z be the set of integers. If A={x∈Z:2(x+2)(x2−5x+6)=1}A = \{x \in Z : 2^{(x+2)(x^{2}-5x+6)} = 1\}A={x∈Z:2(x+2)(x2−5x+6)=1} and B={x∈Z:−3<2x−1<9}B = \{x \in Z : -3 < 2x - 1 < 9\}B={x∈Z:−3<2x−1<9}, then the number of subsets of the set A ×\times× B, is :
  1. (A)2152^{15}215
  2. (B)2182^{18}218
  3. (C)2122^{12}212
  4. (D)2102^{10}210

Correct answer: (A)

Step-by-step solution →
Q303·MathematicsSingle correctJEE Main 2019
Let f:R→Rf:R\rightarrow Rf:R→R be defined by f(x)=x1+x2f(x)=\dfrac{x}{1+x^{2}}f(x)=1+x2x​, x∈Rx\in Rx∈R. Then the range of f is:
  1. (A)[−12,12]\left[-\dfrac{1}{2},\dfrac{1}{2}\right][−21​,21​]
  2. (B)R−[−1,1]R-[-1,1]R−[−1,1]
  3. (C)R−[−12,12]R-\left[-\dfrac{1}{2},\dfrac{1}{2}\right]R−[−21​,21​]
  4. (D)(−1,1)−{0}(-1,1)-\{0\}(−1,1)−{0}

Correct answer: (A)

Step-by-step solution →
Q304·MathematicsSingle correctJEE Main 2019
In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is:
  1. (A)102
  2. (B)42
  3. (C)1
  4. (D)38

Correct answer: (D)

Step-by-step solution →
Q305·MathematicsSingle correctJEE Main 2019
If the Boolean expression (p⊕q)∧(∼p⊖q)(p\oplus q)\wedge(\sim p\ominus q)(p⊕q)∧(∼p⊖q) is equivalent to p∧qp\wedge qp∧q, where ⊕,⊖∈{∧,∨}\oplus,\ominus\in\{\wedge,\vee\}⊕,⊖∈{∧,∨}, then the ordered pair (⊕,⊖)(\oplus,\ominus)(⊕,⊖) is :
  1. (A)(∨,∧)(\vee,\wedge)(∨,∧)
  2. (B)(∨,∨)(\vee,\vee)(∨,∨)
  3. (C)(∧,∨)(\wedge,\vee)(∧,∨)
  4. (D)(∧,∧)(\wedge,\wedge)(∧,∧)

Correct answer: (C)

Step-by-step solution →
Q306·MathematicsSingle correctJEE Main 2019
For x∈R−[0,1]x \in R - [0,1]x∈R−[0,1], let f1(x)=1xf_1(x) = \dfrac{1}{x}f1​(x)=x1​, f2(x)=1−xf_2(x) = 1-xf2​(x)=1−x and f3(x)=11−xf_3(x) = \dfrac{1}{1-x}f3​(x)=1−x1​ be three given functions. If a function, J(x)J(x)J(x) satisfies (f2∘J∘f1)(x)=f3(x)(f_2 \circ J \circ f_1)(x) = f_3(x)(f2​∘J∘f1​)(x)=f3​(x) then J(x)J(x)J(x) is equal to:
  1. (A)f3(x)f_3(x)f3​(x)
  2. (B)1xf3(x)\dfrac{1}{x}f_3(x)x1​f3​(x)
  3. (C)f2(x)f_2(x)f2​(x)
  4. (D)f1(x)f_1(x)f1​(x)

Correct answer: (A)

Step-by-step solution →
Q307·MathematicsSingle correctJEE Main 2019
Let A={x∈R:x is not a positive integer}A = \{x \in R : x \text{ is not a positive integer}\}A={x∈R:x is not a positive integer}. Define a function f:A→Rf : A \rightarrow Rf:A→R as f(x)=2xx−1f(x) = \frac{2x}{x - 1}f(x)=x−12x​ then f is
  1. (A)injective but nor surjective
  2. (B)not injective
  3. (C)surjective but not injective
  4. (D)neither injective nor surjective

Correct answer: (A)

Step-by-step solution →
Q308·MathematicsSingle correctJEE Advanced 2018
Let E1={x∈R:x≠1 and xx−1>0}E_{1} = \left\{x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0\right\}E1​={x∈R:x=1 and x−1x​>0} and E2={x∈E1:sin⁡−1(log⁡e(xx−1)) is a real number}E_{2} = \left\{x \in E_{1} : \sin^{-1}\left(\log_{e}\left(\frac{x}{x-1}\right)\right) \text{ is a real number}\right\}E2​={x∈E1​:sin−1(loge​(x−1x​)) is a real number} (Here, the inverse trigonometric function sin⁡−1x assumes values in [−π2,π2].)\left(\text{Here, the inverse trigonometric function } \sin^{-1}x \text{ assumes values in } \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].\right)(Here, the inverse trigonometric function sin−1x assumes values in [−2π​,2π​].) Let f:E1→Rf : E_{1} \to Rf:E1​→R be the function defined by f(x)=log⁡e(xx−1)f(x) = \log_{e}\left(\frac{x}{x-1}\right)f(x)=loge​(x−1x​) and g:E2→Rg : E_{2} \to Rg:E2​→R be the function defined by g(x)=sin⁡−1(log⁡e(xx−1))g(x) = \sin^{-1}\left(\log_{e}\left(\frac{x}{x-1}\right)\right)g(x)=sin−1(loge​(x−1x​)). The correct option is :
LIST-ILIST-II
P.The range of fff is1.(−∞,11−e]∪[ee−1,∞)\left(-\infty, \frac{1}{1-e}\right] \cup \left[\frac{e}{e-1}, \infty\right)(−∞,1−e1​]∪[e−1e​,∞)
Q.The range of ggg contains2.(0,1)(0, 1)(0,1)
R.The domain of fff contains3.[−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right][−21​,21​]
S.The domain of ggg is4.(−∞,0)∪(0,∞)(-\infty, 0) \cup (0, \infty)(−∞,0)∪(0,∞)
5.(−∞,ee−1]\left(-\infty, \frac{e}{e-1}\right](−∞,e−1e​]
6.(−∞,0)∪(12,ee−1](-\infty, 0) \cup \left(\frac{1}{2}, \frac{e}{e-1}\right](−∞,0)∪(21​,e−1e​]
  1. (A)P →\to→ 4; Q →\to→ 2; R →\to→ 1; S →\to→ 1
  2. (B)P →\to→ 3; Q →\to→ 3; R →\to→ 6; S →\to→ 5
  3. (C)P →\to→ 4; Q →\to→ 2; R →\to→ 1; S →\to→ 6
  4. (D)P →\to→ 4; Q →\to→ 3; R →\to→ 6; S →\to→ 5

Correct answer: (A)

Step-by-step solution →
Q309·MathematicsNumericalJEE Advanced 2018
The value of ((log⁡29)2)1log⁡2(log⁡29)×(7)1log⁡47\left(\left(\log_{2} 9\right)^{2}\right)^{\frac{1}{\log_{2}\left(\log_{2} 9\right)}} \times \left(\sqrt{7}\right)^{\frac{1}{\log_{4} 7}}((log2​9)2)log2​(log2​9)1​×(7​)log4​71​ is ______ .

Correct answer: 8

Step-by-step solution →
Q310·MathematicsMultiple correctJEE Advanced 2015
Let f(x)=sin⁡(π6sin⁡(π2sin⁡x))f(x) = \sin\left(\dfrac{\pi}{6}\sin\left(\dfrac{\pi}{2}\sin x\right)\right)f(x)=sin(6π​sin(2π​sinx)) for all x∈Rx \in \mathbb{R}x∈R and g(x)=π2sin⁡xg(x) = \dfrac{\pi}{2}\sin xg(x)=2π​sinx for all x∈Rx \in \mathbb{R}x∈R. Let (f∘g)(x)(f \circ g)(x)(f∘g)(x) denote f(g(x))f(g(x))f(g(x)) and (g∘f)(x)(g \circ f)(x)(g∘f)(x) denote g(f(x))g(f(x))g(f(x)). Then which of the following is (are) true ?
  1. (A)Range of fff is [−12,12]\left[-\dfrac{1}{2}, \dfrac{1}{2}\right][−21​,21​]
  2. (B)Range of f∘gf \circ gf∘g is [−12,12]\left[-\dfrac{1}{2}, \dfrac{1}{2}\right][−21​,21​]
  3. (C)lim⁡x→0f(x)g(x)=π6\displaystyle\lim_{x \to 0} \dfrac{f(x)}{g(x)} = \dfrac{\pi}{6}x→0lim​g(x)f(x)​=6π​
  4. (D)There is an x∈Rx \in \mathbb{R}x∈R such that (g∘f)(x)=1(g \circ f)(x) = 1(g∘f)(x)=1

Correct answer: (A), (B), (C)

Step-by-step solution →
Q311·MathematicsSingle correctJEE Advanced 2014
Let f1:R→Rf_{1} : \mathbb{R} \to \mathbb{R}f1​:R→R, f2:[0,∞)→Rf_{2} : [0, \infty) \to \mathbb{R}f2​:[0,∞)→R, f3:R→Rf_{3} : \mathbb{R} \to \mathbb{R}f3​:R→R and f4:R→[0,∞)f_{4} : \mathbb{R} \to [0, \infty)f4​:R→[0,∞) be defined by f1(x)={∣x∣if x<0exif x≥0f_{1}(x) = \begin{cases} |x| & \text{if } x < 0 \\ e^{x} & \text{if } x \ge 0 \end{cases}f1​(x)={∣x∣ex​if x<0if x≥0​ ; f2(x)=x2f_{2}(x) = x^{2}f2​(x)=x2 ; f3(x)={sin⁡xif x<0xif x≥0f_{3}(x) = \begin{cases} \sin x & \text{if } x < 0 \\ x & \text{if } x \ge 0 \end{cases}f3​(x)={sinxx​if x<0if x≥0​ and f4(x)={f2(f1(x))if x<0f2(f1(x))−1if x≥0f_{4}(x) = \begin{cases} f_{2}(f_{1}(x)) & \text{if } x < 0 \\ f_{2}(f_{1}(x)) - 1 & \text{if } x \ge 0 \end{cases}f4​(x)={f2​(f1​(x))f2​(f1​(x))−1​if x<0if x≥0​
List – IList – II
P.f4f_{4}f4​ is1.onto but not one-one
Q.f3f_{3}f3​ is2.neither continuous nor one-one
R.f2∘f1f_{2} \circ f_{1}f2​∘f1​ is3.differentiable but not one-one
S.f2f_{2}f2​ is4.continuous and one-one
  1. (A)P-3, Q-1, R-4, S-2
  2. (B)P-1, Q-3, R-4, S-2
  3. (C)P-3, Q-1, R-2, S-4
  4. (D)P-1, Q-3, R-2, S-4

Correct answer: (D)

Step-by-step solution →
Q312·MathematicsMultiple correctJEE Advanced 2014
Let f:(−π2,π2)→Rf: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}f:(−2π​,2π​)→R be given by f(x)=(log⁡(sec⁡x+tan⁡x))3f(x) = (\log(\sec x + \tan x))^3f(x)=(log(secx+tanx))3. Then
  1. (A)f(x)f(x)f(x) is an odd function
  2. (B)f(x)f(x)f(x) is a one-one function
  3. (C)f(x)f(x)f(x) is an onto function
  4. (D)f(x)f(x)f(x) is an even function

Correct answer: (A), (B), (C)

Step-by-step solution →
Q313·MathematicsMultiple correctJEE Advanced 2013
If 3x=4x−13^{x} = 4^{x-1}3x=4x−1, then x=x =x=
  1. (A)2log⁡322log⁡32−1\frac{2\log_{3} 2}{2\log_{3} 2 - 1}2log3​2−12log3​2​
  2. (B)22−log⁡23\frac{2}{2 - \log_{2} 3}2−log2​32​
  3. (C)11−log⁡43\frac{1}{1 - \log_{4} 3}1−log4​31​
  4. (D)2log⁡232log⁡23−1\frac{2\log_{2} 3}{2\log_{2} 3 - 1}2log2​3−12log2​3​

Correct answer: (A), (B), (C)

Step-by-step solution →

Sets, Relations and Functions — frequently asked

How many questions from Sets, Relations and Functions appear in JEE?

Sets, Relations and Functions has appeared in 165 of the last 186 JEE Main and JEE Advanced papers — about 89% of them — contributing 313 questions in total across those papers.

Is Sets, Relations and Functions an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 89% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Sets, Relations and Functions questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226
  • Permutations and Combinations 220

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