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Limits and Continuity — JEE Previous Year Questions

Every Limits and Continuity question asked in JEE Main and JEE Advanced across the last 186 papers — 206 questions, each with its correct answer. Free to read, no account needed.

Questions

206

Papers it appeared in

149/186

Appearance rate

80%

All 206 Limits and Continuity questions

Most recent papers first.

Q1·MathematicsNumericalJEE Advanced 2026
For a real number α\alphaα, let [α][\alpha][α] denote the greatest integer less than or equal to α\alphaα. For a finite set SSS, let ∣S∣|S|∣S∣ denote the number of elements in the set SSS. Consider the functions f:(−3,3)→(−∞,∞)f : (-3, 3) \to (-\infty, \infty)f:(−3,3)→(−∞,∞) and g:(−3,3)→(−∞,∞)g : (-3, 3) \to (-\infty, \infty)g:(−3,3)→(−∞,∞) defined by f(x)=[x3]log⁡e(1+sin⁡2(π(x−[x])))f(x) = [x^{3}] \log_{e}(1 + \sin^{2}(\pi(x - [x])))f(x)=[x3]loge​(1+sin2(π(x−[x]))) and g(x)=x3sin⁡2(πlog⁡e(1+x−[x]))g(x) = x^{3} \sin^{2}(\pi \log_{e}(1 + x - [x]))g(x)=x3sin2(πloge​(1+x−[x])). Let A={x∈(−3,3):fA = \{x \in (-3, 3) : fA={x∈(−3,3):f is discontinuous at x}x\}x} and B={x∈(−3,3):gB = \{x \in (-3, 3) : gB={x∈(−3,3):g is discontinuous at x}x\}x}. Then the value of ∣A∣+2∣B∣−∣A∩B∣|A| + 2|B| - |A \cap B|∣A∣+2∣B∣−∣A∩B∣ is _____.

Correct answer: 56

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
Let f(x)={13,x≤π/2b(1−sin⁡x)(π−2x)2,x>π/2f(x) = \begin{cases} \dfrac{1}{3}, & x \le \pi/2 \\ \dfrac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \pi/2 \end{cases}f(x)=⎩⎨⎧​31​,(π−2x)2b(1−sinx)​,​x≤π/2x>π/2​. If fff is continuous at x=π/2x = \pi/2x=π/2, then the value of ∫03b−6∣x2+2x−3∣ dx\displaystyle\int_{0}^{3b-6} |x^2 + 2x - 3|\,dx∫03b−6​∣x2+2x−3∣dx is:
  1. (A)555
  2. (B)222
  3. (C)333
  4. (D)444

Correct answer: (D)

Step-by-step solution →
Q3·MathematicsNumericalJEE Main 2026
Let f(x)={x3+8;x<0x2−4;x≥0f(x) = \begin{cases} x^{3} + 8; & x < 0 \\ x^{2} - 4; & x \ge 0 \end{cases}f(x)={x3+8;x2−4;​x<0x≥0​ and g(x)={(x−8)1/3;x<0(x+4)1/2;x≥0g(x) = \begin{cases} (x - 8)^{1/3}; & x < 0 \\ (x + 4)^{1/2}; & x \ge 0 \end{cases}g(x)={(x−8)1/3;(x+4)1/2;​x<0x≥0​. Then the number of points, where the function g∘fg \circ fg∘f is discontinuous, is __________.

Correct answer: 3

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
The value of lim⁡x→0(x2sin⁡2xx2−sin⁡2x)\lim_{x \to 0}\left(\frac{x^2 \sin^2 x}{x^2 - \sin^2 x}\right)limx→0​(x2−sin2xx2sin2x​) is:
  1. (A)2
  2. (B)3
  3. (C)4
  4. (D)6

Correct answer: (B)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let f(x)=lim⁡y→0(1−cos⁡(xy))tan⁡(xy)y3f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^{3}}f(x)=limy→0​y3(1−cos(xy))tan(xy)​. Then the number of solutions of the equation f(x) = sin x, x ∈ ℝ is :
  1. (A)0
  2. (B)2
  3. (C)3
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
The product of all possible values of α\alphaα, for which lim⁡x→0(1−cos⁡(αx)cos⁡((α+1)x)cos⁡((α+2)x)sin⁡2((α+1)x))=2\lim\limits_{x \to 0} \left( \frac{1 - \cos(\alpha x)\cos((\alpha + 1)x)\cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} \right) = 2x→0lim​(sin2((α+1)x)1−cos(αx)cos((α+1)x)cos((α+2)x)​)=2, is:
  1. (A)−2-2−2
  2. (B)1
  3. (C)−1-1−1
  4. (D)54\frac{5}{4}45​

Correct answer: (C)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
If lim⁡x→2sin⁡(x3−5x2+ax+b)(x−1−1)log⁡e(x−1)=m\lim_{x \to 2} \frac{\sin\left(x^{3} - 5x^{2} + ax + b\right)}{\left(\sqrt{x-1} - 1\right)\log_{e}(x-1)} = mlimx→2​(x−1​−1)loge​(x−1)sin(x3−5x2+ax+b)​=m, then a+b+ma + b + ma+b+m is equal to :
  1. (A)5
  2. (B)6
  3. (C)8
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q8·MathematicsNumericalJEE Main 2026
The number of points in the interval [2,4][2, 4][2,4], at which the function f(x)=[x2−x−12]f(x) = \left[x^2 - x - \frac{1}{2}\right]f(x)=[x2−x−21​], where [⋅][\cdot][⋅] denotes the greatest integer function, is discontinuous, is ________.

Correct answer: 10

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let f(x)=lim⁡θ→0(cos⁡πx−x(2θ)sin⁡(x−1)1+x(2θ)(x−1)),x∈Rf(x) = \lim\limits_{\theta \to 0}\left(\dfrac{\cos \pi x - x^{\left(\frac{2}{\theta}\right)}\sin(x-1)}{1 + x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in Rf(x)=θ→0lim​(1+x(θ2​)(x−1)cosπx−x(θ2​)sin(x−1)​),x∈R. Consider the following two statements : (I) f(x) is discontinous at x = 1. (II) f(x) is continous at x = - 1. Then,
  1. (A)Neither (I) nor (II) is True
  2. (B)Both (I) and (II) are True
  3. (C)Only (II) is True
  4. (D)Only (I) is True

Correct answer: (A)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
The value of lim⁡x→0log⁡e(sec⁡(ex).sec⁡(e2x)⋅...⋅sec⁡(e10x))e2−e2cos⁡x\lim_{x \to 0} \dfrac{\log_e\left(\sec(ex).\sec\left(e^{2}x\right)\cdot ... \cdot \sec\left(e^{10}x\right)\right)}{e^{2} - e^{2\cos x}}limx→0​e2−e2cosxloge​(sec(ex).sec(e2x)⋅...⋅sec(e10x))​ is equal to
  1. (A)(e10−1)2e2(e2−1)\dfrac{(e^{10}-1)}{2e^{2}(e^{2}-1)}2e2(e2−1)(e10−1)​
  2. (B)(e20−1)2e2(e2−1)\dfrac{(e^{20}-1)}{2e^{2}(e^{2}-1)}2e2(e2−1)(e20−1)​
  3. (C)(e20−1)2(e2−1)\dfrac{(e^{20}-1)}{2(e^{2}-1)}2(e2−1)(e20−1)​
  4. (D)(e10−1)2(e2−1)\dfrac{(e^{10}-1)}{2(e^{2}-1)}2(e2−1)(e10−1)​

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let [t] denote the greatest integer less than or equal to t. If the function f(x)={b2sin⁡(π2[π2(cos⁡x+sin⁡x)cos⁡x]),x<0sin⁡x−12sin⁡2xx3,x>0a,x=0f(x) = \begin{cases} b^{2} \sin\left( \dfrac{\pi}{2} \left[ \dfrac{\pi}{2} (\cos x + \sin x) \cos x \right] \right) & , x < 0 \\ \dfrac{\sin x - \dfrac{1}{2}\sin 2x}{x^{3}} & , x > 0 \\ a & , x = 0 \end{cases}f(x)=⎩⎨⎧​b2sin(2π​[2π​(cosx+sinx)cosx])x3sinx−21​sin2x​a​,x<0,x>0,x=0​ is continuous at x=0x = 0x=0, then a2+b2a^{2} + b^{2}a2+b2 is equal to
  1. (A)58\dfrac{5}{8}85​
  2. (B)916\dfrac{9}{16}169​
  3. (C)34\dfrac{3}{4}43​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
If the function f(x) =ex(etan⁡x−x−1)+log⁡e(sec⁡x+tan⁡x)−xtan⁡x−x= \frac{e^{x}\left(e^{\tan x-x}-1\right)+\log_{e}(\sec x+\tan x)-x}{\tan x-x}=tanx−xex(etanx−x−1)+loge​(secx+tanx)−x​ is Continuous at x = 0, then the value of f(0) is equal to
  1. (A)222
  2. (B)23\frac{2}{3}32​
  3. (C)12\frac{1}{2}21​
  4. (D)32\frac{3}{2}23​

Correct answer: (D)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
If f(x)={a∣x∣+x2−2(sin⁡∣x∣)(cos⁡∣x∣)x, x≠0b, x=0f(x)=\begin{cases} \dfrac{a|x|+x^2-2(\sin|x|)(\cos|x|)}{x} & ,\ x \neq 0 \\[6pt] b & ,\ x = 0 \end{cases}f(x)=⎩⎨⎧​xa∣x∣+x2−2(sin∣x∣)(cos∣x∣)​b​, x=0, x=0​ is continuous at x=0x = 0x=0, then a+ba + ba+b is equal to :
  1. (A)1
  2. (B)2
  3. (C)0
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let f(x)={ax2+2ax+34x2+4x−3,x≠−32,12b,x=−32,12f(x) = \begin{cases} \frac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \ne -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}f(x)={4x2+4x−3ax2+2ax+3​,b,​x=−23​,21​x=−23​,21​​ be continuous at x=−32x = -\frac{3}{2}x=−23​. If fof(x)=75fof(x) = \frac{7}{5}fof(x)=57​, then x is equal to :
  1. (A)2
  2. (B)1
  3. (C)0
  4. (D)1.4

Correct answer: (B)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim\limits_{x \to 0} \frac{e^{(a-1)x} + 2\cos bx + (c-2)e^{-x}}{x \cos x - \log_{e}(1+x)} = 2x→0lim​xcosx−loge​(1+x)e(a−1)x+2cosbx+(c−2)e−x​=2, then a2+b2+c2a^{2} + b^{2} + c^{2}a2+b2+c2 is equal to :
  1. (A)5
  2. (B)3
  3. (C)7
  4. (D)9

Correct answer: (C)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
Let f:R→(0,∞)f : R \to (0, \infty)f:R→(0,∞) be a twice differentiable function such that f(3)=18f(3) = 18f(3)=18, f′(3)=0f'(3) = 0f′(3)=0 and f′′(3)=4f''(3) = 4f′′(3)=4. Then lim⁡x→1(log⁡e(f(2+x)f(3))18(x−1)2)\lim_{x \to 1}\left( \log_{e}\left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right)limx→1​(loge​(f(3)f(2+x)​)(x−1)218​) is equal to :
  1. (A)1
  2. (B)9
  3. (C)2
  4. (D)18

Correct answer: (C)

Step-by-step solution →
Q17·MathematicsNumericalJEE Main 2026
Let [⋅][\cdot][⋅] denote the greatest integer function and f(x)=lim⁡n→∞1n3∑k=1n[k23x]f(x) = \lim_{n \to \infty} \frac{1}{n^{3}} \sum_{k=1}^{n} \left[\frac{k^{2}}{3^{x}}\right]f(x)=limn→∞​n31​∑k=1n​[3xk2​]. Then 12∑j=1∞f(j)12\sum_{j=1}^{\infty} f(j)12∑j=1∞​f(j) is equal to _______.

Correct answer: 2

Step-by-step solution →
Q18·MathematicsSingle correctJEE Advanced 2025
Let x0x_0x0​ be the real number such that ex0+x0=0e^{x_0} + x_0 = 0ex0​+x0​=0. For a given real number α, define g(x)=3xex+3x−αex−αx3(ex+1)g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3\left(e^x + 1\right)}g(x)=3(ex+1)3xex+3x−αex−αx​ for all real numbers x. Then which one of the following statements is TRUE ?
  1. (A)For α = 2, lim⁡x→x0∣g(x)+ex0x−x0∣=0\lim_{x \rightarrow x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0limx→x0​​​x−x0​g(x)+ex0​​​=0
  2. (B)For α = 2, lim⁡x→x0∣g(x)+ex0x−x0∣=1\lim_{x \rightarrow x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1limx→x0​​​x−x0​g(x)+ex0​​​=1
  3. (C)For α = 3, lim⁡x→x0∣g(x)+ex0x−x0∣=0\lim_{x \rightarrow x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0limx→x0​​​x−x0​g(x)+ex0​​​=0
  4. (D)For α = 3, lim⁡x→x0∣g(x)+ex0x−x0∣=23\lim_{x \rightarrow x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}limx→x0​​​x−x0​g(x)+ex0​​​=32​

Correct answer: (C)

Step-by-step solution →
Q19·MathematicsNumericalJEE Advanced 2025
Let α and β be the real numbers such that lim⁡x→01x3(α2∫0x11−t2dt+βxcos⁡x)=2\lim_{x \to 0} \frac{1}{x^3}\left( \frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} dt + \beta x \cos x \right) = 2limx→0​x31​(2α​∫0x​1−t21​dt+βxcosx)=2. Then the value of α + β is _______ .

Correct answer: 2.4

Step-by-step solution →
Q20·MathematicsSingle correctJEE Advanced 2025
Let R denote the set of all real numbers. For a real number x, let [x] denote the greatest integer less than or equal to x. Let n denote a natural number. Match each entry in List-I to the correct entries in List-II and choose the correct option. The correct option is:
List-IList-II
P.The minimum value of n for which the function f(x)=[10x3−45x2+60x+35n]f(x) = \left[\frac{10x^3 - 45x^2 + 60x + 35}{n}\right]f(x)=[n10x3−45x2+60x+35​] is continuous on the interval [1, 2], is1.8
Q.The minimum value of n for which g(x)=(2n2−13n−15)(x3+3x)g(x) = (2n^2 - 13n - 15)(x^3 + 3x)g(x)=(2n2−13n−15)(x3+3x), x∈Rx \in Rx∈R, is an increasing function on R, is2.9
R.The smallest natural number n which is greater than 5, such that x=3x = 3x=3 is a point of local minima of h(x)=(x2−9)n(x2+2x+3)h(x) = (x^2 - 9)^n(x^2 + 2x + 3)h(x)=(x2−9)n(x2+2x+3), is3.5
S.Number of x0∈Rx_0 \in Rx0​∈R such that l(x)=∑k=04(sin⁡∣x−k∣+cos⁡∣x−k+12∣)l(x) = \sum_{k=0}^{4}\left( \sin|x - k| + \cos\left|x - k + \frac{1}{2}\right| \right)l(x)=∑k=04​(sin∣x−k∣+cos​x−k+21​​), x∈Rx \in Rx∈R, In NOT differentiable at x0x_0x0​ is4.6
5.10
  1. (A)(P) → (1), (Q) → (3), (R) → (2), (S) → (5)
  2. (B)(P) → (2), (Q) → (1), (R) → (4), (S) → (3)
  3. (C)(P) → (5), (Q) → (1), (R) → (4), (S) → (3)
  4. (D)(P) → (2), (Q) → (3), (R) → (1), (S) → (5)

Correct answer: (B)

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2025
Given below are two statements: Statement I: lim⁡x→0tan⁡−1x+log⁡e1+x1−x−2xx5=25\lim\limits_{x\to0}\dfrac{\tan^{-1}x+\log_e\sqrt{\dfrac{1+x}{1-x}}-2x}{x^5}=\dfrac{2}{5}x→0lim​x5tan−1x+loge​1−x1+x​​−2x​=52​. Statement II: lim⁡x→1x21−x=1e2\lim\limits_{x\to1}x^{\frac{2}{1-x}}=\dfrac{1}{e^2}x→1lim​x1−x2​=e21​. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are false
  4. (D)Both Statement I and Statement II are true

Correct answer: (D)

Step-by-step solution →
Q22·MathematicsIntegerJEE Main 2025
The number of points of discontinuity of the function f(x)=[x22]−[x]f(x)=\left[\dfrac{x^2}{2}\right]-\left[\sqrt{x}\right]f(x)=[2x2​]−[x​], x∈[0,4]x\in[0,4]x∈[0,4], where [⋅][\cdot][⋅] denotes the greatest integer function is ______.

Correct answer: 8

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2025
lim⁡x→0+tan⁡(5(x)1/3)log⁡e(1+3x2)(tan⁡−13x)2(e5(x)4/3−1)\displaystyle\lim_{x\to0^+}\dfrac{\tan\left(5(x)^{1/3}\right)\log_e(1+3x^2)}{\left(\tan^{-1}3\sqrt{x}\right)^2\left(e^{5(x)^{4/3}}-1\right)}x→0+lim​(tan−13x​)2(e5(x)4/3−1)tan(5(x)1/3)loge​(1+3x2)​ is equal to
  1. (A)115\dfrac{1}{15}151​
  2. (B)1
  3. (C)13\dfrac{1}{3}31​
  4. (D)53\dfrac{5}{3}35​

Correct answer: (C)

Step-by-step solution →
Q24·MathematicsIntegerJEE Main 2025
If the function f(x)=tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(x)=\dfrac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}f(x)=tanx−sinxtan(tanx)−sin(sinx)​ is continuous at x=0x=0x=0, then f(0)f(0)f(0) is equal to __________.

Correct answer: 2

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2025
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a polynomial function of degree four having extreme values at x=4x=4x=4 and x=5x=5x=5. If lim⁡x→0f(x)x2=5\lim\limits_{x\to 0}\dfrac{f(x)}{x^2}=5x→0lim​x2f(x)​=5, then f(2)f(2)f(2) is equal to:
  1. (A)12
  2. (B)10
  3. (C)8
  4. (D)14

Correct answer: (B)

Step-by-step solution →
Q26·MathematicsIntegerJEE Main 2025
For t>−1t>-1t>−1, let αt\alpha_tαt​ and βt\beta_tβt​ be the roots of the equation ((t+2)1/7−1)x2+((t+2)1/6−1)x+((t+2)1/21−1)=0\left((t+2)^{1/7}-1\right)x^2+\left((t+2)^{1/6}-1\right)x+\left((t+2)^{1/21}-1\right)=0((t+2)1/7−1)x2+((t+2)1/6−1)x+((t+2)1/21−1)=0. If lim⁡t→−1+αt=a\lim\limits_{t\to -1^+}\alpha_t=at→−1+lim​αt​=a and lim⁡t→−1+βt=b\lim\limits_{t\to -1^+}\beta_t=bt→−1+lim​βt​=b, then 72(a+b)272(a+b)^272(a+b)2 is equal to __________.

Correct answer: 98

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
If lim⁡x→1(x−1)(6+λcos⁡(x−1))+μsin⁡(1−x)(x−1)3=−1\displaystyle\lim_{x\to1}\dfrac{(x-1)(6+\lambda\cos(x-1))+\mu\sin(1-x)}{(x-1)^3}=-1x→1lim​(x−1)3(x−1)(6+λcos(x−1))+μsin(1−x)​=−1, where λ,μ∈R\lambda,\mu\in\mathbb{R}λ,μ∈R, then λ+μ\lambda+\muλ+μ is equal to
  1. (A)18
  2. (B)20
  3. (C)19
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2025
Let fff be a differentiable function on R\mathbb{R}R such that f(2)=1f(2)=1f(2)=1, f′(2)=4f'(2)=4f′(2)=4. Let lim⁡x→0(f(2+x)f(2))3/x=eα\lim_{x\to0}\left(\dfrac{f(2+x)}{f(2)}\right)^{3/x}=e^{\alpha}limx→0​(f(2)f(2+x)​)3/x=eα. Then the number of times the curve y=4x3−4x2−4(α−7)x−αy=4x^3-4x^2-4(\alpha-7)x-\alphay=4x3−4x2−4(α−7)x−α meets x-axis is:
  1. (A)2
  2. (B)1
  3. (C)0
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q29·MathematicsIntegerJEE Main 2025
Let mmm and nnn be the number of points at which the function f(x)=max⁡{x,x3,x5,…,x21}f(x)=\max\{x,x^3,x^5,\ldots,x^{21}\}f(x)=max{x,x3,x5,…,x21}, x∈Rx\in\mathbb{R}x∈R, is not differentiable and not continuous, respectively. Then m+nm+nm+n is equal to ______.

Correct answer: 3

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2025
Let f(x)={(1+ax)1/x,x<01+b,x=0(x+4)1/2−2(x+c)1/3−2,x>0f(x)=\begin{cases}(1+ax)^{1/x}, & x<0\\ 1+b, & x=0\\ \dfrac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2}, & x>0\end{cases}f(x)=⎩⎨⎧​(1+ax)1/x,1+b,(x+c)1/3−2(x+4)1/2−2​,​x<0x=0x>0​ be continuous at x=0x=0x=0. Then ea⋅b⋅ce^a\cdot b\cdot cea⋅b⋅c is equal to:
  1. (A)64
  2. (B)72
  3. (C)48
  4. (D)36

Correct answer: (C)

Step-by-step solution →
Q31·MathematicsIntegerJEE Main 2025
If lim⁡x→0(tan⁡xx)1/x2=p\displaystyle\lim_{x\to 0}\left(\dfrac{\tan x}{x}\right)^{1/x^2}=px→0lim​(xtanx​)1/x2=p, then 96log⁡ep96\log_e p96loge​p is equal to __________.

Correct answer: 32

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
For α,β,γ∈R\alpha,\beta,\gamma\in Rα,β,γ∈R, if lim⁡x→0x2sin⁡αx+(γ−1)ex2sin⁡2x−βx=3\displaystyle\lim_{x\to0}\dfrac{x^2\sin\alpha x+(\gamma-1)e^{x^2}}{\sin 2x-\beta x}=3x→0lim​sin2x−βxx2sinαx+(γ−1)ex2​=3, then β+γ−α\beta+\gamma-\alphaβ+γ−α is equal to:
  1. (A)7
  2. (B)4
  3. (C)6
  4. (D)−1-1−1

Correct answer: (A)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
If lim⁡x→0cos⁡(2x)+acos⁡(4x)−bx4\displaystyle\lim_{x\to0}\dfrac{\cos(2x)+a\cos(4x)-b}{x^4}x→0lim​x4cos(2x)+acos(4x)−b​ is finite, then (a+b)(a+b)(a+b) is equal to:
  1. (A)12\dfrac{1}{2}21​
  2. (B)0
  3. (C)34\dfrac{3}{4}43​
  4. (D)−1-1−1

Correct answer: (A)

Step-by-step solution →
Q34·MathematicsIntegerJEE Main 2025
Let [t][t][t] be the greatest integer less than or equal to t. Then the least value of p∈Np\in Np∈N for which lim⁡x→0+(x([1x]+[2x]+⋯+[px])−x2([12x2]+[22x2]+⋯+[92x2]))≥1\displaystyle\lim_{x\to 0^+}\left(x\left(\left[\dfrac{1}{x}\right]+\left[\dfrac{2}{x}\right]+\cdots+\left[\dfrac{p}{x}\right]\right)-x^2\left(\left[\dfrac{1^2}{x^2}\right]+\left[\dfrac{2^2}{x^2}\right]+\cdots+\left[\dfrac{9^2}{x^2}\right]\right)\right)\ge 1x→0+lim​(x([x1​]+[x2​]+⋯+[xp​])−x2([x212​]+[x222​]+⋯+[x292​]))≥1 is equal to ______.

Correct answer: 24

Step-by-step solution →
Q35·MathematicsIntegerJEE Main 2025
If lim⁡t→0(∫01(3x+5)t dx)1t=α5e(85)23\displaystyle\lim_{t\to 0}\left(\int_0^1 (3x+5)^t\,dx\right)^{\frac{1}{t}}=\dfrac{\alpha}{5e}\left(\dfrac{8}{5}\right)^{\frac{2}{3}}t→0lim​(∫01​(3x+5)tdx)t1​=5eα​(58​)32​, then α\alphaα is equal to ______.

Correct answer: 64

Step-by-step solution →
Q36·MathematicsIntegerJEE Main 2025
Let f(x)=lim⁡n→∞∑r=0n(tan⁡(x/2r+1)+tan⁡3(x/2r+1)1−tan⁡2(x/2r+1))f(x)=\lim_{n\to\infty}\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1})+\tan^3(x/2^{r+1})}{1-\tan^2(x/2^{r+1})}\right)f(x)=limn→∞​∑r=0n​(1−tan2(x/2r+1)tan(x/2r+1)+tan3(x/2r+1)​). Then lim⁡x→0ex−ef(x)x−f(x)\lim_{x\to0}\frac{e^x-e^{f(x)}}{x-f(x)}limx→0​x−f(x)ex−ef(x)​ is equal to ______.

Correct answer: 1

Step-by-step solution →
Q37·MathematicsIntegerJEE Main 2025
Let f(x)={3x,x<0min⁡{1+x+[x],x+2[x]},0≤x≤25,x>2f(x)=\begin{cases}3x, & x<0\\ \min\{1+x+[x], x+2[x]\}, & 0\le x\le 2\\ 5, & x>2\end{cases}f(x)=⎩⎨⎧​3x,min{1+x+[x],x+2[x]},5,​x<00≤x≤2x>2​ where [.][.][.] denotes greatest integer function. If α\alphaα and β\betaβ are the number of points, where f is not continuous and is not differentiable, respectively, then α+β\alpha+\betaα+β equals____.

Correct answer: 5

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
Consider the region R={(x,y):x≤y≤9−113x2, x≥0}R=\left\{(x,y):x\le y\le 9-\dfrac{11}{3}x^2,\ x\ge 0\right\}R={(x,y):x≤y≤9−311​x2, x≥0}. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is :
  1. (A)625111\dfrac{625}{111}111625​
  2. (B)730119\dfrac{730}{119}119730​
  3. (C)567121\dfrac{567}{121}121567​
  4. (D)821123\dfrac{821}{123}123821​

Correct answer: (C)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
Let f:R−{0}→Rf:\mathbb{R}-\{0\}\to\mathbb{R}f:R−{0}→R be a function such that f(x)−6f(1x)=353x−52f(x)-6f\left(\dfrac{1}{x}\right)=\dfrac{35}{3x}-\dfrac{5}{2}f(x)−6f(x1​)=3x35​−25​. If lim⁡x→0(1αx+f(x))=β\lim_{x\to 0}\left(\dfrac{1}{\alpha x}+f(x)\right)=\betalimx→0​(αx1​+f(x))=β; α,β∈R\alpha,\beta\in\mathbb{R}α,β∈R, then α+2β\alpha+2\betaα+2β is equal to
  1. (A)333
  2. (B)555
  3. (C)444
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
lim⁡x→0cosec⁡x(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)\lim_{x\to 0}\operatorname{cosec}x\left(\sqrt{2\cos^2 x+3\cos x}-\sqrt{\cos^2 x+\sin x+4}\right)limx→0​cosecx(2cos2x+3cosx​−cos2x+sinx+4​) is
  1. (A)000
  2. (B)125\dfrac{1}{2\sqrt5}25​1​
  3. (C)115\dfrac{1}{\sqrt{15}}15​1​
  4. (D)−125-\dfrac{1}{2\sqrt5}−25​1​

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
Let [x][x][x] denote the greatest integer function, and let mmm and nnn respectively be the numbers of the points, where the function f(x)=[x]+∣x−2∣f(x)=[x]+|x-2|f(x)=[x]+∣x−2∣, −2<x<3-2<x<3−2<x<3, is not continuous and not differentiable. Then m+nm+nm+n is equal to:
  1. (A)6
  2. (B)9
  3. (C)8
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
If the function f(x)={2x{sin⁡(k1+1)x+sin⁡(k2−1)x},x<04,x=02xlog⁡e(2+k1x2+k2x),x>0f(x)=\begin{cases}\dfrac{2}{x}\{\sin(k_1+1)x+\sin(k_2-1)x\}, & x<0\\[2mm] 4, & x=0\\[2mm] \dfrac{2}{x}\log_e\left(\dfrac{2+k_1x}{2+k_2x}\right), & x>0\end{cases}f(x)=⎩⎨⎧​x2​{sin(k1​+1)x+sin(k2​−1)x},4,x2​loge​(2+k2​x2+k1​x​),​x<0x=0x>0​ is continuous at x=0x=0x=0, then k12+k22k_1^2+k_2^2k12​+k22​ is equal to
  1. (A)888
  2. (B)202020
  3. (C)555
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q43·MathematicsIntegerJEE Main 2025
If the set of all values of aaa, for which the equation 5x3−15x−a=05x^3-15x-a=05x3−15x−a=0 has three distinct real roots, is the interval (α,β)(\alpha,\beta)(α,β), then β−2α\beta-2\alphaβ−2α is equal to _______

Correct answer: 30

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
lim⁡x→∞(2x2−3x+5)(3x−1)x/2(3x2+5x+4)(3x+2)x\displaystyle\lim_{x\to\infty}\dfrac{\left(2x^2-3x+5\right)\left(3x-1\right)^{x/2}}{\left(3x^2+5x+4\right)\sqrt{\left(3x+2\right)^x}}x→∞lim​(3x2+5x+4)(3x+2)x​(2x2−3x+5)(3x−1)x/2​ is equal to :
  1. (A)23e\dfrac{2}{\sqrt{3e}}3e​2​
  2. (B)2e3\dfrac{2e}{\sqrt{3}}3​2e​
  3. (C)2e3\dfrac{2e}{3}32e​
  4. (D)23e\dfrac{2}{3\sqrt{e}}3e​2​

Correct answer: (D)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2025
A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 111 cm, the ice-cream melts at the rate of 81 cm3/min81\,\text{cm}^3/\text{min}81cm3/min and the thickness of the ice-cream layer decreases at the rate of 14π\dfrac{1}{4\pi}4π1​ cm/min. The surface area (in cm2\text{cm}^2cm2) of the chocolate ball (without the ice-cream layer) is :
  1. (A)225π225\pi225π
  2. (B)128π128\pi128π
  3. (C)196π196\pi196π
  4. (D)256π256\pi256π

Correct answer: (D)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2025
If lim⁡x→∞(e1−e(1e−x1+x))x=α\displaystyle\lim_{x\to\infty}\left(\dfrac{e}{1-e}\left(\dfrac{1}{e}-\dfrac{x}{1+x}\right)\right)^x=\alphax→∞lim​(1−ee​(e1​−1+xx​))x=α, then the value of log⁡eα1+log⁡eα\dfrac{\log_e\alpha}{1+\log_e\alpha}1+loge​αloge​α​ equals:
  1. (A)e
  2. (B)e−2e^{-2}e−2
  3. (C)e2e^{2}e2
  4. (D)e−1e^{-1}e−1

Correct answer: (A)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Advanced 2024
Let k∈Rk \in \mathbb{R}k∈R. If lim⁡x→0+(sin⁡(sin⁡kx)+cos⁡x+x)2x=e6\lim_{x \to 0^+} \left(\sin(\sin kx) + \cos x + x\right)^{\frac{2}{x}} = e^6limx→0+​(sin(sinkx)+cosx+x)x2​=e6, then the value of kkk is
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q48·MathematicsMultiple correctJEE Advanced 2024
Let SSS be the set of all (α,β)∈R×R(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}(α,β)∈R×R such that lim⁡x→∞sin⁡(x2)(log⁡ex)αsin⁡(1x2)xαβ(log⁡e(1+x))β=0\lim_{x \to \infty} \frac{\sin\left(x^2\right)\left(\log_e x\right)^{\alpha} \sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta}\left(\log_e (1+x)\right)^{\beta}} = 0limx→∞​xαβ(loge​(1+x))βsin(x2)(loge​x)αsin(x21​)​=0. Then which of the following is(are) correct ?
  1. (A)(−1,3)∈S(-1, 3) \in S(−1,3)∈S
  2. (B)(−1,1)∈S(-1, 1) \in S(−1,1)∈S
  3. (C)(1,−1)∈S(1, -1) \in S(1,−1)∈S
  4. (D)(1,−2)∈S(1, -2) \in S(1,−2)∈S

Correct answer: (B), (C)

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Q49·MathematicsNumericalJEE Main 2024
Let f:(0,π)→Rf:(0, \pi) \to \mathbb{R}f:(0,π)→R be a function given by f(x)={(87)tan⁡8xtan⁡7x,0<x<π2a−8,x=π2(1+∣cot⁡x∣)ba∣tan⁡x∣,π2<x<πf(x) = \begin{cases} \left(\frac{8}{7}\right)^{\frac{\tan 8x}{\tan 7x}}, & 0 < x < \frac{\pi}{2} \\ a - 8, & x = \frac{\pi}{2} \\ (1 + |\cot x|)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi \end{cases}f(x)=⎩⎨⎧​(78​)tan7xtan8x​,a−8,(1+∣cotx∣)ab​∣tanx∣,​0<x<2π​x=2π​2π​<x<π​ Where a,b∈Za, b \in \mathbb{Z}a,b∈Z. If fff is continuous at x=π2x = \frac{\pi}{2}x=2π​, then a2+b2a^2 + b^2a2+b2 is equal to ________.

Correct answer: 81

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2024
lim⁡x→π2∫x3(π/2)3(sin⁡(2t1/3)+cos⁡(t1/3))dt(x−π2)2\displaystyle\lim_{x\to\frac{\pi}{2}}\dfrac{\int_{x^{3}}^{(\pi/2)^{3}}\left(\sin\left(2t^{1/3}\right)+\cos\left(t^{1/3}\right)\right)dt}{\left(x-\dfrac{\pi}{2}\right)^{2}}x→2π​lim​(x−2π​)2∫x3(π/2)3​(sin(2t1/3)+cos(t1/3))dt​ is equal to:
  1. (A)9π28\dfrac{9\pi^{2}}{8}89π2​
  2. (B)11π210\dfrac{11\pi^{2}}{10}1011π2​
  3. (C)3π22\dfrac{3\pi^{2}}{2}23π2​
  4. (D)4π29\dfrac{4\pi^{2}}{9}94π2​

Correct answer: (A)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2024
lim⁡x→0e−(1+2x)12xx\displaystyle\lim_{x\to 0}\dfrac{e-(1+2x)^{\frac{1}{2x}}}{x}x→0lim​xe−(1+2x)2x1​​ is equal to:
  1. (A)eee
  2. (B)−2e\dfrac{-2}{e}e−2​
  3. (C)000
  4. (D)−e2-e^{2}−e2

Correct answer: (A)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2024
For a,b>0a,b>0a,b>0, let f(x)={tan⁡((a+1)x)+btan⁡xx, x<03, x=0ax+b2x2−axba xx, x>0f(x)=\begin{cases}\dfrac{\tan((a+1)x)+b\tan x}{x} & ,\ x<0\\[2pt] 3 & ,\ x=0\\[2pt] \dfrac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt{a}\,x\sqrt{x}} & ,\ x>0\end{cases}f(x)=⎩⎨⎧​xtan((a+1)x)+btanx​3ba​xx​ax+b2x2​−ax​​​, x<0, x=0, x>0​ be a continuous function at x=0x=0x=0. Then ba\dfrac{b}{a}ab​ is equal to
  1. (A)555
  2. (B)444
  3. (C)888
  4. (D)666

Correct answer: (D)

Step-by-step solution →
Q53·MathematicsNumericalJEE Main 2024
If α=lim⁡x→0+(etan⁡x−extan⁡x−x)\alpha=\displaystyle\lim_{x\to 0^+}\left(\dfrac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)α=x→0+lim​(tanx​−x​etanx​−ex​​) and β=lim⁡x→0(1+sin⁡x)12cot⁡x\beta=\displaystyle\lim_{x\to 0}(1+\sin x)^{\frac{1}{2}\cot x}β=x→0lim​(1+sinx)21​cotx are the roots of the quadratic equation ax2+bx−e=0ax^2+bx-\sqrt{e}=0ax2+bx−e​=0, then 12log⁡e(a+b)12\log_e(a+b)12loge​(a+b) is equal to _______ .

Correct answer: 6

Step-by-step solution →
Q54·MathematicsNumericalJEE Main 2024
The value of lim⁡x→02(1−cos⁡2x cos⁡3x3⋯cos⁡10x10x2)\displaystyle\lim_{x\to0}2\left(\dfrac{1-\sqrt{\cos 2x}\,\sqrt[3]{\cos 3x}\cdots\sqrt[10]{\cos 10x}}{x^2}\right)x→0lim​2(x21−cos2x​3cos3x​⋯10cos10x​​) is ___

Correct answer: 55

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2024
lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1(13+23+⋯+n3)−(12+22+⋯+n2)\lim_{n\to\infty} \frac{(1^2-1)(n-1) + (2^2-2)(n-2) + \cdots + ((n-1)^2-(n-1))\cdot 1}{(1^3 + 2^3 + \cdots + n^3) - (1^2 + 2^2 + \cdots + n^2)}limn→∞​(13+23+⋯+n3)−(12+22+⋯+n2)(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1​ is equal to:
  1. (A)23\tfrac{2}{3}32​
  2. (B)13\tfrac{1}{3}31​
  3. (C)34\tfrac{3}{4}43​
  4. (D)12\tfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2024
If the function f(x)=sin⁡3x+αsin⁡x−βcos⁡3xx3f(x) = \dfrac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}f(x)=x3sin3x+αsinx−βcos3x​, x∈Rx \in \mathbb{R}x∈R, is continuous at x=0x = 0x=0, then f(0)f(0)f(0) is equal to:
  1. (A)222
  2. (B)−2-2−2
  3. (C)444
  4. (D)−4-4−4

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2024
Let f:[−1,2]→Rf:[-1,2]\to\mathbb{R}f:[−1,2]→R be given by f(x)=2x2+x+[x2]−[x]f(x)=2x^2+x+[x^2]-[x]f(x)=2x2+x+[x2]−[x], where [t][t][t] denotes the greatest integer less than or equal to ttt. The number of points where fff is not continuous is:
  1. (A)6
  2. (B)3
  3. (C)4
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsNumericalJEE Main 2024
Let a>0a>0a>0 be a root of the equation 2x2+x−2=02x^2+x-2=02x2+x−2=0. If lim⁡x→1a16(1−cos⁡(2+x−2x2))(1−ax)2=α+β17\displaystyle\lim_{x\to\frac{1}{a}}\dfrac{16\left(1-\cos(2+x-2x^2)\right)}{(1-ax)^2}=\alpha+\beta\sqrt{17}x→a1​lim​(1−ax)216(1−cos(2+x−2x2))​=α+β17​, where α,β∈Z\alpha,\beta\in\mathbb{Z}α,β∈Z, then α+β\alpha+\betaα+β is equal to __________.

Correct answer: 170

Step-by-step solution →
Q59·MathematicsNumericalJEE Main 2024
If lim⁡x→1(5x+1)1/3−(x+5)1/3(2x+3)1/2−(x+4)1/2=m5n(2n)2/3\displaystyle\lim_{x\to1}\dfrac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}}=\dfrac{m\sqrt5}{n(2n)^{2/3}}x→1lim​(2x+3)1/2−(x+4)1/2(5x+1)1/3−(x+5)1/3​=n(2n)2/3m5​​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then 8m+12n8m+12n8m+12n is equal to ___

Correct answer: 100

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2024
Let f(x)=72x−9x−8x+12−1+cos⁡xf(x)=\dfrac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}f(x)=2​−1+cosx​72x−9x−8x+1​ for x≠0x\ne 0x=0, and f(0)=a log⁡e2 log⁡e3f(0)=a\,\log_e 2\,\log_e 3f(0)=aloge​2loge​3. If fff is continuous at x=0x=0x=0, then the value of a2a^2a2 is equal to
  1. (A)768
  2. (B)1152
  3. (C)746
  4. (D)1250

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2024
Let f(x)=∫0x(t+sin⁡(1−et))dtf(x)=\displaystyle\int_0^x\left(t+\sin\left(1-e^t\right)\right)dtf(x)=∫0x​(t+sin(1−et))dt, x∈Rx\in\mathbb{R}x∈R. Then lim⁡x→0f(x)x3\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^3}x→0lim​x3f(x)​ is equal to
  1. (A)16\tfrac{1}{6}61​
  2. (B)−16-\tfrac{1}{6}−61​
  3. (C)−23-\tfrac{2}{3}−32​
  4. (D)23\tfrac{2}{3}32​

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2024
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a function given by f(x)={1−cos⁡2xx2, x<0α, x=0β1−cos⁡xx, x>0f(x)=\begin{cases}\dfrac{1-\cos 2x}{x^2} & ,\ x<0\\ \alpha & ,\ x=0\\ \dfrac{\beta\sqrt{1-\cos x}}{x} & ,\ x>0\end{cases}f(x)=⎩⎨⎧​x21−cos2x​αxβ1−cosx​​​, x<0, x=0, x>0​ where α,β∈R\alpha,\beta\in\mathbb{R}α,β∈R. If fff is continuous at x=0x=0x=0, then α2+β2\alpha^2+\beta^2α2+β2 is equal to:
  1. (A)484848
  2. (B)121212
  3. (C)333
  4. (D)666

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2024
Let {x}\{x\}{x} denote the fractional part of xxx and f(x)=cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3f(x)=\dfrac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}f(x)={x}−{x}3cos−1(1−{x}2)sin−1(1−{x})​, x≠0x\ne0x=0. If LLL and RRR respectively denote the left hand limit and the right hand limit of f(x)f(x)f(x) at x=0x=0x=0, then 32π2(L2+R2)\dfrac{32}{\pi^2}(L^2+R^2)π232​(L2+R2) is equal to ___

Correct answer: 18

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2024
Let f(x)=∣ 2x2+5∣x∣−3 ∣f(x)=\big|\,2x^2+5|x|-3\,\big|f(x)=​2x2+5∣x∣−3​, x∈Rx\in\mathbb Rx∈R. If mmm and nnn denote the number of points where fff is not continuous and not differentiable respectively, then m+nm+nm+n is equal to:
  1. (A)5
  2. (B)2
  3. (C)0
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsNumericalJEE Main 2024
If y=(x+1)(x2−x)xx+x+x+115(3cos⁡2x−5)cos⁡3xy=\dfrac{(\sqrt x+1)(x^2-\sqrt x)}{x\sqrt x+x+\sqrt x}+\dfrac{1}{15}(3\cos^2 x-5)\cos^3 xy=xx​+x+x​(x​+1)(x2−x​)​+151​(3cos2x−5)cos3x, then 96 y′ ⁣(π6)96\,y'\!\left(\dfrac{\pi}{6}\right)96y′(6π​) is equal to __________.

Correct answer: 105

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2024
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be defined as f(x)={a−bcos⁡2xx2, x<0x2+cx+2, 0≤x≤12x+1, x>1f(x)=\begin{cases}\dfrac{a-b\cos 2x}{x^2} & ,\ x<0\\ x^2+cx+2 & ,\ 0\le x\le1\\ 2x+1 & ,\ x>1\end{cases}f(x)=⎩⎨⎧​x2a−bcos2x​x2+cx+22x+1​, x<0, 0≤x≤1, x>1​. If fff is continuous everywhere in R\mathbb{R}R and mmm is the number of points where fff is NOT differentiable then m+a+b+cm+a+b+cm+a+b+c equals:
  1. (A)111
  2. (B)444
  3. (C)333
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2024
Let f(x)={x−1,x is even,2x,x is odd,  x∈Nf(x)=\begin{cases} x-1,& x\text{ is even},\\ 2x,& x\text{ is odd},\end{cases}\;x\in\mathbb Nf(x)={x−1,2x,​x is even,x is odd,​x∈N. If for some a∈Na\in\mathbb Na∈N, f(f(f(a)))=21f(f(f(a)))=21f(f(f(a)))=21, then lim⁡x→a−{∣x∣3a−[xa]}\displaystyle\lim_{x\to a^-}\left\{\dfrac{|x|^3}{a}-\left[\dfrac{x}{a}\right]\right\}x→a−lim​{a∣x∣3​−[ax​]}, where [t][t][t] denotes the greatest integer less than or equal to ttt, is equal to:
  1. (A)121
  2. (B)144
  3. (C)169
  4. (D)225

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2024
Let g(x)g(x)g(x) be a linear function and f(x)={g(x),x≤0(1+x2+x)1x,x>0f(x)=\begin{cases}g(x), & x\le 0\\ \left(\dfrac{1+x}{2+x}\right)^{\frac{1}{x}}, & x>0\end{cases}f(x)=⎩⎨⎧​g(x),(2+x1+x​)x1​,​x≤0x>0​, is continuous at x=0x=0x=0. If f′(1)=f(−1)f'(1)=f(-1)f′(1)=f(−1), then the value of g(3)g(3)g(3) is
  1. (A)13log⁡e(49e1/3)\dfrac{1}{3}\log_e\left(\dfrac{4}{9e^{1/3}}\right)31​loge​(9e1/34​)
  2. (B)13log⁡e(49)+1\dfrac{1}{3}\log_e\left(\dfrac{4}{9}\right)+131​loge​(94​)+1
  3. (C)log⁡e(49)−1\log_e\left(\dfrac{4}{9}\right)-1loge​(94​)−1
  4. (D)log⁡e(49e1/3)\log_e\left(\dfrac{4}{9e^{1/3}}\right)loge​(9e1/34​)

Correct answer: (D)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2024
Consider the function f:(0,∞)→Rf:(0,\infty)\to Rf:(0,∞)→R defined by f(x)=e−∣log⁡ex∣f(x)=e^{-\left|\log_e x\right|}f(x)=e−∣loge​x∣. If mmm and nnn be respectively the number of points at which fff is not continuous and fff is not differentiable, then m+nm+nm+n is
  1. (A)0
  2. (B)3
  3. (C)1
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2024
lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1x2\displaystyle\lim_{x\to 0}\dfrac{e^{2|\sin x|}-2|\sin x|-1}{x^2}x→0lim​x2e2∣sinx∣−2∣sinx∣−1​
  1. (A)is equal to −1-1−1
  2. (B)does not exist
  3. (C)is equal to 111
  4. (D)is equal to 222

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsNumericalJEE Main 2024
If lim⁡x→0ax2ex−blog⁡e(1+x)+cxe−xx2sin⁡x=1\displaystyle\lim_{x\to0}\dfrac{ax^2e^x-b\log_e(1+x)+cxe^{-x}}{x^2\sin x}=1x→0lim​x2sinxax2ex−bloge​(1+x)+cxe−x​=1, then 16(a2+b2+c2)16(a^2+b^2+c^2)16(a2+b2+c2) is equal to ______.

Correct answer: 81

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2024
If the function f:(−∞,−1]→(a,b]f:(-\infty,-1]\to(a,b]f:(−∞,−1]→(a,b] defined by f(x)=ex3−3x+1f(x)=e^{x^3-3x+1}f(x)=ex3−3x+1 is one-one and onto, then the distance of the point P(2b+4, a+2)P(2b+4,\ a+2)P(2b+4, a+2) from the line x+e−3y=4x+e^{-3}y=4x+e−3y=4 is
  1. (A)21+e62\sqrt{1+e^6}21+e6​
  2. (B)41+e64\sqrt{1+e^6}41+e6​
  3. (C)31+e63\sqrt{1+e^6}31+e6​
  4. (D)1+e6\sqrt{1+e^6}1+e6​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correctJEE Main 2024
Let f:R→(0,∞)f:R\to(0,\infty)f:R→(0,∞) be strictly increasing function such that lim⁡x→∞f(7x)f(x)=1\displaystyle\lim_{x\to\infty}\dfrac{f(7x)}{f(x)}=1x→∞lim​f(x)f(7x)​=1. Then the value of lim⁡x→∞[f(5x)f(x)−1]\displaystyle\lim_{x\to\infty}\left[\dfrac{f(5x)}{f(x)}-1\right]x→∞lim​[f(x)f(5x)​−1] is equal to
  1. (A)4
  2. (B)0
  3. (C)75\dfrac{7}{5}57​
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2024
Let f:R−{0}→Rf:R-\{0\}\to Rf:R−{0}→R be a function satisfying f(xy)=f(x)f(y)f\left(\dfrac{x}{y}\right)=\dfrac{f(x)}{f(y)}f(yx​)=f(y)f(x)​ for all x,yx,yx,y, f(y)≠0f(y)\ne 0f(y)=0. If f′(1)=2024f'(1)=2024f′(1)=2024, then:
  1. (A)xf′(x)−2024f(x)=0xf'(x)-2024f(x)=0xf′(x)−2024f(x)=0
  2. (B)xf′(x)+2024f(x)=0xf'(x)+2024f(x)=0xf′(x)+2024f(x)=0
  3. (C)xf′(x)+f(x)=2024xf'(x)+f(x)=2024xf′(x)+f(x)=2024
  4. (D)xf′(x)−2023f(x)=0xf'(x)-2023f(x)=0xf′(x)−2023f(x)=0

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2024
Let f:[−π2,π2]→Rf:\left[-\dfrac\pi2,\dfrac\pi2\right]\to\mathbb{R}f:[−2π​,2π​]→R be a differentiable function such that f(0)=12f(0)=\dfrac12f(0)=21​. If the lim⁡x→0x∫0xf(t) dtex2−1=α\displaystyle\lim_{x\to0}\dfrac{x\displaystyle\int_0^x f(t)\,dt}{e^{x^2}-1}=\alphax→0lim​ex2−1x∫0x​f(t)dt​=α, then 8α28\alpha^28α2 is equal to:
  1. (A)161616
  2. (B)222
  3. (C)111
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2024
The function f(x)=xx2−6x−16, x∈R−{−2,8}f(x)=\dfrac{x}{x^2-6x-16},\ x\in\mathbb{R}-\{-2,8\}f(x)=x2−6x−16x​, x∈R−{−2,8}:
  1. (A)increasing in (−2,8)(-2,8)(−2,8) and decreasing in (−∞,−2)∪(8,∞)(-\infty,-2)\cup(8,\infty)(−∞,−2)∪(8,∞)
  2. (B)decreasing in (−∞,−2)∪(−2,8)(-\infty,-2)\cup(-2,8)(−∞,−2)∪(−2,8) and increasing in (8,∞)(8,\infty)(8,∞)
  3. (C)decreasing in (−∞,−2)∪(8,∞)(-\infty,-2)\cup(8,\infty)(−∞,−2)∪(8,∞) and increasing in (−2,8)(-2,8)(−2,8)
  4. (D)decreasing in R−{−2,8}\mathbb{R}-\{-2,8\}R−{−2,8}

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2024
The function f(x)=2x+3(x)23, x∈Rf(x)=2x+3(x)^{\frac{2}{3}},\ x\in\mathbb{R}f(x)=2x+3(x)32​, x∈R, has:
  1. (A)exactly one point of local minima and no point of local maxima
  2. (B)exactly one point of local maxima and no point of local minima
  3. (C)exactly one point of local maxima and exactly one point of local minima
  4. (D)exactly two points of local maxima and exactly one point of local minima

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2024
Let f(x)=lim⁡r→x{2r2[(f(r))2−f(x)f(r)]r2−x2−r3ef(r)r}f(x)=\sqrt{\displaystyle\lim_{r\to x}\left\{\dfrac{2r^2[(f(r))^2-f(x)f(r)]}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right\}}f(x)=r→xlim​{r2−x22r2[(f(r))2−f(x)f(r)]​−r3erf(r)​}​ be differentiable in (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty)(−∞,0)∪(0,∞) and f(1)=1f(1)=1f(1)=1. Then the value of eaeaea, such that f(a)=0f(a)=0f(a)=0, is equal to ___.

Correct answer: 2

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2024
Let y=log⁡e(1−x21+x2), −1<x<1y=\log_e\left(\dfrac{1-x^2}{1+x^2}\right),\ -1<x<1y=loge​(1+x21−x2​), −1<x<1. Then at x=12x=\dfrac{1}{2}x=21​, the value of 225(y′−y′′)225(y'-y'')225(y′−y′′) is equal to:
  1. (A)732
  2. (B)746
  3. (C)742
  4. (D)736

Correct answer: (D)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2024
lim⁡x→π2(1(x−π2)2∫x3(π2)3cos⁡(t1/3)dt)\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\dfrac{1}{\left(x-\dfrac{\pi}{2}\right)^2}\int_{x^3}^{\left(\frac{\pi}{2}\right)^3}\cos\left(t^{1/3}\right)dt\right)x→2π​lim​​(x−2π​)21​∫x3(2π​)3​cos(t1/3)dt​ is equal to
  1. (A)3π8\dfrac{3\pi}{8}83π​
  2. (B)3π24\dfrac{3\pi^2}{4}43π2​
  3. (C)3π28\dfrac{3\pi^2}{8}83π2​
  4. (D)3π4\dfrac{3\pi}{4}43π​

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsNumericalJEE Main 2024
Let the slope of the line 45x+5y+3=045x+5y+3=045x+5y+3=0 be 27r1+9r2227r_1+\dfrac{9r_2}{2}27r1​+29r2​​ for some r1,r2∈Rr_1,r_2\in\mathbb{R}r1​,r2​∈R. Then lim⁡x→3(∫3x8t23r2x2−r2x2−r1x3−3x dt)\displaystyle\lim_{x\to 3}\left(\int_3^x\dfrac{8t^2}{\dfrac{3r_2 x}{2}-r_2 x^2-r_1 x^3-3x}\,dt\right)x→3lim​​∫3x​23r2​x​−r2​x2−r1​x3−3x8t2​dt​ is equal to ___.

Correct answer: 12

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2024
Suppose f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f(x)=\dfrac{\left(2^x+2^{-x}\right)\tan x\sqrt{\tan^{-1}\left(x^2-x+1\right)}}{\left(7x^2+3x+1\right)^3}f(x)=(7x2+3x+1)3(2x+2−x)tanxtan−1(x2−x+1)​​. Then the value of f′(0)f'(0)f′(0) is equal to
  1. (A)π\piπ
  2. (B)000
  3. (C)π\sqrt{\pi}π​
  4. (D)π2\dfrac{\pi}{2}2π​

Correct answer: (C)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2024
Consider the function f:[12,1]→Rf:\left[\dfrac{1}{2},1\right]\to Rf:[21​,1]→R defined by f(x)=42x3−32x−1f(x)=4\sqrt{2}x^3-3\sqrt{2}x-1f(x)=42​x3−32​x−1. Consider the statements (I) The curve y=f(x)y=f(x)y=f(x) intersects the x-axis exactly at one point (II) The curve y=f(x)y=f(x)y=f(x) intersects the x-axis at x=cos⁡π12x=\cos\dfrac{\pi}{12}x=cos12π​. Then
  1. (A)Only (II) is correct
  2. (B)Both (I) and (II) are incorrect
  3. (C)Only (I) is correct
  4. (D)Both (I) and (II) are correct

Correct answer: (D)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2024
If lim⁡x→03+αsin⁡x+βcos⁡x+log⁡e(1−x)3tan⁡2x=13\displaystyle\lim_{x\to 0}\dfrac{3+\alpha\sin x+\beta\cos x+\log_e(1-x)}{3\tan^2 x}=\dfrac13x→0lim​3tan2x3+αsinx+βcosx+loge​(1−x)​=31​, then 2α−β2\alpha-\beta2α−β is equal to :
  1. (A)2
  2. (B)7
  3. (C)5
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q85·MathematicsNumericalJEE Main 2024
Let f(x)=x3+x2f′(1)+x f′′(2)+f′′′(3)f(x)=x^3+x^2 f'(1)+x\,f''(2)+f'''(3)f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3), x∈Rx\in\mathbb Rx∈R. Then f′(10)f'(10)f′(10) is equal to __________.

Correct answer: 202

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2024
Consider the function f:(0,2)→Rf:(0,2)\to\mathbb{R}f:(0,2)→R defined by f(x)=x2+2xf(x)=\dfrac{x}{2}+\dfrac{2}{x}f(x)=2x​+x2​ and the function g(x)g(x)g(x) defined by g(x)={min⁡{f(t)}, 0<t≤x and 0<x≤132+x, 1<x<2g(x)=\begin{cases}\min\{f(t)\},\ 0<t\le x\text{ and }0<x\le 1 \\ \dfrac{3}{2}+x,\ 1<x<2\end{cases}g(x)=⎩⎨⎧​min{f(t)}, 0<t≤x and 0<x≤123​+x, 1<x<2​. Then
  1. (A)ggg is continuous but not differentiable at x=1x=1x=1
  2. (B)ggg is not continuous for all x∈(0,2)x\in(0,2)x∈(0,2)
  3. (C)ggg is neither continuous nor differentiable at x=1x=1x=1
  4. (D)ggg is continuous and differentiable for all x∈(0,2)x\in(0,2)x∈(0,2)

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2024
If a=lim⁡x→01+1+x4−2x4a=\displaystyle\lim_{x\to 0}\dfrac{\sqrt{1+\sqrt{1+x^4}}-\sqrt2}{x^4}a=x→0lim​x41+1+x4​​−2​​ and b=lim⁡x→0sin⁡2x2−1+cos⁡xb=\displaystyle\lim_{x\to 0}\dfrac{\sin^2 x}{\sqrt2-\sqrt{1+\cos x}}b=x→0lim​2​−1+cosx​sin2x​, then the value of ab3ab^3ab3 is:
  1. (A)36
  2. (B)32
  3. (C)25
  4. (D)30

Correct answer: (B)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2024
Consider the function f(x)={a(7x−12−x2)b ∣x2−7x+12∣,x<32sin⁡(x−3)/(x−[x]),x>3b,x=3f(x)=\begin{cases}\dfrac{a(7x-12-x^2)}{b\,|x^2-7x+12|}, & x<3\\ 2^{\sin(x-3)/(x-[x])}, & x>3\\ b, & x=3\end{cases}f(x)=⎩⎨⎧​b∣x2−7x+12∣a(7x−12−x2)​,2sin(x−3)/(x−[x]),b,​x<3x>3x=3​, where [x][x][x] denotes the greatest integer less than or equal to xxx. If SSS denotes the set of all ordered pairs (a,b)(a,b)(a,b) such that f(x)f(x)f(x) is continuous at x=3x=3x=3, then the number of elements in SSS is:
  1. (A)2
  2. (B)Infinitely many
  3. (C)4
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Advanced 2023
Let f:(0,1)→Rf : (0, 1) \rightarrow Rf:(0,1)→R be the function defined as f(x)=nf(x) = \sqrt{n}f(x)=n​ if x∈[1n+1,1n)x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)x∈[n+11​,n1​) where n∈Nn \in Nn∈N. Let g:(0,1)→Rg : (0, 1) \rightarrow Rg:(0,1)→R be a function such that ∫x2x1−tt dt<g(x)<2x\int_{x^2}^{x} \sqrt{\frac{1-t}{t}}\,dt < g(x) < 2\sqrt{x}∫x2x​t1−t​​dt<g(x)<2x​ for all x∈(0,1)x \in (0, 1)x∈(0,1) . Then lim⁡x→0f(x)g(x)\lim_{x \rightarrow 0} f(x)g(x)limx→0​f(x)g(x)
  1. (A)does NOT exist
  2. (B)is equal to 1
  3. (C)is equal to 2
  4. (D)is equal to 3

Correct answer: (C)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2023
If lim⁡x→0eax−cos⁡(bx)−cxe−cx21−cos⁡(2x)=17\lim\limits_{x \to 0} \dfrac{e^{ax} - \cos(bx) - \dfrac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17x→0lim​1−cos(2x)eax−cos(bx)−2cxe−cx​​=17, then 5a2+b25a^2 + b^25a2+b2 is
  1. (A)72
  2. (B)76
  3. (C)68
  4. (D)64

Correct answer: (C)

Step-by-step solution →
Q91·MathematicsNumericalJEE Main 2023
Let [x][x][x] be the greatest integer ≤x\le x≤x. Then the number of points in the interval (−2,1)(-2,1)(−2,1), where the function f(x)=∣[x]∣+x−[x]f(x)=|[x]|+\sqrt{x-[x]}f(x)=∣[x]∣+x−[x]​ is discontinuous is _________.

Correct answer: 2

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2023
Let ggg and fff be two functions defined as f(x)={x+1,x<0∣x−1∣,x≥0f(x)=\begin{cases} x+1, & x<0 \\ |x-1|, & x\ge 0 \end{cases}f(x)={x+1,∣x−1∣,​x<0x≥0​ and g(x)={x+1,x<01,x≥0g(x)=\begin{cases} x+1, & x<0 \\ 1, & x\ge 0 \end{cases}g(x)={x+1,1,​x<0x≥0​. Then (g∘f)(x)(g\circ f)(x)(g∘f)(x) is
  1. (A)differentiable everywhere
  2. (B)continuous everywhere but not differentiable exactly at one point
  3. (C)not continuous at x=−1x=-1x=−1
  4. (D)continuous everywhere but not differentiable at x=1x=1x=1

Correct answer: (B)

Step-by-step solution →
Q93·MathematicsNumericalJEE Main 2023
Let f:(−2,2)→Rf:(-2,2)\to\mathbb{R}f:(−2,2)→R be defined by f(x)={x[x],−2<x<0(x−1)[x],0≤x<2f(x)=\begin{cases}x[x] & ,-2<x<0\\(x-1)[x] & ,0\le x<2\end{cases}f(x)={x[x](x−1)[x]​,−2<x<0,0≤x<2​ where [x][x][x] denotes the greatest integer function. If m and n respectively are the number of points in (−2,2)(-2,2)(−2,2) at which y=∣f(x)∣y=|f(x)|y=∣f(x)∣ is not continuous and not differentiable, then m+nm+nm+n is equal to _________.

Correct answer: 4

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2023
If α>β>0\alpha>\beta>0α>β>0 are the roots of the equation ax2+bx+1=0ax^{2}+bx+1=0ax2+bx+1=0, and lim⁡x→1α(1−cos⁡(x2+bx+a)2(1−αx)2)12=1k(1β−1α)\displaystyle\lim_{x\to\frac{1}{\alpha}}\left(\dfrac{1-\cos(x^{2}+bx+a)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\dfrac{1}{k}\left(\dfrac{1}{\beta}-\dfrac{1}{\alpha}\right)x→α1​lim​(2(1−αx)21−cos(x2+bx+a)​)21​=k1​(β1​−α1​), then k is equal to
  1. (A)2β2\beta2β
  2. (B)2α2\alpha2α
  3. (C)α\alphaα
  4. (D)β\betaβ

Correct answer: (B)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2023
lim⁡x→0((1−cos⁡2(3x)cos⁡3(4x))(sin⁡3(4x)(log⁡e(2x+1))5))\lim_{x\to 0}\left(\left(\frac{1-\cos^{2}(3x)}{\cos^{3}(4x)}\right)\left(\frac{\sin^{3}(4x)}{(\log_{e}(2x+1))^{5}}\right)\right)limx→0​((cos3(4x)1−cos2(3x)​)((loge​(2x+1))5sin3(4x)​)) is equal to
  1. (A)9
  2. (B)18
  3. (C)15
  4. (D)24

Correct answer: (B)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2023
lim⁡n→∞{(212−213)(212−215)⋯(212−212n+1)}\lim_{n\to\infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)\cdots\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}limn→∞​{(221​−231​)(221​−251​)⋯(221​−22n+11​)} is equal to
  1. (A)12\frac{1}{\sqrt{2}}2​1​
  2. (B)1
  3. (C)2\sqrt{2}2​
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2023
Let a∈Za\in\mathbb{Z}a∈Z and [t][t][t] be the greatest integer ≤t\le t≤t. Then the number of points, where the function f(x)=[a+13sin⁡x], x∈(0,π)f(x)=[a+13\sin x],\,x\in(0,\pi)f(x)=[a+13sinx],x∈(0,π) is not differentiable, is _____.

Correct answer: 25

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2023
lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\displaystyle\lim_{x\to\infty}\dfrac{(\sqrt{3x+1}+\sqrt{3x-1})^6+(\sqrt{3x+1}-\sqrt{3x-1})^6}{(x+\sqrt{x^2-1})^6+(x-\sqrt{x^2-1})^6}x^3x→∞lim​(x+x2−1​)6+(x−x2−1​)6(3x+1​+3x−1​)6+(3x+1​−3x−1​)6​x3
  1. (A)does not exist
  2. (B)is equal to 272727
  3. (C)is equal to 272\dfrac{27}{2}227​
  4. (D)is equal to 999

Correct answer: (B)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2023
Let f,gf, gf,g and hhh be the real valued functions defined on R\mathbb{R}R as f(x)={x∣x∣,x≠01,x=0f(x)=\begin{cases} \dfrac{x}{|x|}, & x \ne 0 \\ 1, & x=0 \end{cases}f(x)=⎩⎨⎧​∣x∣x​,1,​x=0x=0​, g(x)={sin⁡(x+1)(x+1),x≠−11,x=−1g(x)=\begin{cases} \dfrac{\sin(x + 1)}{(x + 1)}, & x \ne -1 \\ 1, & x=-1 \end{cases}g(x)=⎩⎨⎧​(x+1)sin(x+1)​,1,​x=−1x=−1​ and h(x)=2[x]−f(x)h(x)=2[x] - f(x)h(x)=2[x]−f(x), where [x][x][x] is the greatest integer ≤x\le x≤x. Then the value of lim⁡x→1g(h(x−1))\displaystyle\lim_{x \to 1} g(h(x - 1))x→1lim​g(h(x−1)) is:
  1. (A)−1-1−1
  2. (B)000
  3. (C)sin⁡(1)\sin(1)sin(1)
  4. (D)111

Correct answer: (D)

Step-by-step solution →
Q100·MathematicsNumericalJEE Main 2023
lim⁡x→048x4∫0xt3t4+1 dt\lim_{x\to0}\dfrac{48}{x^4}\int_0^x\dfrac{t^3}{t^4+1}\,dtlimx→0​x448​∫0x​t4+1t3​dt is equal to

Correct answer: 12

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2023
Let x=2x = 2x=2 be a root of the equation x2+px+q=0x^2 + px + q = 0x2+px+q=0 and f(x)={1−cos⁡(x2−4px+q2+8q+16)(x−2p)4,x≠2p0,x=2pf(x) = \begin{cases} \dfrac{1 - \cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4}, & x \neq 2p \\ 0, & x = 2p \end{cases}f(x)=⎩⎨⎧​(x−2p)41−cos(x2−4px+q2+8q+16)​,0,​x=2px=2p​. Then lim⁡x→2p+[f(x)]\lim_{x \to 2p^+} [f(x)]limx→2p+​[f(x)], where [⋅][\cdot][⋅] denotes greatest integer function, is
  1. (A)0
  2. (B)−1-1−1
  3. (C)2
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2023
If the function f(x)={(1+∣cos⁡x∣)λ∣cos⁡x∣,0<x<π2μ,x=π2cot⁡6xecot⁡4x,π2<x<πf(x) = \begin{cases} (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}}, & 0 < x < \dfrac{\pi}{2} \\ \mu, & x = \dfrac{\pi}{2} \\ \dfrac{\cot 6x}{e^{\cot 4x}}, & \dfrac{\pi}{2} < x < \pi \end{cases}f(x)=⎩⎨⎧​(1+∣cosx∣)∣cosx∣λ​,μ,ecot4xcot6x​,​0<x<2π​x=2π​2π​<x<π​ is continuous at x=π2x = \dfrac{\pi}{2}x=2π​, then 9λ+6log⁡eμ+μ6−e6λ9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}9λ+6loge​μ+μ6−e6λ is equal to
  1. (A)10
  2. (B)2e4+82e^4 + 82e4+8
  3. (C)11
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2023
The value of lim⁡n→∞1+2−3+4+5−6+⋯+(3n−2)+(3n−1)−3n2n4+4n+3−n4+5n+4\displaystyle\lim_{n\to\infty}\dfrac{1+2-3+4+5-6+\cdots+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}n→∞lim​2n4+4n+3​−n4+5n+4​1+2−3+4+5−6+⋯+(3n−2)+(3n−1)−3n​ is:
  1. (A)32(2+1)\dfrac{3}{2}(\sqrt{2}+1)23​(2​+1)
  2. (B)322\dfrac{3}{2\sqrt{2}}22​3​
  3. (C)2+12\dfrac{\sqrt{2}+1}{2}22​+1​
  4. (D)3(2+1)3(\sqrt{2}+1)3(2​+1)

Correct answer: (A)

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
The set of all values of aaa for which lim⁡x→a([x−5]−[2x+2])=0\lim_{x\to a}([x-5]-[2x+2])=0limx→a​([x−5]−[2x+2])=0, where [α][\alpha][α] denotes the greatest integer less than or equal to α\alphaα, is equal to
  1. (A)[−7.5,−6.5)[-7.5,-6.5)[−7.5,−6.5)
  2. (B)[−7.5,−6.5][-7.5,-6.5][−7.5,−6.5]
  3. (C)(−7.5,−6.5](-7.5,-6.5](−7.5,−6.5]
  4. (D)(−7.5,−6.5)(-7.5,-6.5)(−7.5,−6.5)

Correct answer: (D)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2023
lim⁡t→0(11sin⁡2t+21sin⁡2t+⋯+n1sin⁡2t)sin⁡2t\lim\limits_{t\to 0}\left(1^{\frac{1}{\sin^2 t}}+2^{\frac{1}{\sin^2 t}}+\cdots+n^{\frac{1}{\sin^2 t}}\right)^{\sin^2 t}t→0lim​(1sin2t1​+2sin2t1​+⋯+nsin2t1​)sin2t is equal to
  1. (A)n2n^2n2
  2. (B)n(n+1)2\tfrac{n(n+1)}{2}2n(n+1)​
  3. (C)nnn
  4. (D)n2+nn^2+nn2+n

Correct answer: (C)

Step-by-step solution →
Q106·MathematicsNumericalJEE Advanced 2022
Let α\alphaα be a positive real number. Let f:R→Rf : R \rightarrow Rf:R→R and g:(α,∞)→Rg : (\alpha, \infty) \rightarrow Rg:(α,∞)→R be the functions defined by f(x)=sin⁡(πx12)f(x) = \sin\left(\frac{\pi x}{12}\right)f(x)=sin(12πx​) and g(x)=2log⁡e(x−α)log⁡e(ex−eα)g(x) = \frac{2\log_{e}\left(\sqrt{x} - \sqrt{\alpha}\right)}{\log_{e}\left(e^{\sqrt{x}} - e^{\sqrt{\alpha}}\right)}g(x)=loge​(ex​−eα​)2loge​(x​−α​)​. Then the value of lim⁡x→α+f(g(x))\lim_{x \rightarrow \alpha^{+}} f\left(g(x)\right)limx→α+​f(g(x)) is __________.

Correct answer: 0.5

Step-by-step solution →
Q107·MathematicsIntegerJEE Advanced 2022
If β=lim⁡x→0ex3−(1−x3)1/3+((1−x2)1/2−1)sin⁡xxsin⁡2x\beta = \lim_{x \to 0} \frac{e^{x^{3}} - \left(1 - x^{3}\right)^{1/3} + \left(\left(1 - x^{2}\right)^{1/2} - 1\right)\sin x}{x\sin^{2}x}β=limx→0​xsin2xex3−(1−x3)1/3+((1−x2)1/2−1)sinx​, then the value of 6β is ________.

Correct answer: 5

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2022
Let the function f(x)={log⁡e(1+5x)−log⁡e(1+αx)x;if x≠010;if x=0f(x) = \begin{cases} \dfrac{\log_e (1 + 5x) - \log_e (1 + \alpha x)}{x} & ; \text{if } x \neq 0 \\ 10 & ; \text{if } x = 0 \end{cases}f(x)=⎩⎨⎧​xloge​(1+5x)−loge​(1+αx)​10​;if x=0;if x=0​ be continuous at x=0x = 0x=0. The α\alphaα is equal to :
  1. (A)101010
  2. (B)−10-10−10
  3. (C)555
  4. (D)−5-5−5

Correct answer: (D)

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2022
If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\lim_{x \to 0} \frac{\alpha e^{x} + \beta e^{-x} + \gamma \sin x}{x \sin^{2} x} = \frac{2}{3}limx→0​xsin2xαex+βe−x+γsinx​=32​, where α,β,γ∈R\alpha, \beta, \gamma \in Rα,β,γ∈R, then which of the following is NOT correct ?
  1. (A)α2+β2+γ2=6\alpha^{2} + \beta^{2} + \gamma^{2} = 6α2+β2+γ2=6
  2. (B)αβ+βγ+γα+1=0\alpha\beta + \beta\gamma + \gamma\alpha + 1 = 0αβ+βγ+γα+1=0
  3. (C)αβ2+βγ2+γα2+3=0\alpha\beta^{2} + \beta\gamma^{2} + \gamma\alpha^{2} + 3 = 0αβ2+βγ2+γα2+3=0
  4. (D)α2−β2+γ2=4\alpha^{2} - \beta^{2} + \gamma^{2} = 4α2−β2+γ2=4

Correct answer: (C)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2022
The function f:R→Rf: R \rightarrow Rf:R→R defined by f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2nf(x)=\lim_{n \rightarrow \infty} \frac{\cos(2\pi x)-x^{2n}\sin(x-1)}{1+x^{2n+1}-x^{2n}}f(x)=limn→∞​1+x2n+1−x2ncos(2πx)−x2nsin(x−1)​ is continuous for all x in
  1. (A)R−{−1}R-\{-1\}R−{−1}
  2. (B)R−{−1,1}R-\{-1,1\}R−{−1,1}
  3. (C)R−{1}R-\{1\}R−{1}
  4. (D)R−{0}R-\{0\}R−{0}

Correct answer: (B)

Step-by-step solution →
Q111·MathematicsNumericalJEE Main 2022
lim⁡x→0((x+2cos⁡x)3+2(x+2cos⁡x)2+3sin⁡(x+2cos⁡x)(x+2)3+2(x+2)2+3sin⁡(x+2))100x\lim_{x \to 0} \left( \frac{\left(x + 2\cos x\right)^{3} + 2\left(x + 2\cos x\right)^{2} + 3\sin\left(x + 2\cos x\right)}{\left(x + 2\right)^{3} + 2\left(x + 2\right)^{2} + 3\sin\left(x + 2\right)} \right)^{\frac{100}{x}}limx→0​((x+2)3+2(x+2)2+3sin(x+2)(x+2cosx)3+2(x+2cosx)2+3sin(x+2cosx)​)x100​ is equal to ___________.

Correct answer: 1

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2022
Let a function f:R→Rf:\mathbb{R}\rightarrow\mathbb{R}f:R→R be defined as : f(x)={∫0x(5−∣t−3∣)dt,x>4x2+bx,x≤4f\left(x\right)=\begin{cases}\int_{0}^{x}\left(5-\left|t-3\right|\right)dt, & x>4\\ x^{2}+bx, & x\le 4\end{cases}f(x)={∫0x​(5−∣t−3∣)dt,x2+bx,​x>4x≤4​ where b∈Rb\in\mathbb{R}b∈R. If fff is continuous at x = 4, then which of the following statements is NOT true ?
  1. (A)fff is not differentiable at x = 4
  2. (B)f ′(3)+f ′(5)=354f\,'\left(3\right)+f\,'\left(5\right)=\frac{35}{4}f′(3)+f′(5)=435​
  3. (C)fff is increasing in (−∞,18)∪(8,∞)\left(-\infty,\frac{1}{8}\right)\cup\left(8,\infty\right)(−∞,81​)∪(8,∞)
  4. (D)fff has a local minima at x=18x=\frac{1}{8}x=81​

Correct answer: (C)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2022
If for p≠q≠0p \neq q \neq 0p=q=0, then function f(x)=p(729+x)7−3729+qx3−9f(x) = \frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}f(x)=3729+qx​−97p(729+x)​−3​ is continuous at x=0x = 0x=0, then:
  1. (A)7pq f(0)−1=07pq\,f(0)-1=07pqf(0)−1=0
  2. (B)63q f(0)−p2=063q\,f(0)-p^{2}=063qf(0)−p2=0
  3. (C)21q f(0)−p2=021q\,f(0)-p^{2}=021qf(0)−p2=0
  4. (D)7pq f(0)−9=07pq\,f(0)-9=07pqf(0)−9=0

Correct answer: (B)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2022
If f(x)={x+a,x≤0∣x−4∣,x>0f(x) = \begin{cases} x + a, & x \le 0 \\ |x - 4|, & x > 0 \end{cases}f(x)={x+a,∣x−4∣,​x≤0x>0​ and g(x)={x+1,x<0(x−4)2+b,x≥0g(x) = \begin{cases} x + 1, & x < 0 \\ (x - 4)^2 + b, & x \ge 0 \end{cases}g(x)={x+1,(x−4)2+b,​x<0x≥0​ are continuous on R, then (gof) (2) + (fog) (−2) is equal to :
  1. (A)−10
  2. (B)10
  3. (C)8
  4. (D)−8

Correct answer: (D)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2022
If the function f(x)={log⁡e(1−x+x2)+log⁡e(1+x+x2)sec⁡x−cos⁡x,x∈(−π2,π2)−{0}k,x=0f(x) = \begin{cases} \dfrac{\log_e(1 - x + x^2) + \log_e(1 + x + x^2)}{\sec x - \cos x}, & x \in \left( \dfrac{-\pi}{2}, \dfrac{\pi}{2} \right) - \{0\} \\ k, & x = 0 \end{cases}f(x)=⎩⎨⎧​secx−cosxloge​(1−x+x2)+loge​(1+x+x2)​,k,​x∈(2−π​,2π​)−{0}x=0​ is continuous at x = 0, then k is equal to :
  1. (A)1
  2. (B)−1
  3. (C)eee
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2022
Let β=lim⁡x→0αx−(e3x−1)αx(e3x−1)\beta=\lim\limits_{x\to 0}\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}β=x→0lim​αx(e3x−1)αx−(e3x−1)​ for some α∈R\alpha\in\mathbb{R}α∈R. Then the value of α+β\alpha+\betaα+β is :
  1. (A)145\frac{14}{5}514​
  2. (B)32\frac{3}{2}23​
  3. (C)52\frac{5}{2}25​
  4. (D)72\frac{7}{2}27​

Correct answer: (C)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2022
If lim⁡n→∞(n2−n−1+nα+β)=0\lim_{n \to \infty} \left( \sqrt{n^{2} - n - 1} + n\alpha + \beta \right) = 0limn→∞​(n2−n−1​+nα+β)=0 then 8(α+β)8(\alpha + \beta)8(α+β) is equal to :
  1. (A)444
  2. (B)−8-8−8
  3. (C)−4-4−4
  4. (D)888

Correct answer: (C)

Step-by-step solution →
Q118·MathematicsNumericalJEE Main 2022
Let f(x)={∣4x2−8x+5∣,if 8x2−6x+1≥0[4x2−8x+5],if 8x2−6x+1<0f(x) = \begin{cases} |4x^{2} - 8x + 5|, & \text{if } 8x^{2} - 6x + 1 \ge 0 \\ [4x^{2} - 8x + 5], & \text{if } 8x^{2} - 6x + 1 < 0 \end{cases}f(x)={∣4x2−8x+5∣,[4x2−8x+5],​if 8x2−6x+1≥0if 8x2−6x+1<0​, where [α][\alpha][α] denotes the greatest integer less than or equal to α\alphaα. Then the number of points in R where f is not differentiable is ________.

Correct answer: 3

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2022
The value of lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\lim_{x \to 1} \frac{\left(x^{2} - 1\right)\sin^{2}\left(\pi x\right)}{x^{4} - 2x^{3} + 2x - 1}limx→1​x4−2x3+2x−1(x2−1)sin2(πx)​ is equal to:
  1. (A)π26\frac{\pi^{2}}{6}6π2​
  2. (B)π23\frac{\pi^{2}}{3}3π2​
  3. (C)π22\frac{\pi^{2}}{2}2π2​
  4. (D)π2\pi^{2}π2

Correct answer: (D)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2022
Let f,g:R→Rf, g : \mathbb{R} \to \mathbb{R}f,g:R→R be functions defined by f(x)={[x],x<0∣1−x∣,x≥0f(x) = \begin{cases} [x] & , \quad x < 0 \\ |1 - x| & , \quad x \geq 0 \end{cases}f(x)={[x]∣1−x∣​,x<0,x≥0​ and g(x)={ex−x,x<0(x−1)2−1,x≥0g(x) = \begin{cases} e^{x} - x & , \quad x < 0 \\ (x-1)^{2} - 1 & , \quad x \geq 0 \end{cases}g(x)={ex−x(x−1)2−1​,x<0,x≥0​ where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :
  1. (A)one point
  2. (B)two points
  3. (C)three points
  4. (D)four points

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsNumericalJEE Main 2022
If lim⁡x→1sin⁡(3x2−4x+1)−x2+12x3−7x2+ax+b=−2\lim\limits_{x\to 1} \frac{\sin\left(3x^2 - 4x + 1\right) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2x→1lim​2x3−7x2+ax+bsin(3x2−4x+1)−x2+1​=−2, then the value of (a−b)(a - b)(a−b) is equal to

Correct answer: 11

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2022
Let f:R→R\mathrm{f}:\mathbb{R} \to \mathbb{R}f:R→R be defined as f(x)={[ex],x<0aex+[x−1],0≤x<1b+[sin⁡(πx)],1≤x<2[e−x]−c,x≥2\mathrm{f(x)}=\begin{cases} [e^{x}], & x<0 \\ ae^{x}+[x-1], & 0 \le x < 1 \\ b+[\sin(\pi x)], & 1 \le x < 2 \\ [e^{-x}]-c, & x \ge 2 \end{cases}f(x)=⎩⎨⎧​[ex],aex+[x−1],b+[sin(πx)],[e−x]−c,​x<00≤x<11≤x<2x≥2​ where a,b,c∈Ra, b, c \in \mathbb{R}a,b,c∈R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true ?
  1. (A)There exists a,b,c∈Ra, b, c \in \mathbb{R}a,b,c∈R such that f is continuous of R\mathbb{R}R.
  2. (B)If f is discontinuous at exactly one point, then a+b+c=1a + b + c = 1a+b+c=1.
  3. (C)If f is discontinuous at exactly one point, then a+b+c≠1a + b + c \ne 1a+b+c=1.
  4. (D)f is discontinuous at atleast two points, for any values of a, b and c.

Correct answer: (C)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2022
Let a be an integer such that lim⁡x→718−[1−x][x−3a]\lim_{x \to 7} \frac{18 - \left[1 - x\right]}{\left[x - 3a\right]}limx→7​[x−3a]18−[1−x]​ exists, where [t] is greatest integer ≤\leq≤ t. Then a is equal to :
  1. (A)-6
  2. (B)-2
  3. (C)2
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q124·MathematicsNumericalJEE Main 2022
Let [t] denote the greatest integer ≤\le≤ t and {t} denote the fractional part of t. Then integral value of α\alphaα for which the left hand limit of the function f(x)=[1+x]+α2[x]+{x}+[x]−12[x]+{x}f(x)=[1+x]+\dfrac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}f(x)=[1+x]+2[x]+{x}α2[x]+{x}+[x]−1​ at x = 0 is equal to α−43\alpha-\dfrac{4}{3}α−34​ is ______

Correct answer: 3

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2022
Let f (x) = min {1, 1 + x sin x}, 0 ≤ x ≤ 2π. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair (m, n) is equal to
  1. (A)(2, 0)
  2. (B)(1, 0)
  3. (C)(1, 1)
  4. (D)(2, 1)

Correct answer: (B)

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2022
lim⁡x→0cos⁡(sin⁡x)−cos⁡xx4\lim_{x\to 0}\frac{\cos(\sin x)-\cos x}{x^4}limx→0​x4cos(sinx)−cosx​ is equal to :
  1. (A)13\frac{1}{3}31​
  2. (B)14\frac{1}{4}41​
  3. (C)16\frac{1}{6}61​
  4. (D)112\frac{1}{12}121​

Correct answer: (C)

Step-by-step solution →
Q127·MathematicsNumericalJEE Main 2022
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R satisfy f(x+y)=2xf(y)+4yf(x)f(x+y)=2^{x}f(y)+4^{y}f(x)f(x+y)=2xf(y)+4yf(x), ∀x, y ∈ R\mathbb{R}R. If f(2)=3f(2)=3f(2)=3, then 14⋅f′(4)f′(2)14\cdot\frac{f'(4)}{f'(2)}14⋅f′(2)f′(4)​ is equal to ______.

Correct answer: 248

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2022
Let f, g : R →\to→ R be two real valued functions defined as f(x)={−∣x+3∣,x<0ex,x≥0f(x) = \begin{cases} -|x+3| & , & x < 0 \\ e^{x} & , & x \geq 0 \end{cases}f(x)={−∣x+3∣ex​,,​x<0x≥0​ and g(x)={x2+k1x,x<04x+k2,x≥0g(x) = \begin{cases} x^{2} + k_{1}x & , & x < 0 \\ 4x + k_{2} & , & x \geq 0 \end{cases}g(x)={x2+k1​x4x+k2​​,,​x<0x≥0​, where k1k_{1}k1​ and k2k_{2}k2​ are real constants. If (gof) is differentiable at x = 0, then (gof) (-4) + (gof) (4) is equal to :
  1. (A)4(e4+1)4(e^{4} + 1)4(e4+1)
  2. (B)2(2e4+1)2(2e^{4} + 1)2(2e4+1)
  3. (C)4e44e^{4}4e4
  4. (D)2(2e4−1)2(2e^{4} - 1)2(2e4−1)

Correct answer: (D)

Step-by-step solution →
Q129·MathematicsSingle correctJEE Main 2022
lim⁡x→12sin⁡(cos⁡−1x)−x1−tan⁡(cos⁡−1x)\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin(\cos^{-1} x) - x}{1 - \tan(\cos^{-1} x)}limx→2​1​​1−tan(cos−1x)sin(cos−1x)−x​ is equal to :
  1. (A)2\sqrt{2}2​
  2. (B)−2-\sqrt{2}−2​
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)−12-\frac{1}{\sqrt{2}}−2​1​

Correct answer: (D)

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2022
Let f(x)f(x)f(x) be a polynomial function such that f(x)+f′(x)+f′′(x)=x5+64f(x) + f'(x) + f''(x) = x^5 + 64f(x)+f′(x)+f′′(x)=x5+64. Then, the value of lim⁡x→1f(x)x−1\lim_{x \to 1} \frac{f(x)}{x - 1}limx→1​x−1f(x)​
  1. (A)−15-15−15
  2. (B)−60-60−60
  3. (C)606060
  4. (D)151515

Correct answer: (A)

Step-by-step solution →
Q131·MathematicsNumericalJEE Main 2022
Let f(x)=[2x2+1]f(x) = [2x^{2} + 1]f(x)=[2x2+1] and g(x)={2x−3,x<02x+3,x≥0g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}g(x)={2x−3,2x+3,​x<0x≥0​, where [t] is the greatest integer ≤t\leq t≤t. Then, in the open interval (−1,1)(-1, 1)(−1,1), the number of points where fog is discontinuous is equal to ______

Correct answer: 62

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2022
lim⁡x→π2(tan⁡2x((2sin⁡2x+3sin⁡x+4)12−(sin⁡2x+6sin⁡x+2)12))\lim_{x \to \frac{\pi}{2}}\left( \tan^2 x \left( \left(2\sin^2 x + 3\sin x + 4\right)^{\frac{1}{2}} - \left(\sin^2 x + 6\sin x + 2\right)^{\frac{1}{2}} \right) \right)limx→2π​​(tan2x((2sin2x+3sinx+4)21​−(sin2x+6sinx+2)21​)) is equal to
  1. (A)112\frac{1}{12}121​
  2. (B)−118-\frac{1}{18}−181​
  3. (C)−112-\frac{1}{12}−121​
  4. (D)−16-\frac{1}{6}−61​

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsNumericalJEE Main 2022
The number of points where the function f(x)={∣2x2−3x−7∣if x≤−1[4x2−1]if −1<x<1∣x+1∣+∣x−2∣if x≥1f(x)=\begin{cases}|2x^{2}-3x-7| & \text{if } x\le -1\\ [4x^{2}-1] & \text{if } -1<x<1\\ |x+1|+|x-2| & \text{if } x\ge 1\end{cases}f(x)=⎩⎨⎧​∣2x2−3x−7∣[4x2−1]∣x+1∣+∣x−2∣​if x≤−1if −1<x<1if x≥1​ [t][t][t] denotes the greatest integer ≤t\le t≤t, is discontinuous is ______ .

Correct answer: 7

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2022
Let f(x)={sin⁡(x−[x])x−[x],x∈(−2,−1)max⁡{2x,3[∣x∣]},∣x∣<11,otherwisef(x) = \begin{cases} \dfrac{\sin\left(x-[x]\right)}{x-[x]} &,& x \in (-2,-1) \\ \max\left\{2x, 3\left[|x|\right]\right\} &,& |x| < 1 \\ 1 &,& \text{otherwise} \end{cases}f(x)=⎩⎨⎧​x−[x]sin(x−[x])​max{2x,3[∣x∣]}1​,,,​x∈(−2,−1)∣x∣<1otherwise​ where [t][t][t] denotes greatest integer ≤t\le t≤t. If m is the number of points where fff is not continuous and n is the number of points where fff is not differentiable, then the ordered pair (m, n) is :
  1. (A)(3, 3)
  2. (B)(2, 4)
  3. (C)(2, 3)
  4. (D)(3, 4)

Correct answer: (C)

Step-by-step solution →
Q135·MathematicsNumericalJEE Main 2021
Let f(x)=x6+2x4+x3+2x+3f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3f(x)=x6+2x4+x3+2x+3, x ∈ ℝ. Then the natural number n for which lim⁡x→1xnf(1)−f(x)x−1=44\lim_{x \to 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44limx→1​x−1xnf(1)−f(x)​=44 is ________ .

Correct answer: 7

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2021
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a continuous function. Then lim⁡x→π4π4∫2sec⁡2xf(x) dxx2−π216\lim_{x \to \frac{\pi}{4}} \frac{\frac{\pi}{4}\int_{2}^{\sec^{2} x} f(x)\,dx}{x^{2} - \frac{\pi^{2}}{16}}limx→4π​​x2−16π2​4π​∫2sec2x​f(x)dx​ is equal to :
  1. (A)f(2)f(2)f(2)
  2. (B)2f(2)2f(2)2f(2)
  3. (C)2f(2)2f\left(\sqrt{2}\right)2f(2​)
  4. (D)4f(2)4f(2)4f(2)

Correct answer: (B)

Step-by-step solution →
Q137·MathematicsNumericalJEE Main 2021
Let [t] denote the greatest integer ≤ t. The number of points where the function f(x)=[x]∣x2−1∣+sin⁡(π[x]+3)−[x+1],x∈(−2,2)f(x) = [x]\left|x^{2} - 1\right| + \sin\left(\frac{\pi}{[x] + 3}\right) - [x + 1], x \in (-2, 2)f(x)=[x]​x2−1​+sin([x]+3π​)−[x+1],x∈(−2,2) is not continuous is ______ .

Correct answer: 2

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Q138·MathematicsSingle correctJEE Main 2021
lim⁡x→0sin⁡2(πcos⁡4x)x4\lim_{x \to 0} \frac{\sin^{2}\left(\pi \cos^{4} x\right)}{x^{4}}limx→0​x4sin2(πcos4x)​ is equal to :
  1. (A)π2\pi^{2}π2
  2. (B)2π22\pi^{2}2π2
  3. (C)4π24\pi^{2}4π2
  4. (D)4π4\pi4π

Correct answer: (C)

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2021
If α=lim⁡x→π/4tan⁡3x−tan⁡xcos⁡(x+π4)\alpha = \lim_{x \to \pi/4} \frac{\tan^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)}α=limx→π/4​cos(x+4π​)tan3x−tanx​ and β=lim⁡x→0(cos⁡x)cot⁡x\beta = \lim_{x \to 0} (\cos x)^{\cot x}β=limx→0​(cosx)cotx are the roots of the equation, ax2+bx−4=0ax^2 + bx - 4 = 0ax2+bx−4=0, then the ordered pair (a,b)(a, b)(a,b) is :
  1. (A)(1,−3)(1, -3)(1,−3)
  2. (B)(−1,3)(-1, 3)(−1,3)
  3. (C)(−1,−3)(-1, -3)(−1,−3)
  4. (D)(1,3)(1, 3)(1,3)

Correct answer: (D)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2021
If the function f(x)={1xlog⁡e(1+xa1−xb),x<0k,x=0cos⁡2x−sin⁡2x−1x2+1−1,x>0f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2} + 1} - 1}, & x > 0 \end{cases}f(x)=⎩⎨⎧​x1​loge​(1−bx​1+ax​​),k,x2+1​−1cos2x−sin2x−1​,​x<0x=0x>0​ is continuous at x=0x = 0x=0, then 1a+1b+4k\frac{1}{a} + \frac{1}{b} + \frac{4}{k}a1​+b1​+k4​ is equal to :
  1. (A)−5-5−5
  2. (B)5
  3. (C)−4-4−4
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2021
If α,β\alpha, \betaα,β are the distinct roots of x2+bx+c=0x^2 + bx + c = 0x2+bx+c=0, then lim⁡x→βe2(x2+bx+c)−1−2(x2+bx+c)(x−β)2\lim_{x \to \beta} \frac{e^{2\left(x^2+bx+c\right)} - 1 - 2\left(x^2 + bx + c\right)}{\left(x - \beta\right)^2}limx→β​(x−β)2e2(x2+bx+c)−1−2(x2+bx+c)​ is equal to:
  1. (A)b2+4cb^2 + 4cb2+4c
  2. (B)2(b2+4c)2(b^2 + 4c)2(b2+4c)
  3. (C)2(b2−4c)2(b^2 - 4c)2(b2−4c)
  4. (D)b2−4cb^2 - 4cb2−4c

Correct answer: (C)

Step-by-step solution →
Q142·MathematicsSingle correctJEE Main 2021
If lim⁡x→∞(x2−x+1−ax)=b\lim_{x \to \infty}\left(\sqrt{x^{2} - x + 1} - ax\right) = blimx→∞​(x2−x+1​−ax)=b, then the ordered pair (a, b) is:
  1. (A)(1,12)\left(1, \frac{1}{2}\right)(1,21​)
  2. (B)(1,−12)\left(1, -\frac{1}{2}\right)(1,−21​)
  3. (C)(−1,12)\left(-1, \frac{1}{2}\right)(−1,21​)
  4. (D)(−1,−12)\left(-1, -\frac{1}{2}\right)(−1,−21​)

Correct answer: (B)

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2021
If Un=(1+1n2)(1+22n2)2....(1+n2n2)nU_n = \left(1 + \frac{1}{n^2}\right)\left(1 + \frac{2^2}{n^2}\right)^2 .... \left(1 + \frac{n^2}{n^2}\right)^nUn​=(1+n21​)(1+n222​)2....(1+n2n2​)n, then lim⁡n→∞(Un)−4n2\lim_{n \to \infty} (U_n)^{\frac{-4}{n^2}}limn→∞​(Un​)n2−4​ is equal to :
  1. (A)e216\frac{e^2}{16}16e2​
  2. (B)4e\frac{4}{e}e4​
  3. (C)16e2\frac{16}{e^2}e216​
  4. (D)4e2\frac{4}{e^2}e24​

Correct answer: (A)

Step-by-step solution →
Q144·MathematicsNumericalJEE Main 2021
Let a, b ∈\in∈ R, b≠0b \neq 0b=0, Define a function f(x)={asin⁡π2(x−1),for x≤0tan⁡2x−sin⁡2xbx3,for x>0.f(x) = \begin{cases} a\sin\frac{\pi}{2}(x-1), & \text{for } x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^{3}}, & \text{for } x > 0. \end{cases}f(x)={asin2π​(x−1),bx3tan2x−sin2x​,​for x≤0for x>0.​ If f is continuous at x=0x = 0x=0, then 10−ab10 - ab10−ab is equal to ________.

Correct answer: 14

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2021
lim⁡x→2(∑n=19xn(n+1)x2+2(2n+1)x+4)\lim_{x \to 2}\left(\sum_{n=1}^{9} \frac{x}{n(n + 1)x^{2} + 2(2n + 1)x + 4}\right)limx→2​(∑n=19​n(n+1)x2+2(2n+1)x+4x​) is equal to :
  1. (A)944\frac{9}{44}449​
  2. (B)524\frac{5}{24}245​
  3. (C)15\frac{1}{5}51​
  4. (D)736\frac{7}{36}367​

Correct answer: (A)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2021
The value of lim⁡x→0(x1−sin⁡x8−1+sin⁡x8)\lim_{x \to 0} \left( \frac{x}{\sqrt[8]{1 - \sin x} - \sqrt[8]{1 + \sin x}} \right)limx→0​(81−sinx​−81+sinx​x​) is equal to :
  1. (A)4
  2. (B)0
  3. (C)−4
  4. (D)−1

Correct answer: (C)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2021
Let f:R→Rf:R \to Rf:R→R be a function such that f(2)=4f\left(2\right)=4f(2)=4 and f′(2)=1f'\left(2\right)=1f′(2)=1. Then the value of lim⁡x→2x2f(2)−4f(x)x−2\lim_{x \to 2}\frac{x^2 f\left(2\right)-4f\left(x\right)}{x-2}limx→2​x−2x2f(2)−4f(x)​ is equal to :
  1. (A)12
  2. (B)4
  3. (C)16
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q148·MathematicsNumericalJEE Main 2021
Let f:[0,3]→Rf:\left[0,3\right] \to Rf:[0,3]→R be defined by f(x)=min⁡{x−[x],1+[x]−x}f\left(x\right)=\min\left\{x-\left[x\right],1+\left[x\right]-x\right\}f(x)=min{x−[x],1+[x]−x} Where [x]\left[x\right][x] is the greatest integer less than or equal to x. Let P denote the set containing all x∈[0,3]x \in \left[0,3\right]x∈[0,3] where f is discontinuous, and Q denote the set containing all x∈(0,3)x \in \left(0,3\right)x∈(0,3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to ____.

Correct answer: 5

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2021
Let f:(−π4,π4)→Rf : \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \to Rf:(−4π​,4π​)→R be defined as f(x)={(1+∣sin⁡x∣)3a∣sin⁡x∣, −π4<x<0b, x=0ecot⁡4x/cot⁡2x, 0<x<π4f(x) = \begin{cases} \left(1 + |\sin x|\right)^{\frac{3a}{|\sin x|}} & , \ -\frac{\pi}{4} < x < 0 \\ b & , \ x = 0 \\ e^{\cot 4x / \cot 2x} & , \ 0 < x < \frac{\pi}{4} \end{cases}f(x)=⎩⎨⎧​(1+∣sinx∣)∣sinx∣3a​becot4x/cot2x​, −4π​<x<0, x=0, 0<x<4π​​ If f is continuous at x = 0, then the value of 6a+b26a + b^26a+b2 is equal to :
  1. (A)1−e1 - e1−e
  2. (B)e
  3. (C)1+e1 + e1+e
  4. (D)e−1e - 1e−1

Correct answer: (C)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2021
Let f : R → R be defined as f(x)={λ∣x2−5x+6∣μ(5x−x2−6),x<2etan⁡(x−2)x−[x],x>2μ,x=2f(x) = \begin{cases} \dfrac{\lambda\left|x^{2} - 5x + 6\right|}{\mu\left(5x - x^{2} - 6\right)}, & x < 2 \\ e^{\frac{\tan(x-2)}{x-[x]}}, & x > 2 \\ \mu, & x = 2 \end{cases}f(x)=⎩⎨⎧​μ(5x−x2−6)λ​x2−5x+6​​,ex−[x]tan(x−2)​,μ,​x<2x>2x=2​ where [x] is the greatest integer less than or equal to x. If f is continuous at x = 2, then λ+μ\lambda + \muλ+μ is equal to :
  1. (A)e(−e+1)e(-e + 1)e(−e+1)
  2. (B)e(e−2)e(e - 2)e(e−2)
  3. (C)2e−12e - 12e−1
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2021
If [x] be the greatest integer less than or equal to x, then ∑n=8100[(−1)nn2]\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right]∑n=8100​[2(−1)nn​] is equal to :
  1. (A)4
  2. (B)-2
  3. (C)2
  4. (D)0

Correct answer: (A)

Step-by-step solution →
Q152·MathematicsNumericalJEE Main 2021
Consider the function f(x)=P(x)sin⁡(x−2)f(x) = \dfrac{P(x)}{\sin(x-2)}f(x)=sin(x−2)P(x)​ , x≠2x \neq 2x=2 =7= 7=7 , x=2x = 2x=2 where P(x) is a polynomial such that P''(x) is always a constant and P(3) = 9. If (x) is continuous at x = 2, then P(5) is equal to.......

Correct answer: 39

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2021
The lowest integer which is greater than (1+110100)10100\left(1 + \frac{1}{10^{100}}\right)^{10^{100}}(1+101001​)10100 is..........
  1. (A)3
  2. (B)2
  3. (C)4
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2021
Let f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)={x3(1−cos⁡2x)2log⁡e(1+2xe−2x(1−xe−x)2),x≠0α,x=0f(x) = \begin{cases} \dfrac{x^{3}}{(1-\cos 2x)^{2}} \log_{e}\left(\dfrac{1+2xe^{-2x}}{\left(1-xe^{-x}\right)^{2}}\right) , & x \neq 0 \\ \alpha , & x = 0 \end{cases}f(x)=⎩⎨⎧​(1−cos2x)2x3​loge​((1−xe−x)21+2xe−2x​),α,​x=0x=0​ . If fff is continuous at x = 0, then α\alphaα is equal to :
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q155·MathematicsNumericalJEE Main 2021
If the value of lim⁡x→0(2−cos⁡xcos⁡2x)(x+2x2)\lim_{x \to 0} \left(2 - \cos x \sqrt{\cos 2x}\right)^{\left(\frac{x + 2}{x^2}\right)}limx→0​(2−cosxcos2x​)(x2x+2​) is equal to eae^aea, then a is equal to.......

Correct answer: 3

Step-by-step solution →
Q156·MathematicsSingle correctJEE Main 2021
Let a function f : R → R be defined as f(x)={sin⁡x−exif x≤0a+[−x]if 0<x<12x−bif x≥1f(x) = \begin{cases} \sin x - e^x & \text{if } x \leq 0 \\ a + [-x] & \text{if } 0 < x < 1 \\ 2x - b & \text{if } x \geq 1 \end{cases}f(x)=⎩⎨⎧​sinx−exa+[−x]2x−b​if x≤0if 0<x<1if x≥1​ where [x] is the greatest integer less than or equal to x. If f is continuous on R, then (a+b)\left(a+b\right)(a+b) is equal to :
  1. (A)2
  2. (B)5
  3. (C)3
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2021
Let f : R → R be a function defined as f(x)={sin⁡(a+1)x+sin⁡2x2x, if x<0b, if x=0x+bx3−xbx5/2, if x>0f(x) = \begin{cases} \dfrac{\sin(a+1)x + \sin 2x}{2x} & ,\ \text{if } x < 0 \\ b & ,\ \text{if } x = 0 \\ \dfrac{\sqrt{x + bx^{3}} - \sqrt{x}}{bx^{5/2}} & ,\ \text{if } x > 0 \end{cases}f(x)=⎩⎨⎧​2xsin(a+1)x+sin2x​bbx5/2x+bx3​−x​​​, if x<0, if x=0, if x>0​ If f is continuous at x = 0, then the value of a + b is equal to :
  1. (A)−52-\frac{5}{2}−25​
  2. (B)−2
  3. (C)−3
  4. (D)−32-\frac{3}{2}−23​

Correct answer: (D)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2021
If lim⁡x→0sin⁡−1x−tan⁡−1x3x3\lim_{x \to 0} \dfrac{\sin^{-1}x - \tan^{-1}x}{3x^{3}}limx→0​3x3sin−1x−tan−1x​ is equal to L, then the value of (6L+1)(6L + 1)(6L+1) is
  1. (A)16\dfrac{1}{6}61​
  2. (B)12\dfrac{1}{2}21​
  3. (C)666
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q159·MathematicsNumericalJEE Main 2021
If the function f(x)=cos⁡(sin⁡x)−cos⁡xx4f(x) = \frac{\cos(\sin x) - \cos x}{x^4}f(x)=x4cos(sinx)−cosx​ is continuous at each point in its domain and f(0)=1kf(0) = \frac{1}{k}f(0)=k1​, then k is _________ .

Correct answer: 6

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2021
The value of lim⁡n→∞[r]+[2r]+.....+[nr]n2\lim_{n\rightarrow\infty}\frac{[r]+[2r]+.....+[nr]}{n^{2}}limn→∞​n2[r]+[2r]+.....+[nr]​, where rrr is non-zero real number and [r][r][r] denotes the greatest integer less than or equal to rrr, is equal to :
  1. (A)r2\frac{r}{2}2r​
  2. (B)rrr
  3. (C)2r2r2r
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2021
The value of the limit lim⁡θ→0tan⁡(πcos⁡2θ)sin⁡(2πsin⁡2θ)\lim_{\theta\rightarrow 0}\frac{\tan\left(\pi\cos^{2}\theta\right)}{\sin\left(2\pi\sin^{2}\theta\right)}limθ→0​sin(2πsin2θ)tan(πcos2θ)​ is equal to :
  1. (A)−12-\frac{1}{2}−21​
  2. (B)−14-\frac{1}{4}−41​
  3. (C)000
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2021
The value of lim⁡x→0+cos⁡−1(x−[x]2)⋅sin⁡−1(x−[x]2)x−x3\lim_{x \to 0^{+}} \frac{\cos^{-1}\left(x - [x]^2\right) \cdot \sin^{-1}\left(x - [x]^2\right)}{x - x^3}limx→0+​x−x3cos−1(x−[x]2)⋅sin−1(x−[x]2)​, where [x][x][x] denotes the greatest integer ≤x\leq x≤x is :
  1. (A)π\piπ
  2. (B)0
  3. (C)π4\frac{\pi}{4}4π​
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (D)

Step-by-step solution →
Q163·MathematicsNumericalJEE Main 2021
Let f:R→Rf : R \rightarrow Rf:R→R and g:R→Rg : R \rightarrow Rg:R→R be defined as f(x)={x+a,x<0∣x−1∣,x≥0f(x)=\begin{cases} x+a, & x<0 \\ |x-1|, & x \ge 0 \end{cases}f(x)={x+a,∣x−1∣,​x<0x≥0​ and g(x)={x+1,x<0(x−1)2+b,x≥0g(x)=\begin{cases} x+1, & x<0 \\ (x-1)^{2}+b, & x \ge 0 \end{cases}g(x)={x+1,(x−1)2+b,​x<0x≥0​ where a, b are non-negative real numbers. If (gof)(x)(gof)(x)(gof)(x) is continuous for all x∈Rx \in Rx∈R, then a+ba + ba+b is equal to ___________.

Correct answer: 1

Step-by-step solution →
Q164·MathematicsNumericalJEE Main 2021
If lim⁡x→0aex−bcos⁡x+ce−xxsin⁡x=2\lim_{x \to 0}\frac{ae^{x} - b\cos x + ce^{-x}}{x\sin x} = 2limx→0​xsinxaex−bcosx+ce−x​=2, then a + b + c is equal to _______.

Correct answer: 4

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2021
Let α∈R\alpha \in Rα∈R be such that the function f(x)={cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3,x≠0α,x=0f(x)=\begin{cases} \dfrac{\cos^{-1}(1-\{x\}^{2})\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{cases}f(x)=⎩⎨⎧​{x}−{x}3cos−1(1−{x}2)sin−1(1−{x})​,α,​x=0x=0​ is continuous at x=0x = 0x=0, where {x}=x−[x]\{x\} = x - [x]{x}=x−[x], [x][x][x] is the greatest integer less than or equal to x. Then :
  1. (A)α=π2\alpha = \frac{\pi}{\sqrt{2}}α=2​π​
  2. (B)α=0\alpha = 0α=0
  3. (C)no such α\alphaα exists
  4. (D)α=π4\alpha = \frac{\pi}{4}α=4π​

Correct answer: (C)

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2021
Let f(x) be a differentiable function at x = a with f′(a)=2f'(a) = 2f′(a)=2 and f(a) = 4. Then lim⁡x→axf(a)−af(x)x−a\lim_{x \to a} \frac{xf(a) - af(x)}{x - a}limx→a​x−axf(a)−af(x)​ equals :
  1. (A)2a + 4
  2. (B)2a − 4
  3. (C)4 − 2a
  4. (D)a + 4

Correct answer: (C)

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2021
Let f : R → R be defined as f(x)={2sin⁡(−πx2),if x<−1∣ax2+x+b∣,if −1≤x≤1sin⁡(πx)if x>1f(x) = \begin{cases} 2\sin\left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^{2} + x + b|, & \text{if } -1 \le x \le 1 \\ \sin(\pi x) & \text{if } x > 1 \end{cases}f(x)=⎩⎨⎧​2sin(−2πx​),∣ax2+x+b∣,sin(πx)​if x<−1if −1≤x≤1if x>1​ If f(x) is continuous on R, then a + b equals :
  1. (A)3
  2. (B)−1
  3. (C)−3
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2021
The value of lim⁡h→0{3 sin⁡(π6+h)−cos⁡(π6+h)3h(3 cos⁡h−sin⁡h)}\lim_{h \to 0}\left\{\frac{\sqrt{3}\,\sin\left(\frac{\pi}{6}+h\right)-\cos\left(\frac{\pi}{6}+h\right)}{\sqrt{3}h\left(\sqrt{3}\,\cos h-\sin h\right)}\right\}limh→0​{3​h(3​cosh−sinh)3​sin(6π​+h)−cos(6π​+h)​} is:
  1. (A)34\frac{3}{4}43​
  2. (B)23\frac{2}{\sqrt{3}}3​2​
  3. (C)43\frac{4}{3}34​
  4. (D)23\frac{2}{3}32​

Correct answer: (C)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2021
lim⁡n→∞(1+1+12+⋯+1nn2)n\lim\limits_{n \to \infty} \left( 1 + \dfrac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n^2} \right)^{n}n→∞lim​(1+n21+21​+⋯+n1​​)n is equal to :
  1. (A)12\frac{1}{2}21​
  2. (B)1e\frac{1}{e}e1​
  3. (C)111
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q170·MathematicsNumericalJEE Main 2021
If x_{x}x​lim_{→}_{0}axax− (e(e4x4x_{4x}^{4x}4x4x​−−11)) exists and is equal to b, then the value of a – 2b is ______.

Correct answer: 5

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2021
If f:R→Rf : R \rightarrow Rf:R→R is a function defined by f(x)=[x−1]cos⁡(2x−12)πf(x) = [x-1] \cos\left(\frac{2x - 1}{2}\right)\pif(x)=[x−1]cos(22x−1​)π, where [.][.][.] denotes the greatest integer function, then fff is :
  1. (A)discontinuous only at x=1x = 1x=1
  2. (B)discontinuous at all integral values of xxx except at x=1x = 1x=1
  3. (C)continuous only at x=1x = 1x=1
  4. (D)continuous for every real xxx

Correct answer: (D)

Step-by-step solution →
Q172·MathematicsNumericalJEE Advanced 2020
Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit lim⁡x→0+(1−x)1x−e−1xa\lim_{x \to 0^{+}} \frac{(1 - x)^{\frac{1}{x}} - e^{-1}}{x^{a}}limx→0+​xa(1−x)x1​−e−1​ is equal to a nonzero real number, is ______.

Correct answer: 1.00

Step-by-step solution →
Q173·MathematicsIntegerJEE Advanced 2020
The value of the limit lim⁡x→π242(sin⁡3x+sin⁡x)(2sin⁡2xsin⁡3x2+cos⁡5x2)−(2+2cos⁡2x+cos⁡3x2)\lim_{x \to \frac{\pi}{2}} \frac{4\sqrt{2}(\sin 3x + \sin x)}{\left( 2\sin 2x \sin\frac{3x}{2} + \cos\frac{5x}{2} \right) - \left( \sqrt{2} + \sqrt{2}\cos 2x + \cos\frac{3x}{2} \right)}limx→2π​​(2sin2xsin23x​+cos25x​)−(2​+2​cos2x+cos23x​)42​(sin3x+sinx)​ is ________

Correct answer: 8

Step-by-step solution →
Q174·MathematicsNumericalJEE Main 2020
Let f(x)=x.[x2]f(x)=x.\left[\dfrac{x}{2}\right]f(x)=x.[2x​], for −10<x<10-10<x<10−10<x<10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to ___

Correct answer: 08.00

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2020
lim⁡x→0x(e(1+x2+x4−1)/x−1)1+x2+x4−1\displaystyle\lim_{x\to 0}\dfrac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1}x→0lim​1+x2+x4​−1x(e(1+x2+x4​−1)/x−1)​
  1. (A)does not exist
  2. (B)is equal to 111
  3. (C)is equal to e\sqrt{e}e​
  4. (D)is equal to 000

Correct answer: (B)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2020
lim⁡x→a(a+2x)13−(3x)13(3a+x)13−(4x)13(a≠0)\lim_{x \to a} \frac{\left( a + 2x \right)^{\frac{1}{3}} - \left( 3x \right)^{\frac{1}{3}}}{\left( 3a + x \right)^{\frac{1}{3}} - \left( 4x \right)^{\frac{1}{3}}} \left( a \ne 0 \right)limx→a​(3a+x)31​−(4x)31​(a+2x)31​−(3x)31​​(a=0) is equal to:
  1. (A)(23)43\left( \frac{2}{3} \right)^{\frac{4}{3}}(32​)34​
  2. (B)(29)43\left( \frac{2}{9} \right)^{\frac{4}{3}}(92​)34​
  3. (C)(29)(23)13\left( \frac{2}{9} \right) \left( \frac{2}{3} \right)^{\frac{1}{3}}(92​)(32​)31​
  4. (D)(23)(29)13\left( \frac{2}{3} \right) \left( \frac{2}{9} \right)^{\frac{1}{3}}(32​)(92​)31​

Correct answer: (D)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2020
Let [t][t][t] denote the greatest integer ≤t\leq t≤t. If for some λ∈R−{0,1}\lambda \in R - \{0, 1\}λ∈R−{0,1}, lim⁡x→0∣1−x+∣x∣λ−x+[x]∣=L\lim\limits_{x \to 0} \left| \frac{1 - x + |x|}{\lambda - x + [x]} \right| = Lx→0lim​​λ−x+[x]1−x+∣x∣​​=L, then L is equal to:
  1. (A)0
  2. (B)1
  3. (C)12\frac{1}{2}21​
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q178·MathematicsNumericalJEE Main 2020
If lim⁡x→0{1x8(1−cos⁡x22−cos⁡x24+cos⁡x22cos⁡x24)}=2−k\lim\limits_{x \to 0} \left\{ \frac{1}{x^{8}} \left( 1 - \cos\frac{x^{2}}{2} - \cos\frac{x^{2}}{4} + \cos\frac{x^{2}}{2}\cos\frac{x^{2}}{4} \right) \right\} = 2^{-k}x→0lim​{x81​(1−cos2x2​−cos4x2​+cos2x2​cos4x2​)}=2−k, then the value of k is __________.

Correct answer: 8

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2020
Let [t][t][t] denote the greatest integer ≤t\leq t≤t and lim⁡x→0x[4x]=A\lim\limits_{x \rightarrow 0} x\left[\dfrac{4}{x}\right] = Ax→0lim​x[x4​]=A. Then the function, f(x)=[x2]sin⁡(πx)f(x) = \left[x^{2}\right]\sin(\pi x)f(x)=[x2]sin(πx) is discontinuous, when x is equal to:
  1. (A)A+21\sqrt{A + 21}A+21​
  2. (B)A+5\sqrt{A + 5}A+5​
  3. (C)A\sqrt{A}A​
  4. (D)A+1\sqrt{A + 1}A+1​

Correct answer: (D)

Step-by-step solution →
Q180·MathematicsSingle correctJEE Main 2020
lim⁡x→0∫0xtsin⁡(10t) dtx\displaystyle\lim_{x\to 0}\dfrac{\int_{0}^{x} t\sin(10t)\,dt}{x}x→0lim​x∫0x​tsin(10t)dt​ is equal to:
  1. (A)0
  2. (B)110\dfrac{1}{10}101​
  3. (C)−15-\dfrac{1}{5}−51​
  4. (D)−110-\dfrac{1}{10}−101​

Correct answer: (A)

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2020
lim⁡x→0(3x2+27x2+2)1/x2\displaystyle\lim_{x\to 0}\left(\dfrac{3x^{2}+2}{7x^{2}+2}\right)^{1/x^{2}}x→0lim​(7x2+23x2+2​)1/x2 is equal to:
  1. (A)1e\dfrac{1}{e}e1​
  2. (B)eee
  3. (C)e2e^{2}e2
  4. (D)1e2\dfrac{1}{e^{2}}e21​

Correct answer: (D)

Step-by-step solution →
Q182·MathematicsNumericalJEE Main 2020
If the function f defined on (−13,13)\left(-\dfrac{1}{3},\dfrac{1}{3}\right)(−31​,31​) by f(x)={1xlog⁡e(1+3x1−2x),when x≠0k,when x=0f(x)=\begin{cases}\dfrac{1}{x}\log_{e}\left(\dfrac{1+3x}{1-2x}\right), & \text{when } x \ne 0 \\ k, & \text{when } x=0\end{cases}f(x)=⎩⎨⎧​x1​loge​(1−2x1+3x​),k,​when x=0when x=0​ is continuous, then k is equal to __________.

Correct answer: 5

Step-by-step solution →
Q183·MathematicsNumericalJEE Main 2020
lim⁡x→23x+33−x−123−x/2−31−x\lim\limits_{x \to 2} \frac{3^{x}+3^{3-x}-12}{3^{-x/2}-3^{1-x}}x→2lim​3−x/2−31−x3x+33−x−12​ is equal to _________

Correct answer: 36

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2019
Let f(x)=5−∣x−2∣f(x) = 5 - |x-2|f(x)=5−∣x−2∣ and g(x)=∣x+1∣g(x) = |x+1|g(x)=∣x+1∣, x∈Rx \in Rx∈R. If f(x)f(x)f(x) attains maximum value at α\alphaα and g(x)g(x)g(x) attains minimum value at β\betaβ, then lim⁡x→−αβ(x−1)(x2−5x+6)x2−6x+8\displaystyle\lim_{x \to -\alpha\beta} \frac{(x-1)(x^{2}-5x+6)}{x^{2}-6x+8}x→−αβlim​x2−6x+8(x−1)(x2−5x+6)​ is equal to :
  1. (A)32\dfrac{3}{2}23​
  2. (B)−32\dfrac{-3}{2}2−3​
  3. (C)12\dfrac{1}{2}21​
  4. (D)−12\dfrac{-1}{2}2−1​

Correct answer: (C)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2019
lim⁡x→0x+2sin⁡xx2+2sin⁡x+1−sin⁡2x−x+1\displaystyle\lim_{x\to 0} \dfrac{x + 2\sin x}{\sqrt{x^{2}+2\sin x+1} - \sqrt{\sin^{2}x - x + 1}}x→0lim​x2+2sinx+1​−sin2x−x+1​x+2sinx​ is:
  1. (A)2
  2. (B)6
  3. (C)3
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2019
If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\displaystyle\lim_{x\to 1}\dfrac{x^{4}-1}{x-1}=\lim_{x\to k}\dfrac{x^{3}-k^{3}}{x^{2}-k^{2}}x→1lim​x−1x4−1​=x→klim​x2−k2x3−k3​, then k is
  1. (A)38\dfrac{3}{8}83​
  2. (B)83\dfrac{8}{3}38​
  3. (C)43\dfrac{4}{3}34​
  4. (D)32\dfrac{3}{2}23​

Correct answer: (B)

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2019
If the function f(x)={a∣π−x∣+1,x≤5b∣π−x∣+3,x>5f\left(x\right)=\begin{cases} a\left|\pi-x\right|+1, & x\leq 5 \\ b\left|\pi-x\right|+3, & x>5 \end{cases}f(x)={a∣π−x∣+1,b∣π−x∣+3,​x≤5x>5​ is continuous at x=5x=5x=5, then the value of a−ba-ba−b is
  1. (A)25−π\dfrac{2}{5-\pi}5−π2​
  2. (B)2π−5\dfrac{2}{\pi-5}π−52​
  3. (C)2π+5\dfrac{2}{\pi+5}π+52​
  4. (D)−2π+5\dfrac{-2}{\pi+5}π+5−2​

Correct answer: (A)

Step-by-step solution →
Q188·MathematicsSingle correctJEE Main 2019
If f(x)=[x]−[x4],x∈Rf\left(x\right)=\left[x\right]-\left[\dfrac{x}{4}\right],x\in Rf(x)=[x]−[4x​],x∈R, where [x] denotes the greatest integer function, then:
  1. (A)Both lim⁡x→4−f(x)\lim\limits_{x\to4-}f\left(x\right)x→4−lim​f(x) and lim⁡x→4+f(x)\lim\limits_{x\to4+}f\left(x\right)x→4+lim​f(x) exist but are not equal
  2. (B)lim⁡x→4−f(x)\lim\limits_{x\to4-}f\left(x\right)x→4−lim​f(x) exists but lim⁡x→4+f(x)\lim\limits_{x\to4+}f\left(x\right)x→4+lim​f(x) does not exist
  3. (C)lim⁡x→4+f(x)\lim\limits_{x\to4+}f\left(x\right)x→4+lim​f(x) exists but lim⁡x→4−f(x)\lim\limits_{x\to4-}f\left(x\right)x→4−lim​f(x) does not exist
  4. (D)f is continuous at x = 4

Correct answer: (D)

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2019
If the function f defined on (π6,π3)\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)(6π​,3π​) by f(x)={2cos⁡x−1cot⁡x−1,x≠π4k,x=π4f(x)=\begin{cases}\dfrac{\sqrt{2}\cos x-1}{\cot x-1}, & x\neq \dfrac{\pi}{4}\\ k, & x=\dfrac{\pi}{4}\end{cases}f(x)=⎩⎨⎧​cotx−12​cosx−1​,k,​x=4π​x=4π​​ is continuous, then k is equal to:
  1. (A)1
  2. (B)2
  3. (C)12\dfrac{1}{2}21​
  4. (D)12\dfrac{1}{\sqrt{2}}2​1​

Correct answer: (C)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2019
If f:R→Rf:R \rightarrow Rf:R→R is a differentiable function and f(2)=6f(2)=6f(2)=6, then lim⁡x→2∫6f(x)2t dt(x−2)\displaystyle\lim_{x\to 2}\int_{6}^{f(x)}\dfrac{2t\,dt}{(x-2)}x→2lim​∫6f(x)​(x−2)2tdt​ is:
  1. (A)000
  2. (B)2f′(2)2f'(2)2f′(2)
  3. (C)12f′(2)12f'(2)12f′(2)
  4. (D)24f′(2)24f'(2)24f′(2)

Correct answer: (C)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2019
Let f:R→Rf:R\rightarrow Rf:R→R be a differentiable function satisfying f′′(3)+f′(2)=0f''(3)+f'(2)=0f′′(3)+f′(2)=0. Then lim⁡x→∞(1+f(3+x)−f(3)1+f(2−x)−f(2))1x\lim_{x\rightarrow\infty}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac{1}{x}}limx→∞​(1+f(2−x)−f(2)1+f(3+x)−f(3)​)x1​ is equal to:
  1. (A)e2e^{2}e2
  2. (B)1
  3. (C)e
  4. (D)e−1e^{-1}e−1

Correct answer: (B)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2019
lim⁡x→0sin⁡2x2−1+cos⁡x\displaystyle\lim_{x\to 0}\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}x→0lim​2​−1+cosx​sin2x​ equals
  1. (A)2\sqrt{2}2​
  2. (B)424\sqrt{2}42​
  3. (C)444
  4. (D)222\sqrt{2}22​

Correct answer: (B)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2019
lim⁡x→1−π−2sin⁡−1x1−x\lim_{x\to 1^{-}} \dfrac{\sqrt{\pi} - \sqrt{2\sin^{-1} x}}{\sqrt{1-x}}limx→1−​1−x​π​−2sin−1x​​ is equal to :
  1. (A)12π\dfrac{1}{\sqrt{2\pi}}2π​1​
  2. (B)2π\sqrt{\dfrac{2}{\pi}}π2​​
  3. (C)π2\sqrt{\dfrac{\pi}{2}}2π​​
  4. (D)π\sqrt{\pi}π​

Correct answer: (B)

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2019
lim⁡x→π/4cot⁡3x−tan⁡xcos⁡(x+π/4)\lim_{x\to \pi/4}\frac{\cot^{3}x-\tan x}{\cos\left(x+\pi/4\right)}limx→π/4​cos(x+π/4)cot3x−tanx​ is:
  1. (A)4
  2. (B)424\sqrt{2}42​
  3. (C)828\sqrt{2}82​
  4. (D)8

Correct answer: (D)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2019
lim⁡x→0xcot⁡(4x)sin⁡2xcot⁡2(2x)\displaystyle\lim_{x\to 0}\frac{x\cot(4x)}{\sin^{2}x\cot^{2}(2x)}x→0lim​sin2xcot2(2x)xcot(4x)​ is equal to:
  1. (A)0
  2. (B)2
  3. (C)4
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2019
Let [x] denote the greatest integer less than or equal to x. Then: lim⁡x→0tan⁡(πsin⁡2x)+(∣x∣−sin⁡(x[x]))2x2\lim_{x\to 0}\frac{\tan\left(\pi\sin^{2}x\right)+\left(|x|-\sin\left(x[x]\right)\right)^{2}}{x^{2}}limx→0​x2tan(πsin2x)+(∣x∣−sin(x[x]))2​
  1. (A)does not exist
  2. (B)equals π
  3. (C)equal π + 1
  4. (D)equals 0

Correct answer: (A)

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2019
lim⁡y→01+1+y4−2y4=\displaystyle\lim_{y \to 0} \dfrac{\sqrt{1+\sqrt{1+y^4}}-\sqrt2}{y^4} =y→0lim​y41+1+y4​​−2​​=
  1. (A)exists and equals 142\dfrac{1}{4\sqrt2}42​1​
  2. (B)exists and equals 122(2+1)\dfrac{1}{2\sqrt2(\sqrt2+1)}22​(2​+1)1​
  3. (C)exists and equals 122\dfrac{1}{2\sqrt2}22​1​
  4. (D)does not exist

Correct answer: (A)

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2019
Let f:R→Rf:R \rightarrow Rf:R→R be a function defined as f(x)={5,if x≤1a+bx,if 1<x<3b+5x,if 3≤x<530,if x≥5f(x)=\begin{cases} 5, & \text{if } x \le 1 \\ a+bx, & \text{if } 1 < x < 3 \\ b+5x, & \text{if } 3 \le x < 5 \\ 30, & \text{if } x \ge 5 \end{cases}f(x)=⎩⎨⎧​5,a+bx,b+5x,30,​if x≤1if 1<x<3if 3≤x<5if x≥5​. Then fff is:
  1. (A)continuous if a=5a = 5a=5 and b=5b = 5b=5
  2. (B)continuous if a=5a = 5a=5 and b=10b = 10b=10
  3. (C)continuous if a=0a = 0a=0 and b=5b = 5b=5
  4. (D)not continuous for any values of aaa and bbb

Correct answer: (D)

Step-by-step solution →
Q199·MathematicsNumericalJEE Advanced 2018
For each positive integer n, let yn=1n((n+1)(n+2)....(n+n))1/ny_{n} = \frac{1}{n}\left((n+1)(n+2)....(n+n)\right)^{1/n}yn​=n1​((n+1)(n+2)....(n+n))1/n. For x∈Rx \in Rx∈R, let [x] be the greatest integer less than or equal to x. If lim⁡n→∞yn=L\lim_{n \to \infty} y_{n} = Llimn→∞​yn​=L, then the value of [L] is ______ .

Correct answer: 1

Step-by-step solution →
Q200·MathematicsMultiple correctJEE Advanced 2017
Let [x][x][x] be the greatest integer less than or equals to xxx. Then, at which of the following point(s) the function f(x)=xcos⁡(π(x+[x]))f(x) = x \cos(\pi(x + [x]))f(x)=xcos(π(x+[x])) is discontinuous ?
  1. (A)x=−1x = -1x=−1
  2. (B)x=0x = 0x=0
  3. (C)x=2x = 2x=2
  4. (D)x=1x = 1x=1

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

Step-by-step solution →
Q201·MathematicsMultiple correctJEE Advanced 2017
Let f(x)=1−x(1+∣1−x∣)∣1−x∣cos⁡(11−x)f(x) = \dfrac{1 - x\left(1 + |1-x|\right)}{|1-x|}\cos\left(\dfrac{1}{1-x}\right)f(x)=∣1−x∣1−x(1+∣1−x∣)​cos(1−x1​) for x≠1x \ne 1x=1. Then
  1. (A)lim⁡x→1−f(x)=0\lim\limits_{x \to 1^{-}} f(x) = 0x→1−lim​f(x)=0
  2. (B)lim⁡x→1−f(x)\lim\limits_{x \to 1^{-}} f(x)x→1−lim​f(x) does not exist
  3. (C)lim⁡x→1+f(x)=0\lim\limits_{x \to 1^{+}} f(x) = 0x→1+lim​f(x)=0
  4. (D)lim⁡x→1+f(x)\lim\limits_{x \to 1^{+}} f(x)x→1+lim​f(x) does not exist

Correct answer: (A), (D)

Step-by-step solution →
Q202·MathematicsIntegerJEE Advanced 2016
Let α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R be such that lim⁡x→0x2sin⁡(βx)αx−sin⁡x=1\lim_{x \to 0} \frac{x^{2}\sin(\beta x)}{\alpha x - \sin x} = 1limx→0​αx−sinxx2sin(βx)​=1. Then 6(α+β)6(\alpha + \beta)6(α+β) equals

Correct answer: 7

Step-by-step solution →
Q203·MathematicsIntegerJEE Advanced 2015
Let m and n be two positive integers greater than 1. If lim⁡α→0(ecos⁡(αn)−eαm)=−(e2)\displaystyle\lim_{\alpha \to 0}\left(\dfrac{e^{\cos(\alpha^{n})} - e}{\alpha^{m}}\right) = -\left(\dfrac{e}{2}\right)α→0lim​(αmecos(αn)−e​)=−(2e​) then the value of mn\dfrac{m}{n}nm​ is

Correct answer: 2

Step-by-step solution →
Q204·MathematicsMultiple correctJEE Advanced 2014
For every pair of continuous functions f,g:[0,1]→Rf, g: [0, 1] \to \mathbb{R}f,g:[0,1]→R such that max⁡{f(x):x∈[0,1]}=max⁡{g(x):x∈[0,1]}\max\{ f(x) : x \in [0, 1]\} = \max\{g(x) : x \in [0, 1]\}max{f(x):x∈[0,1]}=max{g(x):x∈[0,1]}, the correct statement(s) is(are)
  1. (A)(f(c))2+3f(c)=(g(c))2+3g(c)(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)(f(c))2+3f(c)=(g(c))2+3g(c) for some c∈[0,1]c \in [0, 1]c∈[0,1]
  2. (B)(f(c))2+f(c)=(g(c))2+3g(c)(f(c))^2 + f(c) = (g(c))^2 + 3g(c)(f(c))2+f(c)=(g(c))2+3g(c) for some c∈[0,1]c \in [0, 1]c∈[0,1]
  3. (C)(f(c))2+3f(c)=(g(c))2+g(c)(f(c))^2 + 3f(c) = (g(c))^2 + g(c)(f(c))2+3f(c)=(g(c))2+g(c) for some c∈[0,1]c \in [0, 1]c∈[0,1]
  4. (D)(f(c))2=(g(c))2(f(c))^2 = (g(c))^2(f(c))2=(g(c))2 for some c∈[0,1]c \in [0, 1]c∈[0,1]

Correct answer: (A), (D)

Step-by-step solution →
Q205·MathematicsIntegerJEE Advanced 2014
The largest value of the non-negative integer aaa for which lim⁡x→1{−ax+sin⁡(x−1)+ax+sin⁡(x−1)−1}1−x1−x=14\lim_{x \to 1} \left\{\frac{-ax + \sin(x - 1) + a}{x + \sin(x - 1) - 1}\right\}^{\frac{1-x}{1-\sqrt{x}}} = \frac{1}{4}limx→1​{x+sin(x−1)−1−ax+sin(x−1)+a​}1−x​1−x​=41​ is __________

Correct answer: 2

Step-by-step solution →
Q206·MathematicsMultiple correctJEE Advanced 2013
For a∈Ra \in \mathbb{R}a∈R (the set of all real numbers), a≠−1a \neq -1a=−1, lim⁡n→∞(1a+2a+…+na)(n+1)a−1[(na+1)+(na+2)+…+(na+n)]=160\lim_{n \to \infty} \frac{\left(1^{a} + 2^{a} + \ldots + n^{a}\right)}{(n+1)^{a-1}\left[(na+1) + (na+2) + \ldots + (na+n)\right]} = \frac{1}{60}limn→∞​(n+1)a−1[(na+1)+(na+2)+…+(na+n)](1a+2a+…+na)​=601​. Then a=a =a=
  1. (A)555
  2. (B)777
  3. (C)−152\frac{-15}{2}2−15​
  4. (D)−172\frac{-17}{2}2−17​

Correct answer: (B), (D)

Step-by-step solution →

Limits and Continuity — frequently asked

How many questions from Limits and Continuity appear in JEE?

Limits and Continuity has appeared in 149 of the last 186 JEE Main and JEE Advanced papers — about 80% of them — contributing 206 questions in total across those papers.

Is Limits and Continuity an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 80% 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 Limits and Continuity 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
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  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

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