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Differentiability — JEE Previous Year Questions

Every Differentiability question asked in JEE Main and JEE Advanced across the last 186 papers — 105 questions, each with its correct answer. Free to read, no account needed.

Questions

105

Papers it appeared in

91/186

Appearance rate

49%

All 105 Differentiability questions

Most recent papers first.

Q1·MathematicsMultiple correctJEE Advanced 2026
Let R\mathbb{R}R denote the set of all real numbers. Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be an arbitrary function and let g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be the function defined by g(x)=xf(x)g(x) = x f(x)g(x)=xf(x), for all x∈Rx \in \mathbb{R}x∈R. Then which of the following statements is (are) TRUE ?
  1. (A)The function ggg is always continuous at x=0x = 0x=0
  2. (B)If fff is continuous at x=0x = 0x=0, then ggg is differentiable at x=0x = 0x=0
  3. (C)If ggg is differentiable at x=0x = 0x=0, then fff is continuous at x=0x = 0x=0
  4. (D)If ggg is differentiable at x=0x = 0x=0, then lim⁡x→0f(x)\lim_{x \to 0} f(x)limx→0​f(x) exists

Correct answer: (B), (D)

Step-by-step solution →
Q2·MathematicsIntegerJEE Advanced 2026
Consider the function f:(−π2,π2)→(−∞,∞)f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to (-\infty, \infty)f:(−2π​,2π​)→(−∞,∞) defined by f(x)=(∣x∣+∣x−1∣)sin⁡x+[xsin⁡x]f(x) = (|x| + |x - 1|) \sin x + [x \sin x]f(x)=(∣x∣+∣x−1∣)sinx+[xsinx], where [xsin⁡x][x \sin x][xsinx] is the greatest integer less than or equal to xsin⁡xx \sin xxsinx. Let α\alphaα be the total number of points in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(−2π​,2π​) at which fff is NOT continuous, and let β\betaβ be the total number of points in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(−2π​,2π​) at which fff is NOT differentiable. Then the value of α+β\alpha + \betaα+β is ______.

Correct answer: 5

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
For the function f(x)=esin⁡∣x∣−∣x∣f(x) = e^{\sin|x|} - |x|f(x)=esin∣x∣−∣x∣, x∈Rx \in \mathbb{R}x∈R, consider the following statements: Statement I: fff is differentiable for all x∈Rx \in \mathbb{R}x∈R. Statement II: fff is increasing in (−π,−π2)\left(-\pi, -\frac{\pi}{2}\right)(−π,−2π​). In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q4·MathematicsNumericalJEE Main 2026
Let f(x)={ex−1,x<0x2−5x+6,x≥0f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^{2} - 5x + 6, & x \geq 0 \end{cases}f(x)={ex−1,x2−5x+6,​x<0x≥0​ and g(x)=f(∣x∣)+∣f(x)∣g(x) = f(|x|) + |f(x)|g(x)=f(∣x∣)+∣f(x)∣. If the number of points where ggg is not continuous and is not differentiable are α and β respectively, then α+β\alpha + \betaα+β is equal to ______

Correct answer: 4

Step-by-step solution →
Q5·MathematicsNumericalJEE Main 2026
The number of points, at which the function f(x)=max⁡{6x,2+3x2}+∣x−1∣∣cos⁡∣x2−14∣∣f(x) = \max\{6x, 2 + 3x^2\} + |x - 1|\left|\cos\left|x^2 - \frac{1}{4}\right|\right|f(x)=max{6x,2+3x2}+∣x−1∣​cos​x2−41​​​, x∈(−π,π)x \in (-\pi, \pi)x∈(−π,π), is not differentiable, is _______.

Correct answer: 9

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let α, β ∈ R\mathbb{R}R be such that the function f(x)={2α(x2−2)+2βx,x<1(α+3)x+(α−β),x≥1f(\mathrm{x}) = \begin{cases} 2\alpha(\mathrm{x}^{2} - 2) + 2\beta \mathrm{x} & ,\mathrm{x} < 1 \\ (\alpha + 3)\mathrm{x} + (\alpha - \beta) & ,\mathrm{x} \geq 1 \end{cases}f(x)={2α(x2−2)+2βx(α+3)x+(α−β)​,x<1,x≥1​ be differentiable at all x∈R\mathbb{R}R. Then 34(α + β) is equal to
  1. (A)84
  2. (B)48
  3. (C)36
  4. (D)24

Correct answer: (B)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Let [•] denote the greatest integer function, and let f(x)=min⁡{2x,x2}f(x) = \min\left\{\sqrt{2}x, x^2\right\}f(x)=min{2​x,x2}. Let S = {x ∈ (–2, 2) : the function g(x)=∣x∣[x2]g(x) = |x|[x^2]g(x)=∣x∣[x2] is discontinuous at x}. Then ∑x∈Sf(x)\sum_{x \in S} f(x)∑x∈S​f(x) equals :
  1. (A)2−22 - \sqrt{2}2−2​
  2. (B)26−322\sqrt{6} - 3\sqrt{2}26​−32​
  3. (C)1−21 - \sqrt{2}1−2​
  4. (D)6−22\sqrt{6} - 2\sqrt{2}6​−22​

Correct answer: (C)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3)f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3)f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3), x ∈ R. Then the value of f′(5)f'(5)f′(5) is :
  1. (A)625\frac{62}{5}562​
  2. (B)6575\frac{657}{5}5657​
  3. (C)25\frac{2}{5}52​
  4. (D)1175\frac{117}{5}5117​

Correct answer: (D)

Step-by-step solution →
Q9·MathematicsNumericalJEE Advanced 2025
Let R denote the set of all real numbers. Let f:R→Rf : R \rightarrow Rf:R→R and g:R→(0,4)g : R \rightarrow (0, 4)g:R→(0,4) be functions defined by f(x)=log⁡e(x2+2x+4)f(x) = \log_e\left(x^2 + 2x + 4\right)f(x)=loge​(x2+2x+4), and g(x)=41+e−2xg(x) = \frac{4}{1 + e^{-2x}}g(x)=1+e−2x4​. Define the composite function f∘g−1f \circ g^{-1}f∘g−1 by (f∘g−1)(x)=f(g−1(x))\left(f \circ g^{-1}\right)(x) = f\left(g^{-1}(x)\right)(f∘g−1)(x)=f(g−1(x)), where g−1g^{-1}g−1 is the inverse of the function g. Then the value of the derivative of the composite function f∘g−1f \circ g^{-1}f∘g−1 at x = 2 is ______

Correct answer: 0.25

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2025
Let the function f(x)=(x2−1)∣x2−ax+2∣+cos⁡∣x∣f(x)=(x^2-1)|x^2-ax+2|+\cos|x|f(x)=(x2−1)∣x2−ax+2∣+cos∣x∣ be not differentiable at the two points x=α=2x=\alpha=2x=α=2 and x=βx=\betax=β. Then the distance of the point (α,β)(\alpha,\beta)(α,β) from the line 12x+5y+10=012x+5y+10=012x+5y+10=0 is equal to:
  1. (A)3
  2. (B)4
  3. (C)2
  4. (D)5

Correct answer: (A)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2025
Let fff be differentiable with f(0)=1f(0)=1f(0)=1 and f(x+y)=f(x)f(y)f(x+y)=f(x)f(y)f(x+y)=f(x)f(y). Under the given conditions, ∑n=1100f′(n)\displaystyle\sum_{n=1}^{100}f'(n)n=1∑100​f′(n) equals:
  1. (A)2384
  2. (B)2525
  3. (C)5220
  4. (D)2406

Correct answer: (B)

Step-by-step solution →
Q12·MathematicsIntegerJEE Main 2025
Let f(x)={−3ax2−2,x<1a2+bx,x≥1f(x)=\begin{cases}-3ax^2-2,&x<1\\ a^2+bx,&x\ge1\end{cases}f(x)={−3ax2−2,a2+bx,​x<1x≥1​ be differentiable for all x∈Rx\in\mathbb{R}x∈R (a>1a>1a>1). With the region enclosed by y=f(x)y=f(x)y=f(x) and the given line equal to β3\beta\sqrt3β3​ (α,β∈Z\alpha,\beta\in\mathbb{Z}α,β∈Z), find α+β\alpha+\betaα+β.

Correct answer: 34

Step-by-step solution →
Q13·MathematicsSingle correctJEE Advanced 2024
Let f:R→Rf : R \to Rf:R→R and g:R→Rg : R \to Rg:R→R be functions defined by f(x)={x∣x∣sin⁡(1x),x≠0,0,x=0,f(x) = \begin{cases} x|x|\sin\left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x = 0, \end{cases}f(x)={x∣x∣sin(x1​),0,​x=0,x=0,​ and g(x)={1−2x,0≤x≤12,0,otherwise.g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise.} \end{cases}g(x)={1−2x,0,​0≤x≤21​,otherwise.​ Let a,b,c,d∈Ra, b, c, d \in Ra,b,c,d∈R. Define the function h:R→Rh : R \to Rh:R→R by h(x)=af(x)+b(g(x)+g(12−x))+c(x−g(x))+d g(x),x∈Rh(x) = af(x) + b\left(g(x) + g\left(\frac{1}{2} - x\right)\right) + c(x - g(x)) + d\,g(x), x \in Rh(x)=af(x)+b(g(x)+g(21​−x))+c(x−g(x))+dg(x),x∈R Match each entry in List-I to the correct entries in List-II. The correct option is:
List-IList-II
P.If a=0a = 0a=0, b=1b = 1b=1, c=0c = 0c=0 and d=0d = 0d=0, then1.hhh is one-one.
Q.If a=1a = 1a=1, b=0b = 0b=0, c=0c = 0c=0 and d=0d = 0d=0, then2.hhh is onto.
R.If a=0a = 0a=0, b=0b = 0b=0, c=1c = 1c=1 and d=0d = 0d=0, then3.hhh is differentiable on RRR.
S.If a=0a = 0a=0, b=0b = 0b=0, c=0c = 0c=0 and d=1d = 1d=1, then4.the range of hhh is [0,1][0, 1][0,1].
5.the range of hhh is {0,1}\{0, 1\}{0,1}.
  1. (A)(P) →\to→ (4) (Q) →\to→ (3) (R) →\to→ (1) (S) →\to→ (2)
  2. (B)(P) →\to→ (5) (Q) →\to→ (2) (R) →\to→ (4) (S) →\to→ (3)
  3. (C)(P) →\to→ (5) (Q) →\to→ (3) (R) →\to→ (2) (S) →\to→ (4)
  4. (D)(P) →\to→ (4) (Q) →\to→ (2) (R) →\to→ (1) (S) →\to→ (3)

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2024
If log⁡ey=3sin⁡−1x\log_{e}y=3\sin^{-1}xloge​y=3sin−1x, then (1−x2)y′′−xy′(1-x^{2})y''-xy'(1−x2)y′′−xy′ at x=12x=\dfrac{1}{2}x=21​ is equal to:
  1. (A)9e5π/69e^{5\pi/6}9e5π/6
  2. (B)3e5π/63e^{5\pi/6}3e5π/6
  3. (C)3eπ/23e^{\pi/2}3eπ/2
  4. (D)9eπ/29e^{\pi/2}9eπ/2

Correct answer: (D)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2024
Let f(x)=ax3+bx2+cx+41f(x) = ax^3 + bx^2 + cx + 41f(x)=ax3+bx2+cx+41 be such that f(1)=40f(1) = 40f(1)=40, f′(1)=2f'(1) = 2f′(1)=2 and f′′(1)=4f''(1) = 4f′′(1)=4. Then a2+b2+c2a^2 + b^2 + c^2a2+b2+c2 is equal to:
  1. (A)62
  2. (B)73
  3. (C)54
  4. (D)51

Correct answer: (D)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2024
Let f:(−∞,∞)−{0}→Rf:(-\infty,\infty)-\{0\}\to\mathbb{R}f:(−∞,∞)−{0}→R be a differentiable function such that f′(1)=lim⁡a→∞a2f(1a)f'(1)=\displaystyle\lim_{a\to\infty}a^{2}f\left(\tfrac{1}{a}\right)f′(1)=a→∞lim​a2f(a1​). Then lim⁡a→∞a(a+1)2tan⁡−1(1a)+a2−2log⁡ea\displaystyle\lim_{a\to\infty}\frac{a(a+1)}{2}\tan^{-1}\left(\tfrac{1}{a}\right)+a^{2}-2\log_{e}aa→∞lim​2a(a+1)​tan−1(a1​)+a2−2loge​a is equal to
  1. (A)32+π4\tfrac{3}{2}+\tfrac{\pi}{4}23​+4π​
  2. (B)38+π4\tfrac{3}{8}+\tfrac{\pi}{4}83​+4π​
  3. (C)52+π8\tfrac{5}{2}+\tfrac{\pi}{8}25​+8π​
  4. (D)34+π8\tfrac{3}{4}+\tfrac{\pi}{8}43​+8π​

Correct answer: (C)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2024
Suppose for a differentiable function h, h(0)=0h(0) = 0h(0)=0, h(1)=1h(1) = 1h(1)=1 and h′(0)=h′(1)=2h'(0) = h'(1) = 2h′(0)=h′(1)=2. If g(x)=h(ex) eh(x)g(x) = h(e^x)\,e^{h(x)}g(x)=h(ex)eh(x), then g′(0)g'(0)g′(0) is equal to:
  1. (A)5
  2. (B)3
  3. (C)8
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2024
If f(x)=x3sin⁡(1x)f(x)=x^{3}\sin\left(\tfrac{1}{x}\right)f(x)=x3sin(x1​) for x≠0x\neq 0x=0 and f(0)=0f(0)=0f(0)=0, then
  1. (A)f′′(0)=1f''(0)=1f′′(0)=1
  2. (B)f′′(2π)=24−π22πf''\left(\tfrac{2}{\pi}\right)=\tfrac{24-\pi^{2}}{2\pi}f′′(π2​)=2π24−π2​
  3. (C)f′′(2π)=12−π22πf''\left(\tfrac{2}{\pi}\right)=\tfrac{12-\pi^{2}}{2\pi}f′′(π2​)=2π12−π2​
  4. (D)f′′(0)=0f''(0)=0f′′(0)=0

Correct answer: (B)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2024
Let f(x)=x5+2x3+3x+1f(x) = x^5 + 2x^3 + 3x + 1f(x)=x5+2x3+3x+1, x∈Rx \in \mathbb{R}x∈R, and g(x)g(x)g(x) be a function such that g(f(x))=xg(f(x)) = xg(f(x))=x for all x∈Rx \in \mathbb{R}x∈R. Then g(7)g′(7)\dfrac{g(7)}{g'(7)}g′(7)g(7)​ is equal to:
  1. (A)777
  2. (B)424242
  3. (C)111
  4. (D)141414

Correct answer: (D)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2024
If y(θ)=2cos⁡θ+cos⁡2θcos⁡3θ+4cos⁡2θ+5cos⁡θ+2y(\theta)=\dfrac{2\cos\theta+\cos 2\theta}{\cos 3\theta+4\cos 2\theta+5\cos\theta+2}y(θ)=cos3θ+4cos2θ+5cosθ+22cosθ+cos2θ​, then at θ=π2\theta=\dfrac{\pi}{2}θ=2π​, y′′+y′+yy''+y'+yy′′+y′+y is equal to:
  1. (A)π2\dfrac{\pi}{2}2π​
  2. (B)1
  3. (C)12\dfrac{1}{2}21​
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q21·MathematicsNumericalJEE Main 2024
Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a thrice differentiable function such that f(0)=0f(0)=0f(0)=0, f(1)=1f(1)=1f(1)=1, f(2)=−1f(2)=-1f(2)=−1, f(3)=2f(3)=2f(3)=2 and f(4)=−2f(4)=-2f(4)=−2. Then, the minimum number of zeros of (3f′f′′+ff′′′)(x)\left(3f'f''+ff'''\right)(x)(3f′f′′+ff′′′)(x) is

Correct answer: 5

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2024
Let f(x)=x5+2ex/4f(x)=x^5+2e^{x/4}f(x)=x5+2ex/4 for all x∈Rx\in\mathbb{R}x∈R. Consider a function g(x)g(x)g(x) such that (g∘f)(x)=x(g\circ f)(x)=x(g∘f)(x)=x for all x∈Rx\in\mathbb{R}x∈R. Then the value of 8g′(2)8g'(2)8g′(2) is:
  1. (A)161616
  2. (B)444
  3. (C)888
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2024
Let aaa and bbb be real constants such that the function fff defined by f(x)={x2+3x+a,x≤1bx+2,x>1f(x)=\begin{cases}x^2+3x+a, & x\le 1\\ bx+2, & x>1\end{cases}f(x)={x2+3x+a,bx+2,​x≤1x>1​ be differentiable on RRR. Then the value of ∫−22f(x) dx\displaystyle\int_{-2}^{2}f(x)\,dx∫−22​f(x)dx equals:
  1. (A)156\dfrac{15}{6}615​
  2. (B)196\dfrac{19}{6}619​
  3. (C)212121
  4. (D)171717

Correct answer: (D)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2024
Let g:R→Rg:\mathbb{R}\to\mathbb{R}g:R→R be a non constant twice differentiable function such that g′(12)=g′(32)g'\left(\dfrac12\right)=g'\left(\dfrac32\right)g′(21​)=g′(23​). If a real valued function fff is defined as f(x)=12[g(x)+g(2−x)]f(x)=\dfrac12[g(x)+g(2-x)]f(x)=21​[g(x)+g(2−x)], then:
  1. (A)f′(x)=0f'(x)=0f′(x)=0 for atleast two xxx in (0,2)(0,2)(0,2)
  2. (B)f′(x)=0f'(x)=0f′(x)=0 for exactly one xxx in (0,1)(0,1)(0,1)
  3. (C)f′(x)=0f'(x)=0f′(x)=0 for no xxx in (0,1)(0,1)(0,1)
  4. (D)f′(32)+f′(12)=1f'\left(\dfrac32\right)+f'\left(\dfrac12\right)=1f′(23​)+f′(21​)=1

Correct answer: (A)

Step-by-step solution →
Q25·MathematicsNumericalJEE Main 2024
If the function f(x)={1∣x∣, ∣x∣≥2ax2+2b, ∣x∣<2f(x)=\begin{cases}\dfrac{1}{|x|} & ,\ |x|\ge2\\ ax^2+2b & ,\ |x|<2\end{cases}f(x)=⎩⎨⎧​∣x∣1​ax2+2b​, ∣x∣≥2, ∣x∣<2​ is differentiable on R\mathbb{R}R, then 48(a+b)48(a+b)48(a+b) is equal to ___

Correct answer: 15

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2024
Let f(x)=2x−x2f(x)=2^x-x^2f(x)=2x−x2, x∈Rx\in Rx∈R. If m and n are respectively the number of points at which the curves y=f(x)y=f(x)y=f(x) and y=f′(x)y=f'(x)y=f′(x) intersects the x-axis, then the value of m+nm+nm+n is ______.

Correct answer: 5

Step-by-step solution →
Q27·MathematicsMultiple correctJEE Advanced 2023
Let f:(0,1)→Rf : (0, 1) \to \mathbb{R}f:(0,1)→R be the function defined as f(x)=[4x](x−14)2(x−12)f(x) = [4x]\left(x - \frac{1}{4}\right)^{2}\left(x - \frac{1}{2}\right)f(x)=[4x](x−41​)2(x−21​), where [x][x][x] denotes the greatest integer less than or equal to x . Then which of the following statements is(are) true?
  1. (A)The function fff is discontinuous exactly at one point in (0,1)(0, 1)(0,1)
  2. (B)There is exactly one point in (0,1)(0, 1)(0,1) at which the function fff is continuous but NOT differentiable
  3. (C)The function f is NOT differentiable at more than three points in (0,1)(0, 1)(0,1)
  4. (D)The minimum value of the function f is −1512-\frac{1}{512}−5121​

Correct answer: (A), (B)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2023
For the differentiable function f:R∖{0}→Rf:\mathbb{R}\setminus\{0\}\to\mathbb{R}f:R∖{0}→R, if 3f(x)+2f(1x)=1x−103f(x)+2f\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-103f(x)+2f(x1​)=x1​−10, then ∣f(3)+f ′ ⁣(14)∣\left|f(3)+f\,'\!\left(\dfrac14\right)\right|​f(3)+f′(41​)​ is equal to:
  1. (A)777
  2. (B)335\dfrac{33}{5}533​
  3. (C)295\dfrac{29}{5}529​
  4. (D)131313

Correct answer: (D)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2023
Let f(x)=x2−x+∣−x+[x]∣f(x) = x^2 - x + |-x + [x]|f(x)=x2−x+∣−x+[x]∣, where x∈Rx \in \mathbb{R}x∈R and [t][t][t] denotes the greatest integer less than or equal to ttt. Then, fff is
  1. (A)continuous at x=0x = 0x=0, but not continuous at x=1x = 1x=1
  2. (B)continuous at x=0x = 0x=0 and x=1x = 1x=1
  3. (C)not continuous at x=0x = 0x=0 and x=1x = 1x=1
  4. (D)continuous at x=1x = 1x=1, but not continuous at x=0x = 0x=0

Correct answer: (D)

Step-by-step solution →
Q30·MathematicsNumericalJEE Main 2023
Let k and m be positive real numbers such that the function f(x)={3x2+kx+1,0<x<1mx2+k2,x≥1f(x)=\begin{cases}3x^{2}+k\sqrt{x+1},&0<x<1\\mx^{2}+k^{2},&x\geq1\end{cases}f(x)={3x2+kx+1​,mx2+k2,​0<x<1x≥1​ is differentiable for all x>0x>0x>0. Then 8f′(8)f′(18)\dfrac{8f'(8)}{f'\left(\dfrac{1}{8}\right)}f′(81​)8f′(8)​ is equal to

Correct answer: 309

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2023
If 2xy+3yx=202^{xy}+3^{yx}=202xy+3yx=20, then dydx\dfrac{dy}{dx}dxdy​ at (2,2)(2,2)(2,2) is equal to:
  1. (A)−(3+log⁡e82+log⁡e4)-\left(\dfrac{3+\log_e 8}{2+\log_e 4}\right)−(2+loge​43+loge​8​)
  2. (B)−(2+log⁡e83+log⁡e4)-\left(\dfrac{2+\log_e 8}{3+\log_e 4}\right)−(3+loge​42+loge​8​)
  3. (C)−(3+log⁡e164+log⁡e8)-\left(\dfrac{3+\log_e 16}{4+\log_e 8}\right)−(4+loge​83+loge​16​)
  4. (D)−(3+log⁡e42+log⁡e8)-\left(\dfrac{3+\log_e 4}{2+\log_e 8}\right)−(2+loge​83+loge​4​)

Correct answer: (B)

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Q32·MathematicsSingle correctJEE Main 2023
If y(x)=xx, x>0y(x)=x^x,\ x>0y(x)=xx, x>0, then y′′(2)−2y′(2)y''(2)-2y'(2)y′′(2)−2y′(2) is equal to
  1. (A)4log⁡e2+24\log_e 2+24loge​2+2
  2. (B)8log⁡e2−28\log_e 2-28loge​2−2
  3. (C)4(log⁡e2)2+24(\log_e 2)^2+24(loge​2)2+2
  4. (D)4(log⁡e2)2−24(\log_e 2)^2-24(loge​2)2−2

Correct answer: (D)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2023
Let y=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)y=(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})y=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′y'-y''y′−y′′ at x=−1x=-1x=−1 is equal to:
  1. (A)976976976
  2. (B)944944944
  3. (C)464464464
  4. (D)496496496

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2023
If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3), x∈Rf(x)=x^3-x^2f'(1)+xf''(2)-f'''(3),\ x\in\mathbb{R}f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3), x∈R then
  1. (A)f(1)+f(2)+f(3)=f(0)f(1)+f(2)+f(3)=f(0)f(1)+f(2)+f(3)=f(0)
  2. (B)2f(0)−f(1)+f(3)=f(2)2f(0)-f(1)+f(3)=f(2)2f(0)−f(1)+f(3)=f(2)
  3. (C)3f(1)+f(2)=f(3)3f(1)+f(2)=f(3)3f(1)+f(2)=f(3)
  4. (D)f(3)−f(2)=f(1)f(3)-f(2)=f(1)f(3)−f(2)=f(1)

Correct answer: (B)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2023
Let f(x)={x2sin⁡ ⁣(1x),x≠00,x=0f(x)=\begin{cases}x^2\sin\!\left(\tfrac{1}{x}\right), & x\ne 0\\ 0, & x=0\end{cases}f(x)={x2sin(x1​),0,​x=0x=0​. Then at x=0x=0x=0
  1. (A)fff is continuous but not differentiable
  2. (B)fff and f′f'f′ both are continuous
  3. (C)f′f'f′ is continuous but not differentiable
  4. (D)fff is continuous but f′f'f′ is not continuous

Correct answer: (D)

Step-by-step solution →
Q36·MathematicsNumericalJEE Main 2022
If [t] denotes the greatest integer ≤t\le t≤t, then number of points, at which the function f(x)=4∣2x+3∣+9[x+12]−12[x+20]f(x) = 4|2x + 3| + 9\left[x + \frac{1}{2}\right] - 12[x + 20]f(x)=4∣2x+3∣+9[x+21​]−12[x+20] is not differentiable in the open interval (−20,20)(-20, 20)(−20,20), is___.

Correct answer: 79

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2022
The number of points, where the function f:R→Rf : \mathbf{R} \rightarrow \mathbf{R}f:R→R, f(x)=∣x−1∣cos⁡∣x−2∣sin⁡∣x−1∣+(x−3)∣x2−5x+4∣f(x) = |x-1|\cos|x-2|\sin|x-1| + (x-3)\left|x^{2}-5x+4\right|f(x)=∣x−1∣cos∣x−2∣sin∣x−1∣+(x−3)​x2−5x+4​, is NOT differentiable, is :
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2022
Let x(t)=22cos⁡tsin⁡2tx(t)=2\sqrt{2}\cos t\sqrt{\sin 2t}x(t)=22​costsin2t​ and y(t)=22sin⁡tsin⁡2ty(t)=2\sqrt{2}\sin t\sqrt{\sin 2t}y(t)=22​sintsin2t​, t∈(0,π2)t \in \left(0,\frac{\pi}{2}\right)t∈(0,2π​). Then 1+(dydx)2d2ydx2\frac{1+\left(\frac{dy}{dx}\right)^{2}}{\frac{d^{2}y}{dx^{2}}}dx2d2y​1+(dxdy​)2​ at t=π4t=\frac{\pi}{4}t=4π​ is equal to
  1. (A)−223\frac{-2\sqrt{2}}{3}3−22​​
  2. (B)23\frac{2}{3}32​
  3. (C)13\frac{1}{3}31​
  4. (D)−23\frac{-2}{3}3−2​

Correct answer: (D)

Step-by-step solution →
Q39·MathematicsNumericalJEE Main 2022
For the curve C:(x2+y2−3)+(x2−y2−1)5=0C : (x^{2} + y^{2} - 3) + (x^{2} - y^{2} - 1)^{5} = 0C:(x2+y2−3)+(x2−y2−1)5=0, the value of 3y′−y3y′′3y' - y^{3}y''3y′−y3y′′, at the point (α,α)(\alpha, \alpha)(α,α), α>0\alpha > 0α>0, on C, is equal to ________.

Correct answer: 16

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2022
Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣f(x) = 2 + |x| - |x-1| + |x+1|f(x)=2+∣x∣−∣x−1∣+∣x+1∣, x∈Rx \in \mathbf{R}x∈R. Consider (S1):f′(−32)+f′(−12)+f′(12)+f′(32)=2(S1) : f'\left(-\frac{3}{2}\right) + f'\left(-\frac{1}{2}\right) + f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right) = 2(S1):f′(−23​)+f′(−21​)+f′(21​)+f′(23​)=2 (S2):∫−22f(x)dx=12(S2) : \int_{-2}^{2} f(x)dx = 12(S2):∫−22​f(x)dx=12 Then,
  1. (A)both (S1) and (S2) are correct
  2. (B)both (S1) and (S2) are wrong
  3. (C)only (S1) is correct
  4. (D)only (S2) is correct

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2022
The value of log⁡e2 ddx(log⁡cos⁡xcosec⁡x)\log_{e}2\,\frac{d}{dx}\left(\log_{\cos x}\operatorname{cosec}x\right)loge​2dxd​(logcosx​cosecx) at x=π4x=\frac{\pi}{4}x=4π​ is
  1. (A)−22-2\sqrt{2}−22​
  2. (B)222\sqrt{2}22​
  3. (C)-4
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q42·MathematicsNumericalJEE Main 2022
Let fff and g be twice differentiable even functions on (−2,2)(-2, 2)(−2,2) such that f(14)=0,f(12)=0,f(1)=1f\left(\frac{1}{4}\right) = 0, f\left(\frac{1}{2}\right) = 0, f(1) = 1f(41​)=0,f(21​)=0,f(1)=1 and g(34)=0,g(1)=2g\left(\frac{3}{4}\right) = 0, g(1) = 2g(43​)=0,g(1)=2 Then, the minimum number of solutions of f(x) g′′(x)+f′(x)g′(x)=0f(x)\ g''(x) + f'(x)g'(x) = 0f(x) g′′(x)+f′(x)g′(x)=0 in (−2,2)(-2,2)(−2,2) is equal to____.

Correct answer: 4

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : R \to Rf:R→R be a function defined by : f(x)={max⁡t≤x{t3−3t} ;x≤2x2+2x−6 ;2<x<3[x−3]+9 ;3≤x≤52x+1 ;x>5f(x) = \begin{cases} \max\limits_{t \le x}\{t^{3} - 3t\}\ ; & x \le 2 \\ x^{2} + 2x - 6\ ; & 2 < x < 3 \\ [x - 3] + 9\ ; & 3 \le x \le 5 \\ 2x + 1\ ; & x > 5 \end{cases}f(x)=⎩⎨⎧​t≤xmax​{t3−3t} ;x2+2x−6 ;[x−3]+9 ;2x+1 ;​x≤22<x<33≤x≤5x>5​ Where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and I=∫−22f(x)dxI = \int_{-2}^{2} f(x)dxI=∫−22​f(x)dx. Then the ordered pair (m, I) is equal to :
  1. (A)(3,274)\left(3, \frac{27}{4}\right)(3,427​)
  2. (B)(3,234)\left(3, \frac{23}{4}\right)(3,423​)
  3. (C)(4,274)\left(4, \frac{27}{4}\right)(4,427​)
  4. (D)(4,234)\left(4, \frac{23}{4}\right)(4,423​)

Correct answer: (C)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2022
If cos⁡−1(y2)=log⁡e(x5)5,∣y∣<2\cos^{-1}\left(\frac{y}{2}\right) = \log_{e}\left(\frac{x}{5}\right)^{5}, |y| < 2cos−1(2y​)=loge​(5x​)5,∣y∣<2, then :
  1. (A)x2y′′+xy′−25y=0x^{2}y'' + xy' - 25y = 0x2y′′+xy′−25y=0
  2. (B)x2y′′−xy′−25y=0x^{2}y'' - xy' - 25y = 0x2y′′−xy′−25y=0
  3. (C)x2y′′−xy′+25y=0x^{2}y'' - xy' + 25y = 0x2y′′−xy′+25y=0
  4. (D)x2y′′+xy′+25y=0x^{2}y'' + xy' + 25y = 0x2y′′+xy′+25y=0

Correct answer: (D)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : R \to Rf:R→R be defined as f(x)=x3+x−5f(x) = x^3 + x - 5f(x)=x3+x−5. If g(x)g(x)g(x) is a function such that f(g(x))=xf(g(x)) = xf(g(x))=x, ∀ x∈Rx \in Rx∈R, then g′(63)g'(63)g′(63) is equal to ______.
  1. (A)149\frac{1}{49}491​
  2. (B)349\frac{3}{49}493​
  3. (C)4349\frac{43}{49}4943​
  4. (D)9149\frac{91}{49}4991​

Correct answer: (A)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2022
If y=tan⁡−1(sec⁡x3−tan⁡x3)y = \tan^{-1}(\sec x^3 - \tan x^3)y=tan−1(secx3−tanx3). π2<x3<3π2\dfrac{\pi}{2} < x^3 < \dfrac{3\pi}{2}2π​<x3<23π​, then
  1. (A)xy′′+2y′=0xy'' + 2y' = 0xy′′+2y′=0
  2. (B)x2y′′−6y+3π2=0x^2 y'' - 6y + \dfrac{3\pi}{2} = 0x2y′′−6y+23π​=0
  3. (C)x2y′′−6y+3π=0x^2 y'' - 6y + 3\pi = 0x2y′′−6y+3π=0
  4. (D)xy′′−4y′=0xy'' - 4y' = 0xy′′−4y′=0

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2021
Let fff be any continuous function on [0,2][0, 2][0,2] and twice differentiable on (0,2)(0, 2)(0,2). If f(0)=0f(0) = 0f(0)=0, f(1)=1f(1) = 1f(1)=1 and f(2)=2f(2) = 2f(2)=2, then
  1. (A)f′′(x)=0f''(x) = 0f′′(x)=0 for all x∈(0,2)x \in (0, 2)x∈(0,2)
  2. (B)f′′(x)=0f''(x) = 0f′′(x)=0 for some x∈(0,2)x \in (0, 2)x∈(0,2)
  3. (C)f′(x)=0f'(x) = 0f′(x)=0 for some x∈[0,2]x \in [0, 2]x∈[0,2]
  4. (D)f′′(x)>0f''(x) > 0f′′(x)>0 for all x∈(0,2)x \in (0, 2)x∈(0,2)

Correct answer: (B)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2021
The function f(x)=∣x2−2x−3∣⋅e∣9x2−12x+4∣f(x) = \left|x^{2} - 2x - 3\right| \cdot e^{\left|9x^{2} - 12x + 4\right|}f(x)=​x2−2x−3​⋅e∣9x2−12x+4∣ is not differentiable at exactly :
  1. (A)four points
  2. (B)three points
  3. (C)two points
  4. (D)one point

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsNumericalJEE Main 2021
If y1/4+y−1/4=2xy^{1/4} + y^{-1/4} = 2xy1/4+y−1/4=2x, and (x2−1)d2ydx2+αxdydx+βy=0\left(x^{2} - 1\right)\frac{d^{2}y}{dx^{2}} + \alpha x\frac{dy}{dx} + \beta y = 0(x2−1)dx2d2y​+αxdxdy​+βy=0, then ∣α−β∣|\alpha - \beta|∣α−β∣ is equal to ___________.

Correct answer: 17

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2021
Let [t] denote the greatest integer less than or equal to t. Let f(x)=x−[x]f(x) = x - [x]f(x)=x−[x], g(x)=1−x+[x]g(x) = 1 - x + [x]g(x)=1−x+[x], and h(x)=min⁡{f(x),g(x)}h(x) = \min\{f(x), g(x)\}h(x)=min{f(x),g(x)}, x∈[−2,2]x \in [-2, 2]x∈[−2,2]. Then h is :
  1. (A)continuous in [−2,2][-2, 2][−2,2] but not differentiable at more than four points in (−2,2)(-2, 2)(−2,2)
  2. (B)not continuous at exactly three points in [−2,2][-2, 2][−2,2]
  3. (C)continuous in [−2,2][-2, 2][−2,2] but not differentiable at exactly three points in (−2,2)(-2, 2)(−2,2)
  4. (D)not continuous at exactly four points in [−2,2][-2, 2][−2,2]

Correct answer: (A)

Step-by-step solution →
Q51·MathematicsNumericalJEE Main 2021
If y=y(x)y = y(x)y=y(x) is an implicit function of x such that log⁡e(x+y)=4xy\log_{e}(x + y) = 4xyloge​(x+y)=4xy, then d2ydx2\frac{d^{2}y}{dx^{2}}dx2d2y​ at x=0x = 0x=0 is equal to ________.

Correct answer: 40

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2021
Let f : [0, ∞) → [0, 3] be a function defined by f(x) = {max⁡{sin⁡t:0≤t≤x},0≤x≤π2+cos⁡x,x>π\begin{cases} \max\left\{ \sin t : 0 \le t \le x \right\}, & 0 \le x \le \pi \\ 2 + \cos x, & x > \pi \end{cases}{max{sint:0≤t≤x},2+cosx,​0≤x≤πx>π​ Then which of the following is true?
  1. (A)f is differentiable everywhere in (0,∞)\left( 0, \infty \right)(0,∞)
  2. (B)f is not continuous exactly at two points in (0,∞)\left( 0, \infty \right)(0,∞)
  3. (C)f is continuous everywhere but not differentiable exactly at two points in (0,∞)\left( 0, \infty \right)(0,∞)
  4. (D)f is continuous everywhere but not differentiable exactly at one point in (0,∞)\left( 0, \infty \right)(0,∞)

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2021
Let f:[0,∞)→[0,∞)f:[0,\infty) \rightarrow [0,\infty)f:[0,∞)→[0,∞) be defined as f(x)=∫0x[y] dyf(x)=\int_{0}^{x}[y]\,dyf(x)=∫0x​[y]dy where [x] is the greatest integer less than or equal to x. Which of the following is true?
  1. (A)f is differentiable at every point in [0,∞)[0,\infty)[0,∞)
  2. (B)f is continuous everywhere except at the integer points in [0,∞)[0,\infty)[0,∞).
  3. (C)f is continuous at every point in [0,∞)[0,\infty)[0,∞) and differentiable except at the integer points.
  4. (D)f is both continuous and differentiable except at the integer points in [0,∞)[0,\infty)[0,∞).

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2021
If f(x)={∫0x(5+∣1−t∣)dt,x>25x+1,x≤2f\left(x\right) = \begin{cases} \int\limits_{0}^{x} \left(5 + \left|1 - t\right|\right)dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}f(x)=⎩⎨⎧​0∫x​(5+∣1−t∣)dt,5x+1,​x>2x≤2​, then
  1. (A)f(x) is not continuous at x = 2
  2. (B)f(x) is not differentiable at x = 1
  3. (C)f(x) is continuous but not differentiable at x = 2
  4. (D)f(x) is everywhere differentiable

Correct answer: (C)

Step-by-step solution →
Q55·MathematicsNumericalJEE Main 2021
Let f:R→Rf : R \rightarrow Rf:R→R be a function defined as f(x)={3(1−∣x∣2)if ∣x∣≤20if ∣x∣>2f(x) = \begin{cases} 3\left(1-\dfrac{|x|}{2}\right) & \text{if } |x| \leq 2 \\ 0 & \text{if } |x| > 2 \end{cases}f(x)=⎩⎨⎧​3(1−2∣x∣​)0​if ∣x∣≤2if ∣x∣>2​ Let g:R→Rg : R \rightarrow Rg:R→R be given by g(x)=f(x+2)−f(x−2)g(x) = f(x+2) - f(x-2)g(x)=f(x+2)−f(x−2). If n and m denote the number of points in R where g is the continuous and not differentiable, respectively, then n + m is equal to............

Correct answer: 4

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2021
Let a function g:[0,4]→Rg:[0,4]\to Rg:[0,4]→R be defined as g(x)={max⁡0≤t≤x{t3−6t2+9t−3},0≤x≤34−x,3<x≤4g(x)=\begin{cases}\max\limits_{0\le t\le x}\{t^{3}-6t^{2}+9t-3\}, & 0\le x\le 3 \\ 4-x, & 3<x\le 4\end{cases}g(x)={0≤t≤xmax​{t3−6t2+9t−3},4−x,​0≤x≤33<x≤4​, then the number of points in the interval (0,4)(0,4)(0,4) where g(x)g(x)g(x) is NOT differentiable, is…….

Correct answer: 1

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2021
If f(x)={1∣x∣;    ∣x∣≥1ax2+b;    ∣x∣<1f\left(x\right) = \begin{cases} \dfrac{1}{\left|x\right|} & ; \;\; \left|x\right| \geq 1 \\ ax^{2} + b & ; \;\; \left|x\right| < 1 \end{cases}f(x)=⎩⎨⎧​∣x∣1​ax2+b​;∣x∣≥1;∣x∣<1​ is differentiable at every point of the domain, then the values of a and b are respectively :
  1. (A)12,12\dfrac{1}{2}, \dfrac{1}{2}21​,21​
  2. (B)12,−32\dfrac{1}{2}, -\dfrac{3}{2}21​,−23​
  3. (C)52,−32\dfrac{5}{2}, -\dfrac{3}{2}25​,−23​
  4. (D)−12,32-\dfrac{1}{2}, \dfrac{3}{2}−21​,23​

Correct answer: (D)

Step-by-step solution →
Q58·MathematicsNumericalJEE Main 2021
Let fff : R → R satisfy the equation f(x+y)=f(x)⋅f(y)f(x + y) = f(x) \cdot f(y)f(x+y)=f(x)⋅f(y) for all x, y ∈ R and f(x) ≠ 0 for any x ∈ R. If the function fff is differentiable at x = 0 and f′(0)=3f'(0) = 3f′(0)=3, then lim⁡h→01h(f(h)−1)\lim_{h \to 0} \frac{1}{h}\left(f(h) - 1\right)limh→0​h1​(f(h)−1) is equal to ________ .

Correct answer: 3

Step-by-step solution →
Q59·MathematicsNumericalJEE Main 2021
If f(x)=sin⁡(cos⁡−1(1−22x1+22x))f(x) = \sin\left(\cos^{-1}\left(\frac{1 - 2^{2x}}{1 + 2^{2x}}\right)\right)f(x)=sin(cos−1(1+22x1−22x​)) and its first derivative with respect to x is −balog⁡e2-\frac{b}{a}\log_e 2−ab​loge​2 when x=1x = 1x=1, where a and b are integers, then the minimum value of ∣a2−b2∣|a^2 - b^2|∣a2−b2∣ is _______ .

Correct answer: 481

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2021
Let f:S→Sf : S \rightarrow Sf:S→S where S=(0,∞)S = (0, \infty)S=(0,∞) be a twice differentiable function such that f(x+1)=xf(x)f(x + 1) = x f(x)f(x+1)=xf(x). If g:S→Rg : S \rightarrow Rg:S→R be defined as g(x)=log⁡ef(x)g(x)=\log_{e}f(x)g(x)=loge​f(x), then the value of ∣g′′(5)−g′′(1)∣\left|g''(5)-g''(1)\right|∣g′′(5)−g′′(1)∣ is equal to :
  1. (A)205144\frac{205}{144}144205​
  2. (B)197144\frac{197}{144}144197​
  3. (C)187144\frac{187}{144}144187​
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2021
Let the functions f : ℝ → ℝ and g : ℝ → ℝ be defined as : f(x)={x+2,x<0x2,x≥0f(x) = \begin{cases} x + 2, & x < 0 \\ x^{2}, & x \ge 0 \end{cases}f(x)={x+2,x2,​x<0x≥0​ and g(x)={x3,x<13x−2,x≥1g(x) = \begin{cases} x^{3}, & x < 1 \\ 3x - 2, & x \ge 1 \end{cases}g(x)={x3,3x−2,​x<1x≥1​ Then, the number of points in ℝ where (fog)(x) is NOT differentiable is equal to :
  1. (A)3
  2. (B)1
  3. (C)0
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2021
Let f be any function defined on R and let it satisfy the condition : (∣f(x)−f(y)∣≤∣(x−y)2∣,∀(x,y)∈R(\left|f(x) - f(y)\right| \le \left|(x - y)^{2}\right|, \forall (x, y) \in R(∣f(x)−f(y)∣≤​(x−y)2​,∀(x,y)∈R If f(0) = 1, then:
  1. (A)f(x) < 0, ∀x ∈ R
  2. (B)f(x) can take any value in R
  3. (C)f(x) = 0, ∀ x ∈ R
  4. (D)f(x) >0, ∀ x ∈ R

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2021
A function f is defined on [–3,3] as f ()x={min {| x |,2−x2^{2}2}, −2≤x≤2 [| x |],2<| x |≤3 where [x] denotes the greatest integer ≤ x. The number of points, where f is not differentiable in (–3,3) is ______.

Correct answer: 5

Step-by-step solution →
Q64·MathematicsNumericalJEE Main 2021
The number of points, at which the function f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣, x∈Rx \in Rx∈R is not differentiable, is __________.

Correct answer: 2

Step-by-step solution →
Q65·MathematicsMultiple correctJEE Advanced 2020
Let the function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=x3−x2+(x−1)sin⁡xf(x) = x^{3} - x^{2} + (x - 1)\sin xf(x)=x3−x2+(x−1)sinx and let g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be an arbitrary function. Let fg :R→R: \mathbb{R} \to \mathbb{R}:R→R be the product function defined by (fg)(x)=f(x) g(x)(f g)(x) = f(x)\, g(x)(fg)(x)=f(x)g(x). Then which of the following statements is/are TRUE ?
  1. (A)If g is continuous at x=1x = 1x=1, then fg is differentiable at x=1x = 1x=1
  2. (B)If fg is differentiable at x=1x = 1x=1, then g is continuous at x=1x = 1x=1
  3. (C)If g is differentiable at x=1x = 1x=1, then fg is differentiable at x=1x = 1x=1
  4. (D)If fg is differentiable at x=1x = 1x=1, then g is differentiable at x=1x = 1x=1

Correct answer: (A), (C)

Step-by-step solution →
Q66·MathematicsIntegerJEE Advanced 2020
Let the functions f:(−1,1)→Rf : (-1,1) \to \mathbb{R}f:(−1,1)→R and g:(−1,1)→(−1,1)g : (-1,1) \to (-1,1)g:(−1,1)→(−1,1) be defined by f(x)=∣2x−1∣+∣2x+1∣f(x) = |2x-1| + |2x+1|f(x)=∣2x−1∣+∣2x+1∣ and g(x)=x−[x]g(x) = x - [x]g(x)=x−[x], where [x][x][x] denotes the greatest integer less than or equal to xxx. Let f∘g:(−1,1)→Rf \circ g : (-1,1) \to \mathbb{R}f∘g:(−1,1)→R be the composite function defined by (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))(f∘g)(x)=f(g(x)). Suppose ccc is the number of points in the interval (−1,1)(-1,1)(−1,1) at which f∘gf \circ gf∘g is NOT continuous, and suppose ddd is the number of points in the interval (−1,1)(-1,1)(−1,1) at which f∘gf \circ gf∘g is NOT differentiable. Then the value of c+dc + dc+d is ________

Correct answer: 4

Step-by-step solution →
Q67·MathematicsMultiple correctJEE Advanced 2020
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R and g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be functions satisfying f(x+y)=f(x)+f(y)+f(x)f(y)f(x+y) = f(x) + f(y) + f(x)f(y)f(x+y)=f(x)+f(y)+f(x)f(y) and f(x)=xg(x)f(x) = x g(x)f(x)=xg(x) for all x,y∈Rx, y \in \mathbb{R}x,y∈R. If lim⁡x→0g(x)=1\lim_{x \to 0} g(x) = 1limx→0​g(x)=1, then which of the following statements is/are TRUE?
  1. (A)fff is differentiable at every x∈Rx \in \mathbb{R}x∈R
  2. (B)If g(0)=1g(0) = 1g(0)=1, then ggg is differentiable at every x∈Rx \in \mathbb{R}x∈R
  3. (C)The derivative f′(1)f'(1)f′(1) is equal to 1
  4. (D)The derivative f′(0)f'(0)f′(0) is equal to 1

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

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2020
Let f:R→Rf : R \to Rf:R→R be a function defined by f(x)=max⁡{x,x2}f(x) = \max\{x, x^{2}\}f(x)=max{x,x2}. Let S denote the set of all points in R, where f is not differentiable. Then:
  1. (A){0,1}\{0, 1\}{0,1}
  2. (B){0}\{0\}{0}
  3. (C)ϕ\phiϕ(an empty set)
  4. (D){1}\{1\}{1}

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsNumericalJEE Main 2020
Let f:R→Rf:R\to Rf:R→R be defined as f(x)={x5sin⁡(1x)+5x2,x<00,x=0x5cos⁡(1x)+λx2,x>0f(x)=\begin{cases}x^{5}\sin\left(\dfrac{1}{x}\right)+5x^{2}, & x<0\\ 0, & x=0\\ x^{5}\cos\left(\dfrac{1}{x}\right)+\lambda x^{2}, & x>0\end{cases}f(x)=⎩⎨⎧​x5sin(x1​)+5x2,0,x5cos(x1​)+λx2,​x<0x=0x>0​. The value of λ\lambdaλ for which f′′(0)f''(0)f′′(0) exists, is ____.

Correct answer: 05

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2020
The derivation of tan⁡−1(1+x2−1x)\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)tan−1(x1+x2​−1​) with respect to tan⁡−1(2x1−x21−2x2)\tan^{-1}\left(\dfrac{2x\sqrt{1-x^2}}{1-2x^2}\right)tan−1(1−2x22x1−x2​​) at x=12x=\dfrac{1}{2}x=21​ is:
  1. (A)233\dfrac{2\sqrt{3}}{3}323​​
  2. (B)310\dfrac{\sqrt{3}}{10}103​​
  3. (C)312\dfrac{\sqrt{3}}{12}123​​
  4. (D)235\dfrac{2\sqrt{3}}{5}523​​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2020
If the function f(x)={k1(x−π)2−1x≤πk2cos⁡xx>πf(x)=\begin{cases} k_{1}(x-\pi)^{2}-1 & x \le \pi \\ k_{2}\cos x & x > \pi \end{cases}f(x)={k1​(x−π)2−1k2​cosx​x≤πx>π​ is twice differentiable, then the ordered pair (k1,k2)(k_{1},k_{2})(k1​,k2​) is equal to:
  1. (A)(1,0)(1,0)(1,0)
  2. (B)(1,1)(1,1)(1,1)
  3. (C)(12,−1)\left(\dfrac{1}{2},-1\right)(21​,−1)
  4. (D)(12,1)\left(\dfrac{1}{2},1\right)(21​,1)

Correct answer: (D)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2020
If (a+2 bcos⁡x)(a−2 bcos⁡y)=a2−b2\left( a + \sqrt{2}\, b \cos x \right)\left( a - \sqrt{2}\, b \cos y \right) = a^{2} - b^{2}(a+2​bcosx)(a−2​bcosy)=a2−b2, where a>b>0a > b > 0a>b>0, then dxdy\frac{dx}{dy}dydx​ at (π4,π4)\left( \frac{\pi}{4}, \frac{\pi}{4} \right)(4π​,4π​) is:
  1. (A)a−2ba+2b\frac{a - 2b}{a + 2b}a+2ba−2b​
  2. (B)2a+b2a−2\frac{2a + b}{2a - 2}2a−22a+b​
  3. (C)a+ba−b\frac{a + b}{a - b}a−ba+b​
  4. (D)a−ba+b\frac{a - b}{a + b}a+ba−b​

Correct answer: (C)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2020
Suppose a differentiable function f(x)f(x)f(x) satisfies the identity f(x+y)=f(x)+f(y)+xy2+x2yf(x+y) = f(x) + f(y) + xy^{2} + x^{2}yf(x+y)=f(x)+f(y)+xy2+x2y, for all real x and y. If lim⁡x→0f(x)x=1\lim_{x \to 0} \frac{f(x)}{x} = 1limx→0​xf(x)​=1 then f′(3)f'(3)f′(3) is equal to __________.

Correct answer: 10

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2020
The function f(x)={π4+tan⁡−1x,∣x∣≤112(∣x∣−1),∣x∣>1f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}\left(|x| - 1\right), & |x| > 1 \end{cases}f(x)={4π​+tan−1x,21​(∣x∣−1),​∣x∣≤1∣x∣>1​ is:
  1. (A)both continuous and differentiable on R − {1}.
  2. (B)both continuous and differentiable on R − {−1}.
  3. (C)continuous on R − {1} and differentiable on R − {−1, 1}.
  4. (D)continuous on R − {−1} and differentiable on R − {−1, 1}.

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2020
If y2+log⁡e(cos⁡2x)=yy^{2}+\log_{e}\left(\cos^{2}x\right)=yy2+loge​(cos2x)=y, x∈(−π2,π2)x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)x∈(−2π​,2π​), Then:
  1. (A)∣y′(0)∣+∣y′′(0)∣=3\left|y'(0)\right|+\left|y''(0)\right|=3∣y′(0)∣+∣y′′(0)∣=3
  2. (B)∣y′′(0)∣=2\left|y''(0)\right|=2∣y′′(0)∣=2
  3. (C)∣y′(0)∣+∣y′′(0)∣=1\left|y'(0)\right|+\left|y''(0)\right|=1∣y′(0)∣+∣y′′(0)∣=1
  4. (D)∣y′′(0)∣=0\left|y''(0)\right|=0∣y′′(0)∣=0

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2020
Let f and g be differentiable functions on R such that fog is the identity function. If for some a,b∈R,g′(a)=5a, b \in R, g'(a) = 5a,b∈R,g′(a)=5 and g(a)=bg(a) = bg(a)=b, then f′(b)f'(b)f′(b) is equal to:
  1. (A)111
  2. (B)555
  3. (C)15\frac{1}{5}51​
  4. (D)25\frac{2}{5}52​

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2020
Let f(x)=xcos⁡−1(−sin⁡∣x∣),x∈[−π2,π2]f(x)=x\cos^{-1}(-\sin|x|), x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]f(x)=xcos−1(−sin∣x∣),x∈[−2π​,2π​], then which of the following is true?
  1. (A)f′(0)=−π2f'(0)=-\dfrac{\pi}{2}f′(0)=−2π​
  2. (B)f is not differentiable at x = 0
  3. (C)f' is decreasing in (−π2,0)\left(-\dfrac{\pi}{2},0\right)(−2π​,0) and increasing in (0,π2)\left(0,\dfrac{\pi}{2}\right)(0,2π​)
  4. (D)f' is increasing in (−π2,0)\left(-\dfrac{\pi}{2},0\right)(−2π​,0) and decreasing in (0,π2)\left(0,\dfrac{\pi}{2}\right)(0,2π​)

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2020
Let y=y(x)y=y(x)y=y(x) be a function of x satisfying y1−x2=k−x1−y2y\sqrt{1-x^{2}}=k-x\sqrt{1-y^{2}}y1−x2​=k−x1−y2​ where k is a constant and y(12)=−14y\left(\dfrac{1}{2}\right)=-\dfrac{1}{4}y(21​)=−41​. Then dydx\dfrac{dy}{dx}dxdy​ at x=12x=\dfrac{1}{2}x=21​, is equal to:
  1. (A)−54-\dfrac{\sqrt{5}}{4}−45​​
  2. (B)−52-\dfrac{\sqrt{5}}{2}−25​​
  3. (C)25\dfrac{2}{\sqrt{5}}5​2​
  4. (D)52\dfrac{\sqrt{5}}{2}25​​

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsNumericalJEE Main 2020
Let S be the set of points where the function, f(x)=∣2−∣x−3∣∣,x∈Rf(x) = \left|2 - |x-3|\right|, x \in Rf(x)=∣2−∣x−3∣∣,x∈R, is not differentiable. Then ∑x∈Sf(f(x))\sum\limits_{x \in S} f(f(x))x∈S∑​f(f(x)) is equal to _________.

Correct answer: 3

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2020
If y(α)=2(tan⁡α+cot⁡α1+tan⁡2α)+1sin⁡2αy(\alpha)=\sqrt{2\left(\frac{\tan\alpha+\cot\alpha}{1+\tan^{2}\alpha}\right)+\frac{1}{\sin^{2}\alpha}}y(α)=2(1+tan2αtanα+cotα​)+sin2α1​​, α∈(3π4,π)\alpha\in\left(\frac{3\pi}{4},\pi\right)α∈(43π​,π) then dydα\frac{dy}{d\alpha}dαdy​ at α=5π6\alpha=\frac{5\pi}{6}α=65π​ is:
  1. (A)4
  2. (B)−14-\frac{1}{4}−41​
  3. (C)−4
  4. (D)43\frac{4}{3}34​

Correct answer: (A)

Step-by-step solution →
Q81·MathematicsMultiple correctJEE Advanced 2019
Let f:R→Rf : R \to Rf:R→R by given by f(x)={x5+5x4+10x3+10x2+3x+1,x<0;x2−x+1,0≤x<1;23x3−4x2+7x−83,1≤x<3;(x−2)log⁡e(x−2)−x+103,x≥3.f(x) = \begin{cases} x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1, & x < 0; \\ x^2 - x + 1, & 0 \le x < 1; \\ \frac{2}{3}x^3 - 4x^2 + 7x - \frac{8}{3}, & 1 \le x < 3; \\ (x - 2)\log_e (x - 2) - x + \frac{10}{3}, & x \ge 3. \end{cases}f(x)=⎩⎨⎧​x5+5x4+10x3+10x2+3x+1,x2−x+1,32​x3−4x2+7x−38​,(x−2)loge​(x−2)−x+310​,​x<0;0≤x<1;1≤x<3;x≥3.​ Then which of the following options is/are correct?
  1. (A)f′f'f′ has a local maximum at x=1x = 1x=1
  2. (B)f′f'f′ is NOT differentiable at x=1x = 1x=1
  3. (C)f is onto
  4. (D)f is increasing on (−∞,0)(-\infty, 0)(−∞,0)

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

Step-by-step solution →
Q82·MathematicsMultiple correctJEE Advanced 2019
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function. We say that f has PROPERTY 1 if lim⁡h→0f(h)−f(0)∣h∣\lim\limits_{h \to 0} \dfrac{f(h) - f(0)}{\sqrt{|h|}}h→0lim​∣h∣​f(h)−f(0)​ exists and is finite, and PROPERTY 2 if lim⁡h→0f(h)−f(0)h2\lim\limits_{h \to 0} \dfrac{f(h) - f(0)}{h^{2}}h→0lim​h2f(h)−f(0)​ exists and if finite. Then which of the following options is/are correct?
  1. (A)f(x)=x∣x∣f(x) = x|x|f(x)=x∣x∣ has PROPERTY 2
  2. (B)f(x)=sin⁡xf(x) = \sin xf(x)=sinx has PROPERTY 2
  3. (C)f(x)=∣x∣f(x) = |x|f(x)=∣x∣ has PROPERTY 1
  4. (D)f(x)=x2/3f(x) = x^{2/3}f(x)=x2/3 has PROPERTY 1

Correct answer: (C), (D)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2019
The derivative of tan⁡−1(sin⁡x−cos⁡xsin⁡x+cos⁡x)\tan^{-1}\left(\dfrac{\sin x - \cos x}{\sin x + \cos x}\right)tan−1(sinx+cosxsinx−cosx​), with respect to x2\dfrac{x}{2}2x​, where (x∈(0,π2))\left(x \in \left(0,\dfrac{\pi}{2}\right)\right)(x∈(0,2π​)) is:
  1. (A)2
  2. (B)12\dfrac{1}{2}21​
  3. (C)1
  4. (D)23\dfrac{2}{3}32​

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2019
If ey+xy=ee^{y} + xy = eey+xy=e, the ordered pair (dydx,d2ydx2)\left(\dfrac{dy}{dx}, \dfrac{d^{2}y}{dx^{2}}\right)(dxdy​,dx2d2y​) at x = 0 is equal to :
  1. (A)(1e,−1e2)\left(\dfrac{1}{e}, -\dfrac{1}{e^{2}}\right)(e1​,−e21​)
  2. (B)(1e,1e2)\left(\dfrac{1}{e}, \dfrac{1}{e^{2}}\right)(e1​,e21​)
  3. (C)(−1e,1e2)\left(-\dfrac{1}{e}, \dfrac{1}{e^{2}}\right)(−e1​,e21​)
  4. (D)(−1e,−1e2)\left(-\dfrac{1}{e}, -\dfrac{1}{e^{2}}\right)(−e1​,−e21​)

Correct answer: (C)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2019
Let f(x) = loge_ee​(sinx), (0 < x < π) and g(x) = sin⁡−1(e−x)\sin^{-1}(e^{-x})sin−1(e−x), (x ≥ 0). If α is a positive real number such that a = (fog)'(α) and b = (fog)(α), then
  1. (A)aα2+bα−a=2α2a\alpha^{2} + b\alpha - a = 2\alpha^{2}aα2+bα−a=2α2
  2. (B)aα2−bα−a=0a\alpha^{2} - b\alpha - a = 0aα2−bα−a=0
  3. (C)aα2−bα−a=1a\alpha^{2} - b\alpha - a = 1aα2−bα−a=1
  4. (D)aα2+bα+a=0a\alpha^{2} + b\alpha + a = 0aα2+bα+a=0

Correct answer: (C)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2019
Let f: R → R be differentiable at c ∈ R and f(c) = 0. If g(x) = |f(x)|, then at x = c, g is:
  1. (A)differentiable if f′(c) = 0
  2. (B)differentiable if f′(c) ≠ 0
  3. (C)not differentiable
  4. (D)not differentiable if f′(c) = 0

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2019
Let f(x)=15−∣x−10∣;x∈Rf(x)=15-|x-10|; x\in Rf(x)=15−∣x−10∣;x∈R. then the set of all values of x, at which the function, g(x)=f(f(x))g(x)=f(f(x))g(x)=f(f(x)) is not differentiable, is:
  1. (A){5,10,15}\{5,10,15\}{5,10,15}
  2. (B){10}\{10\}{10}
  3. (C){5,10,15,20}\{5,10,15,20\}{5,10,15,20}
  4. (D){10,15}\{10,15\}{10,15}

Correct answer: (A)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2019
If f(1)=1,f′(1)=3f(1)=1, f'(1)=3f(1)=1,f′(1)=3, then the derivative of f(f(f(x)))+(f(x))2f(f(f(x)))+(f(x))^{2}f(f(f(x)))+(f(x))2 at x=1x=1x=1 is:
  1. (A)33
  2. (B)15
  3. (C)9
  4. (D)12

Correct answer: (A)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2019
Let f be a differentiable function such that f(1) = 2 and f'(x) = f(x) for all x∈R. If h(x) = f(f(x)), then h'(1) is equal to :
  1. (A)2e22e^{2}2e2
  2. (B)4e4e4e
  3. (C)2e2e2e
  4. (D)4e24e^{2}4e2

Correct answer: (B)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2019
For x > 1, if (2x)2y=4e2x−2y(2x)^{2y}=4e^{2x-2y}(2x)2y=4e2x−2y, then (1+log⁡e2x)2dydx(1+\log_{e}2x)^{2}\frac{dy}{dx}(1+loge​2x)2dxdy​ is equal to:
  1. (A)xlog⁡e2x−log⁡e2x\frac{x\log_{e}2x-\log_{e}2}{x}xxloge​2x−loge​2​
  2. (B)log⁡e2x\log_{e}2xloge​2x
  3. (C)xlog⁡e2x+log⁡e2x\frac{x\log_{e}2x+\log_{e}2}{x}xxloge​2x+loge​2​
  4. (D)xlog⁡e2xx\log_{e}2xxloge​2x

Correct answer: (A)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2019
Let S be the set of all points in (−π,π)(-\pi, \pi)(−π,π) at which the function, f(x)=min⁡{sin⁡x,cos⁡x}f(x)=\min\{\sin x, \cos x\}f(x)=min{sinx,cosx} is non-differentiable. Then S is a subset of which of the following?
  1. (A){−π4,0,π4}\left\{-\frac{\pi}{4},0,\frac{\pi}{4}\right\}{−4π​,0,4π​}
  2. (B){−3π4,−π4,3π4,π4}\left\{-\frac{3\pi}{4},-\frac{\pi}{4},\frac{3\pi}{4},\frac{\pi}{4}\right\}{−43π​,−4π​,43π​,4π​}
  3. (C){−π2,−π4,π4,π2}\left\{-\frac{\pi}{2},-\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{2}\right\}{−2π​,−4π​,4π​,2π​}
  4. (D){−3π4,−π2,π2,3π4}\left\{-\frac{3\pi}{4},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{4}\right\}{−43π​,−2π​,2π​,43π​}

Correct answer: (B)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2019
If xlog⁡e(log⁡ex)−x2+y2=4 (y>0)x\log_{e}(\log_{e}x)-x^{2}+y^{2}=4\,(y>0)xloge​(loge​x)−x2+y2=4(y>0), then dydx\frac{dy}{dx}dxdy​ at x = e is equal to:
  1. (A)(1+2e)24+e2\frac{(1+2e)}{2\sqrt{4+e^{2}}}24+e2​(1+2e)​
  2. (B)(2e−1)24+e2\frac{(2e-1)}{2\sqrt{4+e^{2}}}24+e2​(2e−1)​
  3. (C)(1+2e)4+e2\frac{(1+2e)}{\sqrt{4+e^{2}}}4+e2​(1+2e)​
  4. (D)e4+e2\frac{e}{\sqrt{4+e^{2}}}4+e2​e​

Correct answer: (B)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2019
Let K be the set of all real values of x where the function f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f(x)=\sin|x|-|x|+2(x-\pi)\cos|x|f(x)=sin∣x∣−∣x∣+2(x−π)cos∣x∣ is not differentiable. Then the set K is equal to:
  1. (A)ϕ\phiϕ (en empty set)
  2. (B){π}\{\pi\}{π}
  3. (C){0}\{0\}{0}
  4. (D){0,π}\{0,\pi\}{0,π}

Correct answer: (A)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2019
Let f(x)={−1,−2≤x<0x2−1,0≤x≤2f(x)=\begin{cases}-1, & -2\le x<0\\ x^{2}-1, & 0\le x\le 2\end{cases}f(x)={−1,x2−1,​−2≤x<00≤x≤2​ and g(x)=∣f(x)∣+f(∣x∣)g(x)=|f(x)|+f(|x|)g(x)=∣f(x)∣+f(∣x∣), Then, in the interval (−2,2)(-2,2)(−2,2), g is:
  1. (A)differentiable at all points
  2. (B)not continuous
  3. (C)not differentiable at two points
  4. (D)not differentiable at one point

Correct answer: (D)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2019
Let f:R→Rf : R \to Rf:R→R be a function such that f(x)=x3+x2f′(1)+xf′′(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 Rx∈R. Then f(2)f(2)f(2) equals:
  1. (A)-4
  2. (B)30
  3. (C)-2
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2019
Let f(x)={max⁡{∣x∣,x2},∣x∣≤28−2∣x∣,2<∣x∣≤4f(x) = \begin{cases} \max\{|x|, x^2\}, & |x| \leq 2 \\ 8 - 2|x|, & 2 < |x| \leq 4 \end{cases}f(x)={max{∣x∣,x2},8−2∣x∣,​∣x∣≤22<∣x∣≤4​. Let S be the set of points in the interval (−4, 4) at which f is not differentiable. Then S:
  1. (A)is an empty set
  2. (B)equals {−2,−1,0,1,2}\{-2, -1, 0, 1, 2\}{−2,−1,0,1,2}
  3. (C)equals {−2,−1,1,2}\{-2, -1, 1, 2\}{−2,−1,1,2}
  4. (D)equals {−2,2}\{-2, 2\}{−2,2}

Correct answer: (B)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2019
If x=3tan⁡tx = 3\tan tx=3tant and y=3sec⁡ty = 3\sec ty=3sect, then the value of d2ydx2\frac{d^{2}y}{dx^{2}}dx2d2y​ at t=π4t = \frac{\pi}{4}t=4π​, is:
  1. (A)322\frac{3}{2\sqrt{2}}22​3​
  2. (B)132\frac{1}{3\sqrt{2}}32​1​
  3. (C)16\frac{1}{6}61​
  4. (D)162\frac{1}{6\sqrt{2}}62​1​

Correct answer: (D)

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2019
Let f be a differentiable function R to R such that ∣f(x)−f(y)∣≤2∣x−y∣32\left| f(x) - f(y) \right| \le 2\left| x - y \right|^{\frac{3}{2}}∣f(x)−f(y)∣≤2∣x−y∣23​, for all x,y∈Rx, y \in Rx,y∈R. If f(0)=1f(0) = 1f(0)=1 then ∫01f2(x)dx\int_{0}^{1} f^{2}(x) dx∫01​f2(x)dx is equal to
  1. (A)000
  2. (B)12\frac{1}{2}21​
  3. (C)222
  4. (D)111

Correct answer: (D)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Advanced 2018
Let f1:R→Rf_{1} : R \to Rf1​:R→R, f2:(−π2,π2)→Rf_{2} : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to Rf2​:(−2π​,2π​)→R, f3:(−1,eπ/2−2)→Rf_{3} : (-1, e^{\pi/2} - 2) \to Rf3​:(−1,eπ/2−2)→R and f4:R→Rf_{4} : R \to Rf4​:R→R be functions defined by (i) f1(x)=sin⁡(1−e−x2)f_{1}(x) = \sin\left(\sqrt{1 - e^{-x^{2}}}\right)f1​(x)=sin(1−e−x2​), (ii) f2(x)={∣sin⁡x∣tan⁡−1xif x≠01if x=0f_{2}(x) = \begin{cases} \frac{|\sin x|}{\tan^{-1} x} & \text{if } x \neq 0 \\\\ 1 & \text{if } x = 0 \end{cases}f2​(x)=⎩⎨⎧​tan−1x∣sinx∣​1​if x=0if x=0​, where the inverse trigonometric function tan⁡−1x\tan^{-1}xtan−1x assumes values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(−2π​,2π​), (iii) f3(x)=[sin⁡(log⁡e(x+2))]f_{3}(x) = [\sin(\log_{e}(x + 2))]f3​(x)=[sin(loge​(x+2))], where, for t∈Rt \in Rt∈R, [t][t][t] denotes the greatest integer less than or equal to ttt, (iv) f4(x)={x2sin⁡(1x)if x≠00if x=0f_{4}(x) = \begin{cases} x^{2}\sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\\\ 0 & \text{if } x = 0 \end{cases}f4​(x)=⎩⎨⎧​x2sin(x1​)0​if x=0if x=0​. The correct option is :
LIST-ILIST-II
P.The function f1f_{1}f1​ is1.NOT continuous at x=0x = 0x=0
Q.The function f2f_{2}f2​ is2.continuous at x=0x = 0x=0 and NOT differentiable at x=0x = 0x=0
R.The function f3f_{3}f3​ is3.differentiable at x=0x = 0x=0 and its derivative is NOT continuous at x=0x = 0x=0
S.The function f4f_{4}f4​ is4.differentiable at x=0x = 0x=0 and its derivative is continuous at x=0x = 0x=0
  1. (A)P →\to→ 2; Q →\to→ 3; R →\to→ 1; S →\to→ 4
  2. (B)P →\to→ 4; Q →\to→ 1; R →\to→ 2; S →\to→ 3
  3. (C)P →\to→ 4; Q →\to→ 2; R →\to→ 1; S →\to→ 3
  4. (D)P →\to→ 2; Q →\to→ 1; R →\to→ 4; S →\to→ 3

Correct answer: (D)

Step-by-step solution →
Q100·MathematicsMultiple correctJEE Advanced 2016
Let a,b∈Ra, b \in \mathbb{R}a,b∈R and f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=acos⁡(∣x3−x∣)+b∣x∣sin⁡(∣x3+x∣)f(x) = a \cos(|x^{3} - x|) + b|x| \sin(|x^{3} + x|)f(x)=acos(∣x3−x∣)+b∣x∣sin(∣x3+x∣). Then f is
  1. (A)differentiable at x=0x = 0x=0 if a=0a = 0a=0 and b=1b = 1b=1
  2. (B)differentiable at x=1x = 1x=1 if a=1a = 1a=1 and b=0b = 0b=0
  3. (C)NOT differentiable at x=0x = 0x=0 if a=1a = 1a=1 and b=0b = 0b=0
  4. (D)NOT differentiable at x=1x = 1x=1 if a=1a = 1a=1 and b=1b = 1b=1

Correct answer: (A), (B)

Step-by-step solution →
Q101·MathematicsMultiple correctJEE Advanced 2016
Let f:[−12,2]→Rf : \left[ -\frac{1}{2}, 2 \right] \to \mathbb{R}f:[−21​,2]→R and g:[−12,2]→Rg : \left[ -\frac{1}{2}, 2 \right] \to \mathbb{R}g:[−21​,2]→R be functions defined by f(x)=[x2−3]f(x) = [x^{2} - 3]f(x)=[x2−3] and g(x)=∣x∣f(x)+∣4x−7∣f(x)g(x) = |x| f(x) + |4x - 7| f(x)g(x)=∣x∣f(x)+∣4x−7∣f(x), where [y][y][y] denotes the greatest integer less than or equal to y for y∈Ry \in \mathbb{R}y∈R. Then
  1. (A)f is discontinuous exactly at three points in [−12,2]\left[ -\frac{1}{2}, 2 \right][−21​,2]
  2. (B)f is discontinuous exactly at four points in [−12,2]\left[ -\frac{1}{2}, 2 \right][−21​,2]
  3. (C)g is NOT differentiable exactly at four points in (−12,2)\left( -\frac{1}{2}, 2 \right)(−21​,2)
  4. (D)g is NOT differentiable exactly at five points in (−12,2)\left( -\frac{1}{2}, 2 \right)(−21​,2)

Correct answer: (B), (C)

Step-by-step solution →
Q102·MathematicsMultiple correctJEE Advanced 2016
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R, g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R and h:R→Rh : \mathbb{R} \to \mathbb{R}h:R→R be differentiable functions such that f(x)=x3+3x+2f(x) = x^{3} + 3x + 2f(x)=x3+3x+2, g(f(x))=xg(f(x)) = xg(f(x))=x and h(g(g(x)))=xh(g(g(x))) = xh(g(g(x)))=x for all x∈Rx \in \mathbb{R}x∈R. Then
  1. (A)g′(2)=115g'(2) = \frac{1}{15}g′(2)=151​
  2. (B)h′(1)=666h'(1) = 666h′(1)=666
  3. (C)h(0)=16h(0) = 16h(0)=16
  4. (D)h(g(3))=36h(g(3)) = 36h(g(3))=36

Correct answer: (B), (C)

Step-by-step solution →
Q103·MathematicsMultiple correctJEE Advanced 2015
Let g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be a differential function with g(0)=0g(0) = 0g(0)=0, g′(0)=0g'(0) = 0g′(0)=0 and g′(1)≠0g'(1) \ne 0g′(1)=0. Let f(x)={x∣x∣g(x),x≠00,x=0f(x) = \begin{cases} \dfrac{x}{|x|}g(x), & x \ne 0 \\ 0, & x = 0 \end{cases}f(x)=⎩⎨⎧​∣x∣x​g(x),0,​x=0x=0​ and h(x)=e∣x∣h(x) = e^{|x|}h(x)=e∣x∣ for all x∈Rx \in \mathbb{R}x∈R. Let (f∘h)(x)(f \circ h)(x)(f∘h)(x) denote f(h(x))f(h(x))f(h(x)) and (h∘f)(x)(h \circ f)(x)(h∘f)(x) denote h(f(x))h(f(x))h(f(x)). Then which of the following is (are) true?
  1. (A)fff is differentiable at x=0x = 0x=0
  2. (B)hhh is differentiable at x=0x = 0x=0
  3. (C)f∘hf \circ hf∘h is differentiable at x=0x = 0x=0
  4. (D)h∘fh \circ fh∘f is differentiable at x=0x = 0x=0

Correct answer: (A), (D)

Step-by-step solution →
Q104·MathematicsIntegerJEE Advanced 2014
Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R and g:R→Rg: \mathbb{R} \to \mathbb{R}g:R→R be respectively given by f(x)=∣x∣+1f(x) = |x| + 1f(x)=∣x∣+1 and g(x)=x2+1g(x) = x^2 + 1g(x)=x2+1. Define h:R→Rh: \mathbb{R} \to \mathbb{R}h:R→R by h(x)={max⁡{f(x),g(x)}if x≤0min⁡{f(x),g(x)}if x>0h(x) = \begin{cases} \max \{f(x), g(x)\} & \text{if } x \le 0 \\ \min \{f(x), g(x)\} & \text{if } x > 0 \end{cases}h(x)={max{f(x),g(x)}min{f(x),g(x)}​if x≤0if x>0​. Then number of points at which h(x)h(x)h(x) is not differentiable is __________

Correct answer: 3

Step-by-step solution →
Q105·MathematicsMultiple correctJEE Advanced 2014
Let f:[a,b]→[1,∞)f: [a, b] \to [1, \infty)f:[a,b]→[1,∞) be a continuous function and let g:R→Rg: \mathbb{R} \to \mathbb{R}g:R→R be defined as g(x)={0if x<a,∫axf(t)dtif a≤x≤b,∫abf(t)dtif x>bg(x) = \begin{cases} 0 & \text{if } x < a, \\ \int_{a}^{x} f(t) dt & \text{if } a \le x \le b, \\ \int_{a}^{b} f(t) dt & \text{if } x > b \end{cases}g(x)=⎩⎨⎧​0∫ax​f(t)dt∫ab​f(t)dt​if x<a,if a≤x≤b,if x>b​ Then
  1. (A)g(x)g(x)g(x) is continuous but not differentiable at aaa
  2. (B)g(x)g(x)g(x) is differentiable on R\mathbb{R}R
  3. (C)g(x)g(x)g(x) is continuous but not differentiable at bbb
  4. (D)g(x)g(x)g(x) is continuous and differentiable at either aaa or bbb but not both

Correct answer: (A), (C)

Step-by-step solution →

Differentiability — frequently asked

How many questions from Differentiability appear in JEE?

Differentiability has appeared in 91 of the last 186 JEE Main and JEE Advanced papers — about 49% of them — contributing 105 questions in total across those papers.

Is Differentiability an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 49% 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 Differentiability questions come from?

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

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

All 26 Mathematics chapters →

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