Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Mathematics PYQs
  3. /Application of Derivatives

Application of Derivatives — JEE Previous Year Questions

Every Application of Derivatives question asked in JEE Main and JEE Advanced across the last 186 papers — 207 questions, each with its correct answer. Free to read, no account needed.

Questions

207

Papers it appeared in

139/186

Appearance rate

75%

All 207 Application of Derivatives questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Advanced 2026
Consider the function f:(0,∞)→(−∞,∞)f : (0, \infty) \to (-\infty, \infty)f:(0,∞)→(−∞,∞) given by f(x)=x log⁡e(x)−x+1f(x) = \sqrt{x}\, \log_e(x) - x + 1f(x)=x​loge​(x)−x+1. Then which one of the following statements is TRUE ?
  1. (A)The derivative of the function fff is decreasing in the interval (0,1)(0, 1)(0,1)
  2. (B)The function fff has a local maximum at some point a∈(0,∞)a \in (0, \infty)a∈(0,∞)
  3. (C)The function fff has a local minimum at some point b∈(0,∞)b \in (0, \infty)b∈(0,∞)
  4. (D)The function fff has NEITHER a point of local maximum NOR a point of local minimum in the interval (0,∞)(0, \infty)(0,∞)

Correct answer: (D)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a differentiable function such that f(x+y3)=f(x)+f(y)3f\left(\frac{x + y}{3}\right) = \frac{f(x) + f(y)}{3}f(3x+y​)=3f(x)+f(y)​ for all x,y∈Rx, y \in \mathbb{R}x,y∈R, and f′(0)=3f'(0) = 3f′(0)=3. Then the minimum value of the function g(x)=3+exf(x)g(x) = 3 + e^x f(x)g(x)=3+exf(x), is:
  1. (A)3(e+1e)3\left(\frac{e + 1}{e}\right)3(ee+1​)
  2. (B)3(e−1e)3\left(\frac{e - 1}{e}\right)3(ee−1​)
  3. (C)3−ee\frac{3 - e}{e}e3−e​
  4. (D)3e3e3e

Correct answer: (B)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
max⁡0≤x≤π(16sin⁡(x2)cos⁡3(x2))\max_{0 \leq x \leq \pi}\left(16\sin\left(\frac{x}{2}\right)\cos^{3}\left(\frac{x}{2}\right)\right)max0≤x≤π​(16sin(2x​)cos3(2x​)) is equal to:
  1. (A)332\frac{3\sqrt{3}}{2}233​​
  2. (B)333\sqrt{3}33​
  3. (C)434\sqrt{3}43​
  4. (D)636\sqrt{3}63​

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
Let f(x)f(x)f(x) be a polynomial of degree 5, and have extrema at x=1x = 1x=1 and x=−1x = -1x=−1. If lim⁡x→0(f(x)x3)=−5\lim_{x \to 0} \left(\frac{f(x)}{x^3}\right) = -5limx→0​(x3f(x)​)=−5, then f(2)−f(−2)f(2) - f(-2)f(2)−f(−2) is equal to:
  1. (A)000
  2. (B)505050
  3. (C)929292
  4. (D)112112112

Correct answer: (D)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
The number of critical points of the function f(x)={∣sin⁡xx∣,x≠01,x=0f(x) = \begin{cases} \left|\frac{\sin x}{x}\right|, & x \neq 0 \\ 1, & x = 0 \end{cases}f(x)={​xsinx​​,1,​x=0x=0​ in the interval (−2π,2π)(-2\pi, 2\pi)(−2π,2π) is equal to :
  1. (A)1
  2. (B)3
  3. (C)5
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsNumericalJEE Main 2026
Let f be a differentiable function satisfying f(x)=1−2x+∫0xe(x−t) f(t) dtf(x) = 1 - 2x + \int_0^x e^{(x-t)}\, f(t)\, dtf(x)=1−2x+∫0x​e(x−t)f(t)dt, x ∈\in∈ R\mathbf{R}R and let g(x)=∫0x(f(t)+2)15 (t−4)6 (t+12)17 dtg(x) = \int_0^x (f(t) + 2)^{15}\, (t - 4)^6\, (t + 12)^{17}\, dtg(x)=∫0x​(f(t)+2)15(t−4)6(t+12)17dt, x ∈\in∈ R\mathbf{R}R. If p and q are respectively the points of local minima and local maxima of g, then the value of ∣p+q∣|p + q|∣p+q∣ is equal to _________ .

Correct answer: 9

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Consider the following three statements for the function f:(0,∞)→Rf : (0, \infty) \to \mathbb{R}f:(0,∞)→R defined by f(x)=∣log⁡ex∣−∣x−1∣f(x) = |\log_{e} x| - |x - 1|f(x)=∣loge​x∣−∣x−1∣ : (I) f is differentiable at all x>0x > 0x>0. (II) f is increasing in (0, 1). (III) f is decreasing in (1,∞)(1, \infty)(1,∞). Then.
  1. (A)All (I), (II) and (III) are TRUE.
  2. (B)Only (I) is TRUE.
  3. (C)Only (II) and (III) are TRUE.
  4. (D)Only (I) and (III) are TRUE.

Correct answer: (D)

Step-by-step solution →
Q8·MathematicsNumericalJEE Main 2026
Let (2α, α) be the largest interval in which the function f(t)=∣t+1∣t2f(\mathrm{t}) = \frac{|\mathrm{t} + 1|}{\mathrm{t}^{2}}f(t)=t2∣t+1∣​, t < 0, is strictly decreasing. Then the local maximum value of the function g(x)=2log⁡e(x−2)+αx2+4x−αg(\mathrm{x}) = 2 \log_{e}(\mathrm{x} - 2) + \alpha \mathrm{x}^{2} + 4\mathrm{x} - \alphag(x)=2loge​(x−2)+αx2+4x−α, x > 2, is _____

Correct answer: 4

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let f(x)=x2025−x2000f(x) = x^{2025} - x^{2000}f(x)=x2025−x2000, x∈[0,1]x \in [0, 1]x∈[0,1] and the minimum value of the function f(x) in the interval [0, 1] be (80)80(n)−81(80)^{80} (n)^{-81}(80)80(n)−81. Then n is equal to
  1. (A)−81
  2. (B)−40
  3. (C)−41
  4. (D)−80

Correct answer: (A)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
Let f:R→Rf : \mathbf{R} \to \mathbf{R}f:R→R be a twice differentiable function such that f′′(x)>0f''(x) > 0f′′(x)>0 for all x∈R\mathbf{R}R and f′(a−1)=0f'(a - 1) = 0f′(a−1)=0, where a is real number. Let g(x)=f(tan⁡2x−2tan⁡x+a)g(x) = f(\tan^2 x - 2\tan x + a)g(x)=f(tan2x−2tanx+a), 0<x<π20 < x < \frac{\pi}{2}0<x<2π​. Consider the following two statements : (I) g is increasing in (0,π4)\left(0, \frac{\pi}{4}\right)(0,4π​) (II) g is decreasing in (π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right)(4π​,2π​) Then,
  1. (A)Neither (I) nor (II) is True
  2. (B)Only (II) is True
  3. (C)Only (I) is True
  4. (D)Both (I) and (II) are True

Correct answer: (A)

Step-by-step solution →
Q11·MathematicsNumericalJEE Main 2026
Let f : R→R be a twice differentiable function such that the quadratic equation f(x)m2−2f′(x)m+f′′(x)=0f(x)m^{2}-2f'(x)m+f''(x)=0f(x)m2−2f′(x)m+f′′(x)=0 in m, has two equal roots for every x∈Rx\in Rx∈R. If f(0)=1f(0)=1f(0)=1, f′(0)=2f'(0)=2f′(0)=2 and (α, β) is the largest interval in which the function f(log⁡ex−x)f(\log_{e}x-x)f(loge​x−x) is increasing, then α + β is equal to

Correct answer: 1

Step-by-step solution →
Q12·MathematicsMultiple correctJEE Advanced 2025
Let R denote the set of all real numbers. Let f:R→Rf : R \rightarrow Rf:R→R be defined by f(x)={6x+sin⁡x2x+sin⁡xif x≠073if x=0f(x) = \begin{cases} \frac{6x + \sin x}{2x + \sin x} & \text{if } x \neq 0 \\ \frac{7}{3} & \text{if } x = 0 \end{cases}f(x)={2x+sinx6x+sinx​37​​if x=0if x=0​ Then which of the following statements is (are) TRUE?
  1. (A)The point x = 0 is a point of local maxima of f
  2. (B)The point x = 0 is a point of local minima of f
  3. (C)Number of points of local maxima of f in the interval [π,6π][\pi, 6\pi][π,6π] is 3
  4. (D)Number of points of local minima of f in the interval [2π,4π][2\pi, 4\pi][2π,4π] is 1

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

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2025
Let the function f(x)=x3+3x+3, x≠0f(x)=\dfrac{x}{3}+\dfrac{3}{x}+3,\ x\ne0f(x)=3x​+x3​+3, x=0 be strictly increasing in (−∞,α1)∪(α2,∞)(-\infty,\alpha_1)\cup(\alpha_2,\infty)(−∞,α1​)∪(α2​,∞) and strictly decreasing in (α3,α4)∪(α4,α5)(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)(α3​,α4​)∪(α4​,α5​). Then ∑i=15αi2\sum_{i=1}^{5}\alpha_i^2∑i=15​αi2​ is equal to:
  1. (A)484848
  2. (B)282828
  3. (C)404040
  4. (D)363636

Correct answer: (D)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2025
Let x=−1x=-1x=−1 and x=2x=2x=2 be the critical points of the function f(x)=x3+ax2+blog⁡e∣x∣+1f(x)=x^3+ax^2+b\log_e|x|+1f(x)=x3+ax2+bloge​∣x∣+1, x≠0x\ne0x=0. Let mmm and MMM respectively be the absolute minimum and the absolute maximum values of fff in the interval [−2,−12]\left[-2,-\dfrac{1}{2}\right][−2,−21​]. Then ∣M+m∣|M+m|∣M+m∣ is equal to (Take log⁡e2=0.7\log_e2=0.7loge​2=0.7):
  1. (A)21.1
  2. (B)19.8
  3. (C)22.1
  4. (D)20.9

Correct answer: (A)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2025
Let a>0a>0a>0. If the function f(x)=6x3−45ax2+108a2x+1f(x)=6x^3-45ax^2+108a^2x+1f(x)=6x3−45ax2+108a2x+1 attains its local maximum and minimum values at the points x1x_1x1​ and x2x_2x2​ respectively such that x1x2=54x_1x_2=54x1​x2​=54, then a+x1+x2a+x_1+x_2a+x1​+x2​ is equal to:
  1. (A)15
  2. (B)18
  3. (C)24
  4. (D)13

Correct answer: (B)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2025
If the function f(x)=2x3−9ax2+12a2x+1f(x)=2x^3-9ax^2+12a^2x+1f(x)=2x3−9ax2+12a2x+1, where a>0a>0a>0, attains its local maximum and minimum values at ppp and qqq, respectively, such that p2=qp^2=qp2=q, then f(3)f(3)f(3) is equal to:
  1. (A)55
  2. (B)10
  3. (C)23
  4. (D)37

Correct answer: (D)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2025
The sum of all local minimum values of the function f(x)={1−2x,x<−113(7+2∣x∣),−1≤x≤21118(x−4)(x−5),x>2f(x)=\begin{cases}1-2x, & x<-1\\ \frac{1}{3}(7+2|x|), & -1\le x\le 2\\ \frac{11}{18}(x-4)(x-5), & x>2\end{cases}f(x)=⎩⎨⎧​1−2x,31​(7+2∣x∣),1811​(x−4)(x−5),​x<−1−1≤x≤2x>2​ is:
  1. (A)17172\frac{171}{72}72171​
  2. (B)13172\frac{131}{72}72131​
  3. (C)15772\frac{157}{72}72157​
  4. (D)16772\frac{167}{72}72167​

Correct answer: (C)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2025
Let (2,3)(2,3)(2,3) be the largest open interval in which the function f(x)=2log⁡e(x−2)−x2+ax+1f(x)=2\log_e(x-2)-x^2+ax+1f(x)=2loge​(x−2)−x2+ax+1 is strictly increasing and (b,c)(b,c)(b,c) be the largest open interval, in which the function g(x)=(x−1)3(x+2−a)2g(x)=(x-1)^3(x+2-a)^2g(x)=(x−1)3(x+2−a)2 is strictly decreasing, then 100(a+b−c)100(a+b-c)100(a+b−c) is equal to:
  1. (A)280
  2. (B)360
  3. (C)420
  4. (D)160

Correct answer: (B)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2025
Let f(x)=∫0x2t2−8t+15et dtf(x)=\displaystyle\int_0^{x^2}\dfrac{t^2-8t+15}{e^t}\,dtf(x)=∫0x2​ett2−8t+15​dt, x∈Rx\in\mathbb{R}x∈R. Then the numbers of local maximum and local minimum points of fff, respectively, are:
  1. (A)2 and 3
  2. (B)3 and 2
  3. (C)1 and 3
  4. (D)2 and 2

Correct answer: (A)

Step-by-step solution →
Q20·MathematicsIntegerJEE Advanced 2024
Let the function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=sin⁡xeπx(x2023+2024x+2025)(x2−x+3)+2eπx(x2023+2024x+2025)(x2−x+3)f(x) = \frac{\sin x}{e^{\pi x}} \frac{\left(x^{2023} + 2024x + 2025\right)}{\left(x^2 - x + 3\right)} + \frac{2}{e^{\pi x}} \frac{\left(x^{2023} + 2024x + 2025\right)}{\left(x^2 - x + 3\right)}f(x)=eπxsinx​(x2−x+3)(x2023+2024x+2025)​+eπx2​(x2−x+3)(x2023+2024x+2025)​. Then the number of solutions of f(x) = 0 in R\mathbb{R}R is ______

Correct answer: 1

Step-by-step solution →
Q21·MathematicsNumericalJEE Main 2024
Let the set of all values of ppp, for which f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+cos⁡21f(x)=(p^{2}-6p+8)\left(\sin^{2}2x-\cos^{2}2x\right)+2(2-p)x+\cos^{2}1f(x)=(p2−6p+8)(sin22x−cos22x)+2(2−p)x+cos21 does not have any critical point, be the interval (a,b)(a, b)(a,b). Then 16ab16ab16ab is equal to _______.

Correct answer: 252

Step-by-step solution →
Q22·MathematicsNumericalJEE Main 2024
Let the set of all positive values of λ\lambdaλ, for which the point of local minimum of the function (1+x(λ2−x2))\left(1 + x(\lambda^2 - x^2)\right)(1+x(λ2−x2)) satisfies x2+x+2x2+5x+6<0\frac{x^2 + x + 2}{x^2 + 5x + 6} < 0x2+5x+6x2+x+2​<0, be (α,β)(\alpha, \beta)(α,β). Then α2+β2\alpha^2 + \beta^2α2+β2 is equal to ________.

Correct answer: 39

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2024
For the function f(x)=cos⁡x−x+1f(x)=\cos x-x+1f(x)=cosx−x+1, x∈Rx\in\mathbb{R}x∈R, between the following two statements: (S1)(S_1)(S1​) f(x)=0f(x)=0f(x)=0 for only one value of xxx in [0,π][0,\pi][0,π]. (S2)(S_2)(S2​) f(x)f(x)f(x) is decreasing in [0,π2]\left[0,\dfrac\pi2\right][0,2π​] and increasing in [π2,π]\left[\dfrac\pi2,\pi\right][2π​,π].
  1. (A)Both (S1)(S_1)(S1​) and (S2)(S_2)(S2​) are correct
  2. (B)Only (S1)(S_1)(S1​) is correct
  3. (C)Both (S1)(S_1)(S1​) and (S2)(S_2)(S2​) are incorrect
  4. (D)Only (S2)(S_2)(S2​) is correct

Correct answer: (B)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2024
If the function f(x)=2x3−9ax2+12a2x+1f(x)=2x^3-9ax^2+12a^2x+1f(x)=2x3−9ax2+12a2x+1, a>0a>0a>0 has a local maximum at x=αx=\alphax=α and a local minimum at x=α2x=\alpha^2x=α2, then α\alphaα and α2\alpha^2α2 are the roots of the equation
  1. (A)x2−6x+8=0x^2-6x+8=0x2−6x+8=0
  2. (B)8x2−6x+8=08x^2-6x+8=08x2−6x+8=0
  3. (C)8x2−6x+1=08x^2-6x+1=08x2−6x+1=0
  4. (D)x2+6x+8=0x^2+6x+8=0x2+6x+8=0

Correct answer: (A)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2024
Let f(x)=4cos⁡3x+33cos⁡2x−10f(x)=4\cos^3 x+3\sqrt3\cos^2 x-10f(x)=4cos3x+33​cos2x−10. The number of points of local maxima of fff in interval (0,2π)(0,2\pi)(0,2π) is:
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2024
Let A be the region enclosed by the parabola y2=2xy^2=2xy2=2x and the line x=24x=24x=24. Then the maximum area of the rectangle inscribed in the region A is _______ .

Correct answer: 128

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2024
The number of critical points of the function f(x)=(x−2)2/3(2x+1)f(x)=(x-2)^{2/3}(2x+1)f(x)=(x−2)2/3(2x+1) is:
  1. (A)222
  2. (B)000
  3. (C)111
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2024
If the function f(x)=(1x)2xf(x) = \left(\frac{1}{x}\right)^{2x}f(x)=(x1​)2x ; x>0x > 0x>0 attains the maximum value at x=1ex = \frac{1}{e}x=e1​ then:
  1. (A)eπ<πee^{\pi} < \pi^{e}eπ<πe
  2. (B)e2π<(2π)ee^{2\pi} < (2\pi)^{e}e2π<(2π)e
  3. (C)eπ>πee^{\pi} > \pi^{e}eπ>πe
  4. (D)(2e)π>π(2e)(2e)^{\pi} > \pi^{(2e)}(2e)π>π(2e)

Correct answer: (C)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2024
The interval in which the function f(x)=xxf(x)=x^{x}f(x)=xx, x>0x>0x>0, is strictly increasing is
  1. (A)(0,1e]\left(0,\tfrac{1}{e}\right](0,e1​]
  2. (B)[1e2,1)\left[\tfrac{1}{e^{2}},1\right)[e21​,1)
  3. (C)(0,∞)(0,\infty)(0,∞)
  4. (D)[1e,∞)\left[\tfrac{1}{e},\infty\right)[e1​,∞)

Correct answer: (D)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2024
For the function f(x)=sin⁡x+3x−2π(x2+x)f(x) = \sin x + 3x - \dfrac{2}{\pi}(x^2 + x)f(x)=sinx+3x−π2​(x2+x), where x∈[0,π2]x \in \left[0, \dfrac{\pi}{2}\right]x∈[0,2π​], consider the following two statements: (I) fff is increasing in (0,π2)\left(0, \dfrac{\pi}{2}\right)(0,2π​). (II) f′f'f′ is decreasing in (0,π2)\left(0, \dfrac{\pi}{2}\right)(0,2π​). Between the above two statements,
  1. (A)only (I) is true.
  2. (B)only (II) is true.
  3. (C)neither (I) nor (II) is true.
  4. (D)both (I) and (II) are true.

Correct answer: (D)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2024
Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let aaa and bbb be the sides of the rectangle PQRS when its area is maximum. Then (a+b)2(a + b)^2(a+b)2 is equal to:
  1. (A)727272
  2. (B)606060
  3. (C)808080
  4. (D)646464

Correct answer: (A)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2024
Let f(x)=3x−2+4−xf(x)=3\sqrt{x-2}+\sqrt{4-x}f(x)=3x−2​+4−x​ be a real valued function. If α\alphaα and β\betaβ are respectively the minimum and the maximum values of fff, then α2+2β2\alpha^2+2\beta^2α2+2β2 is equal to
  1. (A)44
  2. (B)42
  3. (C)24
  4. (D)38

Correct answer: (B)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2024
Let the sum of the maximum and the minimum values of the function f(x)=2x2−3x+82x2+3x+8f(x)=\dfrac{2x^2-3x+8}{2x^2+3x+8}f(x)=2x2+3x+82x2−3x+8​ be mn\dfrac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1. Then m+nm+nm+n is equal to:
  1. (A)182182182
  2. (B)217217217
  3. (C)195195195
  4. (D)201201201

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2024
If 5f(x)+4f(1x)=x2−25f(x)+4f\left(\dfrac1x\right)=x^2-25f(x)+4f(x1​)=x2−2, ∀x≠0\forall x\ne0∀x=0 and y=9x2f(x)y=9x^2 f(x)y=9x2f(x), then yyy is strictly increasing in:
  1. (A)(0,15)\left(0,\dfrac{1}{\sqrt5}\right)(0,5​1​)
  2. (B)(−15,0)∪(15,∞)\left(-\dfrac{1}{\sqrt5},0\right)\cup\left(\dfrac{1}{\sqrt5},\infty\right)(−5​1​,0)∪(5​1​,∞)
  3. (C)(−15,0)∪(0,15)\left(-\dfrac{1}{\sqrt5},0\right)\cup\left(0,\dfrac{1}{\sqrt5}\right)(−5​1​,0)∪(0,5​1​)
  4. (D)(−∞,−15)∪(0,15)\left(-\infty,-\dfrac{1}{\sqrt5}\right)\cup\left(0,\dfrac{1}{\sqrt5}\right)(−∞,−5​1​)∪(0,5​1​)

Correct answer: (B)

Step-by-step solution →
Q35·MathematicsNumericalJEE Main 2024
Let S=(−1,∞)S=(-1,\infty)S=(−1,∞) and f:S→Rf:S\to Rf:S→R be defined as f(x)=∫−1x(et−1)11(2t−1)5(t−2)7(t−3)12(2t−10)61dtf(x)=\displaystyle\int_{-1}^x \left(e^t-1\right)^{11}(2t-1)^5(t-2)^7(t-3)^{12}(2t-10)^{61}dtf(x)=∫−1x​(et−1)11(2t−1)5(t−2)7(t−3)12(2t−10)61dt. Let p = Sum of square of the values of x, where f(x)f(x)f(x) attains local maxima on S, and q = Sum of the values of x, where f(x)f(x)f(x) attains local minima on S. Then, the value of p2+2qp^2+2qp2+2q is ______.

Correct answer: 27

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2024
The maximum area of a triangle whose one vertex is at (0,0)(0,0)(0,0) and the other two vertices are on the curve y=−2x2+54y=-2x^2+54y=−2x2+54 at points (x,y)(x,y)(x,y) and (−x,y)(-x,y)(−x,y) where y>0y>0y>0 is:
  1. (A)888888
  2. (B)122122122
  3. (C)929292
  4. (D)108108108

Correct answer: (D)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2024
Let f(x)=(x+3)2(x−2)3f(x)=(x+3)^2(x-2)^3f(x)=(x+3)2(x−2)3, x∈[−4,4]x\in[-4,4]x∈[−4,4]. If MMM and mmm are the maximum and minimum values of fff, respectively in [−4,4][-4,4][−4,4], then the value of M−mM-mM−m is:
  1. (A)600600600
  2. (B)392392392
  3. (C)608608608
  4. (D)108108108

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2024
Let g(x)=3f(x3)+f(3−x)g(x)=3f\left(\dfrac{x}{3}\right)+f(3-x)g(x)=3f(3x​)+f(3−x) and f′′(x)>0f''(x)>0f′′(x)>0 for all x∈(0,3)x\in(0,3)x∈(0,3). If ggg is decreasing in (0,α)(0,\alpha)(0,α) and increasing in (α,3)(\alpha,3)(α,3), then 8α8\alpha8α is
  1. (A)24
  2. (B)0
  3. (C)18
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q39·MathematicsMultiple correctJEE Advanced 2023
Let SSS be the set of all twice differentiable functions f from R\mathbb{R}R to R\mathbb{R}R such that d2fdx2(x)>0\frac{d^{2}f}{dx^{2}}(x) > 0dx2d2f​(x)>0 for all x∈(−1,1)x \in (-1, 1)x∈(−1,1). For f∈Sf \in Sf∈S, let XfX_{f}Xf​ be the number of points x∈(−1,1)x \in (-1, 1)x∈(−1,1) for which f(x)=xf(x) = xf(x)=x. Then which of the following statements is(are) true?
  1. (A)There exists a function f∈Sf \in Sf∈S such that Xf=0X_{f} = 0Xf​=0
  2. (B)For every function f∈Sf \in Sf∈S, we have Xf≤2X_{f} \le 2Xf​≤2
  3. (C)There exists a function f∈Sf \in Sf∈S such that Xf=2X_{f} = 2Xf​=2
  4. (D)There does NOT exist any function fff in SSS such that Xf=1X_{f} = 1Xf​=1

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

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2023
max⁡0≤x≤π/2{x−2sin⁡xcos⁡x+13sin⁡3x}=\displaystyle\max_{0\le x\le\pi/2}\left\{x-2\sin x\cos x+\dfrac13\sin 3x\right\}=0≤x≤π/2max​{x−2sinxcosx+31​sin3x}=
  1. (A)5π+2+336\dfrac{5\pi+2+3\sqrt3}{6}65π+2+33​​
  2. (B)5π+2−336\dfrac{5\pi+2-3\sqrt3}{6}65π+2−33​​
  3. (C)π\piπ
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2023
If the local maximum value of the function f(x)=(3e2sin⁡x)sin⁡2xf(x)=\left(\dfrac{\sqrt{3e}}{2\sin x}\right)^{\sin^2 x}f(x)=(2sinx3e​​)sin2x, x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π​), is ke\dfrac{k}{e}ek​, then (ke)8+k8e5+k5\left(\dfrac{k}{e}\right)^8+\dfrac{k^8}{e^5}+k^5(ek​)8+e5k8​+k5 is equal to
  1. (A)e5+e6+e11e^5+e^6+e^{11}e5+e6+e11
  2. (B)e3+e6+e11e^3+e^6+e^{11}e3+e6+e11
  3. (C)e3+e5+e11e^3+e^5+e^{11}e3+e5+e11
  4. (D)e5+e6+e10e^5+e^6+e^{10}e5+e6+e10

Correct answer: (C)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2023
Let aaa, bbb, ccc and ddd be positive real numbers such that a+b+c+d=11a+b+c+d=11a+b+c+d=11. If the maximum value of a5b3c2da^5 b^3 c^2 da5b3c2d is 3750β3750\beta3750β, then the value of β\betaβ is
  1. (A)909090
  2. (B)110110110
  3. (C)555555
  4. (D)108108108

Correct answer: (A)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2023
Let f:[2,4]→Rf : [2, 4] \to \mathbb{R}f:[2,4]→R be a differentiable function such that (xlog⁡ex)f′(x)+(log⁡ex)f(x)+f(x)≥1(x \log_e x) f'(x) + (\log_e x) f(x) + f(x) \ge 1(xloge​x)f′(x)+(loge​x)f(x)+f(x)≥1, x∈[2,4]x \in [2, 4]x∈[2,4] with f(2)=12f(2) = \frac{1}{2}f(2)=21​ and f(4)=14f(4) = \frac{1}{4}f(4)=41​. Consider the following two statements: (A): f(x)≤1f(x) \le 1f(x)≤1, for all x∈[2,4]x \in [2, 4]x∈[2,4] (B): f(x)≥18f(x) \ge \frac{1}{8}f(x)≥81​, for all x∈[2,4]x \in [2, 4]x∈[2,4] Then,
  1. (A)Only statement (B) is true
  2. (B)Neither statement (A) nor statement (B) is true
  3. (C)Both the statement (A) and (B) are true
  4. (D)Only statement (A) is true

Correct answer: (C)

Step-by-step solution →
Q44·MathematicsNumericalJEE Main 2023
Let a quadratic curve passing through the point (−1,0)(-1,0)(−1,0) and touching the line y=xy=xy=x at (1,1)(1,1)(1,1) be y=f(x)y=f(x)y=f(x). Then the x-intercept of the normal to the curve at the point (α,α+1)(\alpha,\alpha+1)(α,α+1) in the first quadrant is _______ .

Correct answer: 11

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2023
A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm2^22) is equal to
  1. (A)675675675
  2. (B)102510251025
  3. (C)800800800
  4. (D)900900900

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2023
Let g(x)=f(x)+f(1−x)g(x)=f(x)+f(1-x)g(x)=f(x)+f(1−x) and f′′(x)>0, x∈(0,1)f''(x)>0,\ x\in(0,1)f′′(x)>0, x∈(0,1). If ggg is decreasing in the interval (0,α)(0,\alpha)(0,α) and increasing in the interval (α,1)(\alpha,1)(α,1), then tan⁡−1(2x)+tan⁡−1(1α)+tan⁡−1(α+1α)\tan^{-1}(2x)+\tan^{-1}\left(\frac{1}{\alpha}\right)+\tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)tan−1(2x)+tan−1(α1​)+tan−1(αα+1​) is equal to:
  1. (A)3π2\frac{3\pi}{2}23π​
  2. (B)π\piπ
  3. (C)5π4\frac{5\pi}{4}45π​
  4. (D)3π4\frac{3\pi}{4}43π​

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsNumericalJEE Main 2023
The number of points, where the curve y=x5−20x3+50x+2y=x^5-20x^3+50x+2y=x5−20x3+50x+2 crosses the x-axis, is

Correct answer: 5

Step-by-step solution →
Q48·MathematicsNumericalJEE Main 2023
If f(x)=x2+g′(1)x+g′′(2)f(x)=x^2+g'(1)x+g''(2)f(x)=x2+g′(1)x+g′′(2) and g(x)=f(1)x2+xf′(x)+f′′(x)g(x)=f(1)x^2+xf'(x)+f''(x)g(x)=f(1)x2+xf′(x)+f′′(x), then the value of f(4)−g(4)f(4)-g(4)f(4)−g(4) is equal to _______ .

Correct answer: 14

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2023
The sum of the absolute maximum and minimum values of the function f(x)=∣x2−5x+6∣−3x+2f(x)=|x^2-5x+6|-3x+2f(x)=∣x2−5x+6∣−3x+2 in the interval [−1,3][-1,3][−1,3] is equal to:
  1. (A)121212
  2. (B)131313
  3. (C)101010
  4. (D)242424

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2023
Let f(x)=2x+tan⁡−1xf(x)=2x+\tan^{-1}xf(x)=2x+tan−1x and g(x)=log⁡e ⁣(1+x2+x)g(x)=\log_e\!\left(\sqrt{1+x^2}+x\right)g(x)=loge​(1+x2​+x), x∈[0,3]x\in[0,3]x∈[0,3]. Then
  1. (A)min⁡f′(x)=1+max⁡g′(x)\min f'(x)=1+\max g'(x)minf′(x)=1+maxg′(x)
  2. (B)max⁡f′(x)>max⁡g′(x)\max f'(x)>\max g'(x)maxf′(x)>maxg′(x)
  3. (C)there exist 0<x1<x2<30<x_1<x_2<30<x1​<x2​<3 such that f(x)<g(x), ∀x∈(x1,x2)f(x)<g(x),\ \forall x\in(x_1,x_2)f(x)<g(x), ∀x∈(x1​,x2​)
  4. (D)there exists x^∈[0,3]\hat{x}\in[0,3]x^∈[0,3] such that f′(x^)<g′(x^)f'(\hat{x})<g'(\hat{x})f′(x^)<g′(x^)

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2023
A wire of length 20 m is to be cut into two pieces. A piece of length l1l_1l1​ is bent to make a square of area A1A_1A1​ and the other piece of length l2l_2l2​ is made into a circle of area A2A_2A2​. If 2A1+3A22A_1+3A_22A1​+3A2​ is minimum then (πl1):l2(\pi l_1):l_2(πl1​):l2​ is equal to :
  1. (A)1:61:61:6
  2. (B)6:16:16:1
  3. (C)3:13:13:1
  4. (D)4:14:14:1

Correct answer: (B)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2023
The absolute minimum value, of the function f(x)=∣x2−x+1∣+[x2−x+1]f(x)=|x^2-x+1|+[x^2-x+1]f(x)=∣x2−x+1∣+[x2−x+1], where [t][t][t] denotes the greatest integer function, in the interval [−1,2][-1,2][−1,2], is:
  1. (A)14\dfrac1441​
  2. (B)32\dfrac3223​
  3. (C)54\dfrac5445​
  4. (D)34\dfrac3443​

Correct answer: (D)

Step-by-step solution →
Q53·MathematicsNumericalJEE Main 2023
Let for x∈Rx \in \mathbb{R}x∈R, f(x)=x+∣x∣2f(x) = \frac{x + |x|}{2}f(x)=2x+∣x∣​ and g(x)={x,x<0x2,x≥0g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \ge 0 \end{cases}g(x)={x,x2,​x<0x≥0​. Then area bounded by the curve y=(f∘g)(x)y = (f \circ g)(x)y=(f∘g)(x) and the lines y=0, 2y−x=15y = 0,\ 2y - x = 15y=0, 2y−x=15 is equal to

Correct answer: 72

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2023
If the functions f(x)=x33+2bx+ax22f(x)=\dfrac{x^3}{3} + 2bx + \dfrac{ax^2}{2}f(x)=3x3​+2bx+2ax2​ and g(x)=x33+ax+bx2g(x)=\dfrac{x^3}{3} + ax + bx^2g(x)=3x3​+ax+bx2, a≠2ba \ne 2ba=2b, have a common extreme point, then a+2b+7a + 2b + 7a+2b+7 is equal to:
  1. (A)32\dfrac{3}{2}23​
  2. (B)333
  3. (C)444
  4. (D)666

Correct answer: (D)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2023
The number of points on the curve y=54x5−135x4−70x3+180x2+210xy=54x^5-135x^4-70x^3+180x^2+210xy=54x5−135x4−70x3+180x2+210x at which the normal lines are parallel to x+90y+2=0x+90y+2=0x+90y+2=0 is:
  1. (A)444
  2. (B)222
  3. (C)000
  4. (D)333

Correct answer: (A)

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2023
If the equation of the normal to the curve y=x−a(x+b)(x−2)y=\dfrac{x-a}{(x+b)(x-2)}y=(x+b)(x−2)x−a​ at the point (1,−3)(1,-3)(1,−3) is x−4y=13x-4y=13x−4y=13, then the value of a+ba+ba+b is equal to _____.

Correct answer: -6

Step-by-step solution →
Q57·MathematicsSingle correctJEE Main 2023
Let fff and ggg be twice differentiable functions on R\mathbb{R}R such that f ′ ′(x)=g ′ ′(x)+6x, f ′(1)=4g ′(1)−3=9, f(2)=3g(2)=12f\,'\,'(x)=g\,'\,'(x)+6x,\,f\,'(1)=4g\,'(1)-3=9,\,f(2)=3g(2)=12f′′(x)=g′′(x)+6x,f′(1)=4g′(1)−3=9,f(2)=3g(2)=12. Then which of the following is NOT true?
  1. (A)There exists x0∈(1,3/2)x_0\in(1,3/2)x0​∈(1,3/2) such that f(x0)=g(x0)f(x_0)=g(x_0)f(x0​)=g(x0​)
  2. (B)∣f ′(x)−g ′(x)∣<6⇒−1<x<1|f\,'(x)-g\,'(x)|<6\Rightarrow -1<x<1∣f′(x)−g′(x)∣<6⇒−1<x<1
  3. (C)If −1<x<2-1<x<2−1<x<2, then ∣f(x)−g(x)∣<8|f(x)-g(x)|<8∣f(x)−g(x)∣<8
  4. (D)g(−2)−f(−2)=20g(-2)-f(-2)=20g(−2)−f(−2)=20

Correct answer: (C)

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2023
Let the function f(x)=2x3+(2p−7)x2+3(2p−9)x−6f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6f(x)=2x3+(2p−7)x2+3(2p−9)x−6 have a maxima for some value of x<0x < 0x<0 and a minima for some value of x>0x > 0x>0. Then, the set of all values of ppp is
  1. (A)(0,92)\left(0, \dfrac{9}{2}\right)(0,29​)
  2. (B)(−∞,92)\left(-\infty, \dfrac{9}{2}\right)(−∞,29​)
  3. (C)(−92,92)\left(-\dfrac{9}{2}, \dfrac{9}{2}\right)(−29​,29​)
  4. (D)(92,∞)\left(\dfrac{9}{2}, \infty\right)(29​,∞)

Correct answer: (B)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2023
Let x=2x=2x=2 be a local minima of the function f(x)=2x4−18x2+8x+12f(x)=2x^4-18x^2+8x+12f(x)=2x4−18x2+8x+12, x∈(−4,4)x\in(-4,4)x∈(−4,4). If MMM is the local maximum value of the function fff in (−4,4)(-4,4)(−4,4), then M=M=M=
  1. (A)186−31218\sqrt{6}-\dfrac{31}{2}186​−231​
  2. (B)186+33218\sqrt{6}+\dfrac{33}{2}186​+233​
  3. (C)126−31212\sqrt{6}-\dfrac{31}{2}126​−231​
  4. (D)126−33212\sqrt{6}-\dfrac{33}{2}126​−233​

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2023
Let f:(0,1)→Rf:(0,1)\to\mathbb{R}f:(0,1)→R be a function defined by f(x)=11−e−xf(x)=\dfrac{1}{1-e^{-x}}f(x)=1−e−x1​, and g(x)=(f(−x)−f(x))g(x)=(f(-x)-f(x))g(x)=(f(−x)−f(x)). Consider two statements: (I) ggg is an increasing function in (0,1)(0,1)(0,1); (II) ggg is one-one in (0,1)(0,1)(0,1). Then:
  1. (A)Both (I) and (II) are true
  2. (B)Neither (I) nor (II) is true
  3. (C)Only (I) is true
  4. (D)Only (II) is true

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsMultiple correctJEE Advanced 2022
Let α = ∑k=1∞sin⁡2k(π6)\sum_{k=1}^{\infty} \sin^{2k}\left(\frac{\pi}{6}\right)∑k=1∞​sin2k(6π​). Let g : [0, 1] → R be the function defined by g(x) = 2αx^{αx}αx + 2α(1−x)^{α(1−x)}α(1−x). Then, which of the following statements is/are TRUE?
  1. (A)The minimum value of g(x) is 2762^{\frac{7}{6}}267​
  2. (B)The maximum value of g(x) is 1 + 2132^{\frac{1}{3}}231​
  3. (C)The function g(x) attains its maximum at more than one point
  4. (D)The function g(x) attains its minimum at more than one point

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

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2022
If the tangent to the curve y=x3−x2+xy = x^3 - x^2 + xy=x3−x2+x at the point (a,b)(a, b)(a,b) is also tangent to the curve y=5x2+2x−25y = 5x^2 + 2x - 25y=5x2+2x−25 at the point (2,−1)(2, -1)(2,−1), then ∣2a+9b∣|2a + 9b|∣2a+9b∣ is equal to ______.

Correct answer: 195

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2022
Let f(x)=3(x2−2)3+4f(x) = 3^{\left(x^{2}-2\right)^{3}+4}f(x)=3(x2−2)3+4, x∈Rx \in \mathbf{R}x∈R. Then which of the following statements are true ? P : x=0x = 0x=0 is a point of local minima of f Q : x=2x = \sqrt{2}x=2​ is a point of inflection of f R : f' is increasing for x>2x > \sqrt{2}x>2​
  1. (A)Only P and Q
  2. (B)Only P and R
  3. (C)Only Q and R
  4. (D)All, P, Q and R

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2022
The sum of the absolute maximum and absolute minimum values of the function f(x)=tan⁡−1(sin⁡x−cos⁡x)f(x)=\tan^{-1}(\sin x-\cos x)f(x)=tan−1(sinx−cosx) in the interval [0,π][0,\pi][0,π] is
  1. (A)0
  2. (B)tan⁡−1(12)−π4\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}tan−1(2​1​)−4π​
  3. (C)cos⁡−1(13)−π4\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}cos−1(3​1​)−4π​
  4. (D)−π12\frac{-\pi}{12}12−π​

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2022
If the minimum value of f(x)=5x22+αx5f(x) = \frac{5x^{2}}{2} + \frac{\alpha}{x^{5}}f(x)=25x2​+x5α​, x>0x > 0x>0, is 14, then the value of α\alphaα is equal to :
  1. (A)32
  2. (B)64
  3. (C)128
  4. (D)256

Correct answer: (C)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2022
The function f(x)=xex(1−x)f(x)=xe^{x(1-x)}f(x)=xex(1−x), x∈Rx \in Rx∈R, is
  1. (A)increasing in (−12,1)\left(-\frac{1}{2},1\right)(−21​,1)
  2. (B)decreasing in (12,2)\left(\frac{1}{2},2\right)(21​,2)
  3. (C)increasing in (−1,−12)\left(-1,-\frac{1}{2}\right)(−1,−21​)
  4. (D)decreasing in (−12,12)\left(-\frac{1}{2},\frac{1}{2}\right)(−21​,21​)

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsNumericalJEE Main 2022
Let f:[0,1]→Rf : [0, 1] \to Rf:[0,1]→R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1), then the least number of points x∈(0,1)x \in (0, 1)x∈(0,1), at which f′′(x)=0f''(x) = 0f′′(x)=0, is _______.

Correct answer: 2

Step-by-step solution →
Q68·MathematicsNumericalJEE Main 2022
Let MMM and NNN be the number of points on the curve y5−9xy+2x=0y^{5}-9xy+2x=0y5−9xy+2x=0, where the tangents to the curve are parallel to x-axis and y-axis, respectively. Then the value of M + N equals ________.

Correct answer: 2

Step-by-step solution →
Q69·MathematicsNumericalJEE Main 2022
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is tan⁡−134\tan^{-1}\frac{3}{4}tan−143​. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ________.

Correct answer: 5

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2022
Let P and Q be any points on the curves (x−1)2+(y+1)2=1(x-1)^{2}+(y+1)^{2}=1(x−1)2+(y+1)2=1 and y=x2y = x^{2}y=x2, respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval
  1. (A)(0,14)\left(0,\frac{1}{4}\right)(0,41​)
  2. (B)(12,34)\left(\frac{1}{2},\frac{3}{4}\right)(21​,43​)
  3. (C)(14,12)\left(\frac{1}{4},\frac{1}{2}\right)(41​,21​)
  4. (D)(34,1)\left(\frac{3}{4},1\right)(43​,1)

Correct answer: (C)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2022
Let f(x)={x3−x2+10x−7,x≤1−2x+log⁡2(b2−4),x>1f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & x \le 1 \\ -2x + \log_2(b^2 - 4), & x > 1 \end{cases}f(x)={x3−x2+10x−7,−2x+log2​(b2−4),​x≤1x>1​ Then the set of all values of b, for which f(x) has maximum value at x = 1, is :
  1. (A)(−6,−2)(-6, -2)(−6,−2)
  2. (B)(2,6)(2, 6)(2,6)
  3. (C)[−6,−2)∪(2,6][-6, -2) \cup (2, 6][−6,−2)∪(2,6]
  4. (D)[−6,−2)∪(2,6][-\sqrt{6}, -2) \cup (2, \sqrt{6}][−6​,−2)∪(2,6​]

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2022
If the maximum value of a, for which the function fa(x)=tan⁡−12x−3ax+7f_{a}(x)=\tan^{-1}2x-3ax+7fa​(x)=tan−12x−3ax+7 is non-decreasing in (−π6,π6)\left(-\frac{\pi}{6},\frac{\pi}{6}\right)(−6π​,6π​), is a‾\overline{a}a, then fa‾(π8)f_{\overline{a}}\left(\frac{\pi}{8}\right)fa​(8π​) is equal to
  1. (A)8−9π4(9+π2)8-\frac{9\pi}{4\left(9+\pi^{2}\right)}8−4(9+π2)9π​
  2. (B)8−4π9(4+π2)8-\frac{4\pi}{9\left(4+\pi^{2}\right)}8−9(4+π2)4π​
  3. (C)8(1+π29+π2)8\left(\frac{1+\pi^{2}}{9+\pi^{2}}\right)8(9+π21+π2​)
  4. (D)8−π48-\frac{\pi}{4}8−4π​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsSingle correctJEE Main 2022
The curve y(x)=ax3+bx2+cx+5y(x) = ax^{3} + bx^{2} + cx + 5y(x)=ax3+bx2+cx+5 touches the x-axis at the point P(−2,0)(-2, 0)(−2,0) and cuts the y-axis at the point Q, where y' is equal to 3. Then the local maximum value of y(x) is :
  1. (A)274\frac{27}{4}427​
  2. (B)294\frac{29}{4}429​
  3. (C)374\frac{37}{4}437​
  4. (D)92\frac{9}{2}29​

Correct answer: (A)

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2022
If the absolute maximum value of the function f(x)=(x2−2x+7) e(4x3−12x2−180x+31)f(x) = (x^{2} - 2x + 7)\, e^{(4x^{3} - 12x^{2} - 180x + 31)}f(x)=(x2−2x+7)e(4x3−12x2−180x+31) in the interval [−3,0][-3, 0][−3,0] is f(α)f(\alpha)f(α), then :
  1. (A)α=0\alpha = 0α=0
  2. (B)α=−3\alpha = -3α=−3
  3. (C)α∈(−1,0)\alpha \in (-1, 0)α∈(−1,0)
  4. (D)α∈(−3,−1)\alpha \in (-3, -1)α∈(−3,−1)

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2022
A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :
  1. (A)229+43\frac{22}{9 + 4\sqrt{3}}9+43​22​
  2. (B)669+43\frac{66}{9 + 4\sqrt{3}}9+43​66​
  3. (C)224+93\frac{22}{4 + 9\sqrt{3}}4+93​22​
  4. (D)664+93\frac{66}{4 + 9\sqrt{3}}4+93​66​

Correct answer: (B)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)=(x−3)n1 (x−5)n2f(x) = (x - 3)^{n_{1}}\,(x - 5)^{n_{2}}f(x)=(x−3)n1​(x−5)n2​, n1,n2∈Nn_{1}, n_{2} \in Nn1​,n2​∈N. The, which of the following is NOT true?
  1. (A)For n1=3n_{1} = 3n1​=3, n2=4n_{2} = 4n2​=4, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local maxima.
  2. (B)For n1=4n_{1} = 4n1​=4, n2=3n_{2} = 3n2​=3, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local manima.
  3. (C)For n1=3n_{1} = 3n1​=3, n2=5n_{2} = 5n2​=5, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local maxima.
  4. (D)For n1=4n_{1} = 4n1​=4, n2=6n_{2} = 6n2​=6, there exists α∈(3,5)\alpha \in (3,5)α∈(3,5) where fff attains local maxima.

Correct answer: (C)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2022
The number of real solutions of x7+5x3+3x+1=0x^{7}+5x^{3}+3x+1=0x7+5x3+3x+1=0 is equal to ________.
  1. (A)0
  2. (B)1
  3. (C)3
  4. (D)5

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2022
Let ℓ\ellℓ be a line which is normal to the curve y=2x2+x+2y = 2x^2 + x + 2y=2x2+x+2 at a point P on the curve. If the point Q(6, 4) lies on the line ℓ\ellℓ and O is origin, then the area of the triangle OPQ is equal to ______.

Correct answer: 13

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2022
If m and n respectively are the number of local maximum and local minimum points of the function f(x)=∫0x2t2−5t+42+etdtf(x)=\int_{0}^{x^{2}}\dfrac{t^{2}-5t+4}{2+e^{t}}dtf(x)=∫0x2​2+ett2−5t+4​dt, then the ordered pair (m, n) is equal to
  1. (A)(3, 2)(3,\,2)(3,2)
  2. (B)(2, 3)(2,\,3)(2,3)
  3. (C)(2, 2)(2,\,2)(2,2)
  4. (D)(3, 4)(3,\,4)(3,4)

Correct answer: (B)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2022
The lengths of the sides of a triangle are 10+x210 + x^{2}10+x2, 10+x210 + x^{2}10+x2 and 20−2x220 - 2x^{2}20−2x2. If for x = k, the area of the triangle is maximum, then 3k23k^{2}3k2 is equal to :
  1. (A)5
  2. (B)8
  3. (C)10
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2022
Let S be the set of all the natural numbers, for which the line xa+yb=2\frac{x}{a} + \frac{y}{b} = 2ax​+by​=2 is a tangent to the curve (xa)n+(yb)n=2\left(\frac{x}{a}\right)^{n} + \left(\frac{y}{b}\right)^{n} = 2(ax​)n+(by​)n=2 at the point (a, b), ab≠0ab \neq 0ab=0. Then:
  1. (A)S=ϕS = \phiS=ϕ
  2. (B)n(S)=1n(S) = 1n(S)=1
  3. (C)S={2k:k∈N}S = \{2k : k \in N\}S={2k:k∈N}
  4. (D)S=NS = NS=N

Correct answer: (D)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2022
Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :
  1. (A)2 : 5
  2. (B)19:45
  3. (C)3 : 8
  4. (D)19 : 15

Correct answer: (B)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2022
The sum of the absolute minimum and the absolute maximum values of the function f(x)=∣3x−x2+2∣−xf(x) = |3x - x^{2} + 2| - xf(x)=∣3x−x2+2∣−x in the interval [−1,2][-1, 2][−1,2] is :
  1. (A)17+32\frac{\sqrt{17} + 3}{2}217​+3​
  2. (B)17+52\frac{\sqrt{17} + 5}{2}217​+5​
  3. (C)5
  4. (D)9−172\frac{9 - \sqrt{17}}{2}29−17​​

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsNumericalJEE Main 2022
Let f(x)=∣(x−1)(x2−2x−3)∣+x−3f(x) = \left|(x - 1)(x^{2} - 2x - 3)\right| + x - 3f(x)=​(x−1)(x2−2x−3)​+x−3, x∈Rx \in Rx∈R. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ______

Correct answer: 3

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2022
Let f:R→Rf : R \to Rf:R→R and g:R→Rg : R \to Rg:R→R be two functions defined by f(x)=log⁡e(x2+1)−e−x+1f(x) = \log_e(x^2 + 1) - e^{-x} + 1f(x)=loge​(x2+1)−e−x+1 and g(x)=1−2e2xexg(x) = \frac{1 - 2e^{2x}}{e^x}g(x)=ex1−2e2x​. Then, for which of the following range of α, the inequality f(g((α−1)23))>f(g(α−53))f\left(g\left(\frac{(\alpha - 1)^2}{3}\right)\right) > f\left(g\left(\alpha - \frac{5}{3}\right)\right)f(g(3(α−1)2​))>f(g(α−35​)) holds?
  1. (A)(2,3)(2, 3)(2,3)
  2. (B)(−2,−1)(-2, -1)(−2,−1)
  3. (C)(1,2)(1, 2)(1,2)
  4. (D)(−1,1)(-1, 1)(−1,1)

Correct answer: (A)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2022
If the angle made by the tangent at the point (x0,y0)(x_0, y_0)(x0​,y0​) on the curve x=12(t+sin⁡tcos⁡t)x = 12(t + \sin t \cos t)x=12(t+sintcost), y=12(1+sin⁡t)2,0<t<π2y = 12(1 + \sin t)^2, 0 < t < \frac{\pi}{2}y=12(1+sint)2,0<t<2π​, with the positive x-axis is π3\frac{\pi}{3}3π​, then y0y_0y0​ is equal to
  1. (A)6(3+22)6\left(3 + 2\sqrt{2}\right)6(3+22​)
  2. (B)3(7+43)3\left(7 + 4\sqrt{3}\right)3(7+43​)
  3. (C)27
  4. (D)48

Correct answer: (C)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2022
Water is being filled at the rate of 1 cm3/sec⁡1\,\mathrm{cm}^3/\sec1cm3/sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm2/sec⁡\mathrm{cm}^2/\seccm2/sec) at which the wet conical surface area of the vessel increases is
  1. (A)5
  2. (B)215\frac{\sqrt{21}}{5}521​​
  3. (C)265\frac{\sqrt{26}}{5}526​​
  4. (D)2610\frac{\sqrt{26}}{10}1026​​

Correct answer: (C)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2022
f(x)=4log⁡e(x−1)−2x2+4x+5f(x)=4\log_e(x-1)-2x^{2}+4x+5f(x)=4loge​(x−1)−2x2+4x+5, x>1x>1x>1, which one of the following is NOT correct ?
  1. (A)f is increasing in (1,2)(1,2)(1,2) and decreasing in (2,∞)(2,\infty)(2,∞)
  2. (B)f(x)=−1f(x)=-1f(x)=−1 has exactly two solutions
  3. (C)f′(e)−f′′(2)<0f'(e)-f''(2)<0f′(e)−f′′(2)<0
  4. (D)f(x)=0f(x)=0f(x)=0 has a root in the interval (e,e+1)(e,e+1)(e,e+1)

Correct answer: (C)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2022
The sum of absolute maximum and absolute minimum values of the function f(x)=∣2x2+3x−2∣+sin⁡xcos⁡xf(x)=|2x^{2}+3x-2|+\sin x\cos xf(x)=∣2x2+3x−2∣+sinxcosx in the interval [0,1][0,1][0,1] is :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2022
The number of distinct real roots of the equation x7−7x−2=0x^{7} - 7x - 2 = 0x7−7x−2=0 is
  1. (A)5
  2. (B)7
  3. (C)1
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2022
the tangent at the point (x1, y1)(x_{1},\ y_{1})(x1​, y1​) on the curve y=x3+3x2+5y=x^{3}+3x^{2}+5y=x3+3x2+5 passes through the origin, then (x1,y1)(x_{1},y_{1})(x1​,y1​) does NOT lie on the curve :
  1. (A)x2+y281=2x^{2}+\dfrac{y^{2}}{81}=2x2+81y2​=2
  2. (B)y29−x2=8\dfrac{y^{2}}{9}-x^{2}=89y2​−x2=8
  3. (C)y=4x2+5y=4x^{2}+5y=4x2+5
  4. (D)x3−y2=2\dfrac{x}{3}-y^{2}=23x​−y2=2

Correct answer: (D)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2022
Let λ∗\lambda^*λ∗ be the largest value of λ\lambdaλ for which the function fλ(x)=4λx3−36λx2+36x+48f_{\lambda}(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48fλ​(x)=4λx3−36λx2+36x+48 is increasing for all x∈Rx \in Rx∈R. Then fλ∗(1)+fλ∗(−1)f_{\lambda}*(1) + f_{\lambda}*(-1)fλ​∗(1)+fλ​∗(−1) is equal to :
  1. (A)36
  2. (B)48
  3. (C)64
  4. (D)72

Correct answer: (D)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2022
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds,, it becomes 7 units, then its radius after 9 seconds is :
  1. (A)9
  2. (B)10
  3. (C)11
  4. (D)12

Correct answer: (A)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Advanced 2021
Let ψ1:[0,∞)→R\psi_1 : [0, \infty) \to \mathbb{R}ψ1​:[0,∞)→R, ψ2:[0,∞)→R\psi_2 : [0, \infty) \to \mathbb{R}ψ2​:[0,∞)→R, f:[0,∞)→Rf : [0, \infty) \to \mathbb{R}f:[0,∞)→R and g:[0,∞)→Rg : [0, \infty) \to \mathbb{R}g:[0,∞)→R be functions such that f(0)=g(0)=0,f(0) = g(0) = 0,f(0)=g(0)=0, ψ1(x)=e−x+x,x≥0,\psi_1(x) = e^{-x} + x, \quad x \ge 0,ψ1​(x)=e−x+x,x≥0, ψ2(x)=x2−2x−2e−x+2,x≥0,\psi_2(x) = x^2 - 2x - 2e^{-x} + 2, \quad x \ge 0,ψ2​(x)=x2−2x−2e−x+2,x≥0, f(x)=∫−xx(∣t∣−t2)e−t2 dt,x>0f(x) = \int_{-x}^{x} \left( |t| - t^2 \right) e^{-t^2} \, dt, \quad x > 0f(x)=∫−xx​(∣t∣−t2)e−t2dt,x>0 and g(x)=∫0x2t  e−t dt,x>0g(x) = \int_{0}^{x^2} \sqrt{t} \; e^{-t} \, dt, \quad x > 0g(x)=∫0x2​t​e−tdt,x>0 Which of the following statements is TRUE ?
  1. (A)f(ln⁡3)+g(ln⁡3)=13f\left(\sqrt{\ln 3}\right) + g\left(\sqrt{\ln 3}\right) = \frac{1}{3}f(ln3​)+g(ln3​)=31​
  2. (B)For every x > 1, there exists an α ∈ (1, x) such that ψ1(x)=1+αx\psi_1(x) = 1 + \alpha xψ1​(x)=1+αx
  3. (C)For every x > 0, there exists a β ∈ (0, x) such that ψ2(x)=2x(ψ1(β)−1)\psi_2(x) = 2x(\psi_1(\beta) - 1)ψ2​(x)=2x(ψ1​(β)−1)
  4. (D)fff is an increasing function on the interval [0,32]\left[0, \frac{3}{2}\right][0,23​]

Correct answer: (C)

Step-by-step solution →
Q95·MathematicsNumericalJEE Advanced 2021
Let f1:(0,∞)→Rf_1 : (0, \infty) \to \mathbb{R}f1​:(0,∞)→R and f2:(0,∞)→Rf_2 : (0, \infty) \to \mathbb{R}f2​:(0,∞)→R be defined by f1(x)=∫0x∏j=121(t−j)j dt,x>0f_1(x) = \int_{0}^{x} \prod_{j=1}^{21} (t - j)^j \, dt, \quad x > 0f1​(x)=∫0x​∏j=121​(t−j)jdt,x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0,f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450, \quad x > 0,f2​(x)=98(x−1)50−600(x−1)49+2450,x>0, where, for any positive integer n and real numbers a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​, ∏i=1nai\prod_{i=1}^{n} a_i∏i=1n​ai​ denotes the product of a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​. Let mim_imi​ and nin_ini​, respectively, denote the number of points of local minima and the number of points of local maxima of function fif_ifi​, i=1,2i = 1, 2i=1,2, in the interval (0,∞)(0, \infty)(0,∞) The value of 2m1+3n1+m1n12m_1 + 3n_1 + m_1n_12m1​+3n1​+m1​n1​ is _____.

Correct answer: 57.00

Step-by-step solution →
Q96·MathematicsMultiple correctJEE Advanced 2021
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=x2−3x−6x2+2x+4f(x) = \frac{x^2 - 3x - 6}{x^2 + 2x + 4}f(x)=x2+2x+4x2−3x−6​. Then which of the following statements is (are) TRUE ?
  1. (A)fff is decreasing in the interval (−2,−1)(-2, -1)(−2,−1)
  2. (B)fff is increasing in the interval (1,2)(1, 2)(1,2)
  3. (C)fff is onto
  4. (D)Range of fff is [−32,2]\left[-\frac{3}{2}, 2\right][−23​,2]

Correct answer: (A), (B)

Step-by-step solution →
Q97·MathematicsNumericalJEE Advanced 2021
Let f1:(0,∞)→Rf_1 : (0, \infty) \to \mathbb{R}f1​:(0,∞)→R and f2:(0,∞)→Rf_2 : (0, \infty) \to \mathbb{R}f2​:(0,∞)→R be defined by f1(x)=∫0x∏j=121(t−j)j dt,x>0f_1(x) = \int_{0}^{x} \prod_{j=1}^{21} (t - j)^j \, dt, \quad x > 0f1​(x)=∫0x​∏j=121​(t−j)jdt,x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0,f_2(x) = 98(x - 1)^{50} - 600(x - 1)^{49} + 2450, \quad x > 0,f2​(x)=98(x−1)50−600(x−1)49+2450,x>0, where, for any positive integer n and real numbers a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​, ∏i=1nai\prod_{i=1}^{n} a_i∏i=1n​ai​ denotes the product of a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​. Let mim_imi​ and nin_ini​, respectively, denote the number of points of local minima and the number of points of local maxima of function fif_ifi​, i=1,2i = 1, 2i=1,2, in the interval (0,∞)(0, \infty)(0,∞) The value of 6m2+4n2+8m2n26m_2 + 4n_2 + 8m_2n_26m2​+4n2​+8m2​n2​ is _____.

Correct answer: 6.00

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2021
The function f(x)=x3−6x2+ax+bf(x) = x^{3} - 6x^{2} + ax + bf(x)=x3−6x2+ax+b is such that f(2) = f(4) = 0. Consider two statements. (S1) there exists x1,x2∈(2,4)x_{1}, x_{2} \in (2, 4)x1​,x2​∈(2,4), x1<x2x_{1} < x_{2}x1​<x2​, such that f′(x1)=−1f'(x_{1}) = -1f′(x1​)=−1 and f′(x2)=0f'(x_{2}) = 0f′(x2​)=0. (S2) there exists x3,x4∈(2,4)x_{3}, x_{4} \in (2, 4)x3​,x4​∈(2,4), x3<x4x_{3} < x_{4}x3​<x4​, such that f is decreasing in (2,x4)(2, x_{4})(2,x4​), increasing in (x4,4)(x_{4}, 4)(x4​,4) and 2f′(x3)=3f(x4)2f'(x_{3}) = \sqrt{3}f\left(x_{4}\right)2f′(x3​)=3​f(x4​). Then
  1. (A)both (S1) and (S2) are true
  2. (B)(S1) is false and (S2) is true
  3. (C)both (S1) and (S2) are false
  4. (D)(S1) is true and (S2) is false

Correct answer: (A)

Step-by-step solution →
Q99·MathematicsNumericalJEE Main 2021
If 'R' is the least value of 'a' such that the function f(x)=x2+ax+1f(x) = x^{2} + ax + 1f(x)=x2+ax+1 is increasing on [1,2][1, 2][1,2] and 'S' is the greatest value of 'a' such that the function f(x)=x2+ax+1f(x) = x^{2} + ax + 1f(x)=x2+ax+1 is decreasing on [1,2][1, 2][1,2], then the value of ∣R−S∣\left| R - S \right|∣R−S∣ is ______.

Correct answer: 2

Step-by-step solution →
Q100·MathematicsNumericalJEE Main 2021
Let f(x) be a cubic polynomial with f(1) = -10, f(-1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = -1. Then f(3) is equal to ______.

Correct answer: 22

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2021
An angle of intersection of the curves, x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1a2x2​+b2y2​=1 and x2+y2=abx^2 + y^2 = abx2+y2=ab, a>ba > ba>b, is :
  1. (A)tan⁡−1(a+bab)\tan^{-1}\left(\frac{a + b}{\sqrt{ab}}\right)tan−1(ab​a+b​)
  2. (B)tan⁡−1(a−b2ab)\tan^{-1}\left(\frac{a - b}{2\sqrt{ab}}\right)tan−1(2ab​a−b​)
  3. (C)tan⁡−1(a−bab)\tan^{-1}\left(\frac{a - b}{\sqrt{ab}}\right)tan−1(ab​a−b​)
  4. (D)tan⁡−1(2ab)\tan^{-1}\left(2\sqrt{ab}\right)tan−1(2ab​)

Correct answer: (C)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2021
A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:
  1. (A)52+3\frac{5}{2+\sqrt{3}}2+3​5​
  2. (B)102+33\frac{10}{2+3\sqrt{3}}2+33​10​
  3. (C)53+3\frac{5}{3+\sqrt{3}}3+3​5​
  4. (D)103+23\frac{10}{3+2\sqrt{3}}3+23​10​

Correct answer: (D)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2021
A box open from top is made from a rectangular sheet of dimension a × b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :
  1. (A)a+b−a2+b2−ab12\frac{a + b - \sqrt{a^{2} + b^{2} - ab}}{12}12a+b−a2+b2−ab​​
  2. (B)a+b−a2+b2+ab6\frac{a + b - \sqrt{a^{2} + b^{2} + ab}}{6}6a+b−a2+b2+ab​​
  3. (C)a+b−a2+b2−ab6\frac{a + b - \sqrt{a^{2} + b^{2} - ab}}{6}6a+b−a2+b2−ab​​
  4. (D)a+b+a2+b2−ab6\frac{a + b + \sqrt{a^{2} + b^{2} - ab}}{6}6a+b+a2+b2−ab​​

Correct answer: (C)

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2021
Let M and m respectively be the maximum and minimum values of the function f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x) = \tan^{-1}(\sin x + \cos x)f(x)=tan−1(sinx+cosx) in [0,π2]\left[0, \frac{\pi}{2}\right][0,2π​], Then the value of tan(M − m) is equal to:
  1. (A)2+32 + \sqrt{3}2+3​
  2. (B)2−32 - \sqrt{3}2−3​
  3. (C)3+223 + 2\sqrt{2}3+22​
  4. (D)3−223 - 2\sqrt{2}3−22​

Correct answer: (D)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2021
The local maximum value of the function f(x)=(2x)x2f(x) = \left(\frac{2}{x}\right)^{x^{2}}f(x)=(x2​)x2, x>0x > 0x>0, is
  1. (A)(2e)1e\left(2\sqrt{e}\right)^{\frac{1}{e}}(2e​)e1​
  2. (B)(4e)e4\left(\frac{4}{\sqrt{e}}\right)^{\frac{e}{4}}(e​4​)4e​
  3. (C)(e)2e\left(e\right)^{\frac{2}{e}}(e)e2​
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q106·MathematicsNumericalJEE Main 2021
A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then (4π+1)k\left(\frac{4}{\pi} + 1\right)k(π4​+1)k is equal to ________.

Correct answer: 36

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2021
Let f : (a,b) → R be twice differentiable such that f(x)=∫axg(t)dtf(x) = \int_a^x g(t) dtf(x)=∫ax​g(t)dt for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g′(x)=0g(x)g'(x) = 0g(x)g′(x)=0 has at least :
  1. (A)seven roots in (a, b)
  2. (B)twelve roots in (a, b)
  3. (C)five roots in (a, b)
  4. (D)three roots in (a, b)

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsNumericalJEE Main 2021
If a rectangle is inscribed in an equilateral triangle of side length 222\sqrt{2}22​ as shown in the figure, then the square of the largest area of such a rectangle is .........

Correct answer: 3

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2021
Let f(x)=3sin⁡4x+10sin⁡3x+6sin⁡2x−3f(x)=3\sin^{4}x+10\sin^{3}x+6\sin^{2}x-3f(x)=3sin4x+10sin3x+6sin2x−3, x∈[−π6,π2]x \in \left[-\frac{\pi}{6},\frac{\pi}{2}\right]x∈[−6π​,2π​]. Then, f is :
  1. (A)decreasing in (0,π2)\left(0,\frac{\pi}{2}\right)(0,2π​)
  2. (B)increasing in (−π6,π2)\left(-\frac{\pi}{6},\frac{\pi}{2}\right)(−6π​,2π​)
  3. (C)decreasing in (−π6,0)\left(-\frac{\pi}{6},0\right)(−6π​,0)
  4. (D)increasing in (−π6,0)\left(-\frac{\pi}{6},0\right)(−6π​,0)

Correct answer: (C)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2021
Let f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)={−43x3+2x2+3x,x>03xex,x≤0f(x) = \begin{cases} -\dfrac{4}{3}x^3 + 2x^2 + 3x, & x > 0 \\ 3xe^{x}, & x \leq 0 \end{cases}f(x)=⎩⎨⎧​−34​x3+2x2+3x,3xex,​x>0x≤0​ Then f is increasing function in the interval.
  1. (A)(−12,2)\left(-\dfrac{1}{2}, 2\right)(−21​,2)
  2. (B)(−1,32)\left(-1, \dfrac{3}{2}\right)(−1,23​)
  3. (C)(−3,−1)(-3, -1)(−3,−1)
  4. (D)(0,2)(0, 2)(0,2)

Correct answer: (B)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2021
Let 'a' be a real number such that the function f(x)=ax2+6x−15f(x) = ax^2 + 6x - 15f(x)=ax2+6x−15, x ∈ R is increasing in (−∞,34)\left(-\infty, \frac{3}{4}\right)(−∞,43​) and decreasing in (34,∞)\left(\frac{3}{4}, \infty\right)(43​,∞). Then the function g(x)=ax2−6x+15g(x) = ax^2 - 6x + 15g(x)=ax2−6x+15, x ∈ R has a :
  1. (A)local minimum at x=34x = \frac{3}{4}x=43​
  2. (B)local maximum at x=34x = \frac{3}{4}x=43​
  3. (C)local minimum at x=−34x = -\frac{3}{4}x=−43​
  4. (D)local maximum at x=−34x = -\frac{3}{4}x=−43​

Correct answer: (D)

Step-by-step solution →
Q112·MathematicsNumericalJEE Main 2021
If the point on the curve y2=6xy^{2}=6xy2=6x, nearest to the point (3,32)\left(3,\frac{3}{2}\right)(3,23​) is (α,β)(\alpha,\beta)(α,β), then 2(α+β)2(\alpha+\beta)2(α+β) is equal to………

Correct answer: 9

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2021
The sum of all the local minimum values of the twice differentiable function f:R→Rf:R\to Rf:R→R defined by f(x)=x3−3x2−3f′′(2)2x+f′′(1)f(x)=x^{3}-3x^{2}-\frac{3f''(2)}{2}x+f''(1)f(x)=x3−3x2−23f′′(2)​x+f′′(1) is :
  1. (A)−22-22−22
  2. (B)555
  3. (C)000
  4. (D)−27-27−27

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2021
Let A=[aij]A = \left[a_{ij}\right]A=[aij​] be a 3 x 3 matrix, where aij={1,if i=j−x,if ∣i−j∣=12x+1,otherwisea_{ij} = \begin{cases} 1 , & \text{if } i = j \\ -x , & \text{if } |i - j| = 1 \\ 2x+1 , & \text{otherwise} \end{cases}aij​=⎩⎨⎧​1,−x,2x+1,​if i=jif ∣i−j∣=1otherwise​ Let a function f:R→Rf : R \rightarrow Rf:R→R be defined as f(x)=det⁡(A)f(x) = \det(A)f(x)=det(A). Then the sum of maximum and minimum values of f on R is equal to :
  1. (A)−8827-\frac{88}{27}−2788​
  2. (B)2027\frac{20}{27}2720​
  3. (C)8827\frac{88}{27}2788​
  4. (D)−2027-\frac{20}{27}−2720​

Correct answer: (A)

Step-by-step solution →
Q115·MathematicsNumericalJEE Main 2021
Let P(x) be a real polynomial of degree 3 which vanishes at x = −3. Let P(x) have local minima at x = 1, local maxima at x = −1 and ∫−11P(x) dx=18\int_{-1}^{1} P(x)\,dx = 18∫−11​P(x)dx=18, then the sum of all the coefficients of the polynomial P(x) is equal to ________ .

Correct answer: 8

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2021
Consider the function f:R→Rf : R \rightarrow Rf:R→R defined by f(x)={(2−sin⁡(1x))∣x∣,x≠00,x=0f(x) = \begin{cases} \left(2-\sin\left(\frac{1}{x}\right)\right)|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}f(x)={(2−sin(x1​))∣x∣,0,​x=0x=0​. Then fff is :
  1. (A)monotonic on (−∞,0)∪(0,∞)(-\infty, 0) \cup (0, \infty)(−∞,0)∪(0,∞)
  2. (B)not monotonic on (−∞,0)(-\infty, 0)(−∞,0) and (0,∞)(0, \infty)(0,∞)
  3. (C)monotonic on (0,∞)(0, \infty)(0,∞) only
  4. (D)monotonic on (−∞,0)(-\infty, 0)(−∞,0) only

Correct answer: (B)

Step-by-step solution →
Q117·MathematicsNumericalJEE Main 2021
The maximum value of z in the following equation z=6xy+y2z = 6xy + y^2z=6xy+y2, where 3x+4y≤1003x + 4y \leq 1003x+4y≤100 and 4x+3y≤754x + 3y \leq 754x+3y≤75 for x≥0x \geq 0x≥0 and y≥0y \geq 0y≥0 is _______ .

Correct answer: 904

Step-by-step solution →
Q118·MathematicsNumericalJEE Main 2021
Let f:[−1,1]→Rf : [-1, 1] \rightarrow Rf:[−1,1]→R be defined as f(x)=ax2+bx+cf(x) = ax^{2}+bx+cf(x)=ax2+bx+c for all x∈[−1,1]x \in [-1, 1]x∈[−1,1], where a,b,c∈Ra, b, c \in Ra,b,c∈R such that f(−1)=2f(-1) = 2f(−1)=2, f′(−1)=1f'(-1) = 1f′(−1)=1 and for x∈(−1,1)x \in (-1, 1)x∈(−1,1) the maximum value of f′′(x)f''(x)f′′(x) is 12\frac{1}{2}21​. If f(x)≤αf(x) \le \alphaf(x)≤α, x∈[−1,1]x \in [-1, 1]x∈[−1,1], then the least value of α\alphaα is equal to ______.

Correct answer: 5

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2021
Let f be a real valued function, defined on R−{−1,1}R - \{-1, 1\}R−{−1,1} and given by f(x)=3log⁡e∣x−1x+1∣−2x−1.f(x) = 3\log_{e}\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}.f(x)=3loge​​x+1x−1​​−x−12​. Then in which of the following intervals, function f(x) is increasing?
  1. (A)(−∞,−1)∪([12,∞)−{1})(-\infty,-1)\cup\left(\left[\frac{1}{2},\infty\right)-\{1\}\right)(−∞,−1)∪([21​,∞)−{1})
  2. (B)(−∞,∞)−{−1,1}(-\infty, \infty)-\{-1, 1\}(−∞,∞)−{−1,1}
  3. (C)(−1,12]\left(-1,\frac{1}{2}\right](−1,21​]
  4. (D)(−∞,12]−{−1}\left(-\infty,\frac{1}{2}\right]-\{-1\}(−∞,21​]−{−1}

Correct answer: (A)

Step-by-step solution →
Q120·MathematicsNumericalJEE Main 2021
If the normal to the curve y(x)=∫0x(2t2−15t+10)dty(x) = \int_{0}^{x}(2t^{2} - 15t + 10)dty(x)=∫0x​(2t2−15t+10)dt at a point (a,b) is parallel to the line x + 3y = −5, a > 1, then the value of |a + 6b| is equal to ________ .

Correct answer: 406

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2021
The range of a ∈ ℝ for which the function f(x)=(4a−3)(x+log⁡e5)+2(a−7)cot⁡(x2)sin⁡2(x2)f(x) = (4a - 3)(x + \log_{e} 5) + 2(a - 7)\cot\left(\frac{x}{2}\right)\sin^{2}\left(\frac{x}{2}\right)f(x)=(4a−3)(x+loge​5)+2(a−7)cot(2x​)sin2(2x​), x ≠ 2nπ, n ∈ ℕ, has critical points, is :
  1. (A)(−3, 1)
  2. (B)[−43,2]\left[-\frac{4}{3}, 2\right][−34​,2]
  3. (C)[1, ∞)
  4. (D)(−∞, −1]

Correct answer: (B)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2021
The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:
  1. (A)A right angle triangle having two of its sides of length 2r and r.
  2. (B)An equilateral triangle of height 2r3\frac{2r}{3}32r​.
  3. (C)An isosceles triangle with base equal to 2r.
  4. (D)An equilateral triangle having each of its side of length 3\sqrt{3}3​ r.

Correct answer: (D)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2021
The maximum slope of the curve y=12x4−5x3+18x2−19xy = \frac{1}{2}x^{4} - 5x^{3} + 18x^{2} - 19xy=21​x4−5x3+18x2−19x occurs at the point:
  1. (A)(2, 9)
  2. (B)(2,2)
  3. (C)(3,212)\left( 3, \frac{21}{2} \right)(3,221​)
  4. (D)(0, 0)

Correct answer: (B)

Step-by-step solution →
Q124·MathematicsNumericalJEE Main 2021
Let a be an integer such that all the real roots of the polynomial 2x5+5x4+10x3+10x2+10x+102x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+102x5+5x4+10x3+10x2+10x+10 lie in the interval (a, a + 1). Then, ∣a∣|a|∣a∣ is equal to________________.

Correct answer: 2

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2021
The minimum value of f(x) = a^{a}^{x}+a^{1–a}^{x}, where a, x ∈ R and a > 0, is equal to:
  1. (A)a + 1a
  2. (B)a + 1
  3. (C)2a
  4. (D)2 a

Correct answer: (D)

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2021
If the curves, x2a+y2b=1\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1ax2​+by2​=1 and x2c+y2d=1\frac{x^{2}}{c} + \frac{y^{2}}{d} = 1cx2​+dy2​=1 intersect each other at an angle of 90∘90^{\circ}90∘, then which of the following relations is true ?
  1. (A)a+b=c+da + b = c + da+b=c+d
  2. (B)a−b=c−da - b = c - da−b=c−d
  3. (C)ab=c+da+bab = \frac{c+d}{a+b}ab=a+bc+d​
  4. (D)a−c=b+da - c = b + da−c=b+d

Correct answer: (B)

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2021
If Rolle's theorem holds for the function f(x)=x3−ax2+bx−4f(x) = x^{3} - ax^{2} + bx - 4f(x)=x3−ax2+bx−4, x∈[1,2]x \in [1, 2]x∈[1,2] with f′(43)=0f'\left(\frac{4}{3}\right) = 0f′(34​)=0, then ordered pair (a, b) is equal to :
  1. (A)(−5,8)(-5, 8)(−5,8)
  2. (B)(5,8)(5, 8)(5,8)
  3. (C)(5,−8)(5, -8)(5,−8)
  4. (D)(−5,−8)(-5, -8)(−5,−8)

Correct answer: (B)

Step-by-step solution →
Q128·MathematicsNumericalJEE Main 2021
If the curves x = y4^{4}4 and xy = k cut at right angles, then (4k)6^{6}6 is equal to ______.

Correct answer: 4

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2021
Let f(x)f(x)f(x) be a polynomial of degree 6 in xxx, in which the coefficient of x6x^6x6 is unity and it has extrema at x=−1x = -1x=−1 and x=1x = 1x=1. If lim⁡x→0f(x)x3=1\lim\limits_{x \to 0} \dfrac{f(x)}{x^3} = 1x→0lim​x3f(x)​=1, then 5⋅f(2)5 \cdot f(2)5⋅f(2) is equal to __________

Correct answer: 144

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2021
Let f:R→Rf:\mathbf{R}\to\mathbf{R}f:R→R be defined as f(x)={−55x,if x<−52x3−3x2−120x,if −5≤x≤42x3−3x2−36x−336,if x>4f(x)=\begin{cases} -55x, & \text{if } x<-5 \\ 2x^{3}-3x^{2}-120x, & \text{if } -5\le x\le 4 \\ 2x^{3}-3x^{2}-36x-336, & \text{if } x>4 \end{cases}f(x)=⎩⎨⎧​−55x,2x3−3x2−120x,2x3−3x2−36x−336,​if x<−5if −5≤x≤4if x>4​ Let A={x∈R:f is increasing}A=\{x\in R: f\ \text{is increasing}\}A={x∈R:f is increasing}. Then A is equal to :
  1. (A)(−5,−4)∪(4,∞)\left(-5,-4\right)\cup\left(4,\infty\right)(−5,−4)∪(4,∞)
  2. (B)(−5,∞)\left(-5,\infty\right)(−5,∞)
  3. (C)(−∞,−5)∪(4,∞)\left(-\infty,-5\right)\cup\left(4,\infty\right)(−∞,−5)∪(4,∞)
  4. (D)(−∞,−5)∪(−4,∞)\left(-\infty,-5\right)\cup\left(-4,\infty\right)(−∞,−5)∪(−4,∞)

Correct answer: (A)

Step-by-step solution →
Q131·MathematicsNumericalJEE Main 2021
The minimum value of α\alphaα for which the equation 4sin⁡x+11−sin⁡x=α\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alphasinx4​+1−sinx1​=α has at least one solution in (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​) is ______

Correct answer: 9

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2021
If the tangent to the curve y=x3y = x^3y=x3 at the point P(t,t3)P(t, t^3)P(t,t3) meets the curve again at QQQ, then the ordinate of the point which divides PQPQPQ internally in the ratio 1:21 : 21:2 is :
  1. (A)−2t3-2t^3−2t3
  2. (B)−t3-t^3−t3
  3. (C)000
  4. (D)2t32t^32t3

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2021
If P is a point on the parabola y=x2+4y=x^{2}+4y=x2+4 which is closest to the straight line y=4x−1,y=4x-1,y=4x−1, then the co-ordinates of P are :
  1. (A)(−2,8)(-2, 8)(−2,8)
  2. (B)(1,5)(1, 5)(1,5)
  3. (C)(3,13)(3, 13)(3,13)
  4. (D)(2,8)(2, 8)(2,8)

Correct answer: (D)

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2021
The function f(x)=4x3−3x26−2sin⁡x+(2x−1)cos⁡xf(x) = \frac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos xf(x)=64x3−3x2​−2sinx+(2x−1)cosx :
  1. (A)increases in [12,∞)\left[\frac{1}{2}, \infty\right)[21​,∞)
  2. (B)decreases (−∞,12]\left(-\infty, \frac{1}{2}\right](−∞,21​]
  3. (C)increases in (−∞,12]\left(-\infty, \frac{1}{2}\right](−∞,21​]
  4. (D)decreases [12,∞)\left[\frac{1}{2}, \infty\right)[21​,∞)

Correct answer: (A)

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2021
If the curve y=ax2+bx+c,x∈R,y=ax^{2}+bx+c, x\in R,y=ax2+bx+c,x∈R, passes through the point (1,2) and the tangent line to this curve at origin is y=xy=xy=x, then the possible values of a,b,ca,b,ca,b,c are :
  1. (A)a =1, b=1, c=0
  2. (B)a = −1-1−1, b=1, c =1
  3. (C)a =1, b=0, c =1
  4. (D)a=12,b=12,c=1a=\frac{1}{2}, b=\frac{1}{2}, c=1a=21​,b=21​,c=1

Correct answer: (A)

Step-by-step solution →
Q136·MathematicsNumericalJEE Advanced 2020
For a polynomial g(x) with real coefficient, let mgm_{g}mg​ denote the number of distinct real roots of g(x). Suppose S is the set of polynomials with real coefficient defined by S={(x2−1)2(a0+a1x+a2x2+a3x3):a0,a1,a2,a3∈R}S = \left\{\left(x^{2} - 1\right)^{2}\left(a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3}\right) : a_{0}, a_{1}, a_{2}, a_{3} \in \mathbb{R}\right\}S={(x2−1)2(a0​+a1​x+a2​x2+a3​x3):a0​,a1​,a2​,a3​∈R}. For a polynomial f, let f′f'f′ and f′′f''f′′ denote its first and second order derivatives, respectively. Then the minimum possible value of (mf′+mf′′)\left(m_{f'} + m_{f''}\right)(mf′​+mf′′​), where f∈Sf \in Sf∈S, is ______

Correct answer: 5.00

Step-by-step solution →
Q137·MathematicsSingle correctJEE Advanced 2020
Consider all rectangles lying in the region {(x,y)∈R×R:0≤x≤π2 and 0≤y≤2sin⁡(2x)}\left\{(x, y) \in \mathbb{R} \times \mathbb{R} : 0 \leq x \leq \frac{\pi}{2} \text{ and } 0 \leq y \leq 2\sin(2x)\right\}{(x,y)∈R×R:0≤x≤2π​ and 0≤y≤2sin(2x)} and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is
  1. (A)3π2\frac{3\pi}{2}23π​
  2. (B)π\piπ
  3. (C)π23\frac{\pi}{2\sqrt{3}}23​π​
  4. (D)π32\frac{\pi\sqrt{3}}{2}2π3​​

Correct answer: (C)

Step-by-step solution →
Q138·MathematicsNumericalJEE Advanced 2020
Let the function f:(0,π)→Rf : (0,\pi) \to \mathbb{R}f:(0,π)→R be defined by f(θ)=(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)4f(\theta) = (\sin\theta + \cos\theta)^{2} + (\sin\theta - \cos\theta)^{4}f(θ)=(sinθ+cosθ)2+(sinθ−cosθ)4 Suppose the function fff has a local minimum at θ\thetaθ precisely when θ∈{λ1π,…,λrπ}\theta \in \{\lambda_{1}\pi, \ldots, \lambda_{r}\pi\}θ∈{λ1​π,…,λr​π}, where 0<λ1<⋯<λr<10 < \lambda_{1} < \cdots < \lambda_{r} < 10<λ1​<⋯<λr​<1. Then the value of λ1+⋯+λr\lambda_{1} + \cdots + \lambda_{r}λ1​+⋯+λr​ is ________

Correct answer: 0.50

Step-by-step solution →
Q139·MathematicsSingle correctJEE Main 2020
For all twice differentiable functions f:R→Rf : R \to Rf:R→R, with f(0)=f(1)=f′(0)=0f(0) = f(1) = f'(0) = 0f(0)=f(1)=f′(0)=0,
  1. (A)f′′(x)≠0f''(x) \neq 0f′′(x)=0, at every point x∈(0,1)x \in (0,1)x∈(0,1)
  2. (B)f′′(x)=0f''(x) = 0f′′(x)=0, for some x∈(0,1)x \in (0,1)x∈(0,1)
  3. (C)f′′(0)=0f''(0) = 0f′′(0)=0
  4. (D)f′′(x)=0f''(x) = 0f′′(x)=0, at every point x∈(0,1)x \in (0, 1)x∈(0,1)

Correct answer: (B)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2020
The set of all real values of λ\lambdaλ for which the function f(x)=(1−cos⁡2x)⋅(λ+sin⁡x)f(x) = (1 - \cos^{2} x)\cdot(\lambda + \sin x)f(x)=(1−cos2x)⋅(λ+sinx), x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)x∈(−2π​,2π​), has exactly one maxima and exactly one minima, is
  1. (A)(−12,12)−{0}\left(-\frac{1}{2}, \frac{1}{2}\right) - \{0\}(−21​,21​)−{0}
  2. (B)(−32,32)\left(-\frac{3}{2}, \frac{3}{2}\right)(−23​,23​)
  3. (C)(−12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)(−21​,21​)
  4. (D)(−32,32)−{0}\left(-\frac{3}{2}, \frac{3}{2}\right) - \{0\}(−23​,23​)−{0}

Correct answer: (D)

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2020
The position of a moving car at time t is given by f(t)=at2+bt+c, t>0f(t) = at^2+bt+c,\ t>0f(t)=at2+bt+c, t>0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t1,t2][t_1, t_2][t1​,t2​] is attained at the point:
  1. (A)(t2−t1)2\dfrac{(t_2-t_1)}{2}2(t2​−t1​)​
  2. (B)a(t2−t1)+ba(t_2-t_1)+ba(t2​−t1​)+b
  3. (C)(t1+t2)2\dfrac{(t_1+t_2)}{2}2(t1​+t2​)​
  4. (D)2a(t1+t2)+b2a(t_1+t_2)+b2a(t1​+t2​)+b

Correct answer: (C)

Step-by-step solution →
Q142·MathematicsNumericalJEE Main 2020
Let AD and BC be two vertical poles at A and b respectively on a horizontal ground. If AD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point A such that MD2+MC2MD^{2}+MC^{2}MD2+MC2 is minimum is ____.

Correct answer: 05

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2020
If the tangent to the curve, y=f(x)=xlog⁡exy = f(x) = x\log_{e} xy=f(x)=xloge​x, (x>0)(x > 0)(x>0) at a point (c,f(c))(c, f(c))(c,f(c)) is parallel to the line – segment joining the points (1,0)(1, 0)(1,0) and (e,e)(e, e)(e,e) then c is equal to:
  1. (A)e−1e\frac{e-1}{e}ee−1​
  2. (B)1e−1\frac{1}{e-1}e−11​
  3. (C)e(1e−1)e^{\left(\frac{1}{e-1}\right)}e(e−11​)
  4. (D)e(11−e)e^{\left(\frac{1}{1-e}\right)}e(1−e1​)

Correct answer: (C)

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2020
Which of the following point lies on the tangent to the curve x4ey+2y+1=3x^4e^y+2\sqrt{y+1}=3x4ey+2y+1​=3 at the point (1,0)(1,0)(1,0)?
  1. (A)(2,2)(2,2)(2,2)
  2. (B)(−2,6)(-2,6)(−2,6)
  3. (C)(−2,4)(-2,4)(−2,4)
  4. (D)(2,6)(2,6)(2,6)

Correct answer: (B)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2020
If the point P on the curve, 4x2+5y2=204x^{2}+5y^{2}=204x2+5y2=20 is farthest from the point Q(0, –4) then PQ2PQ^{2}PQ2 is equal to:
  1. (A)21
  2. (B)48
  3. (C)36
  4. (D)29

Correct answer: (C)

Step-by-step solution →
Q146·MathematicsNumericalJEE Main 2020
If the lines x + y = a and x - y = b touch the curve y=x2−3x+2y = x^2 - 3x + 2y=x2−3x+2 at the points where the curve intersects the x-axis, then ab\dfrac{a}{b}ba​ is equal to __________.

Correct answer: 0.50

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2020
Let fff be a twice differentiable function on (1,6)(1, 6)(1,6). If f(2)=8f(2) = 8f(2)=8, f′(2)=5f'(2) = 5f′(2)=5, f′(x)≥1f'(x) \geq 1f′(x)≥1 and f′′(x)≥4f''(x) \geq 4f′′(x)≥4, for all x∈(1,6)x \in (1, 6)x∈(1,6), then:
  1. (A)f(5)+f′(5)≤26f(5) + f'(5) \leq 26f(5)+f′(5)≤26
  2. (B)f(5)≤10f(5) \leq 10f(5)≤10
  3. (C)f(5)+f′(5)≥28f(5) + f'(5) \geq 28f(5)+f′(5)≥28
  4. (D)f′(5)+f′′(5)≤20f'(5) + f''(5) \leq 20f′(5)+f′′(5)≤20

Correct answer: (C)

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2020
The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y=x2−1y = x^{2} - 1y=x2−1 below the x-axis, is:
  1. (A)433\frac{4}{3\sqrt{3}}33​4​
  2. (B)133\frac{1}{3\sqrt{3}}33​1​
  3. (C)43\frac{4}{3}34​
  4. (D)233\frac{2}{3\sqrt{3}}33​2​

Correct answer: (A)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2020
If the surface area of a cube is increasing at a rate of 3.6 cm2cm^2cm2/sec, retaining its shape; then the rate of change of its volume (in cm3cm^3cm3/sec), when the length of a side of the cube is 10 cm, is:
  1. (A)999
  2. (B)181818
  3. (C)101010
  4. (D)202020

Correct answer: (A)

Step-by-step solution →
Q150·MathematicsNumericalJEE Main 2020
If the tangent to the curve, y=exy = e^xy=ex at a point (c,ec)\left( c, e^c \right)(c,ec) and the normal to the parabola, y2=4xy^2 = 4xy2=4x at the point (1,2)(1, 2)(1,2) intersect at the same point on the x −-− axis, then the value of c is __________.

Correct answer: 04.00

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2020
The function, f(x)=(3x−7)x2/3f(x) = (3x - 7)x^{2/3}f(x)=(3x−7)x2/3, x∈Rx \in Rx∈R, is increasing for all x lying in:
  1. (A)(−∞,1415)\left(-\infty, \frac{14}{15}\right)(−∞,1514​)
  2. (B)(−∞,0)∪(37,∞)(-\infty, 0) \cup \left(\frac{3}{7}, \infty\right)(−∞,0)∪(73​,∞)
  3. (C)(−∞,−1415)∪(0,∞)\left(-\infty, -\frac{14}{15}\right) \cup (0, \infty)(−∞,−1514​)∪(0,∞)
  4. (D)(−∞,0)∪(1415,∞)(-\infty, 0) \cup \left(\frac{14}{15}, \infty\right)(−∞,0)∪(1514​,∞)

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2020
Suppose f(x)f(x)f(x) is a polynomial of degree four, having critical points at −1,0,1-1, 0, 1−1,0,1. If T={x∈R | f(x)=f(0)}T = \left\{ x \in R \,\middle|\, f(x) = f(0) \right\}T={x∈R∣f(x)=f(0)}, then the sum of squares of all the elements of T is:
  1. (A)8
  2. (B)6
  3. (C)2
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2020
Let f be any function continuous on [a,b][a,b][a,b] and twice differentiable on (a,b)(a,b)(a,b). If for all x∈(a,b),f′(x)>0x \in (a,b), f'(x)>0x∈(a,b),f′(x)>0 and f′′(x)<0f''(x)<0f′′(x)<0, then for any c∈(a,b)c \in (a,b)c∈(a,b), f(c)−f(a)f(b)−f(c)\dfrac{f(c)-f(a)}{f(b)-f(c)}f(b)−f(c)f(c)−f(a)​ is greater than:
  1. (A)c−ab−c\dfrac{c-a}{b-c}b−cc−a​
  2. (B)b+ab−a\dfrac{b+a}{b-a}b−ab+a​
  3. (C)b−cc−a\dfrac{b-c}{c-a}c−ab−c​
  4. (D)111

Correct answer: (A)

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2020
A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3^{3}3/ min. When the thickness of ice is 5 cm, then the rate (in cm/min) at which of the thickness of ice decreases, is:
  1. (A)56π\dfrac{5}{6\pi}6π5​
  2. (B)136π\dfrac{1}{36\pi}36π1​
  3. (C)118π\dfrac{1}{18\pi}18π1​
  4. (D)154π\dfrac{1}{54\pi}54π1​

Correct answer: (C)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2020
Let a function f:[0,5]→Rf : [0, 5] \rightarrow Rf:[0,5]→R be continuous, f(1)=3f(1) = 3f(1)=3 and F be defined as: F(x)=∫1xt2g(t) dtF(x) = \int\limits_{1}^{x} t^{2} g(t)\,dtF(x)=1∫x​t2g(t)dt, where g(t)=∫1tf(u) dug(t) = \int\limits_{1}^{t} f(u)\,dug(t)=1∫t​f(u)du. Then for the function F, the point x = 1 is:
  1. (A)a point of local minima
  2. (B)a point of inflection.
  3. (C)not a critical point
  4. (D)a point of local maxima

Correct answer: (A)

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2020
Let the normal at a point P on the curve y2−3x2+y+10=0y^{2}-3x^{2}+y+10=0y2−3x2+y+10=0 intersect the y-axis at (0,32)\left(0,\dfrac{3}{2}\right)(0,23​). If m is the slope of the tangent at P to the curve, then ∣m∣|m|∣m∣ is equal to

Correct answer: 4

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2020
The length of the perpendicular from the origin, 0n the normal to the curve, x2+2xy−3y2=0x^{2}+2xy-3y^{2}=0x2+2xy−3y2=0 at the point (2, 2) is:
  1. (A)222\sqrt{2}22​
  2. (B)424\sqrt{2}42​
  3. (C)2\sqrt{2}2​
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2020
If c is a point at which Rolle's theorem holds for the function, f(x)=log⁡e(x2+α7x)f(x)=\log_{e}\left(\dfrac{x^{2}+\alpha}{7x}\right)f(x)=loge​(7xx2+α​) in the interval [3,4][3,4][3,4], where α∈R\alpha\in Rα∈R, then f′′(c)f''(c)f′′(c) is equal to:
  1. (A)−112-\dfrac{1}{12}−121​
  2. (B)−124-\dfrac{1}{24}−241​
  3. (C)37\dfrac{\sqrt{3}}{7}73​​
  4. (D)112\dfrac{1}{12}121​

Correct answer: (D)

Step-by-step solution →
Q159·MathematicsNumericalJEE Main 2020
Let f(x)f(x)f(x) be a polynomial of degree 3 such that f(−1)=10,f(1)=−6f(-1)=10, f(1)=-6f(−1)=10,f(1)=−6, f(x)f(x)f(x) has a critical point at x=−1x=-1x=−1 and f′(x)f'(x)f′(x) has a critical point at x=1x=1x=1. Then f(x)f(x)f(x) has a local minima at x=_______.

Correct answer: 3

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2020
The value of c in the Lagrange's mean value theorem for the function f(x)=x3−4x2+8x+11f(x)=x^{3}-4x^{2}+8x+11f(x)=x3−4x2+8x+11, when x∈[0,1]x\in[0,1]x∈[0,1] is:
  1. (A)7−23\dfrac{\sqrt{7}-2}{3}37​−2​
  2. (B)4−53\dfrac{4-\sqrt{5}}{3}34−5​​
  3. (C)4−73\dfrac{4-\sqrt{7}}{3}34−7​​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (C)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2020
Let f(x)f(x)f(x) be a polynomial of degree 5 such that x=±1x=\pm1x=±1 are its critical points. If lim⁡x→0(2+f(x)x3)=4\displaystyle\lim_{x\to0}\left(2+\dfrac{f(x)}{x^{3}}\right)=4x→0lim​(2+x3f(x)​)=4, then which one of the following is not true?
  1. (A)x = 1 is a point of minima and x = -1 is a point of maxims of f.
  2. (B)x = 1 is a point of maxima and x = -1 is a point of minimum of f
  3. (C)f is an odd function
  4. (D)f(1)−4f(−1)=4f(1)-4f(-1)=4f(1)−4f(−1)=4

Correct answer: (A)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2020
Let the function, f:[−7,0]→Rf:[-7,0]\to Rf:[−7,0]→R be continuous on [−7,0][-7,0][−7,0] and differentiable on (−7,0)(-7,0)(−7,0). If f(−7)=−3f(-7)=-3f(−7)=−3 and f′(x)≤2f'(x)\le 2f′(x)≤2, for all x∈(−7,0)x\in(-7,0)x∈(−7,0), then for all such functions f, f(−1)+f(0)f(-1)+f(0)f(−1)+f(0) lies in the interval:
  1. (A)[−3,11][-3,11][−3,11]
  2. (B)(−∞,20](-\infty, 20](−∞,20]
  3. (C)[−6,20][-6,20][−6,20]
  4. (D)(−∞,11](-\infty,11](−∞,11]

Correct answer: (B)

Step-by-step solution →
Q163·MathematicsMultiple correctJEE Advanced 2019
Let f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R be given by f(x)=(x−1)(x−2)(x−5)f(x) = (x - 1)(x - 2)(x - 5)f(x)=(x−1)(x−2)(x−5). Define F(x)=∫0xf(t) dtF(x) = \displaystyle\int_{0}^{x} f(t)\, dtF(x)=∫0x​f(t)dt, x>0x > 0x>0 Then which of the following options is/are correct?
  1. (A)F(x)≠0F(x) \neq 0F(x)=0 for all x∈(0,5)x \in (0, 5)x∈(0,5)
  2. (B)F has a local minimum at x=1x = 1x=1
  3. (C)F has a local maximum at x=2x = 2x=2
  4. (D)F has two local maxima and one local minimum in (0,∞)(0, \infty)(0,∞)

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

Step-by-step solution →
Q164·MathematicsMultiple correctJEE Advanced 2019
Let f(x)=sin⁡πxx2f(x) = \dfrac{\sin \pi x}{x^{2}}f(x)=x2sinπx​, x>0x > 0x>0 Let x1<x2<x3<...<xn<....x_1 < x_2 < x_3 < ... < x_n < ....x1​<x2​<x3​<...<xn​<.... be all the points of local maximum of f and y1<y2<y3<....<yn<....y_1 < y_2 < y_3 < .... < y_n < ....y1​<y2​<y3​<....<yn​<.... be all the points of local minimum of f. Then which of the following options is/are correct?
  1. (A)x1<y1x_1 < y_1x1​<y1​
  2. (B)∣xn−yn∣>1|x_n - y_n| > 1∣xn​−yn​∣>1 for every n
  3. (C)xn∈(2n, 2n+12)x_n \in \left(2n,\ 2n + \dfrac{1}{2}\right)xn​∈(2n, 2n+21​) for every n
  4. (D)xn+1−xn>2x_{n+1} - x_n > 2xn+1​−xn​>2 for every n

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

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2019
A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/ sec., then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is :
  1. (A)25
  2. (B)253\frac{25}{3}325​
  3. (C)25325\sqrt{3}253​
  4. (D)253\frac{25}{\sqrt{3}}3​25​

Correct answer: (D)

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2019
If m is the minimum value of k for which the function f(x)=xkx−x2f(x)=x\sqrt{kx-x^{2}}f(x)=xkx−x2​ is increasing in the interval [0,3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to :
  1. (A)(5,36)(5, 3\sqrt{6})(5,36​)
  2. (B)(4,32)(4, 3\sqrt{2})(4,32​)
  3. (C)(3,33)(3, 3\sqrt{3})(3,33​)
  4. (D)(4,33)(4, 3\sqrt{3})(4,33​)

Correct answer: (D)

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2019
The tangents to the curve y=(x−2)2−1y = (x-2)^{2} -1y=(x−2)2−1 at its points of intersection with the line x−y=3x - y = 3x−y=3, intersect at the point :
  1. (A)(53,1)\left(\dfrac{5}{3},1\right)(35​,1)
  2. (B)(−52,−1)\left(-\dfrac{5}{2},-1\right)(−25​,−1)
  3. (C)(−52,1)\left(-\dfrac{5}{2},1\right)(−25​,1)
  4. (D)(52,−1)\left(\dfrac{5}{2},-1\right)(25​,−1)

Correct answer: (D)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2019
If the tangent to the curve y=xx2−3y = \dfrac{x}{x^{2} - 3}y=x2−3x​, x∈R,(x≠±3)x \in R, \left(x \neq \pm\sqrt{3}\right)x∈R,(x=±3​) at a point (α, β) ≠ (0, 0) on it is parallel to the line 2x + 6y − 11 = 0 then
  1. (A)∣2α+6β∣=11|2\alpha + 6\beta| = 11∣2α+6β∣=11
  2. (B)∣2α+6β∣=19|2\alpha + 6\beta| = 19∣2α+6β∣=19
  3. (C)∣6α+2β∣=19|6\alpha + 2\beta| = 19∣6α+2β∣=19
  4. (D)∣6α+2β∣=9|6\alpha + 2\beta| = 9∣6α+2β∣=9

Correct answer: (C)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2019
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3cm^{3}cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of ice decreases is
  1. (A)136π\dfrac{1}{36\pi}36π1​
  2. (B)56π\dfrac{5}{6\pi}6π5​
  3. (C)19π\dfrac{1}{9\pi}9π1​
  4. (D)118π\dfrac{1}{18\pi}18π1​

Correct answer: (D)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2019
Let f(x) = ex−xe^{x}-xex−x and g(x) = x2−xx^{2}-xx2−x, ∀\forall∀ x ∈\in∈ R. Then the set of all x ∈\in∈ R, where the function h(x) = (fog) (x) is increasing is:
  1. (A)[0,12]∪[1,∞)\left[0,\dfrac{1}{2}\right] \cup [1,\infty)[0,21​]∪[1,∞)
  2. (B)[1,12]∪[12,∞)\left[1,\dfrac{1}{2}\right] \cup \left[\dfrac{1}{2},\infty\right)[1,21​]∪[21​,∞)
  3. (C)[−12,0]∪[1,∞)\left[\dfrac{-1}{2},0\right] \cup [1,\infty)[2−1​,0]∪[1,∞)
  4. (D)[0,∞)[0,\infty)[0,∞)

Correct answer: (A)

Step-by-step solution →
Q171·MathematicsSingle correctJEE Main 2019
If f(x) is a non-zero polynomial of degree four, having local extreme points at x=−1,0,1x=-1,0,1x=−1,0,1; then the set S={x∈R;f(x)=f(0)}S=\{x\in R; f(x)=f(0)\}S={x∈R;f(x)=f(0)} contains exactly:
  1. (A)four irrational numbers
  2. (B)four rational numbers
  3. (C)two irrational and one rational number
  4. (D)two irrational and two rational numbers

Correct answer: (C)

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2019
A water tank has the shape of an inverted right circular cone, whose semi vertical angle is tan⁡−1(12)\tan^{-1}\left(\dfrac{1}{2}\right)tan−1(21​). Water is poured in at a constant rage of 5 cubic meter per minute. Then the rate (in m/min) at which the level of water is rising at the instant when the depth of water in the tank is 10 m is:
  1. (A)2π\dfrac{2}{\pi}π2​
  2. (B)15π\dfrac{1}{5\pi}5π1​
  3. (C)110π\dfrac{1}{10\pi}10π1​
  4. (D)115π\dfrac{1}{15\pi}15π1​

Correct answer: (B)

Step-by-step solution →
Q173·MathematicsSingle correctJEE Main 2019
Let S be the set of all values of x for which the tangent to the curve y=f(x)=x3−x2−2xy=f(x)=x^{3}-x^{2}-2xy=f(x)=x3−x2−2x at (x,y)(x, y)(x,y) is parallel to the line segment joining the points (1,f(1))(1, f(1))(1,f(1)) and (−1,f(−1))(-1, f(-1))(−1,f(−1)), then S is equal to:
  1. (A){13,−1}\left\{\dfrac{1}{3},-1\right\}{31​,−1}
  2. (B){−13,−1}\left\{-\dfrac{1}{3},-1\right\}{−31​,−1}
  3. (C){13,1}\left\{\dfrac{1}{3},1\right\}{31​,1}
  4. (D){−13,1}\left\{-\dfrac{1}{3},1\right\}{−31​,1}

Correct answer: (D)

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2019
If the tangent to the curve, y=x3+ax−by=x^{3}+ax-by=x3+ax−b at the point (1,−5)(1,-5)(1,−5) is perpendicular to the line, −x+y+4=0-x+y+4=0−x+y+4=0, then which one of the following, points lies on the curve?
  1. (A)(2,−2)(2,-2)(2,−2)
  2. (B)(−2,2)(-2,2)(−2,2)
  3. (C)(−2,1)(-2,1)(−2,1)
  4. (D)(2,−1)(2,-1)(2,−1)

Correct answer: (A)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2019
If S1S_{1}S1​ and S2S_{2}S2​ are respectively the sets of local minimum and local maximum points of the function. f(x)=9x4+12x3−36x2+25,x∈Rf(x)=9x^{4}+12x^{3}-36x^{2}+25, x\in Rf(x)=9x4+12x3−36x2+25,x∈R, then
  1. (A)S1={−2,1};S2={0}S_{1}=\{-2,1\};S_{2}=\{0\}S1​={−2,1};S2​={0}
  2. (B)S1={−2,0};S2={1}S_{1}=\{-2,0\};S_{2}=\{1\}S1​={−2,0};S2​={1}
  3. (C)S1={−2};S2={0,1}S_{1}=\{-2\};S_{2}=\{0,1\}S1​={−2};S2​={0,1}
  4. (D)S1={−1};S2={0,2}S_{1}=\{-1\};S_{2}=\{0,2\}S1​={−1};S2​={0,2}

Correct answer: (A)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2019
The shortest distance between the line y=xy=xy=x and the curve y2=x−2y^{2}=x-2y2=x−2 is:
  1. (A)1142\frac{11}{4\sqrt{2}}42​11​
  2. (B)222
  3. (C)742\frac{7}{4\sqrt{2}}42​7​
  4. (D)78\frac{7}{8}87​

Correct answer: (C)

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2019
Let f:[0,2]→Rf:[0,2]\to Rf:[0,2]→R be a twice differentiable function such that f′′(x)>0f''(x)>0f′′(x)>0, for all x∈(0,2)x\in(0,2)x∈(0,2). If ϕ(x)=f(x)+f(2−x)\phi(x)=f(x)+f(2-x)ϕ(x)=f(x)+f(2−x), then ϕ\phiϕ is:
  1. (A)increasing on (0, 2)
  2. (B)decreasing on (0, 2)
  3. (C)decreasing on (0, 1) and increasing on (1, 2)
  4. (D)increasing on (0, 1) and decreasing on (1, 2)

Correct answer: (C)

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2019
The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:
  1. (A)3\sqrt{3}3​
  2. (B)6\sqrt{6}6​
  3. (C)232\sqrt{3}23​
  4. (D)233\frac{2}{3}\sqrt{3}32​3​

Correct answer: (C)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2019
If the function f given by f(x)=x3−3(a−2)x2+3ax+7f(x) = x^{3} - 3(a-2)x^{2} + 3ax + 7f(x)=x3−3(a−2)x2+3ax+7, for some a∈Ra \in Ra∈R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, f(x)−14(x−1)2=0\frac{f(x) - 14}{(x-1)^{2}} = 0(x−1)2f(x)−14​=0 (x≠1)(x \neq 1)(x=1) is :
  1. (A)−7
  2. (B)5
  3. (C)7
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q180·MathematicsSingle correctJEE Main 2019
The tangent to the curve y=x2−5x+5y = x^{2} - 5x + 5y=x2−5x+5, parallel to the line 2y=4x+12y = 4x + 12y=4x+1, also passes through the point :
  1. (A)(72,14)\left(\dfrac{7}{2}, \dfrac{1}{4}\right)(27​,41​)
  2. (B)(18,−7)\left(\dfrac{1}{8}, -7\right)(81​,−7)
  3. (C)(−18,7)\left(-\dfrac{1}{8}, 7\right)(−81​,7)
  4. (D)(14,72)\left(\dfrac{1}{4}, \dfrac{7}{2}\right)(41​,27​)

Correct answer: (B)

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2019
The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y=12−x2y = 12 - x^{2}y=12−x2 such that the rectangle lies inside the parabola, is:
  1. (A)36
  2. (B)20220\sqrt{2}202​
  3. (C)32
  4. (D)18318\sqrt{3}183​

Correct answer: (C)

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2019
Let f(x)=xa2+x2−d−xb2+(d−x)2,r∈Rff(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, r\in R ff(x)=a2+x2​x​−b2+(d−x)2​d−x​,r∈Rf, where a, b and d are non – zero real constant. Then:
  1. (A)f is an increasing function of x
  2. (B)f is a decreasing function of x
  3. (C)f is not a continuous function of x
  4. (D)f is neither increasing nor decreasing function of x

Correct answer: (A)

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2019
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression xmyn(1+x2m)(1+y2n)\dfrac{x^{m}y^{n}}{\left(1+x^{2m}\right)\left(1+y^{2n}\right)}(1+x2m)(1+y2n)xmyn​ is:
  1. (A)1
  2. (B)12\dfrac{1}{2}21​
  3. (C)14\dfrac{1}{4}41​
  4. (D)m+n6mn\dfrac{m+n}{6mn}6mnm+n​

Correct answer: (C)

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2019
The maximum value of the function f(x)=3x3−18x2+27x−40f(x)=3x^{3}-18x^{2}+27x-40f(x)=3x3−18x2+27x−40 on the set S = {x∈R:x2+30≤11x}\{x\in R : x^{2}+30\le 11x\}{x∈R:x2+30≤11x} is
  1. (A)-122
  2. (B)-222
  3. (C)122
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2019
The shortest distance between the point (32,0)\left(\frac{3}{2},0\right)(23​,0) and the curve y=x,(x>0)y=\sqrt{x},(x>0)y=x​,(x>0), is:
  1. (A)52\frac{\sqrt{5}}{2}25​​
  2. (B)32\frac{\sqrt{3}}{2}23​​
  3. (C)32\frac{3}{2}23​
  4. (D)54\frac{5}{4}45​

Correct answer: (A)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2019
The maximum volume (in cu.m) of the right circular cone having slant height 3 m is
  1. (A)6π6\pi6π
  2. (B)33 π3\sqrt{3}\,\pi33​π
  3. (C)43π\dfrac{4}{3}\pi34​π
  4. (D)23π2\sqrt{3}\pi23​π

Correct answer: (D)

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2019
If θ\thetaθ denotes the acute angle between the curves, y=10−x2y=10-x^{2}y=10−x2 and y=2+x2y=2+x^{2}y=2+x2 at a point of their intersection, then ∣tan⁡θ∣|\tan\theta|∣tanθ∣ is equal to
  1. (A)49\dfrac{4}{9}94​
  2. (B)815\dfrac{8}{15}158​
  3. (C)717\dfrac{7}{17}177​
  4. (D)817\dfrac{8}{17}178​

Correct answer: (B)

Step-by-step solution →
Q188·MathematicsMultiple correctJEE Advanced 2018
For every twice differentiable function f:R→[−2,2]f : R \to [-2, 2]f:R→[−2,2] with (f(0))2+(f′(0))2=85(f(0))^{2} + (f'(0))^{2} = 85(f(0))2+(f′(0))2=85, which of the following statement(s) is (are) TRUE ?
  1. (A)There exist r, s ∈\in∈ R, where r<sr < sr<s, such that f is one-one on the open interval (r, s)
  2. (B)There exists x0∈(−4,0)x_{0} \in (-4, 0)x0​∈(−4,0) such that ∣f′(x0)∣≤1|f'(x_{0})| \le 1∣f′(x0​)∣≤1
  3. (C)lim⁡x→∞f(x)=1\lim_{x \to \infty} f(x) = 1limx→∞​f(x)=1
  4. (D)There exist α∈(−4,4)\alpha \in (-4, 4)α∈(−4,4) such that f(α)+f′′(α)=0f(\alpha) + f''(\alpha) = 0f(α)+f′′(α)=0 and f′(α)≠0f'(\alpha) \ne 0f′(α)=0

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

Step-by-step solution →
Q189·MathematicsSingle correctJEE Advanced 2017
Answer by appropriately matching the information given in the three columns of the following table. Let f(x)=x+log⁡ex−xlog⁡exf(x) = x + \log_{e} x - x\log_{e} xf(x)=x+loge​x−xloge​x, x∈(0,∞)x \in (0, \infty)x∈(0,∞). • Column 1 contains information about zeros of f(x)f(x)f(x), f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x). • Column 2 contains information about the limiting behavior of f(x)f(x)f(x), f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x) at infinity. • Column 3 contains information about increasing/decreasing nature of f(x)f(x)f(x) and f′(x)f'(x)f′(x). Which of the following options is the only CORRECT combination ?
Column 1Column 2Column 3
(I) f(x)=0f(x) = 0f(x)=0 for some x∈(1,e2)x \in (1, e^{2})x∈(1,e2)(i) lim⁡x→∞f(x)=0\lim_{x \to \infty} f(x) = 0limx→∞​f(x)=0(P) fff is increasing in (0,1)(0, 1)(0,1)
(II) f′(x)=0f'(x) = 0f′(x)=0 for some x∈(1,e)x \in (1, e)x∈(1,e)(ii) lim⁡x→∞f(x)=−∞\lim_{x \to \infty} f(x) = -\inftylimx→∞​f(x)=−∞(Q) fff is decreasing in (e,e2)(e, e^{2})(e,e2)
(III) f′(x)=0f'(x) = 0f′(x)=0 for some x∈(0,1)x \in (0, 1)x∈(0,1)(iii) lim⁡x→∞f′(x)=−∞\lim_{x \to \infty} f'(x) = -\inftylimx→∞​f′(x)=−∞(R) f′f'f′ is increasing in (0,1)(0, 1)(0,1)
(IV) f′′(x)=0f''(x) = 0f′′(x)=0 for some x∈(1,e)x \in (1, e)x∈(1,e)(iv) lim⁡x→∞f′′(x)=0\lim_{x \to \infty} f''(x) = 0limx→∞​f′′(x)=0(S) f′f'f′ is decreasing in (e,e2)(e, e^{2})(e,e2)
  1. (A)(III) (iii) (R)
  2. (B)(I) (i) (P)
  3. (C)(IV) (iv) (S)
  4. (D)(II) (ii) (Q)

Correct answer: (D)

Step-by-step solution →
Q190·MathematicsMultiple correctJEE Advanced 2017
If f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a differentiable function such that f′(x)>2f(x)f'(x) > 2f(x)f′(x)>2f(x) for all x∈Rx \in \mathbb{R}x∈R, and f(0)=1f(0) = 1f(0)=1, then
  1. (A)f(x)f(x)f(x) is increasing in (0,∞)(0, \infty)(0,∞)
  2. (B)f(x)f(x)f(x) is decreasing in (0,∞)(0, \infty)(0,∞)
  3. (C)f(x)>e2xf(x) > e^{2x}f(x)>e2x in (0,∞)(0, \infty)(0,∞)
  4. (D)f′(x)<e2xf'(x) < e^{2x}f′(x)<e2x in (0,∞)(0, \infty)(0,∞)

Correct answer: (A), (C)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Advanced 2017
If f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R is a twice differentiable function such that f′′(x)>0f''(x) > 0f′′(x)>0 for all x∈Rx \in \mathbb{R}x∈R, and f(12)=12f\left(\dfrac{1}{2}\right) = \dfrac{1}{2}f(21​)=21​, f(1)=1f(1) = 1f(1)=1, then
  1. (A)f′(1)≤0f'(1) \le 0f′(1)≤0
  2. (B)0<f′(1)≤120 < f'(1) \le \dfrac{1}{2}0<f′(1)≤21​
  3. (C)12<f′(1)≤1\dfrac{1}{2} < f'(1) \le 121​<f′(1)≤1
  4. (D)f′(1)>1f'(1) > 1f′(1)>1

Correct answer: (D)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Advanced 2017
Answer by appropriately matching the information given in the three columns of the following table. Let f(x)=x+log⁡ex−xlog⁡exf(x) = x + \log_{e} x - x\log_{e} xf(x)=x+loge​x−xloge​x, x∈(0,∞)x \in (0, \infty)x∈(0,∞). • Column 1 contains information about zeros of f(x)f(x)f(x), f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x). • Column 2 contains information about the limiting behavior of f(x)f(x)f(x), f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x) at infinity. • Column 3 contains information about increasing/decreasing nature of f(x)f(x)f(x) and f′(x)f'(x)f′(x). Which of the following options is the only INCORRECT combination ?
Column 1Column 2Column 3
(I) f(x)=0f(x) = 0f(x)=0 for some x∈(1,e2)x \in (1, e^{2})x∈(1,e2)(i) lim⁡x→∞f(x)=0\lim_{x \to \infty} f(x) = 0limx→∞​f(x)=0(P) fff is increasing in (0,1)(0, 1)(0,1)
(II) f′(x)=0f'(x) = 0f′(x)=0 for some x∈(1,e)x \in (1, e)x∈(1,e)(ii) lim⁡x→∞f(x)=−∞\lim_{x \to \infty} f(x) = -\inftylimx→∞​f(x)=−∞(Q) fff is decreasing in (e,e2)(e, e^{2})(e,e2)
(III) f′(x)=0f'(x) = 0f′(x)=0 for some x∈(0,1)x \in (0, 1)x∈(0,1)(iii) lim⁡x→∞f′(x)=−∞\lim_{x \to \infty} f'(x) = -\inftylimx→∞​f′(x)=−∞(R) f′f'f′ is increasing in (0,1)(0, 1)(0,1)
(IV) f′′(x)=0f''(x) = 0f′′(x)=0 for some x∈(1,e)x \in (1, e)x∈(1,e)(iv) lim⁡x→∞f′′(x)=0\lim_{x \to \infty} f''(x) = 0limx→∞​f′′(x)=0(S) f′f'f′ is decreasing in (e,e2)(e, e^{2})(e,e2)
  1. (A)(II) (iii) (P)
  2. (B)(II) (iv) (Q)
  3. (C)(I) (iii) (P)
  4. (D)(III) (i) (R)

Correct answer: (D)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Advanced 2017
Answer by appropriately matching the information given in the three columns of the following table. Let f(x)=x+log⁡ex−xlog⁡exf(x) = x + \log_{e} x - x\log_{e} xf(x)=x+loge​x−xloge​x, x∈(0,∞)x \in (0, \infty)x∈(0,∞). • Column 1 contains information about zeros of f(x)f(x)f(x), f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x). • Column 2 contains information about the limiting behavior of f(x)f(x)f(x), f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x) at infinity. • Column 3 contains information about increasing/decreasing nature of f(x)f(x)f(x) and f′(x)f'(x)f′(x). Which of the following options is the only CORRECT combination ?
Column 1Column 2Column 3
(I) f(x)=0f(x) = 0f(x)=0 for some x∈(1,e2)x \in (1, e^{2})x∈(1,e2)(i) lim⁡x→∞f(x)=0\lim_{x \to \infty} f(x) = 0limx→∞​f(x)=0(P) fff is increasing in (0,1)(0, 1)(0,1)
(II) f′(x)=0f'(x) = 0f′(x)=0 for some x∈(1,e)x \in (1, e)x∈(1,e)(ii) lim⁡x→∞f(x)=−∞\lim_{x \to \infty} f(x) = -\inftylimx→∞​f(x)=−∞(Q) fff is decreasing in (e,e2)(e, e^{2})(e,e2)
(III) f′(x)=0f'(x) = 0f′(x)=0 for some x∈(0,1)x \in (0, 1)x∈(0,1)(iii) lim⁡x→∞f′(x)=−∞\lim_{x \to \infty} f'(x) = -\inftylimx→∞​f′(x)=−∞(R) f′f'f′ is increasing in (0,1)(0, 1)(0,1)
(IV) f′′(x)=0f''(x) = 0f′′(x)=0 for some x∈(1,e)x \in (1, e)x∈(1,e)(iv) lim⁡x→∞f′′(x)=0\lim_{x \to \infty} f''(x) = 0limx→∞​f′′(x)=0(S) f′f'f′ is decreasing in (e,e2)(e, e^{2})(e,e2)
  1. (A)(IV) (i) (S)
  2. (B)(I) (ii) (R)
  3. (C)(III) (iv) (P)
  4. (D)(II) (iii) (S)

Correct answer: (D)

Step-by-step solution →
Q194·MathematicsMultiple correctJEE Advanced 2016
Let f(x)=lim⁡n→∞(nn(x+n)(x+n2)…(x+nn)n!(x2+n2)(x2+n24)…(x2+n2n2))xnf(x) = \lim_{n \to \infty} \left( \frac{n^{n} (x + n) \left( x + \frac{n}{2} \right) \ldots \left( x + \frac{n}{n} \right)}{n! \left( x^{2} + n^{2} \right) \left( x^{2} + \frac{n^{2}}{4} \right) \ldots \left( x^{2} + \frac{n^{2}}{n^{2}} \right)} \right)^{\frac{x}{n}}f(x)=limn→∞​(n!(x2+n2)(x2+4n2​)…(x2+n2n2​)nn(x+n)(x+2n​)…(x+nn​)​)nx​, for all x>0x > 0x>0. Then
  1. (A)f(12)≥f(1)f\left( \frac{1}{2} \right) \geq f(1)f(21​)≥f(1)
  2. (B)f(13)≤f(23)f\left( \frac{1}{3} \right) \leq f\left( \frac{2}{3} \right)f(31​)≤f(32​)
  3. (C)f′(2)≤0f'(2) \leq 0f′(2)≤0
  4. (D)f′(3)f(3)≥f′(2)f(2)\frac{f'(3)}{f(3)} \geq \frac{f'(2)}{f(2)}f(3)f′(3)​≥f(2)f′(2)​

Correct answer: (B), (C)

Step-by-step solution →
Q195·MathematicsMultiple correctJEE Advanced 2016
Let f:R→(0,∞)f : \mathbb{R} \to (0, \infty)f:R→(0,∞) and g:R→Rg : \mathbb{R} \to \mathbb{R}g:R→R be twice differentiable functions such that f′′f''f′′ and g′′g''g′′ are continuous functions on R\mathbb{R}R. Suppose f′(2)=g(2)=0f'(2) = g(2) = 0f′(2)=g(2)=0, f′′(2)≠0f''(2) \neq 0f′′(2)=0 and g′(2)≠0g'(2) \neq 0g′(2)=0. If lim⁡x→2f(x)g(x)f′(x)g′(x)=1\lim_{x \to 2} \frac{f(x) g(x)}{f'(x) g'(x)} = 1limx→2​f′(x)g′(x)f(x)g(x)​=1, then
  1. (A)f has a local minimum at x=2x = 2x=2
  2. (B)f has a local maximum at x=2x = 2x=2
  3. (C)f′′(2)>f(2)f''(2) > f(2)f′′(2)>f(2)
  4. (D)f(x)−f′′(x)=0f(x) - f''(x) = 0f(x)−f′′(x)=0 for at least one x∈Rx \in \mathbb{R}x∈R

Correct answer: (A), (D)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Advanced 2016
The least value of α∈R\alpha \in \mathbb{R}α∈R for which 4αx2+1x≥14\alpha x^{2} + \frac{1}{x} \geq 14αx2+x1​≥1, for all x>0x > 0x>0, is
  1. (A)164\frac{1}{64}641​
  2. (B)132\frac{1}{32}321​
  3. (C)127\frac{1}{27}271​
  4. (D)125\frac{1}{25}251​

Correct answer: (C)

Step-by-step solution →
Q197·MathematicsMultiple correctJEE Advanced 2015
Let F:R→RF : \mathbb{R} \to \mathbb{R}F:R→R be a thrice differentiable function. Suppose that F(1)=0F(1) = 0F(1)=0, F(3)=−4F(3) = -4F(3)=−4 and F′(x)<0F'(x) < 0F′(x)<0 for all x∈(1/2,3)x \in (1/2, 3)x∈(1/2,3). Let f(x)=xF(x)f(x) = xF(x)f(x)=xF(x) for all x∈Rx \in \mathbb{R}x∈R. The correct statement(s) is(are)
  1. (A)f′(1)<0f'(1) < 0f′(1)<0
  2. (B)f(2)<0f(2) < 0f(2)<0
  3. (C)f′(x)≠0f'(x) \ne 0f′(x)=0 for any x∈(1,3)x \in (1, 3)x∈(1,3)
  4. (D)f′(x)=0f'(x) = 0f′(x)=0 for some x∈(1,3)x \in (1, 3)x∈(1,3)

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

Step-by-step solution →
Q198·MathematicsMultiple correctJEE Advanced 2015
Let fff, g:[−1,2]→Rg : [-1, 2] \to \mathbb{R}g:[−1,2]→R be continuous functions which are twice differentiable on the interval (−1,2)(-1, 2)(−1,2). Let the values of f and g at the points −1-1−1, 0 and 2 be as given in the following table: In each of the intervals (−1,0)(-1, 0)(−1,0) and (0,2)(0, 2)(0,2) the function (f−3g)′′(f - 3g)''(f−3g)′′ never vanishes. Then the correct statement(s) is(are)
x=−1x = -1x=−1x=0x = 0x=0x=2x = 2x=2
f(x)f(x)f(x)360
g(x)g(x)g(x)01−1-1−1
  1. (A)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly three solutions in (−1,0)∪(0,2)(-1, 0) \cup (0, 2)(−1,0)∪(0,2)
  2. (B)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly one solution in (−1,0)(-1, 0)(−1,0)
  3. (C)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly one solution in (0,2)(0, 2)(0,2)
  4. (D)f′(x)−3g′(x)=0f'(x) - 3g'(x) = 0f′(x)−3g′(x)=0 has exactly two solutions in (−1,0)(-1, 0)(−1,0) and exactly two solutions in (0,2)(0, 2)(0,2)

Correct answer: (B), (C)

Step-by-step solution →
Q199·MathematicsIntegerJEE Advanced 2015
A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of V mm3^{3}3, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of V250π\dfrac{V}{250\pi}250πV​ is

Correct answer: 4

Step-by-step solution →
Q200·MathematicsIntegerJEE Advanced 2014
The slope of the tangent to the curve (y−x5)2=x(1+x2)2(y - x^5)^2 = x(1 + x^2)^2(y−x5)2=x(1+x2)2 at the point (1,3)(1, 3)(1,3) is __________

Correct answer: 8

Step-by-step solution →
Q201·MathematicsSingle correctJEE Advanced 2013
A line L:y=mx+3L : y = mx + 3L:y=mx+3 meets y-axis at E(0,3)E(0, 3)E(0,3) and the arc of the parabola y2=16xy^{2} = 16xy2=16x, 0≤y≤60 \leq y \leq 60≤y≤6 at the point F(x0,y0)F(x_{0}, y_{0})F(x0​,y0​). The tangent to the parabola at F(x0,y0)F(x_{0}, y_{0})F(x0​,y0​) intersects the y-axis at G(0,y1)G(0, y_{1})G(0,y1​). The slope mmm of the line L is chosen such that the area of the triangle EFG has a local maximum. Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.m=m =m=1.12\frac{1}{2}21​
Q.Maximum area of ΔEFG\Delta EFGΔEFG is2.444
R.y0=y_{0} =y0​=3.222
S.y1=y_{1} =y1​=4.111
  1. (A)P-4, Q-1, R-2, S-3
  2. (B)P-3, Q-4, R-1, S-2
  3. (C)P-1, Q-3, R-2, S-4
  4. (D)P-1, Q-3, R-4, S-2

Correct answer: (A)

Step-by-step solution →
Q202·MathematicsSingle correctJEE Advanced 2013
Let f:[0,1]→Rf : [0, 1] \to \mathbb{R}f:[0,1]→R (the set of all real numbers) be a function. Suppose the function fff is twice differentiable, f(0)=f(1)=0f(0) = f(1) = 0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥exf''(x) - 2f'(x) + f(x) \geq e^{x}f′′(x)−2f′(x)+f(x)≥ex, x∈[0,1]x \in [0, 1]x∈[0,1]. Which of the following is true for 0<x<10 < x < 10<x<1 ?
  1. (A)0<f(x)<∞0 < f(x) < \infty0<f(x)<∞
  2. (B)−12<f(x)<12-\frac{1}{2} < f(x) < \frac{1}{2}−21​<f(x)<21​
  3. (C)−14<f(x)<1-\frac{1}{4} < f(x) < 1−41​<f(x)<1
  4. (D)−∞<f(x)<0-\infty < f(x) < 0−∞<f(x)<0

Correct answer: (D)

Step-by-step solution →
Q203·MathematicsSingle correctJEE Advanced 2013
The number of points in (−∞,∞)(-\infty,\infty)(−∞,∞), for which x2−xsin⁡x−cos⁡x=0x^2-x\sin x-\cos x=0x2−xsinx−cosx=0, is
  1. (A)666
  2. (B)444
  3. (C)222
  4. (D)000

Correct answer: (C)

Step-by-step solution →
Q204·MathematicsSingle correctJEE Advanced 2013
Let f:[0,1]→Rf : [0, 1] \to \mathbb{R}f:[0,1]→R (the set of all real numbers) be a function. Suppose the function fff is twice differentiable, f(0)=f(1)=0f(0) = f(1) = 0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥exf''(x) - 2f'(x) + f(x) \geq e^{x}f′′(x)−2f′(x)+f(x)≥ex, x∈[0,1]x \in [0, 1]x∈[0,1]. If the function e−xf(x)e^{-x} f(x)e−xf(x) assumes its minimum in the interval [0,1][0, 1][0,1] at x=14x = \frac{1}{4}x=41​, which of the following is true ?
  1. (A)f′(x)<f(x),14<x<34f'(x) < f(x), \frac{1}{4} < x < \frac{3}{4}f′(x)<f(x),41​<x<43​
  2. (B)f′(x)>f(x),0<x<14f'(x) > f(x), 0 < x < \frac{1}{4}f′(x)>f(x),0<x<41​
  3. (C)f′(x)<f(x),0<x<14f'(x) < f(x), 0 < x < \frac{1}{4}f′(x)<f(x),0<x<41​
  4. (D)f′(x)<f(x),34<x<1f'(x) < f(x), \frac{3}{4} < x < 1f′(x)<f(x),43​<x<1

Correct answer: (C)

Step-by-step solution →
Q205·MathematicsMultiple correctJEE Advanced 2013
Let f(x)=xsin⁡πxf(x)=x\sin\pi xf(x)=xsinπx, x>0x>0x>0. Then for all natural numbers nnn, f′(x)f'(x)f′(x) vanishes at
  1. (A)a unique point in the interval (n,n+12)(n,n+\frac{1}{2})(n,n+21​)
  2. (B)a unique point in the interval (n+12,n+1)(n+\frac{1}{2},n+1)(n+21​,n+1)
  3. (C)a unique point in the interval (n,n+1)(n,n+1)(n,n+1)
  4. (D)two points in the interval (n,n+1)(n,n+1)(n,n+1)

Correct answer: (B), (C)

Step-by-step solution →
Q206·MathematicsMultiple correctJEE Advanced 2013
A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:158:158:15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are
  1. (A)242424
  2. (B)323232
  3. (C)454545
  4. (D)606060

Correct answer: (A), (C)

Step-by-step solution →
Q207·MathematicsMultiple correctJEE Advanced 2013
The function f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣f(x) = 2|x| + |x + 2| - \left||x + 2| - 2|x|\right|f(x)=2∣x∣+∣x+2∣−∣∣x+2∣−2∣x∣∣ has a local minimum or a local maximum at x=x =x=
  1. (A)−2-2−2
  2. (B)−23\frac{-2}{3}3−2​
  3. (C)222
  4. (D)23\frac{2}{3}32​

Correct answer: (A), (B)

Step-by-step solution →

Application of Derivatives — frequently asked

How many questions from Application of Derivatives appear in JEE?

Application of Derivatives has appeared in 139 of the last 186 JEE Main and JEE Advanced papers — about 75% of them — contributing 207 questions in total across those papers.

Is Application of Derivatives an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 75% 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 Application of Derivatives 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 →

Practise Application of Derivatives until it stops costing you marks.

Build a timed test from these 207 questions in one click. Jarvis marks it, names the specific misconception behind each wrong answer, and brings the ones you failed back at the right interval.

Practise Application of Derivatives freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS