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JEE Advanced · Mathematics · Integral Calculus

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1
The value of the definite integral ∫0213x+3 dx\displaystyle\int_0^2 \frac{1}{3^x+3}\,dx is
Not attempted
A
12\frac12
B
13\frac13
C
log⁡e33\dfrac{\log_e3}{3}
D
log⁡e32\dfrac{\log_e3}{2}
2
Consider C1:y=e−xC_1: y=e^{-x} for x∈[0,10π]x\in[0,10\pi] and C2:y=e−x(sin⁡x+cos⁡x)C_2: y=e^{-x}(\sin x+\cos x) for x∈[0,10π]x\in[0,10\pi]. Let nn be the total number of points of intersection of C1,C2C_1,C_2. Find nn.
Not attempted
Your answer: —
Correct answer: 11
3
Using the intersection points α1<α2<⋯<α11\alpha_1<\alpha_2<\cdots<\alpha_{11} from the previous question, let β\beta be the area enclosed between C1,C2C_1,C_2 and the lines x=α1, x=α4x=\alpha_1,\ x=\alpha_4. Find −1πlog⁡e(β−2e−π/2)-\frac1\pi\log_e\left(\beta-2e^{-\pi/2}\right).
Not attempted
Your answer: —
Correct answer: 2.5
4
Let R\mathbb{R} denote the set of all real numbers. Then the area of the region {(x,y)∈R×R:x>0, y>1x, 5x−4y−1>0, 4x+4y−17<0}\left\{(x,y)\in\mathbb{R}\times\mathbb{R} : x>0,\ y>\dfrac{1}{x},\ 5x-4y-1>0,\ 4x+4y-17<0\right\} is
Not attempted
A
1716−log⁡e4\dfrac{17}{16}-\log_e4
B
338−log⁡e4\dfrac{33}{8}-\log_e4
C
578−log⁡e4\dfrac{57}{8}-\log_e4
D
172−log⁡e4\dfrac{17}{2}-\log_e4
5
If α=∫1/22tan⁡−1x2x2−3x+2 dx\alpha = \displaystyle\int_{1/2}^{2}\dfrac{\tan^{-1}x}{2x^2-3x+2}\,dx, then the value of 7tan⁡(2α7π)\sqrt7\tan\left(\dfrac{2\alpha\sqrt7}{\pi}\right) is _____________. (Here, the inverse trigonometric function tan⁡−1x\tan^{-1}x assumes values in (−π2,π2)\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).)
Not attempted
Your answer: —
Correct answer: 21
6
Let S={(x,y)∈R×R:x≥0, y≥0, y2≤4x, y2≤12−2x, and 3y+8 x≤58}S=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}: x\ge 0,\ y\ge 0,\ y^2\le 4x,\ y^2\le 12-2x,\ \text{and } 3y+\sqrt{8}\,x\le 5\sqrt{8}\right\}. If the area of the region SS is α2\alpha\sqrt{2}, then α\alpha is equal to
Not attempted
A
172\dfrac{17}{2}
B
173\dfrac{17}{3}
C
174\dfrac{17}{4}
D
175\dfrac{17}{5}
7
Let the function f:[1,∞)→Rf:[1,\infty)\to\mathbb{R} be defined by f(t)={(−1)n+12,if t=2n−1, n∈N,(2n+1−t)2f(2n−1)+(t−(2n−1))2f(2n+1),if 2n−1<t<2n+1, n∈N.f(t)=\begin{cases}(-1)^{n+1}2, & \text{if } t=2n-1,\ n\in\mathbb{N}, \\[4pt] \dfrac{(2n+1-t)}{2}f(2n-1)+\dfrac{(t-(2n-1))}{2}f(2n+1), & \text{if } 2n-1<t<2n+1,\ n\in\mathbb{N}.\end{cases} Define g(x)=∫1xf(t) dt, x∈(1,∞)g(x)=\displaystyle\int_1^x f(t)\,dt,\ x\in(1,\infty). Let α\alpha denote the number of solutions of the equation g(x)=0g(x)=0 in the interval (1,8](1,8] and β=lim⁡x→1+g(x)x−1\beta=\displaystyle\lim_{x\to1^+}\dfrac{g(x)}{x-1}. Then the value of α+β\alpha+\beta is equal to _____.
Not attempted
Your answer: —
Correct answer: 5
8
Paragraph\textbf{Paragraph}: Let f:[0,π2]→[0,1]f:\left[0,\dfrac{\pi}{2}\right]\to[0,1] be the function defined by f(x)=sin⁡2xf(x)=\sin^2x and let g:[0,π2]→[0,∞)g:\left[0,\dfrac{\pi}{2}\right]\to[0,\infty) be the function defined by g(x)=πx2−x2g(x)=\sqrt{\dfrac{\pi x}{2}-x^2}. The value of 2∫0π/2f(x)g(x) dx−∫0π/2g(x) dx2\displaystyle\int_0^{\pi/2}f(x)g(x)\,dx-\int_0^{\pi/2}g(x)\,dx is __________.
Not attempted
Your answer: —
Correct answer: 0
9
Paragraph\textbf{Paragraph}: Let f:[0,π2]→[0,1]f:\left[0,\dfrac{\pi}{2}\right]\to[0,1] be the function defined by f(x)=sin⁡2xf(x)=\sin^2x and let g:[0,π2]→[0,∞)g:\left[0,\dfrac{\pi}{2}\right]\to[0,\infty) be the function defined by g(x)=πx2−x2g(x)=\sqrt{\dfrac{\pi x}{2}-x^2}. The value of 16π3∫0π/2f(x)g(x) dx\dfrac{16}{\pi^3}\displaystyle\int_0^{\pi/2}f(x)g(x)\,dx is __________.
Not attempted
Your answer: —
Correct answer: 0.25
10
For x∈Rx\in\mathbb{R}, let tan⁡−1(x)∈(−π2,π2)\tan^{-1}(x)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right). Then the minimum value of the function f:R→Rf:\mathbb{R}\to\mathbb{R} defined by f(x)=∫0xtan⁡−1xe(t−cos⁡t)1+t2023 dtf(x)=\displaystyle\int_0^{x\tan^{-1}x}\dfrac{e^{(t-\cos t)}}{1+t^{2023}}\,dt is
Not attempted
Your answer: —
Correct answer: 0
11
Let n≥2n\ge2 be a natural number and f:[0,1]→Rf:[0,1]\to\mathbb{R} be the function defined by f(x)={n(1−2nx)if 0≤x≤12n2n(2nx−1)if 12n≤x≤34n4n(1−nx)if 34n≤x≤1nnn−1(nx−1)if 1n≤x≤1f(x)=\begin{cases}n(1-2nx) & \text{if } 0\le x\le\dfrac{1}{2n}\\ 2n(2nx-1) & \text{if } \dfrac{1}{2n}\le x\le\dfrac{3}{4n}\\ 4n(1-nx) & \text{if } \dfrac{3}{4n}\le x\le\dfrac{1}{n}\\ \dfrac{n}{n-1}(nx-1) & \text{if } \dfrac{1}{n}\le x\le1\end{cases}. If nn is such that the area of the region bounded by the curves x=0x=0, x=1x=1, y=0y=0 and y=f(x)y=f(x) is 44, then the maximum value of the function ff is
Not attempted
Your answer: —
Correct answer: 8
12
Let f:[0,1]→[0,1]f:[0,1]\to[0,1] be the function defined by f(x)=x33−x2+5x9+1736f(x)=\dfrac{x^3}{3}-x^2+\dfrac{5x}{9}+\dfrac{17}{36}. Consider the square region S=[0,1]×[0,1]S=[0,1]\times[0,1]. Let G={(x,y)∈S:y>f(x)}G=\{(x,y)\in S: y>f(x)\} be called the green region and R={(x,y)∈S:y<f(x)}R=\{(x,y)\in S: y<f(x)\} be called the red region. Let Lh={(x,h)∈S:x∈[0,1]}L_h=\{(x,h)\in S: x\in[0,1]\} be the horizontal line drawn at a height h∈[0,1]h\in[0,1]. Then which of the following statements is(are) true?
Not attempted
A
There exists an h∈[14,23]h\in\left[\dfrac14,\dfrac23\right] such that the area of the green region above the line LhL_h equals the area of the green region below the line LhL_h
B
There exists an h∈[14,23]h\in\left[\dfrac14,\dfrac23\right] such that the area of the red region above the line LhL_h equals the area of the red region below the line LhL_h
C
There exists an h∈[14,23]h\in\left[\dfrac14,\dfrac23\right] such that the area of the green region above the line LhL_h equals the area of the red region below the line LhL_h
D
There exists an h∈[14,23]h\in\left[\dfrac14,\dfrac23\right] such that the area of the red region above the line LhL_h equals the area of the green region below the line LhL_h
13
Consider the functions f,g:R→Rf,g:\mathbb{R}\to\mathbb{R} defined by f(x)=x2+512f(x)=x^2+\dfrac{5}{12} and g(x)={2(1−4∣x∣3),∣x∣≤34,0,∣x∣>34.g(x)=\begin{cases}2\left(1-\dfrac{4|x|}{3}\right), & |x|\le\dfrac34,\\[4pt] 0, & |x|>\dfrac34.\end{cases} If α\alpha is the area of the region {(x,y)∈R×R:∣x∣≤34, 0≤y≤min⁡{f(x),g(x)}}\left\{(x,y)\in\mathbb{R}\times\mathbb{R} : |x|\le\dfrac34,\ 0\le y\le\min\{f(x),g(x)\}\right\}, then the value of 9α9\alpha is _______.
Not attempted
Your answer: —
Correct answer: 6
14
The greatest integer less than or equal to ∫12log⁡2(x3+1) dx  +  ∫1log⁡29(2x−1)1/3 dx\displaystyle\int_1^2 \log_2(x^3+1)\,dx \;+\; \int_1^{\log_2 9} (2^x-1)^{1/3}\,dx is _______.
Not attempted
Your answer: —
Correct answer: 5
15
Consider the equation ∫1e(log⁡ex)1/2x(a−(log⁡ex)3/2)2 dx=1,a∈(−∞,0)∪(1,∞)\displaystyle\int_1^e \dfrac{(\log_e x)^{1/2}}{x\left(a-(\log_e x)^{3/2}\right)^2}\,dx = 1, \quad a\in(-\infty,0)\cup(1,\infty). Which of the following statements is/are TRUE?
Not attempted
A
No aa satisfies the above equation
B
An integer aa satisfies the above equation
C
An irrational number aa satisfies the above equation
D
More than one aa satisfy the above equation
16
For any real number xx, let [x][x] denote the largest integer less than or equal to xx. If I=∫010[10xx+1]dxI = \int_0^{10}\left[\sqrt{\frac{10x}{x+1}}\right]dx, then the value of 9I9I is ___.
Not attempted
Your answer: —
Correct answer: 182
17
Let ψ1:[0,∞)→R\psi_1:[0,\infty)\to\mathbb{R}, ψ2:[0,∞)→R\psi_2:[0,\infty)\to\mathbb{R}, f:[0,∞)→Rf:[0,\infty)\to\mathbb{R} and g:[0,∞)→Rg:[0,\infty)\to\mathbb{R} be functions such that f(0)=g(0)=0f(0)=g(0)=0, ψ1(x)=e−x+x\psi_1(x)=e^{-x}+x, x≥0x\ge0, ψ2(x)=x2−2x−2e−x+2\psi_2(x)=x^2-2x-2e^{-x}+2, x≥0x\ge0, f(x)=∫−xx(∣t∣−t2)e−t2 dtf(x) = \int_{-x}^{x}(|t|-t^2)e^{-t^2}\,dt, x>0x>0 and g(x)=∫0x2t e−t dtg(x) = \int_0^{x^2}\sqrt{t}\,e^{-t}\,dt, x>0x>0. Which of the following statements is TRUE?
Not attempted
A
f(ln⁡3)+g(ln⁡3)=13f(\sqrt{\ln3})+g(\sqrt{\ln3}) = \frac13
B
For every x>1x>1, there exists an α∈(1,x)\alpha\in(1,x) such that ψ1(x)=1+αx\psi_1(x)=1+\alpha x
C
For every x>0x>0, there exists a β∈(0,x)\beta\in(0,x) such that ψ2(x)=2x(ψ1(β)−1)\psi_2(x)=2x(\psi_1(\beta)-1)
D
ff is an increasing function on the interval [0,32]\left[0,\frac32\right]
18
Let ψ1:[0,∞)→R\psi_1:[0,\infty)\to\mathbb{R}, ψ2:[0,∞)→R\psi_2:[0,\infty)\to\mathbb{R}, f:[0,∞)→Rf:[0,\infty)\to\mathbb{R} and g:[0,∞)→Rg:[0,\infty)\to\mathbb{R} be functions such that f(0)=g(0)=0f(0)=g(0)=0, ψ1(x)=e−x+x\psi_1(x)=e^{-x}+x, x≥0x\ge0, ψ2(x)=x2−2x−2e−x+2\psi_2(x)=x^2-2x-2e^{-x}+2, x≥0x\ge0, f(x)=∫−xx(∣t∣−t2)e−t2 dtf(x) = \int_{-x}^{x}(|t|-t^2)e^{-t^2}\,dt, x>0x>0 and g(x)=∫0x2t e−t dtg(x) = \int_0^{x^2}\sqrt{t}\,e^{-t}\,dt, x>0x>0. Which of the following statements is TRUE?
Not attempted
A
ψ1(x)≤1\psi_1(x)\le1, for all x>0x>0
B
ψ2(x)≤0\psi_2(x)\le0, for all x>0x>0
C
f(x)≥1−e−x2−23x3+25x5f(x) \ge 1-e^{-x^2}-\frac23x^3+\frac25x^5, for all x∈(0,12)x\in\left(0,\frac12\right)
D
g(x)≤23x3−25x5+17x7g(x) \le \frac23x^3-\frac25x^5+\frac17x^7, for all x∈(0,12)x\in\left(0,\frac12\right)
19
Let gi:[π8,3π8]→Rg_i:\left[\frac{\pi}{8},\frac{3\pi}{8}\right]\to\mathbb{R}, i=1,2i=1,2, and f:[π8,3π8]→Rf:\left[\frac{\pi}{8},\frac{3\pi}{8}\right]\to\mathbb{R} be functions such that g1(x)=1g_1(x)=1, g2(x)=∣4x−π∣g_2(x)=|4x-\pi| and f(x)=sin⁡2xf(x)=\sin^2x, for all x∈[π8,3π8]x\in\left[\frac{\pi}{8},\frac{3\pi}{8}\right]. Define Si=∫π/83π/8f(x)⋅gi(x) dxS_i = \int_{\pi/8}^{3\pi/8} f(x)\cdot g_i(x)\,dx, i=1,2i=1,2. The value of 16S1π\frac{16S_1}{\pi} is ___.
Not attempted
Your answer: —
Correct answer: 2
20
Let gi:[π8,3π8]→Rg_i:\left[\frac{\pi}{8},\frac{3\pi}{8}\right]\to\mathbb{R}, i=1,2i=1,2, and f:[π8,3π8]→Rf:\left[\frac{\pi}{8},\frac{3\pi}{8}\right]\to\mathbb{R} be functions such that g1(x)=1g_1(x)=1, g2(x)=∣4x−π∣g_2(x)=|4x-\pi| and f(x)=sin⁡2xf(x)=\sin^2x, for all x∈[π8,3π8]x\in\left[\frac{\pi}{8},\frac{3\pi}{8}\right]. Define Si=∫π/83π/8f(x)⋅gi(x) dxS_i = \int_{\pi/8}^{3\pi/8} f(x)\cdot g_i(x)\,dx, i=1,2i=1,2. The value of 48S2π2\frac{48S_2}{\pi^2} is ___.
Not attempted
Your answer: —
Correct answer: 1.5
21
Let f:[−π2,π2]→Rf:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\to\mathbb{R} be a continuous function such that f(0)=1f(0)=1 and ∫0π/3f(t) dt=0\int_0^{\pi/3} f(t)\,dt = 0. Then which of the following statements is (are) TRUE?
Not attempted
A
The equation f(x)−3cos⁡3x=0f(x) - 3\cos 3x = 0 has at least one solution in (0,π3)\left(0,\frac{\pi}{3}\right)
B
The equation f(x)−3sin⁡3x=−6πf(x) - 3\sin 3x = -\frac{6}{\pi} has at least one solution in (0,π3)\left(0,\frac{\pi}{3}\right)
C
lim⁡x→0x∫0xf(t) dt1−ex2=−1\lim_{x\to 0} \frac{x\int_0^x f(t)\,dt}{1-e^{x^2}} = -1
D
lim⁡x→0sin⁡x∫0xf(t) dtx2=−1\lim_{x\to 0} \frac{\sin x\int_0^x f(t)\,dt}{x^2} = -1
22
The area of the region {(x,y):0≤x≤94, 0≤y≤1, x≥3y, x+y≥2}\left\{(x,y): 0 \le x \le \frac{9}{4},\ 0 \le y \le 1,\ x \ge 3y,\ x+y \ge 2\right\} is
Not attempted
A
1132\frac{11}{32}
B
3596\frac{35}{96}
C
3796\frac{37}{96}
D
1332\frac{13}{32}
23
Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a differentiable function such that its derivative f′f' is continuous and f(π)=−6f(\pi)=-6. If F:[0,π]→RF:[0,\pi]\to\mathbb{R} is defined by F(x)=∫0xf(t) dtF(x)=\displaystyle\int_0^x f(t)\,dt, and if ∫0π(f′(x)+F(x))cos⁡x dx=2\displaystyle\int_0^\pi \left(f'(x)+F(x)\right)\cos x\,dx = 2, then the value of f(0)f(0) is ____
Not attempted
Your answer: —
Correct answer: 4
24
Which of the following inequalities is/are TRUE?
Not attempted
A
∫01xcos⁡x dx≥38\int_0^1 x\cos x \, dx \geq \frac{3}{8}
B
∫01xsin⁡x dx≥310\int_0^1 x\sin x \, dx \geq \frac{3}{10}
C
∫01x2cos⁡x dx≥12\int_0^1 x^2\cos x \, dx \geq \frac{1}{2}
D
∫01x2sin⁡x dx≥29\int_0^1 x^2\sin x \, dx \geq \frac{2}{9}
25
Let the functions f:R→Rf: \mathbb{R} \to \mathbb{R} and g:R→Rg: \mathbb{R} \to \mathbb{R} be defined by f(x)=ex−1−e−∣x−1∣f(x) = e^{x-1} - e^{-|x-1|} and g(x)=12(ex−1+e1−x)g(x) = \frac{1}{2}(e^{x-1} + e^{1-x}). Then the area of the region in the first quadrant bounded by the curves y=f(x)y=f(x), y=g(x)y=g(x) and x=0x=0 is
Not attempted
A
(2−3)+12(e−e−1)(2-\sqrt{3}) + \frac{1}{2}(e-e^{-1})
B
(2+3)+12(e−e−1)(2+\sqrt{3}) + \frac{1}{2}(e-e^{-1})
C
(2−3)+12(e+e−1)(2-\sqrt{3}) + \frac{1}{2}(e+e^{-1})
D
(2+3)+12(e+e−1)(2+\sqrt{3}) + \frac{1}{2}(e+e^{-1})
26
The value of the integral ∫0π/23cos⁡θ(cos⁡θ+sin⁡θ)5 dθ\displaystyle\int_0^{\pi/2}\dfrac{3\sqrt{\cos\theta}}{\left(\sqrt{\cos\theta}+\sqrt{\sin\theta}\right)^5}\,d\theta equals ____
Not attempted
Your answer: —
Correct answer: 0.5
27
Let f:R→Rf:\mathbb{R}\to\mathbb{R} be given by f(x)=(x−1)(x−2)(x−5)f(x) = (x-1)(x-2)(x-5). Define F(x)=∫0xf(t) dtF(x) = \int_0^x f(t)\,dt for x>0x>0. Then which of the following options is/are correct?
Not attempted
A
FF has a local minimum at x=1x=1
B
FF has a local maximum at x=2x=2
C
F(x)≠0F(x)\neq 0 for all x∈(0,5)x\in(0,5)
D
FF has two local maxima and one local minimum in (0,∞)(0,\infty)
28
For a∈Ra\in\mathbb{R}, ∣a∣>1|a|>1, let lim⁡n→∞(1+23+⋯+n3n7/3(1(an+1)2+1(an+2)2+⋯+1(an+n)2))=54.\lim_{n\to\infty}\left(\dfrac{1+\sqrt[3]{2}+\cdots+\sqrt[3]{n}}{n^{7/3}\left(\dfrac{1}{(an+1)^2}+\dfrac{1}{(an+2)^2}+\cdots+\dfrac{1}{(an+n)^2}\right)}\right) = 54. Then the possible value(s) of aa is/are:
Not attempted
A
88
B
−9-9
C
−6-6
D
77
29
The area of the region {(x,y):xy≤8, 1≤y≤x2}\{(x,y):xy\le8,\ 1\le y\le x^2\} is
Not attempted
A
8log⁡e2−1438\log_e2-\dfrac{14}{3}
B
16log⁡e2−14316\log_e2-\dfrac{14}{3}
C
16log⁡e2−616\log_e2-6
D
8log⁡e2−738\log_e2-\dfrac{7}{3}
30
If I=2π∫−π/4π/4dx(1+esin⁡x)(2−cos⁡2x)I=\dfrac{2}{\pi}\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{dx}{(1+e^{\sin x})(2-\cos2x)} then 27I227I^2 equals _____
Not attempted
Your answer: —
Correct answer: 4
31
The value of the integral ∫01/21+3((x+1)2(1−x)6)1/4 dx\int_{0}^{1/2}\frac{1+\sqrt3}{\left((x+1)^2(1-x)^6\right)^{1/4}}\,dx is _____.
Not attempted
Your answer: —
Correct answer: 2
32
A farmer F1F_1 has a land in the shape of a triangle with vertices at P(0,0)P(0,0), Q(1,1)Q(1,1) and R(2,0)R(2,0). From this land, a neighbouring farmer F2F_2 takes away the region which lies between the side PQPQ and a curve of the form y=xny=x^n (n>1)(n>1). If the area of the region taken away by the farmer F2F_2 is exactly 30% of the area of △PQR\triangle PQR, then the value of nn is ______.
Not attempted
Your answer: —
Correct answer: 4
33
If g(x)=∫sin⁡xsin⁡(2x)sin⁡−1(t) dtg(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t)\, dt, then
Not attempted
A
g′(π2)=−2πg'\left(\frac{\pi}{2}\right) = -2\pi
B
g′(−π2)=2πg'\left(-\frac{\pi}{2}\right) = 2\pi
C
g′(π2)=2πg'\left(\frac{\pi}{2}\right) = 2\pi
D
g′(−π2)=−2πg'\left(-\frac{\pi}{2}\right) = -2\pi
34
If the line x=αx = \alpha divides the area of the region R={(x,y)∈R2:x3≤y≤x, 0≤x≤1}R = \{(x,y) \in \mathbb{R}^2 : x^3 \le y \le x,\ 0 \le x \le 1\} into two equal parts, then
Not attempted
A
0<α≤120 < \alpha \le \frac{1}{2}
B
12<α<1\frac{1}{2} < \alpha < 1
C
2α4−4α2+1=02\alpha^4 - 4\alpha^2 + 1 = 0
D
α4+4α2−1=0\alpha^4 + 4\alpha^2 - 1 = 0
35
If I=∑k=198∫kk+1k+1x(x+1) dxI = \displaystyle\sum_{k=1}^{98} \int_{k}^{k+1} \dfrac{k+1}{x(x+1)}\, dx, then
Not attempted
A
I>log⁡e99I > \log_e 99
B
I<log⁡e99I < \log_e 99
C
I<4950I < \frac{49}{50}
D
I>4950I > \frac{49}{50}
36
Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a differentiable function such that f(0)=0f(0)=0, f(π2)=3f\left(\dfrac{\pi}{2}\right)=3 and f′(0)=1f'(0)=1. If g(x)=∫xπ2[f′(t)csc⁡t−cot⁡t csc⁡t f(t)]dtg(x)=\displaystyle\int_x^{\frac{\pi}{2}}\left[f'(t)\csc t-\cot t\,\csc t\,f(t)\right]dt for x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then lim⁡x→0g(x)=\displaystyle\lim_{x\to0}g(x)=
Not attempted
Your answer: —
Correct answer: 2
37
The value of ∫−π/2π/2x2cos⁡x1+ex dx\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{x^2\cos x}{1+e^x}\,dx is equal to
Not attempted
A
π24−2\dfrac{\pi^2}{4}-2
B
π24+2\dfrac{\pi^2}{4}+2
C
π2−eπ/2\pi^2-e^{\pi/2}
D
π2+eπ/2\pi^2+e^{\pi/2}
38
Area of the region {(x,y)∈R2:y≥∣x+3∣, 5y≤x+9≤15}\{(x,y)\in\mathbb{R}^2: y\ge\sqrt{|x+3|},\ 5y\le x+9\le15\} is equal to
Not attempted
A
16\dfrac{1}{6}
B
43\dfrac{4}{3}
C
32\dfrac{3}{2}
D
53\dfrac{5}{3}
39
The total number of distinct x∈[0,1]x\in[0,1] for which ∫0xt21+t4 dt=2x−1\displaystyle\int_0^x\dfrac{t^2}{1+t^4}\,dt=2x-1 is
Not attempted
Your answer: —
Correct answer: 1
40
Setup (shared with Q.60)\textbf{Setup (shared with Q.60)}: Let F:R→RF:\mathbb R\to\mathbb R be a thrice differentiable function. Suppose that F(1)=0, F(3)=−4F(1)=0,\ F(3)=-4 and F′(x)<0F'(x)<0 for all x∈(12,3)x\in\left(\dfrac12,3\right). Let f(x)=xF(x)f(x)=xF(x) for all x∈Rx\in\mathbb R. If ∫13x2F′(x) dx=−12\displaystyle\int_1^3x^2F'(x)\,dx=-12 and ∫13x3F′′(x) dx=40\displaystyle\int_1^3x^3F''(x)\,dx=40, then the correct expression(s) is(are)
Not attempted
A
9f′(3)+f′(1)−32=09f'(3)+f'(1)-32=0
B
∫13f(x) dx=12\displaystyle\int_1^3f(x)\,dx=12
C
9f′(3)−f′(1)+32=09f'(3)-f'(1)+32=0
D
∫13f(x) dx=−12\displaystyle\int_1^3f(x)\,dx=-12
41
Let f(x)=7tan⁡8x+7tan⁡6x−3tan⁡4x−3tan⁡2xf(x)=7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2x for all x∈(−π2,π2)x\in\left(-\dfrac\pi2,\dfrac\pi2\right). Then the correct expression(s) is(are)
Not attempted
A
∫0π/4xf(x) dx=112\displaystyle\int_0^{\pi/4}xf(x)\,dx=\dfrac1{12}
B
∫0π/4f(x) dx=0\displaystyle\int_0^{\pi/4}f(x)\,dx=0
C
∫0π/4xf(x) dx=16\displaystyle\int_0^{\pi/4}xf(x)\,dx=\dfrac16
D
∫0π/4f(x) dx=1\displaystyle\int_0^{\pi/4}f(x)\,dx=1
42
Let f′(x)=192x32+sin⁡4πxf'(x)=\dfrac{192x^3}{2+\sin^4\pi x} for all x∈Rx\in\mathbb R with f ⁣(12)=0f\!\left(\dfrac12\right)=0. If m≤∫1/21f(x) dx≤Mm\le\displaystyle\int_{1/2}^{1}f(x)\,dx\le M, then the possible values of mm and MM are
Not attempted
A
m=13, M=24m=13,\ M=24
B
m=14, M=12m=\dfrac14,\ M=\dfrac12
C
m=−11, M=0m=-11,\ M=0
D
m=1, M=12m=1,\ M=12
43
The option(s) with the values of aa and LL that satisfy the following equation is(are) ∫04πet(sin⁡6at+cos⁡4at)dt∫0πet(sin⁡6at+cos⁡4at)dt=L\dfrac{\displaystyle\int_0^{4\pi}e^t\left(\sin^6at+\cos^4at\right)dt}{\displaystyle\int_0^{\pi}e^t\left(\sin^6at+\cos^4at\right)dt}=L?
Not attempted
A
a=2, L=e4π−1eπ−1a=2,\ L=\dfrac{e^{4\pi}-1}{e^\pi-1}
B
a=2, L=e4π+1eπ+1a=2,\ L=\dfrac{e^{4\pi}+1}{e^\pi+1}
C
a=4, L=e4π−1eπ−1a=4,\ L=\dfrac{e^{4\pi}-1}{e^\pi-1}
D
a=4, L=e4π+1eπ+1a=4,\ L=\dfrac{e^{4\pi}+1}{e^\pi+1}
44
If α=∫01(e9x+3tan⁡−1x)(12+9x21+x2)dx\alpha=\displaystyle\int_0^1\left(e^{9x+3\tan^{-1}x}\right)\left(\dfrac{12+9x^2}{1+x^2}\right)dx, where tan⁡−1x\tan^{-1}x takes only principal values, then the value of (log⁡e∣1+α∣−3π4)\left(\log_e|1+\alpha|-\dfrac{3\pi}{4}\right) is
Not attempted
Your answer: —
Correct answer: 9
45
Let f:R→Rf:\mathbb R\to\mathbb R be a continuous odd function, which vanishes exactly at one point and f(1)=12f(1)=\dfrac12. Suppose that F(x)=∫−1xf(t) dtF(x)=\displaystyle\int_{-1}^{x}f(t)\,dt for all x∈[−1,2]x\in[-1,2] and G(x)=∫−1xt ∣f(f(t))∣dtG(x)=\displaystyle\int_{-1}^{x}t\,\left|f(f(t))\right|dt for all x∈[−1,2]x\in[-1,2]. If lim⁡x→1F(x)G(x)=114\displaystyle\lim_{x\to1}\dfrac{F(x)}{G(x)}=\dfrac1{14}, then the value of f ⁣(12)f\!\left(\dfrac12\right) is
Not attempted
Your answer: —
Correct answer: 7
46
This question is part of a Column-I to Column-II matching set (JEE Advanced 2015 Section 3, Q.60). Column II lists the values 1,2,3,5,61,2,3,5,6; select every value that correctly matches this Column-I entry. Entry (D)\textbf{Entry (D)}: Suppose that F(α)F(\alpha) denotes the area of the region bounded by x=0, x=2, y2=4xx=0,\ x=2,\ y^2=4x and y=∣αx−1∣+∣αx−2∣+αxy=|\alpha x-1|+|\alpha x-2|+\alpha x, where α∈{0,1}\alpha\in\{0,1\}. Then the value(s) of F(α)+832F(\alpha)+\dfrac83\sqrt2, when α=0\alpha=0 and α=1\alpha=1, is (are)
Not attempted
A
11
B
22
C
33
D
55
E
66
47
Let F(x)=∫xx2+π62cos⁡2t dtF(x)=\displaystyle\int_{x}^{x^2+\frac{\pi}{6}}2\cos^2t\,dt for all x∈Rx\in\mathbb{R} and f:[0,12]→[0,∞)f:\left[0,\dfrac12\right]\to[0,\infty) be a continuous function. For a∈[0,12]a\in\left[0,\dfrac12\right], if F′(a)+2F'(a)+2 is the area of the region bounded by x=0x=0, y=0y=0, y=f(x)y=f(x) and x=ax=a, then f(0)f(0) is
Not attempted
Your answer: —
Correct answer: 3
48
Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a function defined by f(x)={[x],x≤20,x>2f(x)=\begin{cases}[x], & x\le 2\\ 0, & x>2\end{cases}, where [x][x] is the greatest integer less than or equal to xx. If I=∫−12xf(x2)2+f(x+1) dxI=\displaystyle\int_{-1}^{2}\dfrac{xf(x^2)}{2+f(x+1)}\,dx, then the value of (4I−1)(4I-1) is
Not attempted
Your answer: —
Correct answer: 0
49
Match each entry of List I\textbf{List I} to its value in List II\textbf{List II}. List I\textbf{List I} P.\textbf{P.} The number of polynomials f(x)f(x) with non-negative integer coefficients of degree ≤2\le2, satisfying f(0)=0f(0)=0 and ∫01f(x) dx=1\displaystyle\int_0^1 f(x)\,dx=1, is Q.\textbf{Q.} The number of points in the interval [−13,13][-\sqrt{13},\sqrt{13}] at which f(x)=sin⁡(x2)+cos⁡(x2)f(x)=\sin(x^2)+\cos(x^2) attains its maximum value, is R.\textbf{R.} ∫−223x21+ex dx\displaystyle\int_{-2}^{2}\dfrac{3x^2}{1+e^x}\,dx equals S.\textbf{S.} ∫−1/21/2cos⁡2x log⁡ ⁣(1+x1−x)dx∫01/2cos⁡2x log⁡ ⁣(1+x1−x)dx\dfrac{\displaystyle\int_{-1/2}^{1/2}\cos2x\,\log\!\left(\dfrac{1+x}{1-x}\right)dx}{\displaystyle\int_{0}^{1/2}\cos2x\,\log\!\left(\dfrac{1+x}{1-x}\right)dx} equals List II\textbf{List II} 1.\textbf{1.} 88 2.\textbf{2.} 22 3.\textbf{3.} 44 4.\textbf{4.} 00 Which of the following is the correct combination?
Not attempted
A
P→3, Q→2, R→4, S→1P{\to}3,\ Q{\to}2,\ R{\to}4,\ S{\to}1
B
P→2, Q→3, R→4, S→1P{\to}2,\ Q{\to}3,\ R{\to}4,\ S{\to}1
C
P→3, Q→2, R→1, S→4P{\to}3,\ Q{\to}2,\ R{\to}1,\ S{\to}4
D
P→2, Q→3, R→1, S→4P{\to}2,\ Q{\to}3,\ R{\to}1,\ S{\to}4
50
Comprehension\textbf{Comprehension}: Given that for each a∈(0,1)a\in(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\displaystyle\lim_{h\to0^+}\int_h^{1-h}t^{-a}(1-t)^{a-1}\,dt exists. Let this limit be g(a)g(a). It is given that g(a)g(a) is differentiable on (0,1)(0,1). The value of g ⁣(12)g\!\left(\dfrac12\right) is
Not attempted
A
π\pi
B
2π2\pi
C
π2\dfrac{\pi}{2}
D
π4\dfrac{\pi}{4}
51
Comprehension\textbf{Comprehension}: Given that for each a∈(0,1)a\in(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\displaystyle\lim_{h\to0^+}\int_h^{1-h}t^{-a}(1-t)^{a-1}\,dt exists. Let this limit be g(a)g(a). It is given that g(a)g(a) is differentiable on (0,1)(0,1). The value of g′ ⁣(12)g'\!\left(\dfrac12\right) is
Not attempted
A
π2\dfrac{\pi}{2}
B
π\pi
C
−π2-\dfrac{\pi}{2}
D
00
52
Let f:[0,2]→Rf:[0,2]\to\mathbb{R} be a function which is continuous on [0,2][0,2] and differentiable on (0,2)(0,2) with f(0)=1f(0)=1. Let F(x)=∫0x2f(t) dtF(x)=\displaystyle\int_0^{x^2}f(\sqrt t)\,dt for x∈[0,2]x\in[0,2]. If F′(x)=f′(x)F'(x)=f'(x) for all x∈(0,2)x\in(0,2), then F(2)F(2) equals
Not attempted
A
e2−1e^2-1
B
e4−1e^4-1
C
e−1e-1
D
e4e^4
53
The following integral ∫π/4π/2(2 cosec x)17 dx\displaystyle\int_{\pi/4}^{\pi/2}(2\,\text{cosec }x)^{17}\,dx is equal to
Not attempted
A
∫0log⁡(1+2)2(eu+e−u)16 du\displaystyle\int_0^{\log(1+\sqrt2)}2(e^u+e^{-u})^{16}\,du
B
∫0log⁡(1+2)(eu+e−u)17 du\displaystyle\int_0^{\log(1+\sqrt2)}(e^u+e^{-u})^{17}\,du
C
∫0log⁡(1+2)(eu−e−u)17 du\displaystyle\int_0^{\log(1+\sqrt2)}(e^u-e^{-u})^{17}\,du
D
∫0log⁡(1+2)2(eu−e−u)16 du\displaystyle\int_0^{\log(1+\sqrt2)}2(e^u-e^{-u})^{16}\,du
54
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} be given by f(x)=∫1/xxe−(t+1t)dttf(x)=\displaystyle\int_{1/x}^{x} e^{-\left(t+\frac{1}{t}\right)}\frac{dt}{t}. Then:
Not attempted
A
f(x)f(x) is monotonically increasing on [1,∞)[1,\infty).
B
f(x)f(x) is monotonically decreasing on (0,1)(0,1).
C
f(x)+f(1x)=0f(x)+f\left(\frac{1}{x}\right)=0, for all x∈(0,∞)x \in (0,\infty).
D
f(2x)f(2^x) is an odd function of xx on R\mathbb{R}.
55
Let f:[a,b]→[1,∞)f:[a,b]\to[1,\infty) be a continuous function and let g:R→Rg:\mathbb{R}\to\mathbb{R} be defined as g(x)={0if x<a,∫axf(t) dtif a≤x≤b,∫abf(t) dtif x>b.g(x)=\begin{cases}0 & \text{if } x<a,\\ \displaystyle\int_a^x f(t)\,dt & \text{if } a\leq x\leq b,\\ \displaystyle\int_a^b f(t)\,dt & \text{if } x>b.\end{cases} Then:
Not attempted
A
g(x)g(x) is continuous but not differentiable at aa.
B
g(x)g(x) is differentiable on R\mathbb{R}.
C
g(x)g(x) is continuous but not differentiable at bb.
D
g(x)g(x) is continuous and differentiable at either aa or bb but not both.
56
The value of ∫014x3{d2dx2(1−x2)5}dx\displaystyle\int_0^1 4x^3\left\{\frac{d^2}{dx^2}(1-x^2)^5\right\}dx is
Not attempted
Your answer: —
Correct answer: 2
57
For a∈Ra \in \mathbb{R} (the set of all real numbers), a≠−1a \neq -1, lim⁡n→∞(1a+2a+⋯+na)(n+1)a−1[(na+1)+(na+2)+⋯+(na+n)]=160\displaystyle\lim_{n\to\infty}\frac{\left(1^{a}+2^{a}+\cdots+n^{a}\right)}{(n+1)^{a-1}\left[(na+1)+(na+2)+\cdots+(na+n)\right]}=\frac{1}{60}. Then a=a=
Not attempted
A
55
B
77
C
−152-\dfrac{15}{2}
D
−172-\dfrac{17}{2}
58
The area enclosed by the curves y=sin⁡x+cos⁡xy=\sin x+\cos x and y=∣cos⁡x−sin⁡x∣y=|\cos x-\sin x| over the interval [0,π2]\left[0,\dfrac{\pi}{2}\right] is
Not attempted
A
4(2−1)4(\sqrt2-1)
B
22(2−1)2\sqrt2(\sqrt2-1)
C
2(2+1)2(\sqrt2+1)
D
22(2+1)2\sqrt2(\sqrt2+1)
59
Let f:[12,1]→Rf:\left[\dfrac12,1\right]\to\mathbb{R} (the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x)<2f(x)f'(x) < 2f(x) and f(12)=1f\left(\dfrac12\right)=1. Then the value of ∫1/21f(x) dx\displaystyle\int_{1/2}^{1} f(x)\,dx lies in the interval
Not attempted
A
(2e−1,2e)(2e-1,2e)
B
(e−1,2e−1)(e-1,2e-1)
C
(e−12,e−1)\left(\dfrac{e-1}{2},e-1\right)
D
(0,e−12)\left(0,\dfrac{e-1}{2}\right)
Consider g(x)=f(x)e−2xg(x)=f(x)e^{-2x}; the given inequality makes g′(x)<0g'(x)<0, i.e. gg is strictly decreasing, which bounds f(x)f(x) above by e2x−1e^{2x-1} on the interval.
Deconstruction\textbf{Deconstruction}: Given a differential inequality and one boundary value, bound the definite integral of ff. Strategy\textbf{Strategy}: Multiply the inequality by an integrating-factor-like term to make it a clean monotonicity statement, then bound ff pointwise and integrate the bound. Execution\textbf{Execution}: Let g(x)=f(x)e−2xg(x)=f(x)e^{-2x}. Then g′(x)=e−2x(f′(x)−2f(x))<0g'(x)=e^{-2x}\big(f'(x)-2f(x)\big)<0 since f′(x)<2f(x)f'(x)<2f(x). So gg is strictly decreasing on [12,1]\left[\frac12,1\right]. - g(12)=f(12)e−1=e−1g\left(\frac12\right)=f\left(\frac12\right)e^{-1}=e^{-1}. - For x>12x>\frac12: g(x)<g(12)=e−1g(x)<g\left(\frac12\right)=e^{-1}, i.e. f(x)<e2x−1f(x)<e^{2x-1}. - Since ff is given positive, f(x)>0f(x)>0. - Integrating: 0<∫1/21f(x) dx<∫1/21e2x−1 dx=[e2x−12]1/21=e−120<\displaystyle\int_{1/2}^1 f(x)\,dx<\int_{1/2}^1 e^{2x-1}\,dx=\left[\frac{e^{2x-1}}{2}\right]_{1/2}^1=\frac{e-1}{2}. Verification\textbf{Verification}: The upper bound e−12≈0.859\frac{e-1}{2}\approx0.859 is a reasonable size for an integral of a function starting at f(1/2)=1f(1/2)=1 over an interval of length 1/21/2; the strict inequalities throughout (from non-constant ff and strict differential inequality) correctly make both bounds open. Final Answer\textbf{Final Answer}: (D) (0,e−12)\left(0,\dfrac{e-1}{2}\right)