Definite integration – JEE Advanced Previous Year Questions with Solutions

Class 9-10, JEE & NEET

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Q. Let ƒ be a non-negative function defined on the interval $[0,1] .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t$ $0 \leq \mathrm{x} \leq 1,$ and $f(0)=0,$ then $-$ (A) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$ (B) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$ (C) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$ (D) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$ [JEE 2009, 3]

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Sol. (C) $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \mathrm{d} t=\int_{0}^{x} f(t) d t, 0 \leq x \leq 1$ differentiating both the sides & squreing $\Rightarrow 1-\left(f^{\prime}(\mathrm{x})\right)^{2}=f^{2}(\mathrm{x})$ $\Rightarrow \frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$ $\Rightarrow \sin ^{-1} f(\mathrm{x})=\mathrm{x}+\mathrm{c}$ $f(0)=0$ $\Rightarrow f(\mathrm{x})=\sin \mathrm{x}$ $\Rightarrow \because \sin \mathrm{x} \leq \mathrm{x}$ for $\mathrm{x} \in[0,1]$ $\Rightarrow f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

Q. If $\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{d} \mathrm{x}, \mathrm{n}=0,1,2, \ldots, \mathrm{then}-$ (A) $\mathrm{I}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}+2}$ (B) $\sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2 \mathrm{m}+1}=10 \pi$ (C) $\sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2 \mathrm{m}}=0$ (D) $\mathrm{I}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}+1}$ [JEE 2009, 4]

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Sol. (A,B,D) $\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{dx}$ $\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\pi^{\mathrm{x}} \sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{dx}$ $2 \mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\sin \mathrm{x}} \mathrm{dx}$ ..(i) $2 \mathrm{I}_{\mathrm{n}+2}=\int_{-\pi}^{\pi} \frac{\sin (\mathrm{n}+2) \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx} \quad \ldots(\mathrm{i})$ (ii) – (i) $\Rightarrow 2\left(\ln _{+2}-\mathrm{I}_{\mathrm{n}}\right)=\int_{-\pi}^{\pi} \cos (\mathrm{n}+1) \mathrm{x}=0$ $\Rightarrow \quad \mathrm{I}_{\mathrm{n}+2}=\mathrm{I}_{\mathrm{n}}$ $\sum_{m=1}^{10} \mathrm{I}_{2 \mathrm{m}}=10 \sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2}=\frac{10}{2} \int_{-\pi}^{\pi} \frac{\sin 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{d} \mathrm{x}=0$ Put n = 1 in equation (i) $2 \mathrm{I}_{1}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{x} \mathrm{d} \mathrm{x}}{\sin \mathrm{x}}=2 \pi$ $\mathrm{I}_{1}=\pi$ $\sum_{m=1}^{10} I_{2 m+1}=10 \pi$

Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a continuous function which satisfies $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{d} \mathrm{t}$ Then the value of f(ln 5) is…….. [JEE 2009, 4]

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Sol. 0 $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}$ $\mathrm{f}^{\mathrm{l}}(\mathrm{x})=\mathrm{f}(\mathrm{x})$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}$ $\Rightarrow \int \frac{d y}{y}=\int d x$ $\Rightarrow \ln y=x+c$ $\Rightarrow y=e^{x+c}$ $\Rightarrow y=0$ $\left(\begin{array}{c}{\text { at } x=0, y=0} \\ {c \rightarrow-\infty}\end{array}\right)$ $f(x)=0$ $f(\ell n 5)=0$

Q. The value of $\lim _{x \rightarrow 0} \frac{1}{x^{3}} \int_{0}^{x} \frac{t \ell n(1+t)}{t^{4}+4} d t$ is (A) 0 (B) $\frac{1}{12}$ (C) $\frac{1}{24}$ (D) $\frac{1}{64}$ [JEE 2010, 3 (–1)]

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Sol. (B) Applying L-Hospital rule,

Q. The value(s) of $\int_{0}^{1} \frac{\mathrm{x}^{4}(1-\mathrm{x})^{4}}{1+\mathrm{x}^{2}} \mathrm{dx}$ is (are) (A) $\frac{22}{7}-\pi$ (B) $\frac{2}{105}$ (C) 0 (D) $\frac{71}{15}-\frac{3 \pi}{2}$ [JEE 2010, 3]

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Sol. (A)

Q. Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int_{0}^{x} \sqrt{t^{4}+1} d t,$ for all $x \in(-1,1),$ and let $f^{-1}$ be the inverse function of $f$ Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to- (A) 1 (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) $\frac{1}{\mathrm{e}}$ [JEE2010, 5 (–2)]

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Sol. (B) from $(2), f^{-1}(2)=\frac{1}{3}$

Q. For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by \mathrm{f}(\mathrm{x})=\left\{\begin{aligned} \mathrm{x}-[\mathrm{x}] & \text { if }[\mathrm{x}] \text { is odd } \\ 1+[\mathrm{x}]-\mathrm{x} & \text { if }[\mathrm{x}] \text { is even } \end{aligned}\right. Then the value of $\frac{\pi^{2}}{10} \int_{-10}^{10} \mathrm{f}(\mathrm{x}) \cos \pi \mathrm{x} \mathrm{d} \mathrm{x}$ is [JEE 2010, 3]

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Sol. 4

Q. The value of $\int_{\sqrt{\mathrm{in} 2}}^{\sqrt{\mathrm{n} 3}} \frac{\mathrm{x} \sin \mathrm{x}^{2}}{\sin \mathrm{x}^{2}+\sin \left(\mathrm{ln} 6-\mathrm{x}^{2}\right)} \mathrm{dx}$ is (A) $\frac{1}{4} \ln \frac{3}{2}$ (B) $\frac{1}{2} \ln \frac{3}{2}$ (C) $\ln \frac{3}{2}$ (D) $\frac{1}{6} \ln \frac{3}{2}$ [JEE 2011, 3 (–1)]

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Sol. (A)

Q. Let $S$ be the area of the region enclosed by $y=e^{-x^{2}}, y=0, x=0,$ and $x=1 .$ Then (A) $\mathrm{S} \geq \frac{1}{\mathrm{e}}$ (B) $\mathrm{S} \geq 1-\frac{1}{\mathrm{e}}$ (C) $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{\mathrm{e}}}\right)$ (D) $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{\mathrm{e}}}\left(1-\frac{1}{\sqrt{2}}\right)$ [JEE 2012, 4M]

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Sol. (A,B,D)

Q. The value of the integral $\int_{-\pi / 2}^{\pi / 2}\left(\mathrm{x}^{2}+\ln \frac{\pi+\mathrm{x}}{\pi-\mathrm{x}}\right) \cos \mathrm{xd} \mathrm{x}$ is (A) 0 (B) $\frac{\pi^{2}}{2}-4$ (C) $\frac{\pi^{2}}{2}+4$ (D) $\frac{\pi^{2}}{2}$c [JEE 2012, 3M, –1M]

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Sol. (B)

Q. For a $\in \mathrm{R}$ (the set of all real numbers), a\neq-1. $\lim _{n \rightarrow \infty} \frac{\left(1^{a}+2^{a}+\ldots \ldots+n^{a}\right)}{(n+1)^{a-1}[(n a+1)+(n a+2)+\ldots \ldots+(n a+n)]}=\frac{1}{60}$ Then $a=$ (A) 5 (B) 7 (C) $\frac{-15}{2}$ (D) $\frac{-17}{2}$ [JEE(Advanced) 2013, 3, (–1)]

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Sol. (B)

Q. Let $f:[\mathrm{a}, \mathrm{b}] \rightarrow[1, \infty)$ be a continuous function and let $\mathrm{g}: \square \rightarrow \square$ be defined as Then (A) g(x) is continuous but not differentiable at a (B) g(x) is differentiable on  (C) g(x) is continuous but not differentiable at b (D) g(x) is continuous and differentiable at either a or b but not both. [JEE(Advanced)-2014, 3]

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Sol. (A,C)

Q. The value of $\int_{0}^{1} 4 x^{3}\left\{\frac{d^{2}}{d x^{2}}\left(1-x^{2}\right)^{5}\right\} d x$ is [JEE(Advanced)-2014, 3]

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Sol. 2

Q. The following integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \csc x)^{17} d x$ is equal to (A) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{\mathfrak{u}}+e^{-\mathfrak{u}}\right)^{16} \mathrm{d} \mathfrak{u}$ (B) $\int_{0}^{\log (1+\sqrt{2})}\left(\mathrm{e}^{\mathrm{u}}+\mathrm{e}^{-\mathrm{u}}\right)^{17} \mathrm{du}$ (C) $\int_{0}^{\log (1+\sqrt{2})}\left(e^{\mathfrak{u}}-e^{-\mathfrak{u}}\right)^{17} \mathrm{d} \mathfrak{u}$ (D) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{\mathfrak{u}}-e^{-\mathfrak{u}}\right)^{16} d \mathfrak{u}$ [JEE(Advanced)-2014, 3(–1)]

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Sol. (A)

Q. Let $f:[0,2] \rightarrow \square$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1 .$ Let $F(x)=\int_{0}^{x^{2}} f(\sqrt{t}) d t$ for $x \in[0,2] .$ If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$ then $F(2)$ equals $-$ (A) $\mathrm{e}^{2}-1$ (B) $\mathrm{e}^{4}-1$ (C) e – 1 (D) e $^{4}$ [JEE(Advanced)-2014, 3(–1)]

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Sol. (B)

Given that for each a $\in(0,1)$, $\lim _{\mathrm{h} \rightarrow 0^{+}} \int_{\mathrm{h}}^{1-\mathrm{h}} \mathrm{t}^{-\mathrm{a}}(1-\mathrm{t})^{\mathrm{a}-1}$ $\mathrm{dt}$ exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0,1).
Q. The value of $\mathrm{g}\left(\frac{1}{2}\right)$ is – (A) $\pi$ (B) $2 \pi$ (C) $\frac{\pi}{2}$ (D) $\frac{\pi}{4}$ [JEE(Advanced)-2014, 3(–1)]

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Sol. (A)

Q. The value of $\mathrm{g}^{\prime}\left(\frac{1}{2}\right)$ is- (A) $\frac{\pi}{2}$ (B) $\pi$ (C) $-\frac{\pi}{2}$ (D) 0 [JEE(Advanced)-2014, 3(–1)]

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Sol. (D)

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Sol. (C)

Q. Let $f: \square \rightarrow \square$ be a function defined by $f(x)$ $=\left\{\begin{array}{ccc}{[\mathrm{x}]} & {,} & {\mathrm{x} \leq 2} \\ {0} & {,} & {\mathrm{x}>2}\end{array}\right.$ where [x] is the greatest integer less than or equal to x. If $\mathrm{I}=\int_{-1}^{2} \frac{\mathrm{x} f\left(\mathrm{x}^{2}\right)}{2+f(\mathrm{x}+1)} \mathrm{dx}$ , then the value of (4I – 1) is [JEE 2015, 4M, –0M]

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Sol. (A)

Q. If $\alpha=\int_{0}^{1}\left(\mathrm{e}^{9 \mathrm{x}+3 \tan ^{-1} \mathrm{x}}\right)\left(\frac{12+9 \mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$ where $\tan ^{-1} \mathrm{x}$ takes only principal values, then the value of $\left(\log _{\mathrm{e}}|1+\alpha|-\frac{3 \pi}{4}\right)$ is [JEE 2015, 4M, –0M]

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Sol. 0

Q. Let $f: \mathbb{U} \rightarrow \square$ be a continuous odd function, which vanishes exactly at one point and $f(1)=\frac{1}{2}$ Suppose that $\mathrm{F}(\mathrm{x})=\int_{-1}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$ for all $\mathrm{x} \in[-1,2]$ and $\mathrm{G}(\mathrm{x})$ $=\int_{-1}^{x} \mathfrak{t}|f(f(\mathfrak{t}))| d \mathfrak{t}$ for all $x \in[-1,2]$. If $\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14},$ then the value of $f\left(\frac{1}{2}\right)$ is [JEE 2015, 4M, –0M]

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Sol. 9

Q. The option(s) with the values of a and L that satisfy the following equation is(are) (A) $a=2, L=\frac{e^{4 \pi}-1}{e^{\pi}-1}$ (B) $a=2, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$c (C) $a=4, L=\frac{e^{4 \pi}-1}{e^{\pi}-1}$ (D) $a=4, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$ [JEE 2015, 4M, –0M]

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Sol. 7

Q. Let $f(\mathrm{x})=7 \tan ^{8} \mathrm{x}+7 \tan ^{6} \mathrm{x}-3 \tan ^{4} \mathrm{x}-3 \tan ^{2} \mathrm{x}$ for all $\mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$ Then the correct expression(s)is(are) (A) $\int_{0}^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{12}$ (B) $\int_{0}^{\pi / 4} f(\mathrm{x}) \mathrm{d} \mathrm{x}=0$ (C) $\int_{0}^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{6}$ (D) $\int_{0}^{\pi / 4} f(\mathrm{x}) \mathrm{d} \mathrm{x}=1$ [JEE 2015, 4M, –0M]

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Sol. (A,C)

Q. Let $f^{\prime}(x)=\frac{192 x^{3}}{2+\sin ^{4} \pi x}$ for all $\mathrm{x} \in \square$ with $f$ $\left(\frac{1}{2}\right)$ $=0 .$ If $\mathrm{m} \leq \int_{1 / 2}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x} \leq \mathrm{M}$ then the possible values of m and M are (A) m = 13, M = 24 (B) $\quad \mathrm{m}=\frac{1}{4}, \mathrm{M}=\frac{1}{2}$ (C) m = –11, M = 0 (D) m = 1, M = 12 [JEE 2015, 4M, –0M]

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Sol. (A,B)

Let $\mathrm{F}: \mathbb{U} \rightarrow \square$ be a thrice differentiable function. Suppose that $\mathrm{F}(1)=0, \mathrm{F}(3)=-4 \mathrm{F}^{\prime}(\mathrm{x})<$ 0 for all $\mathrm{x} \in(1 / 2,3) .$ Let $f(\mathrm{x})=\mathrm{xF}(\mathrm{x})$ for all $\mathrm{x} \in \mathbb{D}$.
Q. The correct statement(s) is(are) (A) $f^{\prime}(1)<0$ (B) $f(2)<0$ (C) $f^{\prime}(\mathrm{x}) \neq 0$ for any $\mathrm{x} \in(1,3)$ (D) $f^{\prime}(x)=0$ for some $x \in(1,3)$ [JEE 2015, 4M, –0M]

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Sol. (D)

Q. If $\int_{1}^{3} \mathrm{x}^{2} \mathrm{F}^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=-12$ and $\int_{1}^{3} \mathrm{x}^{3} \mathrm{F}^{\prime \prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=40,$ then the correct expression(s) is (are) (A) 9ƒ'(3) + ƒ'(1) – 32 = 0 (B) $\int_{1}^{3} f(\mathrm{x}) \mathrm{d} \mathrm{x}=12$ (C) 9ƒ'(3) – ƒ'(1) + 32 = 0 (D) $\left.\int_{1}^{3} f(x) d x=-12\right]$ [JEE 2015, 4M, –0M]

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Sol. (A,B,C)

Q. The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2} \cos x}{1+e^{x}} d x$ is equal to (A) $\frac{\pi^{2}}{4}-2$ (B) $\frac{\pi^{2}}{4}+2$ (C) $\pi^{2}-\mathrm{e}^{\frac{\pi}{2}}$ (D) $\pi^{2}+\mathrm{e}^{\frac{\pi}{2}}$ [JEE(Advanced)2016]

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Sol. (C,D)

Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a differentiable function such that $\mathrm{f}(0)=0, \mathrm{f}\left(\frac{\pi}{2}\right)=3$ and $\mathrm{f}^{\prime}(0)=1$ If $\mathrm{g}(\mathrm{x})=\int_{\mathrm{x}}^{\frac{\pi}{2}}\left[\mathrm{f}^{\prime}(\mathrm{t}) \csc \mathrm{t}-\cot t \csc \mathrm{t} \mathrm{f}(\mathrm{t})\right] \mathrm{d} \mathrm{t}$ for $\mathrm{x} \in\left(0, \frac{\pi}{2}\right],$ then $\lim _{\mathrm{x} \rightarrow 0} \mathrm{g}(\mathrm{x})=$ [JEE(Advanced)-2017]

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Sol. 2

Q. If $\mathrm{I}=\sum_{\mathrm{k}=1}^{98} \int_{\mathrm{k}}^{\mathrm{k}+1} \frac{\mathrm{k}+1}{\mathrm{x}(\mathrm{x}+1)} \mathrm{d} \mathrm{x},$ then (A) $\mathrm{I}<\frac{49}{50}$ (B) $\mathrm{I}<\log _{\mathrm{e}} 99$ (C) $\mathrm{I}>\frac{49}{50}$ (D) $\mathrm{I}>\log _{\mathrm{e}} 99$ [JEE(Advanced)-2017]

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Sol. (B,C)

Q. If $\mathrm{g}(\mathrm{x})=\int_{\sin \mathrm{x}}^{\sin (2 \mathrm{x})} \sin ^{-1}(\mathrm{t}) \mathrm{dt},$ then (A) $\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=-2 \pi$ (B) $\mathrm{g}^{\prime}\left(-\frac{\pi}{2}\right)=2 \pi$ (C) $\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=2 \pi$ (D) $\mathrm{g}^{\prime}\left(-\frac{\pi}{2}\right)=-2 \pi$ [JEE(Advanced)-2017]

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Sol. (Bonus)

Q. The value of the integral $\int_{0}^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{\frac{1}{4}}} d x$ is [JEE(Advanced)-2018]

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Sol. 2

• February 18, 2021 at 10:18 pm

nice👍

9
• October 10, 2020 at 6:36 am