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1

JEE Advanced 2016 Paper 1 Offline

Numerical
The total number of distinct $$x \in \left[ {0,1} \right]$$ for which

$$\int\limits_0^x {{{{t^2}} \over {1 + {t^4}}}} dt = 2x - 1$$
Your Input ________

Answer

Correct Answer is 1
2

JEE Advanced 2015 Paper 1 Offline

Numerical
Let $$F\left( x \right) = \int\limits_x^{{x^2} + {\pi \over 6}} {2{{\cos }^2}t\left( {dt} \right)} $$ for all $$x \in R$$ and $$f:\left[ {0,{1 \over 2}} \right] \to \left[ {0,\infty } \right]$$ be a continuous function. For $$a \in \left[ {0,{1 \over 2}} \right],\,$$ $$F'(a)+2$$ is the area of the region bounded by $$x=0, y=0, y=f(x)$$ and $$x=a,$$ then $$f(0)$$ is
Your Input ________

Answer

Correct Answer is 3
3

JEE Advanced 2015 Paper 2 Offline

Numerical
Let $$f:R \to R$$ be a continuous odd function, which vanishes exactly at one point and $$f\left( 1 \right) = {1 \over {2.}}$$ Suppose that $$F\left( x \right) = \int\limits_{ - 1}^x {f\left( t \right)dt} $$ for all $$x \in \,\,\left[ { - 1,2} \right]$$ and $$G(x)=$$ $$\int\limits_{ - 1}^x {t\left| {f\left( {f\left( t \right)} \right)} \right|} dt$$ for all $$x \in \,\,\left[ { - 1,2} \right].$$ If $$\mathop {\lim }\limits_{x \to 1} {{F\left( x \right)} \over {G\left( x \right)}} = {1 \over {14}},$$ then the value of $$f\left( {{1 \over 2}} \right)$$ is
Your Input ________

Answer

Correct Answer is 7

Explanation

Here, $$\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$$

$$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = {1 \over {14}}$$ [using L' Hospital's rule] ....... (i)

As $$F(x) = \int_{ - 1}^x {f(t)dt} $$

$$ \Rightarrow F'(x) = f(x)$$ ...... (ii)

and $$G(x) = \int_{ - 1}^x {t|f\{ f(t)\} |dt} $$

$$ \Rightarrow G'(x) = x|f\{ f(x)\} |$$ ...... (iii)

$$\therefore$$ $$\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = \mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x|f\{ f(x)\} |}}$$

$$ = {{f(1)} \over {1|f\{ f(1)\} |}} = {{1/2} \over {|f(1/2)|}}$$ ....... (iv)

Given, $$\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$$

$$\therefore$$ $${{{1 \over 2}} \over {\left| {f\left( {{1 \over 2}} \right)} \right|}} = {1 \over {14}} \Rightarrow \left| {f\left( {{1 \over 2}} \right)} \right| = 7$$

4

JEE Advanced 2015 Paper 2 Offline

Numerical
If $$\alpha = \int\limits_0^1 {\left( {{e^{9x + 3{{\tan }^{ - 1}}x}}} \right)\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)} dx$$ where $${\tan ^{ - 1}}x$$ takes only principal values, then the value of $$\left( {{{\log }_e}\left| {1 + \alpha } \right| - {{3\pi } \over 4}} \right)$$ is
Your Input ________

Answer

Correct Answer is 9

Explanation

$$\alpha = \int\limits_0^1 {{e^{(9x + 3{{\tan }^{ - 1}}x)}}\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)dx} $$

Set $$9x + 3{\tan ^{ - 1}}x = t$$

so that $${{dt} \over {dx}} = 9 + {3 \over {1 + {x^2}}} = {{12 + 9{x^2}} \over {1 + {x^2}}}$$

We have, $$\alpha = \int\limits_0^{9 + {{3\pi } \over 4}} {{e^t}dt = {e^{9 + {{3\pi } \over 4}}} - 1} $$

$$\therefore$$ $$\ln \left| {\alpha + 1} \right| = 9 + {{3\pi } \over 4}$$

Thus $$\ln \left| {\alpha + 1} \right| - {{3\pi } \over 4} = 9$$

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