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1

### 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

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

2

### 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

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

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

3

### JEE Advanced 2015 Paper 1 Offline

Numerical
Let $$f:R \to R$$ be a function defined by
$$f\left( x \right) = \left\{ {\matrix{ {\left[ x \right],} & {x \le 2} \cr {0,} & {x > 2} \cr } } \right.$$ where $$\left[ x \right]$$ is the greatest integer less than or equal to $$x$$, if $$I = \int\limits_{ - 1}^2 {{{xf\left( {{x^2}} \right)} \over {2 + f\left( {x + 1} \right)}}dx,}$$ then the value of $$(4I-1)$$ is

4

### JEE Advanced 2014 Paper 1 Offline

Numerical
The value of $$\int\limits_0^1 {4{x^3}\left\{ {{{{d^2}} \over {d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx}$$ is

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