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1

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

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

2

### 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

3

### 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

4

### IIT-JEE 2010 Paper 1 Offline

Numerical
For any real number $$x,$$ let $$\left[ x \right]$$ denote the largest integer less than or equal to $$x.$$ Let $$f$$ be a real valued function defined on the interval $$\left[ { - 10,10} \right]$$ by $$f\left( x \right) = \left\{ {\matrix{ {x - \left[ x \right]} & {if\left[ x \right]is\,odd,} \cr {1 + \left[ x \right] - x} & {if\left[ x \right]is\,even} \cr } } \right.$$\$

Then the value of $${{{\pi ^2}} \over {10}}\int\limits_{ - 10}^{10} {f\left( x \right)\cos \,\pi x\,dx}$$ is

## Explanation

Given,

$$f(x) = \left\{ {\matrix{ {x - [x]} & {if\,[x]\,is\,odd} \cr {1 + [x] - x} & {if\,[x]\,is\,even} \cr } } \right.$$

f(x) and cos $$\theta$$ x both are periodic with period 2 and both are even.

$$\therefore$$ $$\int\limits_{ - 10}^{10} {f(x)\cos \pi x\,dx}$$ $$= 2\int\limits_0^{10} {f(x)\cos \pi x\,dx}$$

$$= 10\int\limits_0^2 {f(x)\cos \pi x\,dx}$$

Now, $$\int\limits_0^1 {f(x)\cos \pi x\,dx}$$

$$= \int\limits_0^1 {(1 - x)\cos \pi x\,dx = - \int\limits_0^1 {u\cos \pi u\,du} }$$ and $$\int\limits_1^2 {f(x)\cos \pi x\,dx}$$

$$= \int\limits_1^2 {(x - 1)\cos \pi x\,dx = - \int\limits_0^1 {u\cos \pi u\,du} }$$

$$\therefore$$ $$\int\limits_{ - 10}^{10} {f(x)\cos \pi x\,dx}$$

$$= - 20\int\limits_0^1 {u\cos \pi u\,du = {{40} \over {{\pi ^2}}}}$$

$$\Rightarrow {{{\pi ^2}} \over {10}}\int\limits_{ - 10}^{10} {f(x)\cos \pi x\,dx = 4}$$

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