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1

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
Your Input ________

Answer

Correct Answer is 2
2

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

Your Input ________

Answer

Correct Answer is 4

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 $$\int\limits_1^2 {f(x)\cos \pi x\,dx} $$

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

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

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

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

3

IIT-JEE 2009

Numerical
Let $$f:R \to R$$ be a continuous function which satisfies $$$f\left( x \right) = \int\limits_0^x {f\left( t \right)dt.} $$$ Then the value of $$f\left( {\ln \,5} \right)$$ is
Your Input ________

Answer

Correct Answer is 0

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