NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### IIT-JEE 2011 Paper 1 Offline

Numerical

Let $$f:[1,\infty ) \to [2,\infty )$$ be a differentiable function such that $$f(1) = 2$$. If $$6\int\limits_1^x {f(t)dt = 3xf(x) - {x^3} - 5}$$ for all $$x \ge 1$$, then the value of f(2) is ___________.

## Explanation

It is given that

$$6\int\limits_1^x {f(t)dt = 3xf(x) - {x^3}} - 5$$

$$\Rightarrow 6f(x) = 3f(x) + 3xf'(x) - 3{x^2}$$

$$\Rightarrow 3f(x) = 3xf'(x) - 3{x^2} \Rightarrow xf'(x) - f(x) = {x^2}$$

$$\Rightarrow x{{dy} \over {dx}} - y = {x^2} \Rightarrow {{dy} \over {dx}} - {1 \over x}y = x$$ .... (1)

Now, $$I.F. = {e^{\int { - {1 \over x}dx} }} = {e^{ - {{\log }_e}x}}$$

Multiplying Eq. (1) both sides by $${1 \over x}$$, we get

$${1 \over x}{{dy} \over {dx}} - {1 \over {{x^2}}}y = 1 \Rightarrow {d \over {dx}}\left( {y.{1 \over x}} \right) = 1$$

Integrating, we get

$${y \over x} = x + c$$

Substituting x = 1 and y = 2, we get

$$\Rightarrow 2 = 1 + c \Rightarrow c = 1 \Rightarrow y = {x^2} + x$$

$$\Rightarrow f(x) = {x^2} + x \Rightarrow f(2) = 6$$

2

### IIT-JEE 2011 Paper 2 Offline

Numerical
Let $$y'\left( x \right) + y\left( x \right)g'\left( x \right) = g\left( x \right),g'\left( x \right),y\left( 0 \right) = 0,x \in R,$$ where $$f'(x)$$ denotes $${{df\left( x \right)} \over {dx}}$$ and $$g(x)$$ is a given non-constant differentiable function on $$R$$ with $$g(0)=g(2)=0.$$ Then the value of $$y(2)$$ is

## Explanation

It is given that

$$y'(x) + y(x)g'(x) = g(x)g'(x)$$

$$\Rightarrow {e^{g(x)}}y'(x) + {e^{g(x)}}g'(x)y(x) = {e^{g(x)}}g(x)g'(x)$$

$$\Rightarrow {d \over {dx}}(y(x){e^{g(x)}} )= {e^{g(x)}}g(x)g'(x)$$

Therefore, $$y(x) = {e^{g(x)}} = \int {{e^{g(x)}}g(x)g'(x)dx}$$

$$= \int {{e^t}t\,dt}$$ [where g(x) = t]

$$= (t - 1){e^t} + c$$

Therefore, $$y(x){e^{g(x)}} = (g(x) - 1){e^{g(x)}} + c$$

Substituting $$x = 0 \Rightarrow 0 = (0 - 1) \times 1 + c \Rightarrow c = 1$$

Substituting $$x = 2 \Rightarrow y(2) \times 1 = (0 - 1) \times (1) + 1$$

Hence, $$y(2) = 0$$.

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12