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1

### JEE Advanced 2018 Paper 2 Offline

Numerical
Let f : R $$\to$$ R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equation $${{dy} \over {dx}} = (2 + 5y)(5y - 2)$$, then the value of $$\mathop {\lim }\limits_{n \to - \infty } f(x)$$ is ...........

## Explanation

We have,

$${{dy} \over {dx}} = (2 + 5y)(5y - 2)$$

$$\Rightarrow {{dy} \over {25{y^2} - 4}} = dx$$

$$\Rightarrow {1 \over {25}}\left( {{{dy} \over {{y^2} - {4 \over {25}}}}} \right) = dx$$

On integrating both sides, we get

$${1 \over {25}}\int {{{dy} \over {{y^2} - {{\left( {{2 \over 5}} \right)}^2}}} = \int {dx} }$$

$$\Rightarrow {1 \over {25}} \times {1 \over {2 \times 2/5}}\log \left| {{{y - 2/5} \over {y + 2/5}}} \right| = x + C$$

$$\Rightarrow \log \left| {{{5y - 2} \over {5y + 2}}} \right| = 20(x + C)$$

$$\Rightarrow \left| {{{5y - 2} \over {5y + 2}}} \right| = A{e^{20x}}$$ [$$\because$$ e20C = A]

when x = 0 $$\Rightarrow$$ y = 0, then A = 1

$$\therefore$$ $$\left| {{{5y - 2} \over {5y + 2}}} \right| = {e^{20x}}$$

$$\mathop {\lim }\limits_{x \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = \mathop {\lim }\limits_{x \to - \infty } {e^{20x}}$$

$$\Rightarrow \mathop {\lim }\limits_{n \to - \infty } \left| {{{5f(x) - 2} \over {5f(x) + 2}}} \right| = 0$$

$$\Rightarrow \mathop {\lim }\limits_{n \to - \infty } 5f(x) - 2 = 0$$

$$\Rightarrow \mathop {\lim }\limits_{n \to - \infty } f(x) = {2 \over 5} = 0.4$$
2

### 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

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

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

3

### 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

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