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1

### AIEEE 2003

Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,}$$ is
A
$$e + {{{e^2}} \over 2} + {5 \over 2}$$
B
$$e - {{{e^2}} \over 2} - {5 \over 2}$$
C
$$e + {{{e^2}} \over 2} - {3 \over 2}$$
D
$$e - {{{e^2}} \over 2} - {3 \over 2}$$

## Explanation

Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$

Integrating log

$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$

$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$

$$\therefore$$ $$\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$$

$$= \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$$

$$= \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$$

$$= \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$$

$$= e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$$

$$= e - {{{e^2}} \over 2} - {3 \over 2}$$
2

### AIEEE 2003

The area of the region bounded by the curves $$y = \left| {x - 1} \right|$$ and $$y = 3 - \left| x \right|$$ is
A
$$6$$ sq. units
B
$$2$$ sq. units
C
$$3$$ sq. units
D
$$4$$ sq. units

## Explanation

$$A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + }$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + }$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx}$$

$$= \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( {4 - 2x} \right)dx} } }$$

$$= \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2$$

$$= 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right)$$

$$= 1 + 2 + 4 - 3 = 4$$ sq. units
3

### AIEEE 2002

The area bounded by the curves $$y = \ln x,y = \ln \left| x \right|,y = \left| {\ln {\mkern 1mu} x} \right|$$ and $$y = \left| {\ln \left| x \right|} \right|$$ is
A
$$4$$sq. units
B
$$6$$sq. units
C
$$10$$sq. units
D
none of these

## Explanation

First we draw each curve as separate graph

NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.

Clearly the bounded area is as shown in the following figure.

Required area $$= 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$$

$$= - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$$ sq. units
4

### AIEEE 2002

If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx}$$ is
A
$$3/2$$
B
$$1$$
C
$$5/4$$
D
$$-3/4$$

## Explanation

We have $$\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$$

$$\int\limits_0^2 {xf'\left( x \right)} dx$$

$$= x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$

$$= \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$$

$$= 2f\left( 2 \right) - {3 \over 4}$$

$$= 0 - {3 \over 4}$$

$$\left( {} \right.$$ As $$f\left( 2 \right) = 0$$ $$\left. {} \right)$$ $$= - {3 \over 4}.$$

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