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1

WB JEE 2008

MCQ (Single Correct Answer)

The area included between the parabolas y2 = 4x and x2 = 4y is

A
$${8 \over 3}$$ sq. units
B
8 sq. units
C
$${16 \over 3}$$ sq. units
D
12 sq. units

Explanation

Equation of curves are

y2 = 4x ..... (i)

and x2 = 4y .... (ii)

From (i) and (ii)

$${\left( {{{{x^2}} \over 4}} \right)^2} = 4x \Rightarrow {x^4} - 64x = 0$$

$$ \Rightarrow x({x^3} - 64) = 0 \Rightarrow x = 0$$ or $$x = 4$$

$$\therefore$$ Required area $$ = \int\limits_0^4 {\left( {\sqrt {4x} - {{{x^2}} \over 4}} \right)dx = \left[ {2.{{2{x^{3/2}}} \over 3} - {{{x^3}} \over {12}}} \right]_0^4} $$

$$ = {{32} \over 3} - {{64} \over {12}} = {{32} \over 3} - {{16} \over 3} = {{16} \over 3}$$ sq. unit.

2

WB JEE 2008

MCQ (Single Correct Answer)

The line which is parallel to x-axis and crosses the curve y = $$\sqrt x$$ at an angle 45$$^\circ$$ is

A
y = $${1 \over 4}$$
B
y = $${1 \over 2}$$
C
y = 1
D
y = 4

Explanation

Since equation of any line parallel to y-axis is y = k where k is any real constant. We draw a tangent which makes an angle 45$$^\circ$$ with the line parallel to x-axis.

Given equation of curve is

$$y = \sqrt x $$ ..... (1)

Taking derivative

$${{dy} \over {dx}} = {1 \over {2\sqrt x }}$$

Since $${{dy} \over {dx}} = \tan 45^\circ = 1$$

$$\therefore$$ $${1 \over {2\sqrt x }} = 1$$ or $$x = {1 \over 4}$$

Substituting $$x = {1 \over 4}$$ in (1)

$$y = \sqrt {{1 \over 4}} = {1 \over 2}$$

$$\therefore$$ Point $$P\left( {{1 \over 4},{1 \over 2}} \right)$$, it also passes through the line parallel to x-axis.

$$\therefore$$ $$y = {1 \over 2}$$ is required equation of line.

3

WB JEE 2008

MCQ (Single Correct Answer)

The area enclosed between the curve y = 1 + x2, the y-axis, and the straight line y = 5 is given by

A
$${{14} \over 3}$$ square units
B
$${{7} \over 3}$$ square units
C
5 square units
D
$${{16} \over 3}$$ square units

Explanation

Given equation of curve is y = 1 + x2 or x2 = (y $$-$$ 1)

This is equation of parabola having vertex (0, 1)

$$\therefore$$ Required area $$ = \int\limits_1^5 {\sqrt {y - 1} \,dy} $$

$$ = {2 \over 3}\left[ {{{(y - 1)}^{3/2}}} \right]_1^5$$

$$ = {2 \over 3}\left[ {{{(4)}^{3/2}}} \right] = {2 \over 3} \times 8 = {{16} \over 3}$$ sq. units

4

WB JEE 2008

MCQ (Single Correct Answer)

The function f(x) which satisfies $$f(x) = f( - x) = {{f'(x)} \over x}$$ is given by

A
$$f(x) = {1 \over 2}{e^{{x^2}}}$$
B
$$f(x) = {1 \over 2}{e^{ - {x^2}}}$$
C
$$f(x) = {x^2}{e^{{x^2}/2}}$$
D
$$f(x) = {e^{{x^2}/2}}$$

Explanation

$$f(x) = {{f'(x)} \over x} \Rightarrow {{f'(x)} \over {f(x)}} = x$$

Integrating both sides, we get

$$\int {{{f'(x)} \over {f(x)}}dx = \int {x\,dx \Rightarrow \log |f(x)| = {{{x^2}} \over 2} + k} } $$

$$ \Rightarrow |f(x)| = {e^{({x^2}/2) + k}}{e^k}\,.\,{e^{{{{x^2}} \over 2}}}$$

$$ \Rightarrow |f(x)| = c{e^{{x^2}/2}}$$ ($$c = {e^k}$$ is an arbitrary constant)

$$ \Rightarrow f(x) = c{e^{{x^2}/2}}$$

keeping c = 1, $$f(x) = {e^{{x^2}/2}}$$.

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