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1

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

2

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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