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1

WB JEE 2008

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

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

Questions Asked from Application of Integration

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
WB JEE 2022 (3)
WB JEE 2021 (3)
WB JEE 2020 (3)
WB JEE 2019 (1)

Joint Entrance Examination

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