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1

WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
The area of the figure bounded by the parabola $$x = - 2{y^2},\,x = 1 - 3{y^2}$$ is
A
$${1 \over 3}$$sq unit
B
$${4 \over 3}$$sq unit
C
1 sq unit
D
2 sq units

Explanation

Intersecting points of both curves are $$P( - 2,1)$$ and $$Q( - 2, - 1)$$.


$$ \therefore $$ Area of the bounded region

$$ = \int_{ - 1}^1 {((1 - 3{y^2}) - ( - 2{y^2}))\,dy} $$

$$ = 2\int_0^1 {(1 - {y^2})\,dy} $$

$$ = 2\left( {y - {{{y^3}} \over 3}} \right)_0^1$$

$$ = 2\left( {1 - {1 \over 3}} \right) = {4 \over 3}$$ sq unit

$$x = - 2{y^2}$$ এবং $$x = 1 - 3{y^2}$$ অধিবৃত্তদ্বয়ের দ্বারা বেষ্টিত অঞ্চলের ক্ষেত্রফল হবে

A
$${1 \over 3}$$ বর্গ একক
B
$${4 \over 3}$$ বর্গ একক
C
1 বর্গ একক
D
2 বর্গ একক

Explanation

$${y^2} = - {1 \over 2}x$$ ..... (i)

$${y^2} = - {1 \over 3}(x - 1)$$ .... (ii)

$$ - {1 \over 2}x = - {1 \over 3}(x - 1)$$

বা, $$3x = 2x - 2$$

বা, $$x = - 2$$

$$ \therefore $$ $${y^2} = - {1 \over 2}x = 1$$

ক্ষেত্রফল $$ = 2\left[ {\left| {\int\limits_{ - 2}^1 {{1 \over {\sqrt 3 }}\sqrt {1 - x} dx} } \right| - \left| {\int\limits_{ - 2}^0 {{1 \over {\sqrt 2 }}\sqrt { - x} dx} } \right|} \right]$$

$$ = 2\left| {\left[ {{1 \over {\sqrt 3 }}{{{{(1 - x)}^{{3 \over 2}}}} \over { - {3 \over 2}}}} \right]_{ - 2}^2} \right| - 2\left| {{1 \over {\sqrt 2 }}\left[ {{{{{( - x)}^{{3 \over 2}}}} \over { - {3 \over 2}}}} \right]_{ - 2}^0} \right|$$

$$ = 2\left| {{2 \over {3\sqrt 3 }}.3\sqrt 3 } \right| - 2\left| {{1 \over {\sqrt 2 }}.{2 \over 3}.2\sqrt 2 } \right|$$

$$ = 2\left( {2 - {4 \over 3}} \right) = {4 \over 3}$$ বর্গ একক।

2

WB JEE 2019

MCQ (More than One Correct Answer)
English
Bengali
The area bounded by y = x + 1 and y = cos x and the X-axis, is
A
1 sq unit
B
$${3 \over 2}$$ sq unit
C
$${1 \over 4}$$ sq unit
D
$${1 \over 8}$$ sq unit

Explanation

We have,

y = x + 1 and y = cos x


$$ \therefore $$ Required area = $$\int_{ - 1}^0 {(x + 1)dx + \int_0^{\pi /2} {\cos xdx} } $$

$$ = \left[ {{{{x^2}} \over 2} + x} \right]_1^0 + [\sin x]_0^{\pi /2}$$

$$ = \left[ {(0) - \left( {{1 \over 2} - 1} \right)} \right] + [1 - 0]$$

$$ = {1 \over 2} + 1 = {3 \over 2}$$ sq unit.

$$y = x + 1$$ ও y = $$\cos x$$ এবং $$X$$-অক্ষদ্বারা সীমাবদ্ধ অঞ্চলটির ক্ষেত্রফল হল -

A
1 বর্গ একক
B
$${3 \over 2}$$ বর্গ একক
C
$${1 \over 4}$$ বর্গ একক
D
$${1 \over 8}$$ বর্গ একক

Explanation

$$y = x + 1$$ .... (i)

$$y = \cos x$$ ... (ii)

ক্ষেত্রফল $$=\int_{ - 1}^0 {(x + 1)dx + \int_0^{{\pi \over 2}} {\cos xdx} } $$

$$ = \left[ {{{{x^2}} \over 2} + x} \right]_{ - 1}^0 + [\sin x]_0^{{\pi \over 2}}$$

$$ = - \left( {{1 \over 2} - 1} \right) + 1 = {3 \over 2}$$ বর্গ একক

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