Area Under Curves — Page 11 | JEE Mathematics

Q101

If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is -

A

\sqrt 3

B

{{\sqrt 3 } \over 2}

C

{2 \over {\sqrt 3 }}

D

{1 \over {\sqrt 3 }}

Correct Answer

Option D

Solution

Area bounded by y2 = 4ax & x2 = 4by, a, b $\ne$ 0 is

\left| {{{16ab} \over 3}} \right|

by using formula : 4a $=$

{1 \over k} = 4b,k > 0

Area

= \left| {{{16.{1 \over {4k}}.{1 \over {4k}}} \over 3}} \right| = 1

$\Rightarrow$ k2

= {1 \over 3}

$\Rightarrow$ k

= {1 \over {\sqrt 3 }}

Q102

If the area of the region bounded by the curves,

y = {x^2},y = {1 \over x}

and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :

A

{e^{{3 \over 2}}}

B

{4 \over 3}

C

{3 \over 2}

D

{e^{{2 \over 3}}}

Correct Answer

Option D

Solution

Point of intersection of y = x2 and y =

{1 \over x}

. put y =

{1 \over x}

in y = x2, then we get,

{1 \over x} = {x^2}

$\Rightarrow$

\,\,\,

x3 $-$ 1 = 0 $\Rightarrow$

\,\,\,

x = 1

\therefore\,\,\,

y = 1

\therefore\,\,\,

point B = (1, 1) Area of region ABCDA =

\int\limits_0^1 {{x^2}}

dx +

\int\limits_1^t {{1 \over x}}

dx $=$

\left[ {{{{x^3}} \over 3}} \right]_0^1

+

\left[ {\ell n\,x} \right]_1^t

=

{1 \over 3}

+

\ell n\,t

$-$

\ell n

1 =

{1 \over 3}

+

\ell n\,t

[ as

\ell n

1 = 0] given this Area = 1 sq unit.

\therefore\,\,\,

{1 \over 3}

+

\ell n\,t

= 1 $\Rightarrow$

\ell n\,t

=

{2 \over 3}

$\Rightarrow$ t = e

{^{{2 \over 3}}}

Q103

The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :

A

{3 \over 4}

B

{5 \over 4}

C

{7 \over 8}

D

{9 \over 8}

Correct Answer

Option D

Solution

x = 4y $-$ 2 & x2 = 4y $\Rightarrow$ x2 = x + 2 $\Rightarrow$ x2 $-$ x $-$ 2 = 0 x = 2, $-$ 1 So,

\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}}

Q104

The area (in sq. units) bounded by the parabolae y = x2 – 1, the tangent at the point (2, 3) to it and the y-axis is :

A

56\over3

B

32\over3

C

8\over3

D

14\over3

Correct Answer

Option C

Solution

Equation of tangent at (2, 3) on the parabola y = x2 $-$ 1 is

{{y + 3} \over 2} = 2x - 1

$\Rightarrow$ y + 3 = 4x $-$ 2 $\Rightarrow$ y = 4x $-$ 5 When x = 0 then for the tangent y = $-$ 5 $\therefore$ Tangent cuts x y axis at (0, $-$ 5) point.

$\therefore$ Area of the bounded region is =

\int\limits_{ - 5}^3 {{{y + 5} \over 4}} \,\,\,dy - \int\limits_{ - 1}^3 {\sqrt {y + 1} } \,\,\,dy

=

{1 \over 4}\left[ {{{{y^2}} \over 2} + 5y} \right]_{ - 5}^3 - \left[ {{2 \over 3} \times {{\left( {y + 1} \right)}^{{3 \over 2}}}} \right]_{ - 1}^3

{1 \over 4}\left[ {\left( {{9 \over 2} + 15} \right) - \left( {{{25} \over 2} - 25} \right)} \right] - {2 \over 3}{\left( 4 \right)^{{3 \over 2}}}

=

{1 \over 4}\left[ {{{93} \over 2} + {{25} \over 2}} \right] - {2 \over 3} \times 8

=

{1 \over 4} \times {{64} \over 2} - {{16} \over 3}

=

8 - {{16} \over 3}

=

{8 \over 3}

Q105

The area of the region A = {(x, y) : 0

\le

y

\le

x |x| + 1 and

-

1

\le

x

\le

1} in sq. units, is :

A

{2 \over 3}

B 2

C

{4 \over 3}

D

{1 \over 3}

Correct Answer

Option B

Solution

Required area

= \int\limits_{ - 1}^1 {\left( {x\left| x \right| + 1} \right)} dx

= 0 + \left( x \right)_{ - 1}^1 = 2

Q106

The area (in sq. units) of the region {x

\in

R : x

\ge

0, y

\ge

0, y

\ge

x

-

2 and y

\le

\sqrt x

}, is :

A

{{13} \over 3}

B

{{8} \over 3}

C

{{10} \over 3}

D

{{5} \over 3}

Correct Answer

Option C

Solution

y =

\sqrt x

y = x $-$ 2

\therefore\,\,\,

\sqrt x

= x $-$ 2 $\Rightarrow$

\,\,\,

x = x2 $-$ 4x + 4 x2 $-$ 5x + 4 = 0 x2 $-$ 4x $-$ x + 4 = 0 $\Rightarrow$

\,\,\,

x(x $-$ 4) $-$ (x $-$ 4) = 0 $\Rightarrow$

\,\,\,

(x $-$ 4) (x $-$ 1) = 0

\therefore\,\,\,

x = 4, 1 and y = 2, $-$ 1

\therefore\,\,\,

Their point of intersection (4, 2) and (1, $-$ 1) Required area is shown in the shaded figure.

\therefore\,\,\,

Required area =

\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} }

=

\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} }

=

\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4

=

{2 \over 3}\left( 8 \right)

+ 2

\left( {4 - 2} \right)

$-$

{1 \over 2}

(16 $-$ 4) =

{{16} \over 3}

+ 4 $-$ 6 =

{{10} \over 3}

Q107

The area (in sq. units) of the region A = { (x, y)

\in

R × R| 0

\le

x

\le

3, 0

\le

y

\le

4, y

\le

x2 + 3x} is :

A

{{59} \over 6}

B

{{26} \over 3}

C 8

D

{{53} \over 6}

Correct Answer

Option A

Solution

When y = 4 then, x2 + 3x = 4 $\Rightarrow$ x2 + 3x - 4 = 0 $\Rightarrow$ x2 + 4x - x - 4 = 0 $\Rightarrow$ x(x + 4) - (x + 4) = 0 $\Rightarrow$ (x + 4)(x - 1) = 0 $\Rightarrow$ x = 1, - 4 As 0 $\le$ x $\le$ 3, so possible value of x = 1. $\therefore$ y = x2 + 3x parabola cut the line y = 4 at x = 1.

Required area =

\int\limits_0^1 {\left( {{x^2} + 3x} \right)} dx

+ 2 $\times$ 4 =

\left[ {{{{x^3}} \over 3} + 3\left( {{{{x^2}} \over 2}} \right)} \right]_0^1

+ 8 =

{\left( {{1 \over 3} + {3 \over 2}} \right)}

+ 8 =

{{11} \over 6} + 8

=

{{59} \over 6}