Area Under Curves — Page 5 | JEE Mathematics

Q41

Consider a region R = {(x, y)

\in

R : x2

\le

y

\le

2x}. if a line y =

\alpha

divides the area of region R into two equal parts, then which of the following is true?

A 3

\alpha

2 - 8

\alpha

+ 8 = 0

B

\alpha

3 - 6

\alpha

3/2 - 16 = 0

C 3

\alpha

2 - 8

\alpha

3/2 + 8 = 0

D

\alpha

3 - 6

\alpha

2 + 16 = 0

Correct Answer

Option C

Solution

y $\ge$ x2 $\Rightarrow$ upper region of y = x2 y $\le$ 2x $\Rightarrow$ lower region of y = 2x According to question, area of OABC = 2 $\times$ area of OAC $\Rightarrow$

\int\limits^{4}_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy

= 2

\int\limits^{\alpha }_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy

\Rightarrow \left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^4 = 2\left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^\alpha

\Rightarrow {{16} \over 3} - 4 = 2\left[ {{2 \over 3}{{\left( \alpha \right)}^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]

\Rightarrow {4 \over 3} = 2\left[ {{2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]

\Rightarrow {2 \over 3} = {2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}

\Rightarrow 8 = 8{\alpha ^{{3 \over 2}}} - 3{\alpha ^2}

\Rightarrow 3{\alpha ^2} - 8{\alpha ^{{3 \over 2}}} + 8 = 0

Q42

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

\in

R2 : x2

\le

y

\le

3 – 2x}, is :

A

{{34} \over 3}

B

{{29} \over 3}

C

{{31} \over 3}

D

{{32} \over 3}

Correct Answer

Option D

Solution

x2 $\le$ y $\le$ – 2x + 3 $\Rightarrow$ x2 = – 2x + 3 $\Rightarrow$ x2 + 2x – 3 = 0 $\Rightarrow$ (x + 3) (x – 1) = 0 $\Rightarrow$ x = – 3, x = 1 Area =

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

=

\left( { - {x^2} + 3x - {{{x^3}} \over 3}} \right)_{ - 3}^1

=

\left( { - 1 + 3 - {1 \over 3} + 9 + 9 - 9} \right)

= 11 -

{{1 \over 3}}

=

{{{32} \over 3}}

sq. unit

Q43

The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = x and the straight line y = x, is:

A

{1 \over 6}\left( {24\pi - 1} \right)

B

{1 \over 3}\left( {12\pi - 1} \right)

C

{1 \over 3}\left( {6\pi - 1} \right)

D

{1 \over 6}\left( {12\pi - 1} \right)

Correct Answer

Option D

Solution

Required area = Area of circle –

\int\limits_0^1 {\left( {\sqrt x - x} \right)dx}

= $\pi$ r2 -

\left( {{{2{x^{3/2}}} \over 3} - {{{x^2}} \over 2}} \right)_0^1

=

\pi {\left( {\sqrt 2 } \right)^2}

-

{1 \over 6}

=

{1 \over 6}\left( {12\pi - 1} \right)

Q44

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

\in

R2 | 4x2

\le

y

\le

8x + 12} is :

A

{{125} \over 3}

B

{{128} \over 3}

C

{{127} \over 3}

D

{{124} \over 3}

Correct Answer

Option B

Solution

For point of intersection 4x2 = 8x + 12 $\Rightarrow$ x2 - 2x - 3 = 0 $\Rightarrow$ x = –1, 3 Required area = area of the shaded region =

\int\limits_{ - 1}^3 {\left( {8x + 12 - 4{x^2}} \right)dx}

=

4\left[ {2.{{{x^2}} \over 2} + 3x - {{{x^3}} \over 3}} \right]_{ - 1}^3

= (36 + 36 – 36) – (4 – 12 +

{4 \over 3}

) =

{{128} \over 3}

Q45

For a > 0, let the curves C1 : y2 = ax and C2 : x2 = ay intersect at origin O and a point P. Let the line x = b (0 < b < a) intersect the chord OP and the x-axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, C1 and C2, and the area of

\Delta

OQR =

{1 \over 2}

, then 'a' satisfies the equation :

A x6 – 12x3 + 4 = 0

B x6 – 12x3 – 4 = 0

C x6 + 6x3 – 4 = 0

D x6 – 6x3 + 4 = 0

Correct Answer

Option A

Solution

C1 : y2 = ax, C2 : x2 = ay (a > 0) P : intersection point of y2 = ax and x2 = ay $\Rightarrow$ x4 = a2 y2 $\Rightarrow$ x4 = a2ax $\Rightarrow$ x = a, y = a $\therefore$ Point P : (a, a) Line OP : y = x $\Rightarrow$ Point Q = (b, b) Area

\Delta

OQR =

{1 \over 2}

$\Rightarrow$

{1 \over 2} \times b \times b

=

{1 \over 2}

$\Rightarrow$ b = 1 As line x = b bisect the area between curve $\therefore$

{1 \over 2}\int\limits_0^a {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx

=

\int\limits_0^1 {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx

$\Rightarrow$

\left[ {{1 \over 2}\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {1 \over 2}{{{x^3}} \over {3a}}} \right]_0^a

=

\left[ {\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {{{x^3}} \over {3a}}} \right]_0^1

$\Rightarrow$

{{{a^2}} \over 3} - {1 \over 2}{{{a^2}} \over 3}

- 0 =

{{2\sqrt a } \over 3} - {1 \over {3a}}

- 0 $\Rightarrow$

{{{a^2}} \over 2} + {1 \over a} = 2\sqrt a

$\Rightarrow$

{a^3} + 2 = 4a\sqrt a

Squareing both sides, we get $\Rightarrow$

{a^6} + 4{a^3} + 4 = 16{a^3}

$\Rightarrow$

{a^6} - 12{a^3} + 4 =

0 Hence

a

satisfy x6 – 12x3 + 4 = 0.

Q46

The area (in sq. units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is :

A

12\pi - 3\sqrt 3

B

24\pi + 3\sqrt 3

C

24\pi - 3\sqrt 3

D

12\pi + 3\sqrt 3

Correct Answer

Option C

Solution

{x^2} + {y^2} = 36

and

{y^2} = 9x

$\therefore$

{x^2} + 9x - 36 = 0

\Rightarrow x = 3, - 12

Required Area,

A = \pi {{{(6)}^2}} - 2\left[ {{A_1} + {A_2}} \right]

A = \pi {{{(6)}^2} - 2\left[ {\int_0^3 {\sqrt {9x} dx + \int_3^6 {\sqrt {36 - {x^2}} } dx} } \right]}

= 36\pi -

2\left[ {\left[ {3 \times {2 \over 3}{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + {{36} \over 2}{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]

= 36\pi -

2\left[ {\left[ {2{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + 18{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]

= 36\pi -

2\left[ {\left[ {6\sqrt 3 - 0} \right] + \left[ {\left( {0 + 18{{\sin }^{ - 1}}{6 \over 6}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + 18{{\sin }^{ - 1}}{3 \over 6}} \right)} \right]} \right]

= 36\pi -

2\left[ {\left[ {6\sqrt 3 } \right] + \left[ {\left( {{{18\pi } \over 2}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + {{18\pi } \over 6}} \right)} \right]} \right]

= 36\pi -

\left( {12\sqrt 3 + 18\pi - 9\sqrt 3 - 6\pi } \right)

=

24\pi - 3\sqrt 3

Note : (i)

\int {\sqrt {{x^2} + {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} + {a^2}} + {a^2}\log |x + \sqrt {{x^2} + {a^2}} |} \right]} + C

(ii)

\int {\sqrt {{a^2} - {x^2}} dx = {1 \over 2}\left[ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\left( {{x \over a}} \right)} \right]} + C

(iii)

\int {\sqrt {{x^2} - {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} - {a^2}} - {a^2}\log |x + \sqrt {{x^2} - {a^2}} |} \right]} + C

Q47

The area of the region :

R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\}

is :

A

6\sqrt 3

square units

B

12\sqrt 3

square units

C

11\sqrt 3

square units

D

9\sqrt 3

square units

Correct Answer

Option B

Solution

Required area

= 2\int\limits_0^{\sqrt 3 } {\left( {2{x^2} + 9 - 5{x^2}} \right)} dx

= 2\int\limits_0^{\sqrt 3 } {\left( {9 - 3{x^2}} \right)dx}

= 2|\,9x - {x^3}\,|_0^{\sqrt 3 } = 12\sqrt 3

Q48

The area bounded by the curve 4y2 = x2(4

-

x)(x

-

2) is equal to :

A

{\pi \over {16}}

B

{\pi \over {8}}

C

{3\pi \over {2}}

D

{3\pi \over {8}}

Correct Answer

Option C

Solution

Given, 4y2 = x2(4 $-$ x)(x $-$ 2) .....

(1) Here, Left hand side 4y2 is always positive.

So Right hand side should also be positive.

In x

\in

[2, 4] Right hand side is positive.

By putting y = $-$ y in equation (1), equation remains same.

So, graph is symmetric about x axis.

Required Area = 2A1

= 2\int_2^4 y dx

= \int_2^4 {x\sqrt {(x - 2)(4 - x)} } dx

put

x = 4{\sin ^2}\theta + 2{\cos ^2}\theta

\Rightarrow dx = \left[ {4(2\sin \theta \cos \theta - 4\sin \theta \cos \theta )} \right]d\theta

\Rightarrow dx = 4\sin \theta \cos \theta d\theta

When lower limit = 2 then

2 = 4{\sin ^2}\theta + 2{\cos ^2}\theta

\Rightarrow 4(1 - {\cos ^2}\theta ) + 2{\cos ^2}\theta = 2

\Rightarrow 4 - 4{\cos ^2}\theta + 2{\cos ^2}\theta = 2

\Rightarrow 2{\cos ^2}\theta = 2

\Rightarrow \cos \theta = \pm 1

\Rightarrow \theta = 0,\pi {}

When upper limit = 4 then

4 = 4{\sin ^2}\theta + 2{\cos ^2}\theta

\Rightarrow - 2{\cos ^2}\theta = 0

\Rightarrow \theta = {{\pi {} } \over 2}

$\therefore$ Range of $\theta$ = 0 to

{{{\pi {} } \over 2}}

$\therefore$ Area =

\int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )\sqrt {(2{{\sin }^2}\theta )(2{{\cos }^2}\theta )} (4\sin \theta \cos \theta )d\theta }

= \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )8{{\sin }^2}\theta {{\cos }^2}\theta d\theta }

= 32\int_0^{{{\pi {} } \over 2}} {{{\sin }^4}\theta {{\cos }^2}\theta d\theta } + 16\int_0^{{{\pi {} } \over 2}} {{{\sin }^2}\theta {{\cos }^2}\theta d\theta }

Using Wallis formula,

= 32.{{3.1.1} \over {6.4.2}}.{{\pi {} } \over 2} + 16.{{1.3.1} \over {6.4.2}}.{{\pi {} } \over 2}

= \pi {} + {{\pi {} } \over 2}

= {{3\pi {} } \over 2}

Q49

If the area of the bounded region

R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}

is ,

\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma

, then the value of

{(\alpha + \beta - 2\lambda )^2}

is equal to :

A 8

B 2

C 4

D 1

Correct Answer

Option B

Solution

R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}

\int\limits_{{1 \over 2}}^2 {{2^x}dx} - \int\limits_1^2 {\ln xdx}

\Rightarrow \left[ {{{{2^x}} \over {\ln 2}}} \right]_{1/2}^2 - [x\ln x - x]_1^2

\Rightarrow {{({2^2}) - {2^{1/2}}} \over {{{\log }_e}2}} - (2\ln 2 - 1)

\Rightarrow {{\left( {{2^2} - \sqrt 2 } \right)} \over {{{\log }_e}2}} - 2\ln 2 + 1

$\therefore$

\alpha = {2^2} - \sqrt 2

,

\beta = - 2

,

\gamma = 1

\therefore {(\alpha + \beta + 2\gamma )^2}

= {({2^2} - \sqrt 2 - 2 - 2)^2}

= {(\sqrt 2 )^2} = 2

Q50

The area of the region bounded by the parabola (y

-

2)2 = (x

-

1), the tangent to it at the point whose ordinate is 3 and the x-axis is :

A 9

B 10

C 4

D 6

Correct Answer

Option A

Solution

y = 3 $\Rightarrow$ x = 2 Point is (2, 3) Diff. w.r.t x 2 (y $-$ 2) y' = 1 $\Rightarrow$

y' = {1 \over {2(y - 2)}}

\Rightarrow y{'_{(2,3)}} = {1 \over 2}

\Rightarrow {{y - 3} \over {x - 2}} = {1 \over 2} \Rightarrow x - 2y + 4 = 0

Area

= \int\limits_0^3 {\left( {{{(y - 2)}^2} + 1 - (2y - 4)} \right)} \,dy

= 9 sq. units