Utilizing ML Aggarwal Class 12 ISC Solutions Chapter 3 Applications of Integrals MCQs as a study aid can enhance exam preparation.

## ML Aggarwal Class 12 Maths Solutions Section B Chapter 3 Applications of Integrals MCQs

Choose the correct answer from the given four options in questions (1 to 17) :

Question 1.

The area of the region (in square units) bounded by the curve x^{2} = 4y, line x = 2 and x-axis is

(a) 1

(b) 2/3

(c) 4/3

(d) 8/3

Solution:

(b) 2/3

Given eqn. of curve be x^{2} = 4y represents upward parabola with vertex (0, 0).

Now x = 2 meets x^{2} = 4y at y = 1

divide the shaded region into vertical strips with tower end on x-axis and upper end on given parabola.

The strip moves from x = 0 to x = 2.

∴ required area = \(\int_0^2\) y dx

= \(\int_0^2 \frac{x^2}{4}\) dx

= \(\left.\frac{x^3}{12}\right]_0^2\)

= \(\frac{8}{12}=\frac{2}{3}\) sq. units

Question 2.

The area of the region bounded by the curve y = x^{2} and the line y = 16 is

(a) \(\frac{32}{3}\)

(b) \(\frac{256}{3}\)

(c) \(\frac{64}{3}\)

(d) \(\frac{128}{3}\) [NCERT Exemplar]

Solution:

(b) \(\frac{256}{3}\)

Given eqn. of curve be y = x^{2} which represents an upward parabola with vertex (0, 0)

and y = 16 represents a line || to x-axis.

Further, both curves y = x^{2} and line y = 16

intersects when x^{2} = 16

i.e. x = ± 4

i.e. intersects at points (± 4, 16).

∴ Required area of region OCBO = 2 × area of region OABO

= 2 \(\int_0^4\) (16 – x^{2}) dx

= 2 \(\left[16 x-\frac{x^3}{3}\right]_0^1\)

= 2 [64 – \(\frac{64}{3}\)]

= \(\frac{256}{3}\) sq. units

Question 3.

The area (in sq. units) enclosed by the parabola y^{2} = 4ax and its latus-rectum is

(a) \(\frac{4}{3}\) a^{2}

(b) \(\frac{16}{3}\) a^{2}

(c) \(\frac{8}{3}\) a^{2}

(d) none of these

Solution:

(c) \(\frac{8}{3}\) a^{2}

eqn. of given parabola be y^{2} = 4ax

eqn. of latus-rectum be x = a

The end points of latus rectum are (a, 2a) and (a, – 2a)

∴ required area = 2 × Area of region OABA

= 2 \(\int_0^a\) 2√a√x dx

= 4√a \(\left.\frac{x^{3 / 2}}{3 / 2}\right]_0^a\)

= \(\frac{8}{3}\) √a × a^{3/2}

= \(\frac{8}{3}\) a^{2}

Question 4.

The area bounded by the parabola y^{2} = 8x, the x-axis and the latusrectum is

(a) \(\frac{16}{3}\)

(b) \(\frac{23}{3}\)

(c) \(\frac{32}{3}\)

(d) \(\frac{16 \sqrt{2}}{3}\)

Solution:

Given eqn. of parabola be y^{2} = 8x, which is right handed with vertex (0, 0).

On comparing with y^{2} = 4ax

⇒ 4a = 8

⇒ a = 2

∴ eqn. of latus rectum be x = a

i.e. x = 2

Now the line x = 2 intersects the parabola

y^{2} = 8x when y^{2} = 16

i.e y = ± 4

∴ points of intersection are (2, ± 4).

∴ required area = 2 × area of shaded region OABO divide the shaded region into vertical strips.

each vertical strip has lower end on x-axis and upper end on given parabola.

Clearly the rectangle moves from x = 0 to x = 2.

∴ required area = 2 \(\int_0^2 \sqrt{8} \sqrt{x}\)

= 2\(\left.\sqrt{8} \frac{x^{3 / 2}}{3 / 2}\right]_0^2\)

= \(\frac{4 \sqrt{8}}{3} \times 2^{3 / 2}\)

= \(\frac{32}{3}\) sq.unlts

Question 5.

The area of the region bounded by parabola y^{2} = x and the straight line 2y = x ¡s

(a) \(\frac{4}{3}\) sq units

(b) 1 sq units

(c) \(\frac{2}{3}\) sq units

(d) \(\frac{1}{3}\) sq units

Solution:

(a) \(\frac{4}{3}\) sq units

Given parabola y^{2} = x ………………..(1)

is a right handed parabola with vertex (0, 0)

and line 2y = x …………………..(2)

eqn. (1) and (2) intersects when

y^{2} = 2y

⇒ y (y – 2) = 0

⇒ y = o, 2

∴ From (2) ;

x = 0, 4

∴ points of intersection of both curves are (0, 0) and (4, 2)

Divide the region into vertical strips with lower end on line 2y = x

and upper end on parabola y^{2} = x.

∴ required area = \(\int_0^4\left[\sqrt{x}-\frac{x}{2}\right]\)

= \(\left[\frac{2}{3} x^{3 / 2}-\frac{x^2}{4}\right]_0^4\)

= \(\frac{16}{3}-4\)

= \(\frac{4}{3}\) sq. units

Question 6.

The area of the region (in sq. units) bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is

(a) \(\frac{7}{2}\)

(b) \(\frac{9}{2}\)

(c) \(\frac{11}{2}\)

(d) \(\frac{13}{2}\)

Solution:

(a) \(\frac{7}{2}\)

eqn. of given line be

y = x + 1

⇒ – x + y = 1

which meets coordinate axes are (- 1, 0) and (0, 1)

∴ required area of region ABCDA = \(\int_2^3\) (x + 1) dx

= \(\left.\frac{(x+1)^2}{2}\right]_2^3\)

= \(\frac{1}{2}\) [16 – 9]

= \(\frac{7}{2}\) sq. units

Question 7.

The area of the region bounded by the curve x = 2y + 3 and they lines y = 1 and y = – 1 is

(a) 4 sq. units

(b) \(\frac{3}{2}\) sq. units

(c) 6 sq. units

(d) 8 sq. units

Answer:

(c) 6 sq. units

Equation of given curve be x = 2y + 3 which represents a line meets x-axis at (3, 0) and

y-axis at (0, – \(\frac{3}{2}\))

Thus shaded region R is given by

R = {(x, y) ; 0 ≤ x ≤ 2y + 3 ; – 1 ≤ y ≤ 1}

∴ required area = \(\int_{-1}^1\) (2y + 3) dy

= \(\left.\frac{(2 y+3)^2}{2 \times 2}\right]_{-1}^1\)

= \(\frac{1}{4}\) [25 – 1]

= 6 sq. units.

Question 8.

The area bounded by the curve y = log_{e} x and x-axis and the straight line x = e is

(a) e sq. units

(b) 1 sq. units

(c) 1 – \(\frac{1}{e}\) sq.units

(d) 1 + \(\frac{1}{e}\) sq.units

Solution:

(b) 1 sq. units

Given eqn. of curve bey = log_{e} x

When x = e

⇒ y log_{e} e = 1

∴ required area = \(\int_1^e \log x \cdot 1\) dx

= [x log x – x\(]_1^e\)

= [e log e – e – 1 . log 1 + 1]

= 1 sq. units

Question 9.

If the area above the x-axis, bounded by the curves y = 2^{kx} and x = 0, and x = 2 is \(\frac{3}{\log _e 2}\) then the value of k is

(a) 1/2

(b) 1

(c) – 1

(d) 2

Solution:

(b) 1

Given eqn. of curve be y = 2^{kx}

When x = 0

⇒ y = 2^{0} = 1

When x = 2

⇒ y = 2^{2k}

Question 10.

If the area bounded by the curve y^{2} = 16x and the line y = mx is \(\frac{2}{3}\) sq. units, then the value of m is

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

(d) 4

eqns. of given curves are

y^{2} = 16x ………………(1)

and y = mx …………………(2)

eqn. (1) represents right handed parabola.

both curves (1) and (2) intersects

When m^{2}x^{2} = 16x

⇒ x(m^{2}x – 16) = 0

⇒ x = 0, \(\frac{16}{m^2}\)

∴ y = 0, \(\frac{16}{m}\)

Question 11.

The area (in sq. units) enclosed by the circle x^{2} + y^{2} is

(a) 4π

(b) 2√2π

(c) 2π

(d) 4π^{2}

Solution:

(c) 2π

eqn. of given circle be x^{2} + y^{2} = 2

Required area of circle = 4 × area of region OABO

= 4 × \(\int_0^{\sqrt{2}} \sqrt{2-x^2}\) dx

= 4 \(\left[\frac{x \sqrt{2-x^2}}{2}+\frac{2}{2} \sin ^{-1} \frac{x}{\sqrt{2}}\right]_0^{\sqrt{2}}\)

= 4 [0 + sin^{-1} 1 – 0 – 0]

= 4 × \(\frac{\pi}{2}\)

= 2π sq. units

Question 12.

The area enclosed by the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 is equal to

(a) π^{2} ab

(b) π ab

(c) π a^{2}b

(d) π ab^{2} [NCERT Exemplar]

Solution:

(b) π ab

Given eqn. of ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1

R = \(\left\{\begin{array}{r}

(x, y) ; \frac{-b}{a} \sqrt{a^2-x^2} \leq y \leq \frac{b}{a} \sqrt{a^2-x^2} \\

-a \leq x \leq a

\end{array}\right\}\)

Question 13.

The area (in sq. units) enclosed by the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1 is

(a) 20π

(b) 20π^{2}

(c) 16π^{2}

(d) 25π

Solution:

(a) 20π

Given eqn. of ellipse be

Question 14.

The area of the region bounded by the curve y = \(\sqrt{16-x^2}\) and x-axis is

(a) 8π sq. units

(b) 20π sq. units

(c) 16π sq. units

(d) 256π sq. units

Solution:

(a) 8π sq. units

Given equation of curve be

y = \(\sqrt{16-x^2}\)

which represents a circle with centre (0, 0) and radius 4.

∴ shaded region is given by

R = {(x, y) ; 0 ≤ y ≤ \(\sqrt{16-x^2}\) ; – 4 ≤ x ≤ 4}

Thus required area = \(\int_{-4}^4 \sqrt{16-x^2}\) dx

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

= \(\left.\frac{x \sqrt{16-x^2}}{2}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{-4}^4\)

= 0 + 8 sin^{-1} 1 – 0 – 8 sin^{-1} 1

= 16 sin^{-1} 1

= 16 × \(\frac{\pi}{2}\)

= 8π sq. units.

Question 15.

The area bounded by the curve y = sin x and the ordinates x = 0 and x = \(\frac{\pi}{2}\) is

(a) 2 sq. units

(b) 3 sq. units

(c) 4 sq. units

(d) 1 sq. units

Solution:

(d) 1 sq. units

Required area of region = \(\int_0^{\pi / 2}\) y dx

= \(\int_0^{\pi / 2}\) sin x dx

= – cos x\(]_0^{\pi / 2}\)

= – [cos \(\frac{\pi}{2}\) – cos 0]

= – (0 – 1)

= 1 sq. units

Question 16.

The area (in sq. units) bounded by the curve y = cos x and the x-axis between x = 0 and x = π is

(a) 1

(b) 2

(c) 4

(d) 6

Solution:

(b) 2

required area = area of region OABO + area of region AECA

= \(\int_0^{\pi / 2}|\cos x| d x+\int_{\pi / 2}^\pi|\cos x| d x\)

= \(\int_0^{\pi / 2}\) cos x dx – \(\int_{\pi / 2}^\pi\) cos x dx

= \(\left.\sin x]_0^{\pi / 2}-\sin x\right]_{\pi / 2}^\pi\)

= (1 – 0) – (0 – 1)

= 2 sq. units

Question 17.

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x ≤ \(\frac{\pi}{2}\) is

(a) 2 (√2 – 1)

(b) √2 – 1

(c) √2 + 1

(d) √2

Solution:

(b) √2 – 1

eqns. of both curves are;

y = cos x …………….(1)

and y = sin x ………………..(2)

Thus, both curves intersects

when sin x = cos x

⇒ tan x = 1

⇒ x = \(\frac{\pi}{4}\)

∴ From (1) ;

y = \(\frac{1}{\sqrt{2}}\)

Thus both curves intersects at \(\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)\)

Divide the region into vertical strips with lower end on eqn. (2) and upper end on curve y = cos x.

∴ required area = \(\int_0^{\pi / 4}\) [cos x – sin x] dx

= [sin x + cos x\(]_0^{\pi / 4}\)

= [\(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) – 0 – 1]

= [√2 – 1] sq. units