MCQ Questions for Class 12 Maths Chapter 8 Application of Integrals

Application of Integrals Class 12 MCQs Questions with Answers

MCQ On Application Of Integration Class 12 Chapter 8 Question 1.
The area of the y = cos x and y = sin x, 0 ≤ x ≤ \(\frac {π}{2}\) is
(A) \(\sqrt{2}\) sq units
(B) \((\sqrt{2}+1)\) sq units
(C) \((\sqrt{2}-1)\) sq units
(D) \((2 \sqrt{2}-1)\)
Answer:
(C) \((\sqrt{2}-1)\) sq units

Explanation:
We have y = cos x and y = sin x
Where o ≤ x ≤ \(\frac {π}{2}\)
we get cos x = sin π
⇒ x = \(\frac {π}{4}\)
From the figure, area of the shaded region,

A = \(\int_{0}^{\pi / 4}(\cos x+\sin x) d x\)
= \([\sin x+\cos x]_{0}^{\pi / 4}\)
= \(\left[\sin \frac{\pi}{4}+\cos \frac{\pi}{4}-\sin 0-\cos 0\right]\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
= \((\sqrt{2}-1)\) sq units

Application Of Integrals Class 12 MCQ Chapter 8 Question 2.
The area of the region bounded by the curve x² = 4y and the straight-line x = 4y – 2 is
(A) \(\frac{3}{8}\) sq units
(B) \(\frac{5}{8}\) sq units
(C) \(\frac{7}{8}\) sq units
(D) \(\frac{9}{8}\) sq units
Answer:
(D) \(\frac{9}{8}\) sq units

Explanation:
x² = x + 2
x² – x – 2 = 0
(x – 2)(x + 1) = 0
x = -1, 2
For x = -1, y = \(\frac{1}{4}\) and for x = 2, y = 1
Points of intersection are (-1, \(\frac{1}{4}\)) and (2, 1).
Graphs of parabola x²=4y and x =4y – 2 are shown in the following figure:

A = \(\int_{-1}^{2}\left[\frac{x+2}{4}-\frac{x^{2}}{4}\right] d x\)
= \(\frac{1}{4}\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{2}\)
= \(\frac{1}{4}\left[8-\frac{1}{2}-3\right]\)
= \(\frac{1}{4}\left[8-\frac{1}{2}-3\right]\)
= \(\frac{9}{8}\) sq units

Application Of Integration MCQ Maths Class 12 Chapter 8 Question 3.
Area of the region in the first quadrant endosed by the x-axis, the line y = x and the circle x² + y² = 32 is
(A) 16π sq units
(B) 4π sq units
(C) 32π sq units
(D) 24π sq units
Answer:
(B) 4π sq units

Explanation:
We have y = 0,y = x and the circle
x² + y² = 32 in the frist quadrant.
Solving y = x with thedrcle
x² + x² = 32
x² = 16
x = 4 (In the first quadrant)
When x = 4, y = 4 for the point of intersection of the circle with the x-axis.
Put y = 0
x² + 0 = 32
x = \(\pm 4 \sqrt{2}\)
So, the drde intersects the x-axis. at (\(\pm 4 \sqrt{2}\), 0).

From the above figure, area oi the shaded region,
A = \(\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}} \sqrt{(4 \sqrt{2})^{2}-x^{2}} d x\)
= \(\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{x}{2} \sqrt{(4 \sqrt{2})^{2}-x^{2}}+\frac{(4 \sqrt{2})^{2}}{2} \sin ^{-1} \frac{x}{4 \sqrt{2}}\right]_{4}^{4 \sqrt{2}}\)
= \(\left[\frac{16}{2}\right]+\left[\begin{array}{l}
0+16 \sin ^{-1} 1-\frac{4}{2} \sqrt{(4 \sqrt{2})^{2}-16^{2}} \\
-16 \sin ^{-1} \frac{4}{4 \sqrt{2}}
\end{array}\right]\)
= 8 + \(\left[\frac{16 \pi}{2}-2 \sqrt{16}-16 \frac{\pi}{4}\right]\)
= 8 + [8π – 8 – 4π]
= 4π sq units

Application Of Integrals MCQ Class 12 Chapter 8 Question 4.
Area of the region bounded by the curve y = cos x between x = 0 and x = π is
(A) 2 sq units
(B) 4 sq units
(C) 3 sq units
(D) 1 sq unit
Answer:
(A) 2 sq units

Explanation:
We have y = cos x, x = 0, x = π

From the figure, area of the shaded region,
A = \(\int_{0}^{\pi}|\cos x| d x+\int_{0}^{\pi / 2} \cos x d x\)
A = \(2[\sin x]_{0}^{\pi / 2}\)
= 2 sq units

Application Of Integration MCQ Maths Class 12 Question 5.
The area of the region bounded by parabola y² = x and the straight line 2y = x is
(A) sq units
(B) 1 sq unit
(C) sq unit
(D) sq unit
Answer:
(A) sq units

Explanation:
When y = x and 2y = x
Solving we get y’= 2y
= y = 0, 2 and when y = 2,x = 4
So, points of intersection are (0, 0) and (4,2).
Graphs of parabola 9 = x and 2y = x are as shown in the following figure:

From the figure, area of the shaded region,
A = \(\int_{0}^{4}\left[\sqrt{x}-\frac{x}{2}\right] d x\)
= \(\left[\frac{2}{3} x^{3 / 2}-\frac{1}{2} \cdot \frac{x^{2}}{2}\right]_{0}^{4}\)
= \(\frac{2}{3} \cdot(4)^{3 / 2}-\frac{16}{4}-0\)
= \(\frac{16}{3}-4\)
= \(\frac{4}{3}\) sq units

Question 6.
The area of the region bounded by the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is
(A) 20 π sq units
(B) 20 π sq units
(C) 16 π sq units
(D) 25 π sq units
Answer:
(A) 20 π sq units

Explanation:
We have \(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}\) = 1, which is ellipse with its axes as coordinate axes.

\(\frac{y^{2}}{4^{2}}=1-\frac{x^{2}}{5^{2}}\)
y² = \(y^{2}=16\left(1-\frac{x^{2}}{25}\right)\)
y = \(\frac{4}{5} \sqrt{5^{2}-x^{2}}\)
From the figure, area of the shaded region,
A = \(4 \int_{0}^{5} \frac{4}{5} \sqrt{5^{2}-x^{2}} d x\)
= \(\frac{16}{5}\left[\frac{x}{2} \sqrt{5^{2}-x^{2}}-\frac{5^{2}}{2} \sin ^{-1} \frac{x}{5}\right]_{0}^{5}\)
= \(\frac{16}{5}\left[0+\frac{5^{2}}{2} \sin ^{-1} 1-0-0\right]\)
= \(\frac{16}{5} \cdot \frac{25}{2} \cdot \frac{\pi}{2}\)
= 20 π sq units

Question 7.
The area of the region bounded by the circle x² + y² = 1 is
(A) 2π sq units
(B) 7π sq units
(C) 3π sq units
(D) 4π sq units
Answer:
(B) 7π sq units

Explanation:
We have, x² + y² = I, which is a arde having centre at (0, 0) and radius ‘1’ unIt.
⇒ y² = 1 – x²
y = \(\sqrt{1-x^{2}}\)

From the figure, area of the shaded region,
A = \(4 \int_{0}^{1} \sqrt{1^{2}-x^{2}} d x\)
= \(4\left[\frac{x}{2} \sqrt{1^{2}-x^{2}}-\frac{1^{2}}{2} \sin ^{-1} \frac{x}{1}\right]_{0}^{1}\)
= \(4\left[0+\frac{1^{2}}{2} \times \frac{\pi}{2}-0-0\right]\)
= π sq units

Question 8.
The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is
(A) \(\frac {7}{2}\) sq units
(B) \(\frac {9}{2}\) sq units
(C) \(\frac {11}{2}\) sq units
(D) \(\frac {13}{2}\)sq units
= \(\int_{0}^{3} \frac{y^{2}}{4} d y\)
= \(\frac{1}{4}\left[\frac{y^{3}}{4}\right]_{0}^{3}\)
= \(\frac{1}{12} \times 27\)
= \(\frac {9}{4}\) sq units
Answer:
(A) \(\frac {7}{2}\) sq units

Explanation:

From the figure, area of the shaded region,
A = \(\int_{2}^{3}(x+1) d x\)
= \(\left[\frac{x^{2}}{2}+x\right]_{2}^{3}\)
= \(\left[\frac{9}{2}+3-\frac{4}{2}-2\right]\)
= \(\frac {7}{2}\) sq units

Question 9.
The area of the region bounded by the curve x = 2y + 3 and the y lines y = 1 and y = -1 is,
(A) 4 sq units
(B) \(\frac {3}{2}\)
(C) 6 sq units
(D) 8 sq units
Answer:
(C) 6 sq units

Explanation:

From the figure, area of the shaded region,
A = \(\int_{-1}^{1}(2 y+3) d y\)
= \(\left[y^{2}+3 y\right]_{-1}^{1}\)
= [ 1 + 3 – 1 + 3]
= 6 sq units

Question 10.
Area lying in the first quadrant and bounded by circic x² + y² = 4 and the lines x = 0 and x = 2 is
(A) π
(B) \(\frac {π}{2}\)
(C) \(\frac {π}{3}\)
(D) \(\frac {π}{4}\)
Answer:
(A) π

Explanation:
The area bounded by the circle and the lines in the first quadrant is represented as:

A = \(\int_{0}^{2} y d x\)
= \(\int_{0}^{2} \sqrt{4-x^{2}} d x\)
= \(\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}\)
= π sq units

Question 11.
Area of the region hounded by the curve 2 = 4x, ii-axis and the line y = 3 is
(A) 2
(B) \(\frac {9}{4}\)
(C) \(\frac {9}{3}\)
(D) \(\frac {9}{2}\)
Answer:
(B) \(\frac {9}{4}\)

Explanation:
The area bounded by the curve, y² = 4x, y-axis, and y = 3 is represented as:

Area Of OAB = \(\int_{0}^{3} x d y\)
= \(\int_{0}^{3} \frac{y^{2}}{4} d y\)
= \(\frac{1}{4}\left[\frac{y^{3}}{4}\right]_{0}^{3}\)
= \(\frac{1}{12} \times 27\)
= \(\frac {9}{4}\)

Question 12.
Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2
(A) 2(π – 2)
(B) π – 2
(C) 2,π – 1
(D) 2(π + 2)
Answer:
(B) π – 2

Explanation:
The smaller area enclosed by the circle x² + y² = 4 and the line, x + y = 2 is represented by the shaded area ACBA as:

It can be observed that
Area of ACBA = Area of OACBO – Area of MOB
A = \(\int_{0}^{2} \sqrt{4-x^{2}} d x-\int_{0}^{2}(2-x) d x\)
= \(\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}\) – \(\left[2 x-\frac{x}{2}\right]_{0}^{2}\)
= \(\left[2 \times \frac{\pi}{2}\right]-[4-2]\)
= π – 2 sq units

Question 13.
Area lying between the curve y² = 4x and y = 2x
(A) \(\frac {2}{3}\)
(B) \(\frac {1}{3}\)
(C) \(\frac {1}{4}\)
(D) \(\frac {3}{4}\)
Answer:
(B) \(\frac {1}{3}\)

Explanation:
The area lying between the curve y² = 4x and y = 2x is represented by the shaded area OBAO as

The points of intersection of the curves are 0(0, 0) and A(1, 2).
We draw AC perpendicular to x-axis such that coordinate of C is (1,0).
Area of OBAO = Area of SOCA – Area of OCABO
A = \(\int_{0}^{1} 2 x d x-\int_{0}^{1} 2 \sqrt{x} d x\)
= \(2\left[\frac{x^{2}}{2}\right]_{0}^{1}-2\left[\frac{x^{3 / 2}}{\frac{3}{2}}\right]_{0}^{1}\)
= \(\left[\left[1-\frac{4}{3}\right]\right]\)
= \(\left|-\frac{1}{3}\right|\)
= \(\frac {1}{3}\) sq unit

Question 14.
Area bounded by the curve y = x³, the x-axis and the ordinates x = -2 and x = 1 is
(A) – 9
(B) – \(\frac {15}{3}\)
(C) \(\frac {15}{3}\)
(D) \(\frac {17}{4}\)
Answer:
(C) \(\frac {15}{3}\)

Explanation:
Required area,
A = \(\int_{-2}^{1} y d x\)
= \(\int_{-2}^{1} x^{3} d x\)

= \(\left[\frac{x^{4}}{4}\right]_{-2}^{1}\)
= \(\left(\frac{1}{4}-4\right)\)
= \(-\frac{15}{4}\)
∴ Area = \(\left|-\frac{15}{4}\right|\)
= \(\frac{15}{4}\)

Question 15.
The area of the circle x² + y² = 16 exterior to the parabola y² = 6x is
(A) \(\frac{4}{3}(4 \pi-\sqrt{3})\)
(B) \(\frac{4}{3}(4 \pi+\sqrt{3})\)
(C) \(\frac{4}{3}(8 \pi-\sqrt{3})\)
(D) \(\frac{4}{3}(8 \pi+\sqrt{3})\)
Answer:
(C) \(\frac{4}{3}(8 \pi-\sqrt{3})\)

Explanation:

Area bounded by the cirde and parabola
= 2[(area (OAMO) + area (AMBA)]
= 2\(\left[\int_{0}^{2} \sqrt{6 x} d x+\int_{2}^{4} \sqrt{16-x^{2}} d x\right]\)
= 2\(\int_{0}^{2} \sqrt{6 x} d x+2 \int_{2}^{4} \sqrt{16-x^{2}} d x\)
= \(2 \sqrt{6} \int_{0}^{2} \sqrt{x} d x+2 \int_{2}^{4} \sqrt{16-x^{2}} d x\)
= \(2 \sqrt{6} \times \frac{2}{3}\left[x^{3 / 2}\right]_{0}^{2}+2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1}\left(\frac{x}{4}\right)\right]_{2}^{4}\)
= \(2\left[\left\{0+8 \sin ^{-1}(1)\right\}-\left\{2 \sqrt{3}+8 \sin ^{-1}\left(\frac{1}{2}\right)\right\}\right]\)
= \(\frac{16 \sqrt{3}}{3}+2\left[8 \times \frac{\pi}{2}-2 \sqrt{3}-8 \times \frac{\pi}{6}\right]\)
= \(\frac{16 \sqrt{3}}{3}+2\left(4 \pi-2 \sqrt{3}-\frac{4 \pi}{3}\right)\)
= \(\frac{16 \sqrt{3}}{3}+8 \pi-4 \sqrt{3}-\frac{8 \pi}{3}\)
= \(\frac{16 \sqrt{3}+24 \pi-4 \sqrt{3}-8 \pi}{3}\)
= \(\frac{16 \pi+12 \sqrt{3}}{3}\)
= \(\frac{4}{3}[4 \pi+\sqrt{3}]\)

Area of circle = π(r)²
= π(4)² = 16π sq unit

∴ Required area = \(16 \pi-\frac{4}{3}(4 \pi+\sqrt{3})\)
= \(16 \pi-\frac{16 \pi}{3}-\frac{4 \sqrt{3}}{3}\)
= \(\frac{32 \pi}{3}-\frac{4 \sqrt{3}}{3}\)
= \(\frac{4}{3}[8 \pi-\sqrt{3}]\)sq unit

Assertion And Reason Based MCQs (1 Mark each)

Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R ¡s True

Question 1.
Assertion (A):

The area of region PQML = \(\int_{a}^{b} y d x=\int_{a}^{b} f(x) d x\)
Reason (R):

The area A of the region bounded by curve x = g(,y), y-axis and the lines y = c and y = d is given by
A = \(\int_{c}^{d} x d y\)
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Assertion (A) and Reason (R) both are individually correct.

Question 2.
Assertion (A):

Area = \(\left|\int_{a}^{b} f(x) d x\right|\)
Reason (R): If the curve under consideration lies below x-axis, then f(x) < 0 from x = a to x = b, the area bounded by the curve y = f(x) and the ordinates x = a, x = b and x-axis is negative. But, if the numerical value of the area is to be taken into consideration, then
Area = \(\left|\int_{a}^{b} f(x) d x\right|\)
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Assertion (A) and Reason (R) both are correct, Reason (R) is the correct explanation of Assertion (A).

Question 3.
Assertion (A):

Reason (R):
It may happen that some portion of the curve is above the x-axis and some portion is below the x-axis as shown in the figure. Let A₁ be the area below the x-axis and A₂ be the area above the x-axis. Therefore, area bounded by the curve y = f(x), x-axis and the ordinates x = a and x = b is given by
Area = \(\left|A_{1}\right|+\left|A_{2}\right|\)
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Assertion (A) and Reason (R) both are correct, Reason (R) is the correct explanation of Assertion (A).

Question 4.
Assertion (A): The area enclosed by the circle x² + y² = a² is πa²

Reason (R): The area enclosed by the circle
= 4\(\int_{0}^{a} x d y\)
=4\(\int_{0}^{a} \sqrt{a^{2}-y^{2}} d y\)
=4\(\left[\frac{y}{2} \sqrt{a^{2}-y^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{y}{a}\right]_{0}^{a}\)
=4\(\left[\left(\frac{a}{2} \times 0+\frac{a^{2}}{2} \sin ^{-1} 1\right)-0\right]\)
= 4\(\frac{a^{2}}{2} \frac{\pi}{2}\)
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Assertion (A) and Reason (R) both are correct Reason (R) is the correct explanation of Assertion (A)

Question 5.
Assertion (A): The area of the region bounded by the curvey = x² and the line y = 4 is \(\frac {3}{32}\)
Reason (R):

Since the given curve represented by the equation y = x² is a parabola symmetrical about y-axis only, therefore, from figure, the required area of the region AOBA is given by
A = 2\(2 \int_{0}^{4} x d y\)
= 2\(2 \int_{0}^{4} \sqrt{y} d y\)
= 2\(2 \times \frac{2}{3}\left[y^{3 / 2}\right]_{0}^{4}\)
= \(\frac{4}{3} \times 8\)
= \(\frac {3}{32}\)
Answer:
(D) A is false and R ¡s True

Explanation:
Assertion (A) is wrong. Reason (R) is the correct solution of Assertion (A).

Question 6.
Assertion (A): if the two curves y = f(x) and y = g(x) intersect at x = a, x = c and x = b, such that a < b < c

if f(x) > g(x) in [a, c] and g(x) ≤ f(x) in [c, b], then Area of the regions bounded by the curve
= Area of region PACQP + Area of region QDRBQuestion
= \(\int_{a}^{c}|f(x)-g(x)| d x+\int_{c}^{b}|g(x)-f(x)| d x\)

Reason (R):
Let the two curves by y = f(x) and y = as shown in the figure. Suppose these curves intersect at f(x) with width dx.

Area = \(\int_{a}^{b}[f(x)-g(x)] d x\)
= \(\int_{a}^{b} f(x) d x-\int_{a}^{b} g(x) d x\)
= Area bounded by the curve {y = f(x)} – Area bounded by the curve {y = g(x)}
where f(x) > g(x).
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Assertion (A) and Reason (R) both are individually correct.

Case-Based MCQs

Attempt any four sub-parts from each question.
Each sub-part carries 1 mark.

I. Read the following text and answer the following
questions on the basis of the same:

The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road in the middle of the bridge as seen in the figure.

Question 1.
The equation of the parabola designed on the bridge is
(A) x²= 250 y
(B) x² = -250 y
(C) y² = 250 x
(D) y² = 250 y
Answer:
(C) y² = 250 x

Question 2.
The value of the integral \(\int_{-50}^{50} \frac{x^{2}}{250}\)
(A) \(\frac {1000}{3}\)
(B) \(\frac {250}{3}\)
(C) 1200
(D) 0
Answer:
(A) \(\frac {1000}{3}\)

Explanation:
= \(\int_{-50}^{50} \frac{x^{2}}{250}=\frac{1}{250}\left[\frac{x^{3}}{3}\right]_{-50}^{50}\)
= \(\frac{1}{250} \times \frac{1}{3}\left[(50)^{3}-(-50)^{3}\right]\)
= \(\frac{1}{750}[125000+125000]\)
= \(\frac {100}{3}\).

Question 3.
The integrand of the integral \(\int_{-50}^{50} x^{2}\) is …………. function.
(A) Even
(B) Odd
(C) Neither odd nor even
(D) None of these
Answer:
(A) Even

Explanation:
f(x) = x²
f(-x) = x²
∴ f(x) is even function.

Question 4.
The area formed by the curve x² = 250y, x – axis,
y = 0 and y = 10 is
(A) \(\frac{1000 \sqrt{2}}{3}\)
(B) \(\frac{4}{3}\)
(C) \(\frac{1000}{3}\)
(D) 0
Answer:
(C) \(\frac{1000}{3}\)

Explanation:
x² = 250 y
y = \(\frac{1}{250}\)x²
at y = 0 x = 0
at y = 10 x = 50,-50
∴ Area formed by curve
= \(\int_{-50}^{50} \frac{1}{250} x^{2} d x\)
= \(\frac{1}{250} \times \frac{1}{3}\left[x^{3}\right]_{0}^{50}\)
= \(\frac{1}{750}[250,000]\)
= \(\frac{1000}{3}\) sq.units

Question 5.
The area formed between x² = 250y, y-axis, y = 2 and y = 4is
(A) \(\frac{1000}{3}\)
(B) 0
(C) \(\frac{1000 \sqrt{2}}{3}\)
(D) None of these
Answer:
(D) None of these

II. Read the following text and answer the following
questions on the basis of the same:

In figure 0(0, 0) is the center of the circle. The line y = x meets the circle in the first quadrant at point B.

Question 1.
The equation of the circle is …………..
(A) x ² + 9 = 4\(\sqrt{2}\)
(B) x² + y = 16
(C) x² + y² = 32
(D) (x² – 4\(\sqrt{2}\))² +0
Answer:
(C) x² + y² = 32

Explanation:
Centre = (0,0),
r = 4\(\sqrt{2}\)
Equation of circle is
x² + y² = (4)²
x² + y² = 32

Question 2.
The co-ordinates of B are ………….
(A) (1,1)
(B) (2,2)
(C) (4\(\sqrt{2}\), 4 \(\sqrt{2}\);)
(D) (4, 4)
Answer:
(D) (4, 4)

Explanation:
x² + y² = 32 …….(i)
y = x ……..(ii)
Solving (i) and (ii),
⇒ x² + y² = 32
⇒ x² = 16
⇒ x = 4
⇒ y = x = 4
∴B = (4,4)

Question 3.
Area of ∆OBM is ………….. sq. units
(A) 8
(B) 16
(C) 32
(D) 32π
Answer:
(A) 8

Explanation:
Ar (∆OBM) = \(\int_{0}^{4} x d x\)
= \(\left[\frac{x^{2}}{2}\right]_{0}^{4}\)
8 sq. units

Question 4.
Ar (BAMB) = …………… sq. units
(A) 32 π
(B) 4 π
(C) 8
(D) 4π – 8
Answer:
(D) 4π – 8

Explanation:
Ar (BAMB) = \(\int_{4}^{4 \sqrt{2}} \sqrt{32-x^{2} d x}\)
= \(\left[\frac{x}{2} \sqrt{32-x^{2}}+16 \sin ^{-1} \frac{x}{4 \sqrt{2}}\right]_{4}^{4 \sqrt{2}}\)
= (4π – 8) sq. units.

Question 5.
Area of the shaded region is ………… sq. units.
(A) 32π
(B) 4π
(C) 8
(D) 4π – 8
Answer:
(B) 4π

Explanation:
Area of shaded region
= Ar (∆OBM) + Ar(BAMB)
= 8 + 4π – 8
= 4π sq. units

III. Read the following text and answer the following
questions on the basis of the same:

A farmer has a square plot of land. Three of its boundaries are x = 0, y = 0 and y = 4. He wants to divide this land among his three sons A, B and C as shown in figure.

Question 1.
The equation of PQ is …………..
(A) x = 0
(C) x = 4
(B) x = 2
(d) y = 4
Answer:
Option (C) is correct.

Explanation:
Equation of PQ is x = 4.

Question 2.
The co-ordinates of Q are ………..
(A) (2,2)
(B) (4,4)
(C) (1, 1)
(D) (5, 5)
Answer:
(B) (4,4)

Explanation:
Q = (4,4)

Question 3.
Area received by son B is …………. sq. units.
(A) 4
(B) 16
(C) \(\frac{16}{3}\)
(D) \(\frac{8}{3}\)
Answer:
(C) \(\frac{16}{3}\)

Explanation:
Ar (son B) = \(\int_{0}^{4} x d y\)
= \(\int_{0}^{4} \frac{y^{2}}{4} d y\)
= \(\left[\frac{y^{3}}{12}\right]_{0}^{4}\)
= \(\frac{1}{12}\left[4^{3}-0\right]\)
= \(\frac{64}{12}\)
= \(\frac{16}{3}\) sq. units.

Question 4.
Area received by son A is ______ sq. units.
(A) 4
(B) 16
(C) \(\frac{16}{3}\)
(D) \(\frac{8}{3}\)
Answer:
(C) \(\frac{16}{3}\)

Explanation:
Ar(son A) = \(\int_{0}^{4} y d x\)
= \(\int_{0}^{4} \frac{x^{2}}{4} d x\)
= \(\frac{1}{12}\left[x^{3}\right]_{0}^{4}\)
= \(\frac{16}{3}\) sq. units

Question 5.
The total area of the square field is …………. sq. units.
(A) 4’
(B) 16
(C) \(\frac{16}{3}\)
(D) \(\frac{8}{3}\)
Answer:
(B) 16

Explanation:
Total area = 4 x 4 = 16 sq.units

MCQ Questions for Class 12 Maths with Answers