MCQ Questions

Liner Programming Class 12 MCQs Questions with Answers

Linear Programming Class 12 MCQ Chapter 12 Question 1.
The corner points of the feasible region determined by the system of linear constraints are (0,0), (0,40), (20,40), (60,20), (60,0). The objective function is Z = 4x + 3y.
Compare the quantity in Column A and Column B

(A) The quantity in column A is greater.
(B) The quantity in column B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined on the basis of the information supplied.
Answer:
(B) The quantity in column B is greater.

Explanation:

Hence, maximum value of Z = 300 < 325 So, the quantity in column B is greater.

MCQ Questions On Linear Programming Class 12 Question 2.
The feasible solution for a LPP is si iown in given figure. Let Z = 3x – 4y be the objective function Minimum of Z occurs at

(A) (0,0)
(B) (0,8)
(C) (5,0)
(D) (4,10)
Answer:
(B) (0,8)

Explanation:

Hence, the minimum of Z occurs at (0, 8) and its minimum value is (-32).

Linear Programming MCQ Chapter 12 Class 12 Question 3.
The feasible solution for a LPP is si iown in given figure. Let Z = 3x – 4y be the obje .ctive function Minimum of Z, maximum of Z occurs at
(A) (5,0)
(B) (6,5)
(C) (6, 8)
(D) (4,10)
Answer:
(A) (5,0)

Explanation:
Maximum of Z occurs at (5, 0).

Linear Programming MCQ With Answers Pdf Question 4.
The feasible solution for a LPP is si iown in given figure. Let Z = 3x – 4y be the obje .ctive function Minimum of Z , (Maximum va lue of Z + Minimum value of Z) is equal to ……………
(A ) 13
(B) 1
(C) -13
(D) -17
Answer:
(D) -17

Explanation:
Maximum value of Z + Minimum value of Z = 15 – 32 = -17

MCQ Of Linear Programming Class 12 Question 5.
The t easible region for an LPP is shown in the given Figure. Let F = 3x – 4y be the objective function. Maximum value of F is …………..

(A) 0
(B) 8
(C) 12
(D) -18
Answer:
(C) 12

Explanation:
The feasible region as shown in the figure, has objective function F = 3x – 4y.

Hence, the maximum, value of F is 12.

MCQ On Linear Programming Class 12 Question 6.
The teasible region for an LPP is shown in the given Figure. Let F = 3x – 4y be the objective function, minimum value of F is …….
(A) 0
(B) -16
(C) 12
(D) does not exist
Answer:
(B) -16

Explanation:
Minimum value of F is -16 at (0,4).

MCQ On Linear Programming Chapter 12 Question 7.
Corner points of the feasible region for an LPP are. (0,2), (3,0), (6,0), (6,8) and (0, 5). Let F = 4x + 6y be the objective function. The minimum value of F occurs at
(A) (0, 2) only
(B) (3, 0) only
(C) the mid-point of the line segment joining the points (0, 2) and (3, 0) only
(D) any point on the line segment joining the points (0, 2) and (3, 0)
Answer:
(D) any point on the line segment joining the points (0, 2) and (3, 0)

Explanation :
Corner points Corresponding value of F = 4x + 6y

Hence, minimum value of F occurs at any points on the line segment joining the points (0, 2) and (3,0).

Class 12 Maths Chapter 12 MCQ Question 8.
Corner points of the feasible region for an LPP are. (0,2), (3,0), (6,0), (6,8) and (0, 5). Let F = 4x + 6y be the objective function, Maximum of F – Minimum of F =
(A) 60
(B) 48
(C) 42
(D) 18
Answer:
(A) 60

Explanation:
Maximum of F – Minimum of F = 72 – 12 = 60

Linear Programming MCQs Chapter 12 Question 9.
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1,1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3,0) and (1, 1) is ……….
(A) p = 2q
(B) p = \(\frac {q}{2}\)
(C) p = 3q
(D) p = q
Answer:
(B) p = \(\frac {q}{2}\)

Explanation:

So, condition of p and Question so that the minimum of Z occurs at (3,0) and (1,1) is
p+q = 3p
2p = q
p = \(\frac {q}{2}\)

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 but R is True

Linear Programming MCQ Class 12 Chapter 12 Question 1.
Assertion (A): Feasible region is the set of points which satisfy all of the given constraints and objective function too.
Reason (R): The optimal value of the objective function is attained at the points on X-axis only.
Answer:
(C) A is true but R is false

Explanation:
The optimal value of the objective function is attained at the comer points of feasible region.

Question 2.
Assertion (A): The intermediate solutions of constraints must be checked by substituting them back into objective function.
Reason (R):

Here (0,2); (0,0) and (3,0) all are vertices of feasible region.
Answer:
(D) A is false but R is True.

Explanation:
The intermediate solutions of constraints must be checked by substituting them back into constraint equations.

Linear Programming Class 12 MCQ Questions Question 3.
Assertion (A) : For the constraints of linear optimizing function Z = x₁ + x₂ given by x₁ + x₂ ≤ 1, 3x₁ + x₂ ≥ 1, there is no feasible region.
Reason (R): Z = 7x + y, subject to 5x + y ≤ 5, x + y ≥ 3, x ≥ 0, y > 0. Out of the corner points of feasible region (3, 0), (\(\frac {1}{2}\), \(\frac {5}{2}\)), (7, 0) and (0,5), the maximum value of Z occurs at (7, 0).
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Assertion (A) is corred. Clearly from the graph below that there is no feasible region.

Assertion (R) is also correct.

Linear Programming MCQ With Answers Chapter 12 Question 4.
Assertion (A): Z = 20x₁ + 20x₂, subject to x₁ ≥ 0, x₂ ≥ 2, x₁ + 2x₂ ≥ 8, 3x₁ + 2x₂ ≥ 15, 5x₁ 2x₂ ≥ 20. Out, of the corner points of feasible region (8, 0), (\(\frac {5}{2}\),\(\frac {15}{2}\)), (\(\frac {7}{2}\),\(\frac {9}{4}\)) and (0,10), the mini mum value of Z occurs at \(\frac {7}{2}\),\(\frac {9}{4}\).
Reason (R):

Answer:
(A) Both A and R are true and R is the correct explanation of A

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

Question 5.
Assertion (A): For the constraintc of a LPP problem given by x₁ + 2x₂ ≤ 2000, x₁ + x₂ ≤ 1500, x₂ ≤ 600 and x₁, x₂ ≥ 0, the points (1000, 0), (0, 500), (2, 0) lie in the positive bounded region, but point (2000, 0) does not lie in the positive bounded region.
Reason (R):

From the graph, it is clear that the point (2000, 0) is outside.
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 6.
Assertion (A) : The graph of x ≤ 2 and y ≥ 2 will be situated in the first and second quadrants.
Reason (R):

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
It is clear from the graph given in the Reason (R) that Assertion (A) is true.

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:

An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for the executive class. However, at least 4 times as many passengers prefer to travel by economy class, than by executive class. It is given that the number of executive class tickets is x and that of economy class tickets is y.

Question 1.
The maximum value of x + y is ………………
(A) 100
(B) 200
(C) 20
(D) 80
Answer:
(B) 200

Question 2.
The relation between x and y is ………..
(A) x < y (B) y > 80
(C) x ≥ 4y
(D) y ≥ 4x
Answer:
(D) y ≥ 4x

Question 3.
Which among these is not a constraint for this LPP?
(A) x ≥ 0
(B) x + y ≤ 200
(C) x ≥ 80
(D) 4x – y ≤ 0
Answer:
(C) x ≥ 80

Question 4.
The profit when x = 20 and y = 80 is ……….
(A) ₹60,000
(B) ₹68,000
(C) ₹64,000
(D) ₹1,36,000
Answer:
(B) ₹68,000

Question 5.
The maximum profit is ₹ …………..
(A) 1,36,000
(B) 1,28,000
(C) 68,000
(D) 1,20,000
Answer:
(A) 1,36,000

Explanation:
Objective function :
Maximise Z = 1000x + 600y
Constraints:
x + y ≥ 200
y ≥ 20, x > 0
y ≥ 4x

The corner points are 1(20, 180), 8(40, 160), C(20,80)
Evaluating the objective function
Z = 1,000x + 600y at A, B and C
At A(20,180), Z = 1,000 x 20 + 600 x 180
= 20,000 + 1,08,000
= ₹1,28,000
At B(40,160), Z = 1,000 x 40 + 600 x 160
= 40,000 + 96,000
= ₹1,36,000 (max.)
At C(20, 80), Z = 1000 x 20 + 600 x 80
= 20,000 + 48,000
= ₹68,000
or
Z is maximum, when x = 40, y = 160.
or
40 tickets of executive class and 160 tickets of economy class should be sold to get the maximum profit of ₹1,36,000.

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

A dealer in rural area wishes to purchase a number of sewing machines. He has only ₹5,760 to invest and has space for at most 20 items for storage. An electronic sewing machine cost him ₹360 and a manually operated sewing machine ₹240. He can sell an electronic sewing machine at a profit of ₹22 and a manually operated machine at a profit of ₹18. Assume that the electronic sewing machines he can sell is x and that of manually operated machines is y.

Question 1.
The objective function is …………..
(A) Maximize Z = 360x + 240y
(B) Maximize Z = 22x + 18y
(C) Minimize Z = 360x + 240y
(D) Minimize Z = 22x + 18y
Answer:
(B) Maximize Z = 22x + 18y

Question 2.
The maximum value of x + y is ………….
(A) 5760
(B) 18
(C) 22
(D) 20
Answer:
(D) 20

Question 3.
Which of the following is not a constraint ₹
(A) x + y ≥ 20
(B) 360x + 240y ≤ 5,760
(C) x ≥ 0
(D) y ≥ 0
Answer:
(A) x + y ≥ 20

Question 4.
The profit is maximum when (x, y) =
(A) (5,15)
(B) (8,12)
(C) (12,8)
(D) (15,5)
Answer:
(B) (8,12)

Question 5.
The maximum profit is …………..
(A) 5,760
(B) 392
(C) 362
(D) 290
Answer:
(B) 392

Explanation:
Objective function :
Maximize Z = 22x + 18y
Constraints :
x + y ≤ 20
360x + 240y ≤ 5,760
Or
3x + 2y ≤ 48
x ≥ 0, y ≤ 0

Vertices of feasible region are :
A(0, 20), B(8,12), C(16, 0) & O(0,0)
P(A) = 360, P(B) = 392, P(C) = 352
∴ For Maximum P, Electronic machines = 8, and Manual machines = 12. Max. profit ₹392

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

A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

Question 1.
The Objective function to minimise the amount of nitogen added to garden?
(A) Maximise Z = 3x + 4y
(B) Minimise Z = 3x + 3.5y
(C) Maximise Z = 4x + 3.5y
(D) Minimise Z = 3x + 4y
Answer:
(B) Minimise Z = 3x + 3.5y

Question 2.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of brand P should be used?
(A) 40
(B) 50
(C) 100
(D) 60
Answer:
(A) 40

Question 3.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of brand Q should be used?
(A) 40
(B) 50
(C) 100
(D) 60
Answer:
(C) 100

Question 4.
What is the minimum amount of nitrogen added in the garden?
(A) 595 kg
(B) 550 kg
(C) 400 kg
(D) 470 kg
Answer:
(D) 470 kg

Question 5.
What is the total number of bags used by fruit grower to minimise the amount of nitrogen?
(A) 160
(B) 190
(C) 140
(D) 130
Answer:
(C) 140

Explanation:
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows:
Minimise Z = 3x + 3.5y ……(i)
Subject to the constraings,
x + 2y ≥ 240 ……….(ii)
x + 0.5y ≥ 90 ………(iii)
1.5x + 2y ≤ 310 ……(iv)
x,y ≥ 0 …..(v)
The feasible region determined by the system of constraints is as follows:

The corner points are A(240, 50), B(20, 140), and C(40,100)

The minimum value of Z is 470 at (40,100).
Thus, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.
The minimum amount of nitrogen added to the garden is 470 kg.

MCQ Questions for Class 12 Maths with Answers

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