# CBSE Class 12 Maths Question Paper 2021 (Term-I) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Maths with Solutions and CBSE Class 12 Maths Question Paper 2021 (Term-I) to familiarize themselves with the exam format and marking scheme.

## CBSE Class 12 Maths Question Paper 2021 (Term-I) with Solutions

Time Allowed: 3 hours

Maximum Marks: 80

General Instructions:

- This Question paper comprises 50 questions out of which 40 questions are to be attempted as per instructions. All questions carry equal marks.
- This question paper consists three Sections — Section A, B and C.
- Section—A contains 20 questions. Attempt any 16 questions from Q. No. 1 to 20.
- Section—B also contains 20 questions. At tempt anti 16 questions from Q. No. 21 to 40.
- Section — C contains 10 questions including one Case Study. Attempt any 8 from Q. No. 41 to 50.
- There is only one correct option for every Multiple Choice Question (MC Q). Marks will not be awarded for answering more than one option.
- There is no negative marking.

Section – A

In this section, attempt any 16 questions out of Question 1-20. Each question is of one mark.

Question 1.

Differential of log [log(log x^{5})] w.r.t. x is

(a) \(\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)

(b) \(\frac{5}{x \log \left(\log x^5\right)}\)

(c) \(\frac{5 x^4}{\log \left(x^5\right) \log \left(\log x^5\right)}\)

(d) \(\frac{5 x^4}{\log x^5 \log \left(\log x^5\right)}\)

Solution:

(a) \(\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)

Let y = log[log(log x^{5})]

Differentiating both sides w.r.t.x, we have

Question 2.

The number of all possible matrices of order 2 × 3 with each entry 1 or 2 is

(a) 16

(b) 6

(c) 64

(d) 24

Solution:

(c) 64

Order = 2 × 3 = 6

Each entry 1 or 2 i.e., 2 numbers

∴ Total possible matrices = 2^{6} = 64

Question 3.

A function f : R → R is defined as f(x) = x^{3} + 1. Then the function has

(a) no minimum value

(b) no maximum value

(c) both maximum and minimum values

(d) neither maximum value nor minimum value

Solution:

(d) neither maximum value nor minimum value

Question 4.

If sin y = x cos (a + y), then \(\frac{d x}{d y}\) is:

(a) \(\frac{\cos a}{\cos ^2(a+y)}\)

(b) \(\frac{-\cos a}{\cos ^2(a+y)}\)

(c) \(\frac{\cos a}{\sin ^2 y}\)

(d) \(\frac{-\cos a}{\sin ^2 y}\)

Solution:

(a) \(\frac{\cos a}{\cos ^2(a+y)}\)

Question 5.

If radius of a spherical soap bubble is increasing at the rate of 0.2 cnVsec. Find the rate of increase of its surface area, when the radius is 7 cm.

(a) 11.2π cm/sec

(b) 11.2π cm^{2}/sec

(c) 11π cm^{2}/sec

(d) 11π cm^{2}/sec

Solution:

(b) 11.2π cm^{2}/sec

Given, radius, r = 7 cm and

\(\frac{d r}{d t}\) = 0.2 cm/sec

Surface area of Sphere, s = 4πr^{2}

Difference w.r.t, t i.e., time, we have

∴ \(\frac{d s}{d t}\) = 8πr\(\frac{d r}{d t}\) = 8π(7)(0.2) = 11.2 π cm^{2}/sec

Question 6.

Three points F(2x, x + 3), Q(0, x) and R(x + 3, x + 6) are collinear, then x is equal to

(a) 0

(b) 2

(c) 3

(d) 1

Solution:

(d) 1

\(\left|\begin{array}{ccc}

2 x & x+3 & 1 \\

0 & x & 1 \\

x+3 & x+6 & 1

\end{array}\right|\) = 0

[∵ Points P, Q R are collinear.

Expanding along C_{1}, we have

2x(x – x – 6) + (x + 3) (x + 3 – x) = 0

⇒ -12x + 3x + 9 = 0

⇒ -9x = -9 ∴ x = 1

Question 7.

The principal value of cos^{-1}\(\left(\frac{1}{2}\right)\) + sin^{-1}\(\left(-\frac{1}{\sqrt{2}}\right)\) is

(a) \(\frac{\pi}{12}\)

(b) π

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

(d) \(\frac{\pi}{6}\)

Solution:

(a) \(\frac{\pi}{12}\)

Question 8.

If (x^{2} + y^{2})^{2} = xy, then \(\frac{d y}{d x}\) is:

(a) \(\frac{y+4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)

(b) \(\frac{y-4 x\left(x^2+y^2\right)}{x+4\left(x^2+y^2\right)}\)

(c) \(\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)

(d) \(\frac{4 y\left(x^2+y^2\right)-x}{y-4 x\left(x^2+y^2\right)}\)

Solution:

(c) \(\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)

Question 9.

If a matrix A is both symmetric and skew symmetric, then A is necessarily a

(a) Diagonal matrix

(b) Zero square matrix

(c) Square matrix

(d) Identity matrix

Solution:

(b) Zero square matrix

A^{t} = A (∵ A is a symmetric matrix)

A^{t} = -A (∵ A is a skew symmetric matrix)

A = -A

⇒ 2A = 0 ⇒ A = 0

∴ A is a zero square matrix.

Question 10.

Let set X = {1, 2, 3} and a relation R is defined in X as: R = {(1, 3), (2, 2), (3, 2)}, then minimum ordered pairs which should be added in relation R to make it reflexive and symmetric are

(a) {(1,1), (2, 3), (1, 2)}

(b) {(3, 3), (3, 1), (1, 2)}

(c) {(1, 1), (3, 3), (3, 1), (2, 3)}

(d) {(1, 1), (3, 3), (3, 1), (1, 2)}

Solution:

(c) {(1, 1), (3, 3), (3, 1), (2, 3)}

Given. X = {1, 2, 3}

Reflexive : {(1, 1), (2, 2), (3, 3)}

Symmetric : {(1, 3), (3, 1), (3, 2), (2, 3)}

R to makes it reflexive and symmetric then add = {(1, 1), (3, 3), (3, 1), (2, 3)}

Question 11.

A Linear Programming Problem is as follows:

Minimise z = 2x + y

Subject to the constraints x ≥ 3, x ≤ 9, y ≥ 0

x – y ≥ 0, x + y ≤ 14

(a) 5 corner points including (0, 0) and (9, 5)

(b) 5 corner points including (7, 7) and (3, 3)

(c) 5 corner points including (14, 0) and (9, 0)

(d) 5 corner points including (3, 6) and (9, 5)

Solution:

(b) 5 corner points including (7, 7) and (3, 3)

Question 12.

(a) 3

(b) 5

(c) 2

(d) 8

Solution:

(d) 8

Question 13.

If C_{ij} denotes the cofactor of element P of the matrix P = \(\left[\begin{array}{rrr}

1 & -1 & 2 \\

0 & 2 & -3 \\

3 & 2 & 4

\end{array}\right]\), then the value of C_{31}.C_{23} is

(a) 5

(b) 24

(c) -24

(d) -5

Solution:

(d) -5

Cofactor of a_{31} = C_{31} = (-1)^{3+1}\(\left|\begin{array}{rr}

-1 & 2 \\

2 & -3

\end{array}\right|\)

= 3 – 4 = -1

Cofactor of a_{23} = C_{23} = (-1)^{2+3}\(\left|\begin{array}{rr}

1 & -1 \\

3 & 2

\end{array}\right|\)

= -(2 + 3) = -5

∴ C_{31} . C_{23} = (-1) (-5) = 5

Question 14.

If function y = x^{2}e^{-x} is decreasing in the interval

(a) (0, 2)

(b) (2, ∞)

(c) (-∞, 0)

(d) (-∞, 0) ∪ (2, ∞)

Solution:

(d) (-∞, 0) ∪ (2, ∞)

Question 15.

If R = {(x, y); x y ∈ Z, x^{2} + y^{2} ≤ 4) is a relation in set Z, then domain of R is

(a) {0, 1, 2}

(b) {-2, -1, 0, 1, 2}

(c) {0, -1, -2}

(d) {-1, 0, 1}

Solution:

(b) {-2, -1, 0, 1, 2}

Equation of circle with centre (0, 0) and radius = 2

x ∈ Z (integers)

∴ Domain = {-2, -1, 0, 1, 2}

Question 16.

The system of linear equations

5x + ky = 5,

3x + 3y = 5;

will be consistent if

(a) k ≠ -3

(b) k = -5

(c) k = 5

(d) k ≠ 5

Solution:

(d) k ≠ 5

For consistent, | A | ≠ 0

∴ \(\left|\begin{array}{ll}

5 & k \\

3 & 3

\end{array}\right|\) ≠ 0

⇒ 15 – 3k ≠ 0

⇒ -3k ≠ -15 ∴ k ≠ 5

Question 17.

Find the point on the curve y^{2} = 8x for which the abscissa and ordinate change at the same rate

(a) (1, 2)

(b) (2, 3)

(c) (2, 4)

(d) (1, 4)

Solution:

(c) (2, 4)

Question 18.

If \(\left[\begin{array}{cc}

3 c+6 & a-d \\

a+d & 2-3 b

\end{array}\right]\) = \(\left[\begin{array}{rr}

12 & 2 \\

-8 & -4

\end{array}\right]\) are equal, then value ab – cd

(a) 4

(b) 16

(c) -4

(d) -16

Solution:

(a) 4

Question 19.

The principal value of tan^{-1}\(\left(\tan \frac{9 \pi}{8}\right)\) is

(a) \(\frac{\pi}{8}\)

(b) \(\frac{3 \pi}{8}\)

(c) \(-\frac{\pi}{8}\)

(d) \(-\frac{3 \pi}{8}\)

Solution:

(a) \(\frac{\pi}{8}\)

Question 20.

For two matrices P = \(\left[\begin{array}{rr}

3 & 4 \\

-1 & 2 \\

0 & 1

\end{array}\right]\) and Q^{T} = \(\left[\begin{array}{rrr}

-1 & 2 & 1 \\

1 & 2 & 3

\end{array}\right]\) then, P – Q is

(a) \(\left[\begin{array}{rr}

2 & 3 \\

-3 & 0 \\

0 & -3

\end{array}\right]\)

(b) \(\left[\begin{array}{rr}

4 & 3 \\

-3 & 0 \\

-1 & -2

\end{array}\right]\)

(c) \(\left[\begin{array}{rr}

4 & 3 \\

0 & -3 \\

-1 & -2

\end{array}\right]\)

(d) \(\left[\begin{array}{rr}

2 & 3 \\

0 & -3 \\

0 & -3

\end{array}\right]\)

Solution:

(b) \(\left[\begin{array}{rr}

4 & 3 \\

-3 & 0 \\

-1 & -2

\end{array}\right]\)

Section – B

In this section, attempt any 16 questions out of Question 21-40. Each question is of one mark.

Question 21.

The function f(x) = 2x^{3} – 15x^{2} + 36x + 6 is increasing in the interval

(a) (-∞, 2) ∪ (3, ∞)

(b) (-∞, 2)

(c) (-∞, 2) ∪ (3, ∞)

(d) [3, ∞)

Solution:

(a) (-∞, 2) ∪ (3, ∞)

f(x) = 2x^{3} – 15x^{2} + 36x + 6

Differentiating w.r.t. x, we get

f'(x) = 6x^{2} – 30x + 36

= 6(x^{2} – 5x + 6)

= 6(x^{2} – 3x – 2x + 6)

= 6(x(x – 3) – 2(x – 3))

= 6(x – 2) (x – 3)

Question 22.

If x = 2 cos θ – cos 2θ and y = 2 sinθ – sin 2θ, then \(\frac{d y}{d x}\) is

(a) \(\frac{\cos \theta+\cos 2 \theta}{\sin \theta-\sin 2 \theta}\)

(b) \(\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\)

(c) \(\frac{\cos \theta-\cos 2 \theta}{\sin \theta-\sin 2 \theta}\)

(d) \(\frac{\cos 2 \theta-\cos \theta}{\sin 2 \theta+\sin \theta}\)

Solution:

(b) \(\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\)

Question 23.

What is the domain of the function cos^{-1}(2x – 3)?

(a) [-1, 1]

(b) (1, 2)

(c) (-1, 1)

(d) [1, 2]

Solution:

(b) (1, 2)

-1 ≤ 2x – 3 ≤ 1

[Domain of cos^{-1}θ = -1 ≤ θ ≤ 1

-1 + 3 ≤ 2x – 3 + 3 ≤ 1 + 3

2 ≤ 2x ≤ 4

\(\frac{2}{2}\) ≤ \(\frac{2 x}{2}\) ≤ \(\frac{4}{2}\) ≤

1 ≤ x ≤ 2; ∴ Domain x ∈ [1, 2]

Question 24.

A matrix A = [a_{ij}]_{3×3} is defined by

The number of elements in A which are more than 5, is

(a) 3

(b) 4

(c) 5

(d) 6

Solution:

(b) 4

Here, a_{ij} = 2i + 3j, i < j

a_{12} = 2(1) + 3(2) = 2 + 6 = 8 > 5 ………. (i)

a_{13} = 2(1) + 3(3) = 2 + 9 = 11 > 5 ………… (ii)

a_{23} = 2(2) + 3(3) = 4 + 9 = 1 3 > 5 ……….. (iii)

and a_{ij} = 5, i = j

a_{11} = 5; a_{22} = 5; a_{33} = 5

then a_{ij} = 3i – 2j, i > j

a_{21} = 3(2) – 2(1) = 6 – 2 = 4

a_{31} = 3(3) – 2(1) = 9 – 2 = 7 > 5 ……… (iv)

a_{32} = 3(3) – 2(2) = 9 – 2 = 5

Number of elements in A which are more than 5 is 4 i.e., (a_{12}, a_{13}, a_{23}, a_{31}).

Question 25.

If a function f defined by

is continuous at x = \(\frac{\pi}{2}\), then the value of k is

(a) 2

(b) 3

(c) 6

(d) -6

Solution:

(c) 6

Question 26.

For the matrix X = \(\left[\begin{array}{lll}

0 & 1 & 1 \\

1 & 0 & 1 \\

1 & 1 & 0

\end{array}\right]\), (X^{2} — X) is

(a) 21

(b) 31

(c) I

(d) 5I

Solution:

(a) 21

Question 27.

Let X = {x^{2} : x ∈ N} and the function f : N → X is defined by f(x) = x^{2}, x ∈ N. Then this function is

(a) injective only

(b) not bijective

(c) surjective only

(d) bijective

Solution:

(d) bijective

X = {x^{2} : x ∈ N)

Here x = 1, 2, 3, 4, 5,…………

N → X

Domain = N (natural numbers)

Range = X = {1, 4, 9, 16, 25, ………}

f(x) = x^{2}, x ∈ N

one-one, f(x_{1}) = f(x_{2}), x_{1}, x_{2} ∈ N

\(x_1^2\) = \(x_2{ }^2\)

x_{1} = x_{2} ∴ f is one-one

For Onto, When x ∈ N, f(x) = x^{2}

Range = {1, 4, 9, 16, 25,……….}

Co-domain = {1, 4, 9, 16, 25,……….}

∴ Range (f) = co-domain (f)

∴ f is onto. ∴ f is bijective.

Question 28.

The corner points of the feasible region for a Linear Programming problem are P(0, 5), Q(1, 5), R(4, 2) and S(12, 0). The minimum value of objective function Z = 2x + 5y is at

(a) P

(b) Q

(c) R

(d) S

Solution:

(c) R

Question 29.

A kite is flying at a height of 3 m and 5 m of string in out. If the kite in moving away horizontally at the rate of 200 cm/s, then the rate at which string is being released.

(a) 5 m/s

(b) 3 m/s

(c) \(\frac{3}{5}\) m/s

(d) \(\frac{8}{5}\) m/s

Solution:

(d) \(\frac{8}{5}\) m/s

Question 30.

If A is a square matrix of order 3 and |A| = -5, then |adj A| is

(a) 125

(b) -25

(c) 25

(d) ± 25

Solution:

(c) 25

As we know, |adj A| = |A|^{n-1}

= |A|^{2} (∵ order n = 3)

= (-5)^{2} = 25

Question 31.

The simplest form of tan^{-1}\(\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]\) is

(a) \(\frac{\pi}{4}-\frac{x}{2}\)

(b) \(\frac{\pi}{4}+\frac{x}{2}\)

(c) \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)

(d) \(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\)

Solution:

(c) \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)

Question 32.

If for the matrix A = \(\left[\begin{array}{cc}

\alpha & -2 \\

-2 & \alpha

\end{array}\right]\), |A^{3}| = 125, then the value of α is

(a) ± 3

(b) -3

(c) ± 1

(d) 1

Solution:

(a) ± 3

Question 33.

If y = sin(m sin^{-1} x), then which one of the following equations is true?

Solution:

(b) (1 – x^{2})\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) + m^{2}y = 0

Question 34.

The principal value of [tan^{-1}\(\sqrt{3}\) – cot^{-1} (-\(\sqrt{3}\))] is

(a) π

(b) \(-\frac{\pi}{2}\)

(c) 0

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

Solution:

(b) \(-\frac{\pi}{2}\)

tan^{-1}\(\sqrt{3}\) – cot^{-1}\((-\sqrt{3})\)

= \(\frac{\pi}{3}\)(π – cot^{-1}\(\sqrt{3}\)) [∵ cot^{-1}(-A) = π – cot^{-1}A

= \(\frac{\pi}{3}-\pi+\frac{\pi}{6}\) = \(-\frac{\pi}{2}\)

Question 35.

The maximum value of \(\left(\frac{1}{x}\right)^x\) is

(a) e^{1/e}

(b) e

(c) \(\left(\frac{1}{e}\right)^{1 / e}\)

(d) e^{e}

Solution:

(a) e^{1/e}

Let y = \(-\frac{\pi}{2}\)

Taking log on both sides,

log y = x log \(\left(\frac{1}{x}\right)\)

log y = -x log x

Differentiating w.r.t. x, we have

Question 36.

Let matrix X = [x_{ij}] is givne by X = \(\left[\begin{array}{ccc}

1 & -1 & 2 \\

3 & 4 & -5 \\

2 & -1 & 3

\end{array}\right]\). Then the matrix Y = [m<sub.ij], where m_{ij} = Minor of x_{ij}, is

(a) \(\left[\begin{array}{ccc}

7 & -5 & -3 \\

19 & 1 & -11 \\

-11 & 1 & 7

\end{array}\right]\)

(b) \(\left[\begin{array}{ccc}

7 & -19 & -11 \\

5 & -1 & -1 \\

3 & 11 & 7

\end{array}\right]\)

(c) \(\left[\begin{array}{ccc}

7 & 19 & -11 \\

-3 & 11 & 7 \\

-5 & -1 & -1

\end{array}\right]\)

(d) \(\left[\begin{array}{ccc}

7 & 19 & -11 \\

-1 & -1 & 1 \\

-3 & -11 & 7

\end{array}\right]\)

Solution:

(d) \(\left[\begin{array}{ccc}

7 & 19 & -11 \\

-1 & -1 & 1 \\

-3 & -11 & 7

\end{array}\right]\)

M_{11} = 12 – 5 = 7

M_{12} = 9 + 10 = 19

M_{13} = -3 – 8 = -11

M_{21} = -3 + 2 = -1

M_{22} = 3 – 4 = -1

M_{23} = -1 + 2 = 1

M_{31} = 5 – 8 = -3

M_{32} = -5 – 6 = -11

M_{33} = 4 + 3 = 7

∴ Minor of X = \(\left[\begin{array}{rrr}

7 & 19 & -11 \\

-1 & -1 & 1 \\

-3 & -11 & 7

\end{array}\right]\)

Question 37.

A function f : R → R defined by f(x) = 2 + x^{2} is

(a) not one-one

(b) one-one

(c) not onto

(d) neither one-one nor onto

Solution:

(d) neither one-one nor onto

For one-one: f(x_{1}) = f(x_{2})

2 + \(x_1^2\) = 2 + \(x_2^2\)

\(x_1^2\) = ± x_{2} ∴ f is not one-one.

For onto: Let y = f(x)

y = 2 + x^{2}

y – 2 = x^{2}

x = ±\(\sqrt{y-2}\) ∉ R for some y ∈ R.

For y = 0 ∈ R, there is no x ∈ R such that f(x) = y.

∴ 0 ∈ R does not have pre-image in R.

∴ f is not onto.

Hence f is neither one-one nor onto.

Question 38.

A Linear Programming Problem is as follows:

Maximise/Minimise objective function Z = 2x – y + 5

Subject to the constraints:

3x + 4y ≤ 60

x + 3y ≤ 30

x ≥ 0, y ≥ 0

If the corner points of the feasible region are A(0,10), B(12, 6), C(20, 0) and 0(0, 0), then which of the following is true?

(a) Maximum value of Z is 40

(b) Minimum value of Z is -5

(c) Difference of maximum and minimum values of Z is 35

(d) At two corner points, value of Z are equal

Solution:

(b) Minimum value of Z is -5

∴ Minimum value is Z is -5.

Question 39.

If x = -4 is a root of \(\left|\begin{array}{lll}

x & 2 & 3 \\

1 & x & 1 \\

3 & 2 & x

\end{array}\right|\) = 0, then the sum of the other two roots is

(a) 4

(b) -3

(c) 2

(d) 5

Solution:

(a) 4

We have, \(\left|\begin{array}{lll}

x & 2 & 3 \\

1 & x & 1 \\

3 & 2 & x

\end{array}\right|\) = 0

x(x^{2} – 2) -2 (x – 3) + 3(2 – 3x) = 0

⇒ x^{3} – 2x – 2x + 6 + 6 – 9x = 0

x^{3} – 13x + 12 = 0

Here a = 1, b = 0, c = -13, d = 12

Let roots are: α = -4, β, and δ

Sum of roots, α + β + δ = \(\frac{-b}{a}\)

-4 + β + δ = \(\frac{0}{1}\) ∴ β + δ = 4

Question 40.

The absolute maximum value of the function f(x) = 4x – \(\frac{1}{2} x^2\) in the interval [-2, \(\frac{9}{2}\)] is

(a) 8

(b) 9

(c) 6

(d) 10

Solution:

(a) 8

Section – C

Attempt any 8 Questions out of the Questions 41-50. Each question is of one mark.

Question 41.

In a sphere of radius r, a right circular cone of height h having maximum curved surface area is inscribed. The expression for the square of curved surface of cone is

(a) 2π^{2}rh(2rh + h^{2})

(b) π^{2}hr(2rh + h^{2})

(c) 2π^{2}r(2rh^{2} – h^{3})

(d) 2π^{2}r^{2}(2rh – h^{2})

Solution:

(c) 2π^{2}r(2rh^{2} – h^{3})

OB = OC = r = radii of Sphere

Height of cone, AC = h,

∴ OA = h – r

In rt. ∆OAB,

AB^{2} = OB^{2} – OA^{2} ….. [Pythagoras theorem

AB^{2} = r^{2} – (h – r)^{2}

= r^{2} (h^{2} + r^{2} – 2hr)

= 2hr – h^{2} …….. (i)

In rt. ∆CAB,

BC^{2} = AC^{2} + AB^{2} ……[Pythagoras theorem)

= h^{2} + 2hr – h^{2}

= 2hr ……….(ii)

Square of C.S. of cone = (πrl)^{2}

= π^{2} (AB)^{2} (BC)^{2}

= π^{2}(2hr – h^{2}) 2hr ……[From (i) and (ii)

= 2π^{2}r (2rh^{2} – h^{3})

Question 42.

The corner points of the feasible region determined by a set of constraints (linear inequalities) are P(0, 5), Q(3, 5), R(5, 0) and S(4,1) and the objective function is Z = ax + 2by where a, b > 0. The condition on a and b such that the maximum Z occurs at Q and S is

(a) a – 5b = 0

(b) a – 3b – 0

(c) a – 2b = 0

(d) a – 8b = 0

Solution:

(d) a – 8b = 0

Z = ax + 2by

At Q(3, 5),

At S(4, 1),

Z = a(3) + 2b(5)

Z = a(4) + 2b(1)

Maximum Z occurs at Q and S (Given)

∴ 3a + 10b = 4a + 2b

⇒ 0 = 4a + 2b – 3a – 10b

∴ a – 8b = 0

Question 43.

The volume of metal of a hollow sphere is constant. If the inner radius is increasing at the rate of 1 cm/s, find the rate of increase of the outer radius, when the radii are 3 cm and 6 cm respectively.

(a) 0.25 cm/sec

(b) 0.5 cm/sec

(c) 0.75 cm/sec

(d) 2 cm/sec

Solution:

(a) 0.25 cm/sec

Let r and R be the inner and outer radius.

Volume of hollow sphere,

Question 44.

The inverse of matrix X = \(\left[\begin{array}{lll}

2 & 0 & 0 \\

0 & 3 & 0 \\

0 & 0 & 4

\end{array}\right]\) is

Solution:

(d) \(\left[\begin{array}{ccc}

1 / 2 & 0 & 0 \\

0 & 1 / 3 & 0 \\

0 & 0 & 1 / 4

\end{array}\right]\)

Question 45.

For an L.P.P. the objective function is Z = 4x + 3y, and the feasible region determined by a set of constraints (linear inequations) is shown in the graph

Which one of the following statements is true?

(a) Maximum value of Z is at R.

(b) Maximum value of Z is at Q.

(c) Value of Z at R is less than the value at P.

(d) Value of Z at Q is less than the value at R.

Solution:

(a) Maximum value of Z is at R.

∴ Maximum value of Z is at Q(30, 20).

Case Study

In a residential society comprising of 100 houses, there were 60 children between the ages of 10-15 years. They were inspired by their teachers to start composting to ensure that biodegradable waste is recycled. For this purpose, instead of each child doing it for only his/her house, children convinced the Residents welfare association to do it as a society initiative. For this they identified a

square area in the local park. Local authorities charged amount of ₹50 per square metre for space so that there is association takes it seriously. Association charged ₹400 × (depth)^{2}. Association will like to have minimum cost.

Based on this information, answer any four of the following questions.

Question 46.

Let side of square plots is x m and its depth is h metres, then cost c for the pit is

(a) \(\frac{50}{h}\) + 400 h^{2}

(b) \(\frac{12,500}{h}\) + 400h^{2}

(c) \(\frac{250}{h}\) + h^{2}

(d) \(\frac{250}{h}\) + 400h^{2}

Solution:

(b) \(\frac{12,500}{h}\) + 400h^{2}

Total cost, c = (cost of a space) + (cost of labourer)

= 50 × (Area of square) + 400 × (depth)^{2}

= 50 × (x)^{2} + 400 h^{2}

= 50 × \(\left(\frac{250}{h}\right)\) + 400 h^{2} …… [From (i)

∴ c = \(\frac{12,500}{h}\) + 400 h^{2}

Question 47.

Value of h (in m) for which \(\frac{d c}{d h}\) = 0 is

(a) 1.5

(b) 2

(c) 2.5

(d) 3

Solution:

(c) 2.5

From Question 46, c = \(\frac{12,500}{h}\) + 400 h^{2}

Differentiating both sides w.r.f, h, we have

Question 48.

\(\frac{d^2 c}{d h^2}\) is given by

(a) \(\frac{25,000}{h^3}\) + 800

(b) \(\frac{500}{h^3}\) + 800

(c) \(\frac{100}{h^3}\) + 800

(d) \(\frac{500}{h^3}\) + 2

Solution:

(a) \(\frac{25,000}{h^3}\) + 800

\(\frac{d c}{d h}\) = -12,500 h^{2} + 800h

Differentiating w.r.t. h, we get

\(\frac{d^2 c}{d h^2}\) = \(\frac{25,000}{h^3}\) + 800

Question 49.

Value of x (in m) for minimum cost is

(a) 5

(b) 10\(\sqrt{\frac{5}{3}}\)

(c) 5\(\sqrt{5}\)

(d) 10

Solution:

(d) 10

Volume of cuboid = 250 m^{2}

x.x.h = 250

x^{2} = \(\frac{250}{h}\)

⇒ x^{2} = 250 × \(\frac{2}{5}\) …… [From Question 47

⇒ x^{2} = 100 ∴ x = 10

Question 50.

Total minimum cost of digging the pit (in ₹) is

(a) 4,100

(b) 7,500

(c) 7,850

(d) 3,220

Solution:

(b) 7,500

\(\left(\frac{d^2 c}{d h^2}\right)_{h=\frac{5}{2}}\)

= 25,000 × \(\frac{2}{5}\) × \(\frac{2}{5}\) × \(\frac{2}{5}\) + 800 > 0

Cost is minimum at h = \(\frac{5}{2}\)

c = \(\frac{12,500}{h}\) + 400h^{2}

c = 12,500 × \(\frac{2}{5}\) + 400 × \(\frac{5}{2}\) × \(\frac{5}{2}\)

= 5,000 + 2,500 = ₹7,500