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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:

  1. This Question paper comprises 50 questions out of which 40 questions are to be attempted as per instructions. All questions carry equal marks.
  2. This question paper consists three Sections — Section A, B and C.
  3. Section—A contains 20 questions. Attempt any 16 questions from Q. No. 1 to 20.
  4. Section—B also contains 20 questions. At tempt anti 16 questions from Q. No. 21 to 40.
  5. Section — C contains 10 questions including one Case Study. Attempt any 8 from Q. No. 41 to 50.
  6. There is only one correct option for every Multiple Choice Question (MC Q). Marks will not be awarded for answering more than one option.
  7. 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 x5)] 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 x5)]
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 = 26 = 64

Question 3.
A function f : R → R is defined as f(x) = x3 + 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 cm2/sec
(c) 11π cm2/sec
(d) 11π cm2/sec
Solution:
(b) 11.2π cm2/sec

Given, radius, r = 7 cm and
\(\frac{d r}{d t}\) = 0.2 cm/sec
Surface area of Sphere, s = 4πr2
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 cm2/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 C1, 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 (x2 + y2)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

At = A (∵ A is a symmetric matrix)
At = -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 Cij 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 C31.C23 is
(a) 5
(b) 24
(c) -24
(d) -5
Solution:
(d) -5

Cofactor of a31 = C31 = (-1)3+1\(\left|\begin{array}{rr}
-1 & 2 \\
2 & -3
\end{array}\right|\)
= 3 – 4 = -1
Cofactor of a23 = C23 = (-1)2+3\(\left|\begin{array}{rr}
1 & -1 \\
3 & 2
\end{array}\right|\)
= -(2 + 3) = -5
∴ C31 . C23 = (-1) (-5) = 5

Question 14.
If function y = x2e-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, x2 + y2 ≤ 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 y2 = 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 QT = \(\left[\begin{array}{rrr}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\) 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) = 2x3 – 15x2 + 36x + 6 is increasing in the interval
(a) (-∞, 2) ∪ (3, ∞)
(b) (-∞, 2)
(c) (-∞, 2) ∪ (3, ∞)
(d) [3, ∞)
Solution:
(a) (-∞, 2) ∪ (3, ∞)

f(x) = 2x3 – 15x2 + 36x + 6
Differentiating w.r.t. x, we get
f'(x) = 6x2 – 30x + 36
= 6(x2 – 5x + 6)
= 6(x2 – 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-1θ = -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 = [aij]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, aij = 2i + 3j, i < j
a12 = 2(1) + 3(2) = 2 + 6 = 8 > 5 ………. (i)
a13 = 2(1) + 3(3) = 2 + 9 = 11 > 5 ………… (ii)
a23 = 2(2) + 3(3) = 4 + 9 = 1 3 > 5 ……….. (iii)
and aij = 5, i = j
a11 = 5; a22 = 5; a33 = 5
then aij = 3i – 2j, i > j
a21 = 3(2) – 2(1) = 6 – 2 = 4
a31 = 3(3) – 2(1) = 9 – 2 = 7 > 5 ……… (iv)
a32 = 3(3) – 2(2) = 9 – 2 = 5
Number of elements in A which are more than 5 is 4 i.e., (a12, a13, a23, a31).

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 = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\), (X2 — X) is
(a) 21
(b) 31
(c) I
(d) 5I
Solution:
(a) 21

Question 27.
Let X = {x2 : x ∈ N} and the function f : N → X is defined by f(x) = x2, x ∈ N. Then this function is
(a) injective only
(b) not bijective
(c) surjective only
(d) bijective
Solution:
(d) bijective

X = {x2 : x ∈ N)
Here x = 1, 2, 3, 4, 5,…………
N → X
Domain = N (natural numbers)
Range = X = {1, 4, 9, 16, 25, ………}
f(x) = x2, x ∈ N
one-one, f(x1) = f(x2), x1, x2 ∈ N
\(x_1^2\) = \(x_2{ }^2\)
x1 = x2 ∴ f is one-one
For Onto, When x ∈ N, f(x) = x2
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 = \(\left[\begin{array}{cc}
\alpha & -2 \\
-2 & \alpha
\end{array}\right]\), |A3| = 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 – x2)\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) + m2y = 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-1A
= \(\frac{\pi}{3}-\pi+\frac{\pi}{6}\) = \(-\frac{\pi}{2}\)

Question 35.
The maximum value of \(\left(\frac{1}{x}\right)^x\) is
(a) e1/e
(b) e
(c) \(\left(\frac{1}{e}\right)^{1 / e}\)
(d) ee
Solution:
(a) e1/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 = [xij] 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 mij = Minor of xij, 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) \(\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]\)

M11 = 12 – 5 = 7
M12 = 9 + 10 = 19
M13 = -3 – 8 = -11
M21 = -3 + 2 = -1
M22 = 3 – 4 = -1
M23 = -1 + 2 = 1
M31 = 5 – 8 = -3
M32 = -5 – 6 = -11
M33 = 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 + x2 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(x1) = f(x2)
2 + \(x_1^2\) = 2 + \(x_2^2\)
\(x_1^2\) = ± x2 ∴ f is not one-one.
For onto: Let y = f(x)
y = 2 + x2
y – 2 = x2
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(x2 – 2) -2 (x – 3) + 3(2 – 3x) = 0
⇒ x3 – 2x – 2x + 6 + 6 – 9x = 0
x3 – 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) 2π2rh(2rh + h2)
(b) π2hr(2rh + h2)
(c) 2π2r(2rh2 – h3)
(d) 2π2r2(2rh – h2)
Solution:
(c) 2π2r(2rh2 – h3)

OB = OC = r = radii of Sphere
Height of cone, AC = h,
∴ OA = h – r

In rt. ∆OAB,
AB2 = OB2 – OA2 ….. [Pythagoras theorem
AB2 = r2 – (h – r)2
= r2 (h2 + r2 – 2hr)
= 2hr – h2 …….. (i)
In rt. ∆CAB,
BC2 = AC2 + AB2 ……[Pythagoras theorem)
= h2 + 2hr – h2
= 2hr ……….(ii)
Square of C.S. of cone = (πrl)2
= π2 (AB)2 (BC)2
= π2(2hr – h2) 2hr ……[From (i) and (ii)
= 2π2r (2rh2 – h3)

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 h2
(b) \(\frac{12,500}{h}\) + 400h2
(c) \(\frac{250}{h}\) + h2
(d) \(\frac{250}{h}\) + 400h2
Solution:
(b) \(\frac{12,500}{h}\) + 400h2

Total cost, c = (cost of a space) + (cost of labourer)
= 50 × (Area of square) + 400 × (depth)2
= 50 × (x)2 + 400 h2

= 50 × \(\left(\frac{250}{h}\right)\) + 400 h2 …… [From (i)
∴ c = \(\frac{12,500}{h}\) + 400 h2

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 h2
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 h2 + 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 m2
x.x.h = 250
x2 = \(\frac{250}{h}\)
⇒ x2 = 250 × \(\frac{2}{5}\) …… [From Question 47
⇒ x2 = 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 = \(\frac{12,500}{h}\) + 400h2
c = 12,500 × \(\frac{2}{5}\) + 400 × \(\frac{5}{2}\) × \(\frac{5}{2}\)
= 5,000 + 2,500 = ₹7,500


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