Question Set B

cbse 12th Standard Maths Inverse Trigonometric Functions Model Questions Papers Study Material - QB365

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Questions:

Generic 5

2 x 5 = 10

Express matrix A as the sum of a symmetric and a skew-symmetric matrices, where \(A=\left[\begin{array}{rrr}2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2\end{array}\right]\)

Using elementary transformations, find the inverse of the matrix

\(\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] \).

Generic 3

4 x 3 = 12

Prove that \(2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}\)

Find AB, if \(A=\begin{bmatrix} 6 & 9 \\ 2 & 3 \end{bmatrix}\quad \)and \(B=\left[ \begin{matrix} 2 & 6 & 0 \\ 7 & 9 & 8 \end{matrix} \right] \) .

Find the principal value of \({ \cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) \)

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Generic 2

4 x 2 = 8

Evaluate \(3 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}(0)\)

For the matrix \(A=\left[\begin{array}{ll}3 & 1 \\ 7 & 5\end{array}\right]\) find a and b such that \(A^{2}+a I=b A,\) where I is a \(2 \times 2\) identity matrix.

Find the domain of the function \(\cos ^{-1}(2 x-1)\)

10.

If A =\(\left[ \begin{matrix} 0 & 0 \\ -1 & 0 \end{matrix} \right] \)find A⁶

Multiple Choice Question

9 x 1 = 9

11.

If value of \(\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot \theta \operatorname{cosec}^2 \theta d \theta\) is

(a)

\(\frac{1}{2}\)

(b)

- \(\frac{1}{2}\)

(c)

(d)

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

12.

State the function which is continuous for all x \(\in\) R,

(a)

sin x

(b)

\(\frac{x^{2}-25}{x-5}\)

(c)

[x]

(d)

sgn (x)

13.

If \(\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{-1} x,\) then x is equal to

(a)

(b)

\(\frac{\sqrt{5}-4 \sqrt{2}}{9}\)

(c)

\(\frac{\sqrt{5}+4 \sqrt{2}}{9}\)

(d)

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

14.

Let \(\begin{cases} 4x \ >2 \\ ax \ 0\le x\le 2 \\ b \ x<0 \end{cases}\) For what values of a and b, f is a continuous function.

(a)

a = 2, b = 0

(b)

a = 1, b = 0

(c)

a = 0, b = 2

(d)

a = 0, b = 0

15.

If y = x^x-∞, then x(l -y log x)\(\frac { dy }{ dx } \) is equal to

(a)

x²

(b)

y²

(c)

xy²

(d)

x²y

16.

If y = tan^-1 \(\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \), then \(\frac { dy }{ dx } \) is equal to

(a)

\(\frac { 1 }{ 1+{ x }^{ 4 } } \)

(b)

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

(c)

\(\frac { -1 }{ 1+{ x }^{ 4 } } \)

(d)

\(\frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } \)

17.

A function \(f(x)=\begin{cases} \frac { sinx }{ x } +cosx,x\neq 0 \\ 2k\quad \quad \quad \quad ,x=0 \end{cases}\) is continuous at x = 0 for

(a)

k = 1

(b)

k = 2

(c)

K = \(\frac12\)

(d)

k = \(\frac32\)

18.

The diagonal elements of a skew symmetric matrix are

(a)

all zeroes

(b)

are all equal to some scalar k(≠ 0)

(c)

can be any number

(d)

none of these

19.

Total number of possible matrices of order 2 × 3 with each entry 1 or 0 is

(a)

(b)

(c)

(d)

Case Study Questions

1 x 4 = 4

20.

Three car dealers, say A, Band C, deals in three types of cars, namely Hatchback cars, Sedan cars, SUV cars. The sales figure of 2019 and 2020 showed that dealer A sold 120 Hatchback, 50 Sedan, 10 SUV cars in 2019 and 300 Hatchback, 150 Sedan, 20 SUV cars in 2020; dealer B sold 100 Hatchback, 30 Sedan,S SUV cars in 2019 and 200 Hatchback, 50 Sedan, 6 SUV cars in 2020; dealer C sold 90 Hatchback, 40 Sedan, 2 SUV cars in 2019 and 100 Hatchback, 60 Sedan,S SUV cars in 2020.

Based on the above information, answer the following questions.
(i) The matrix summarizing sales data of 2019 is

(ii) The matrix summarizing sales data of 2020 is

(iii) The total number of cars sold in two given years, by each dealer, is given by the matrix

(iv) The increase in sales from 2019 to 2020 is given by the matrix

(v) If each dealer receive profit of Rs. 50000 on sale of a Hatchback, Rs. 100000 on sale of a Sedan and Rs. 200000 on sale of a SUV (v) then amount of profit received in the year 2020 by each dealer is given by the matrix.

Assertion and reason

1 x 1 = 1

21.

Assertion: If \( A=\frac{1}{3}\begin{bmatrix} 1& -2& 2\\ -2& 1& 2\\ -2& -2& -1\\ \end{bmatrix}\), then (A^T)A = I
Reason: For any square matrix, A(A^T)^T = A
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

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Answers Key:

Generic 5

2 x 5 = 10

\(\left[\begin{array}{rrr} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{rrr} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]\)

Use A=lA. Proceed. [Refer 2]

\(\left[ \begin{matrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{matrix} \right] \)

Generic 3

4 x 3 = 12

\(\text {Let } \sin ^{-1} \frac{3}{5}=x . \text { Then, } \sin x=\frac{3}{5}\)
\(\Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}\)
\(\therefore \tan x=\frac{3}{4}\)
\(\therefore x=\tan ^{-1} \frac{3}{4} \Rightarrow \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}\)
Now, we have :
\(\text {L.H.S. }=2 \sin ^{-1} \frac{3}{5}=2 \tan ^{-1} \frac{3}{4} \)
\(=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right) \quad\left[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right] \)
\(=\tan ^{-1}\left(\frac{\frac{3}{2}}{\left.\frac{16-9}{16}\right)}=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)\right. \)
\(=\tan ^{-1} \frac{24}{7}=\mathrm{R} . \mathrm{H.S} . \)

The matrix A has 2 columns which is equal to the number of rows of B.
Hence AB is defined. Now
\(\mathrm{AB}=\left[\begin{array}{lll} 6(2)+9(7) & 6(6)+9(9) & 6(0)+9(8) \\ 2(2)+3(7) & 2(6)+3(9) & 2(0)+3(8) \end{array}\right]\)
\(=\left[\begin{array}{ccc} 12+63 & 36+81 & 0+72 \\ 4+21 & 12+27 & 0+24 \end{array}\right]=\left[\begin{array}{ccc} 75 & 117 & 72 \\ 25 & 39 & 24 \end{array}\right]\)

Let \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =y\), Then \(\cot y=\frac{-1}{\sqrt{3}}=-\cot \left(\frac{\pi}{3}\right)=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \left(\frac{2 \pi}{3}\right)\)
We know that the range of principal value branch of cot^–1 is (0, π) and \(\cot \left(\frac{2 \pi}{3}\right)=\frac{-1}{\sqrt{3}}\)
Hence, principal value of \({ cot }^{ -1 }\left( -\frac { 1 }{ \sqrt { 3 } } \right) =\frac { 2\pi }{ 3 } .\)

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are:
1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3
(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

Generic 2

4 x 2 = 8

\(3 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}(0) \)
\(=3 \times \frac{\pi}{4}+2 \times \frac{\pi}{6}+\frac{\pi}{2} \)
\( =\frac{3 \pi}{4}+\frac{\pi}{3}+\frac{\pi}{2} \)
\( =\frac{9 \pi+4 \pi+6 \pi}{12}=\frac{19 \pi}{12}\)

a = 8, b = 8

We have, f(x) = cos⁻¹(2x−1)
∵ cos⁻¹ ≤ 2x −1 ≤ 1
⇒ 0 ≤ 2x ≤ 2
⇒ 0 ≤ x ≤ 1
∴ x ∈ [0, 1]

10.

\(A^6 = \left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)

Multiple Choice Question

9 x 1 = 9

11.

(a)
\(\frac{1}{2}\)

12.

(a)
sin x

13.

14.

(a)
a = 2, b = 0

15.

As y = xy ⇒ log y = y log x
⇒ \(\frac { 1 }{ y } .{ y }^{ ' }=\frac { y }{ x } +logx.{ y }^{ ' }\)
\(\Rightarrow { y }^{ ' }\left[ \frac { 1 }{ y } -log \ x \right] \)
\(=\frac { y }{ x } \Rightarrow x(1-y \ log \ x){ y }^{ ' }={ y }^{ 2 }\)

16.

y = tan^-1 \(\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \)
= tan^-1 \(\left( \frac { \pi }{ 4 } \right) -{ tan }^{ -1 }{ x }^{ 2 }\)
\({ y }^{ ' }=0-\frac { -1 }{ 1+{ x }^{ 4 } } 2x=\frac { -2x }{ 1+{ x }^{ 4 } } \)

17.

As \(\lim _{ x\rightarrow 0 }{ \left( \frac { sinx }{ x } +cosx \right) } \)
= 1 + 1 = 2 2k
⇒ k = 1

18.

As in skew symmetric matrix, a_ij = -a_ji
⇒ a_ii = – a_ii
⇒ 2a_ii = 0
⇒ a_ii = 0, i.e. diagonal elements are zeroes.

19.

As total elements are 6 and each entry can be done in 2 ways. Hence, total possibilities = 2⁶ = 64.

Case Study Questions

1 x 4 = 4

20.

(i) (b) : In 2019,dealer A sold 120 Hatchback, 50 Sedan and 10 SUV; dealer B sold 100 Hatchback, 30 Sedan and 5 SUV and dealer C sold 90 Hatchback, 40 Sedan and 2 SUV
\(\therefore\) Required matrix, say P, is given by

(ii) (a) : In 2020,
dealer A sold 300 Hatchback, 150 Sedan, 20 SUV
dealer B sold 200 Hatchback, 50 sedan, 6 SUV
dealer C sold 100 Hatchback, 60 sedan,S SUV
\(\therefore\) Required matrix, say Q, is given by

(iii) (c) : Total number of cars sold in two given years, by each dealer, is given by

(iv) (c): The increase in sales from 2019 to 2020 is given by

(v) (c) : The amount of profit in 2020 received by each dealer is given by the matrix

\(\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{c} 15000000+15000000+4000000 \\ 10000000+5000000+1200000 \\ 5000000+6000000+1000000 \end{array}\right]\)
\(\begin{array}{r} A \\ =B \\ C \end{array}\left[\begin{array}{l} 34000000 \\ 16200000 \\ 12000000 \end{array}\right]\)