CBSE Class 12

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-ex-3-2/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.2

Ex 3.2 Class 12 Question 1.
Let
A = \(\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]\)
B = \(\left[\begin{array}{ll} 1 & 3 \\ -2 & 5 \end{array}\right]\)
C = \(\left[\begin{array}{ll} -2 & 5 \\ 3 & 4 \end{array}\right]\)
F ind each of the following:
i. A + B
ii. A – B
iii. 3A – C
iv. AB
v. BA
Solution:

Class 12 Maths 3.2 Question 2.
Compute the following
i. \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\)
ii. \(\left[\begin{array}{cc}
a^{2}+b^{2} & b^{2}+c^{2} \\
a^{2}+c^{2} & a^{2}+b^{2}
\end{array}\right]+\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
iv. \(\left[\begin{array}{cc}
\cos ^{2} x & \sin ^{2} x \\
\sin ^{2} x & \cos ^{2} x
\end{array}\right]+\left[\begin{array}{cc}
\sin ^{2} x & \cos ^{2} x \\
\cos ^{2} x & \sin ^{2} x
\end{array}\right]\)
Solution:

Exercise 3.2 Class 12 Maths Question 3.
Compute the indicated products.

Solution:

iii. Number of columns of first matrix = Number of rows of second matrix.

Class 12 Maths Ncert Solutions Chapter 3 Matrices Exercise 3.2 Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\), B = \(=\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]\) then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A+ B) – C.
Solution:

Class 12 Ex 3.2 Question 5.
If A = \(\left[\begin{array}{ccc}
\frac{2}{\dot{3}} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{array}\right]\), then compute 3A – 5B.
Solution:

Question 6.
Simplify cos θ\(\left[\begin{array}{ll}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sin θ\(\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:

Question 7.
Find X and Y, if
i. X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)

ii. 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) and 3X + 2Y = \(\left[\begin{array}{ll}
2 & -2 \\
-1 & 5
\end{array}\right]\)
Solution:

ii. 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) …… (1)
3X + 2Y = \(\left[\begin{array}{ll}
2 & -2 \\
-1 & 5
\end{array}\right]\) …… (2)

Question 8.
Find X, if Y = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) and 2X + Y = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\).
Solution:

Question 9.
Find x and y, if 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Solution:
2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\) ⇒ \(\left[\begin{array}{ll}
2 & 6 \\
0 & 2x
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Equating the corresponding elements, we get
2 + y = 5 and 2x + 2 = 8
⇒ y = 3 and 2x = 6 ⇒ x = 3
∴ x = 3, y = 3

Question 10.
Solve the equation for x, y, z and t, if 2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]\) + 3\(\left[\begin{array}{ll}
1 & -1 \\
0 & 2
\end{array}\right]\) = 3\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]\) + 3\(\left[\begin{array}{ll}
1 & -1 \\
0 & 2
\end{array}\right]\) = 3\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
2x & 2z \\
2y & 2t
\end{array}\right]\) + \(\left[\begin{array}{cc}
3 & -3 \\
0 & 6
\end{array}\right]\) = \(\left[\begin{array}{ll}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2x+3 & 2z-3 \\
2y & 2t+6
\end{array}\right]\) = \(\left[\begin{array}{ll}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ 2x + 3 = 9, 2z – 3 = 15
2y = 12, 2t + 6 = 18 ⇒ x = 3, y = 6, t = 6, z = 9

Question 11.
If x\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\) + y\(\left[\begin{array}{l}
-1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
10 \\
5
\end{array}\right]\)
Solution:
x\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\) + y\(\left[\begin{array}{l}
-1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
10 \\
5
\end{array}\right]\) ⇒ \(\left[\begin{array}{l}
2 x \\
3 x
\end{array}\right]+\left[\begin{array}{c}
-y \\
y
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\) ⇒ \(\left[\begin{array}{l}
10 \\
5
\end{array}\right]\)
⇒ 2x – y = 10, 3x + y = 5
Solving, we get x = 3, y = – 4

Question 12.
Given
\(3\begin{bmatrix} x & \quad y \\ z & \quad w \end{bmatrix}=\begin{bmatrix} x & \quad 6 \\ -1 & \quad 2w \end{bmatrix}+\begin{bmatrix} 4 & \quad x+y \\ z+w & 3 \end{bmatrix} \)
find the values of x,y,z and w.
Solution:
\(3\begin{bmatrix} x & \quad y \\ z & \quad w \end{bmatrix}=\begin{bmatrix} x & \quad 6 \\ -1 & \quad 2w \end{bmatrix}+\begin{bmatrix} 4 & \quad x+y \\ z+w & 3 \end{bmatrix} \)
⇒ \(\begin{bmatrix} 3x & \quad 3y \\ 3z & \quad 3w \end{bmatrix}=\begin{bmatrix} x+4 & \quad 6+x+y \\ -1+z+w & \quad 2w+3 \end{bmatrix}\)
⇒ 3x = x + 4 ⇒ x = 2
and 3y = 6 + x + y ⇒ y = 4
Also, 3w = 2w + 3 ⇒ w = 3
Again, 3z = – 1 + z + w
⇒ 2z = – 1 + 3
⇒ 2z = 2
⇒ z = 1
Hence x = 2 ,y = 4, z = 1, w = 3.

Question 13.
If F(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\) then show that F(x).F(y) = F(x+y)
Solution:

Question 14.
Show that

Solution:

Question 15.
Find A² – 5A + 6I, if A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Solution:

Question 16.
If A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\) Prove that A³ – 6A² + 7A + 2I = 0
Solution:

Question 17.
If \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find k so that A² = kA – 2I
Solution:

Question 18.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\) and I is the identity matrix of order 2, show that I + A = (I – A)\(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:

Question 19.
A trust has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bond if the trust fund obtains an annual total interest of
(a) Rs 1800
(b) Rs 2000
Solution:
(a) Let ₹ and ₹ y be the amount invested in two bonds such that x + y = 30000 … (1)
Then [x y]\(\left[\begin{array}{l}
5 \% \\
7 \%
\end{array}\right]\) = [1800]
⇒ [x × 5% + y x 7%] = [1800]
⇒ \(\frac{x \times 5}{100}+\frac{y \times 7}{100}\)
⇒ 5x + 7y = 180000 … (2)
(1) x 5 = 5x + 5y = 150000 … (3)
Subtracting (3) from (2), we get
2y = 30000 ⇒ y = 15000
From (1), we get x= 15000
∴ To get an interest of ₹ 1800, the amount to be invested in each bonds are ₹ 15000 and ₹ 15000

(b) Here x + y = 30000 … (4)
[x y]\(\left[\begin{array}{l}
5 \% \\
7 \%
\end{array}\right]\) = 2000
⇒ \(\frac{x \times 5}{100}+\frac{y \times 7}{100}\) = 2000
⇒ 5x + 7y = 200000 … (5)
(4) x 5 ⇒ 5x + 5y = 150000 …. (6)
Subtracting (6) from (5), we get
2y = 50000 ⇒ y = 25000
From (4) we get x = 5000
To get an interest of ₹ 2000, the amount to be invested in each bonds is ₹ 5000 and ₹ 25000.

Question 20.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution:
1 dozen = 12
∴ 10 dozen chemistry books = 10 x 12 = 120
8 dozen physics books = 8 x 12 = 96
10 dozen economics books = 10 x 12 = 120
Total amount the bookshop will receive
= [120 96 120]\(\left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right]\)
= [120 x 80 + 96 x 60 + 120 x 40]
= [9600 + 5760 + 4800] = ₹ 20160
Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k, respectively. Choose the correct answer in Exercises 21 and 22.

Question 21.
The restriction on n, k and p so that PY + WY will be defined are
a. k = 3, p = n
b. k is arbitrary, p = 2
c. p is arbitrary, k = 3
d. k = 2, p = 3
Solution:
Answer:
PY + WY = (P + W)Y
⇒ order of P = order of W
⇒ p k = n x 3
⇒ p = n and k = 3

Question 22.
If n = p, then the order of the matrix 7X – 5Z is
a. p x 2
b. 2 x n
c. n x 3
d. p x n
Solution:
Order of X = 2 x n
Order of Z = 2 x p = 2 x n ∵ (n = p)
∴ Order of 7X – 5Z is order of X or order of Z.

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-ex-3-3/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.3

Ex 3.3 Class 12 NCERT Solutions Question 1.
Find the transpose of each of the following matrices:
(i) \(\left[ \begin{matrix} 5 \\ \frac { 1 }{ 2 } \\ -1 \end{matrix} \right] \)
(ii) \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
(iii) \(\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt { 3 } & 5 & 6 \\ 2 & 3 & -1 \end{matrix} \right] \)
Solution:

Question 2.
If \(A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix} \right] \)
then verify that:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’
Solution:
i. (A + B)

From (1) and (2), (A + B)’ = A’ + B’

ii. (A – B)

From (1) and (2), (A – B)’ = A’ – B’.

Question 3.
If \(A’=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right] \)
then verify that:
(i) (A+B)’ = A’+B’
(ii) (A-B)’ = A’-B’
Solution:

Question 4.
If \(A’=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} ,B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} \) then find (A+2B)’
Solution:

Question 5.
For the matrices A and B, verify that (AB)’ = B’A’, where
\((i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \end{matrix} \right] \)
\((ii)\quad A=\left[ \begin{matrix} 0 \\ 1 \\ 2 \end{matrix} \right] ,B=\left[ \begin{matrix} 1 & 5 & 7 \end{matrix} \right] \)
Solution:
\((i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right] \)
\(A’=\left[ \begin{matrix} 1 & -4 & 3 \end{matrix} \right] \)

Question 6.
If (i) \(A=\begin{bmatrix} cos\alpha & \quad sin\alpha \\ -sin\alpha & \quad cos\alpha \end{bmatrix} \) , the verify that A’A = I
If (ii) \(A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -cos\alpha & \quad sin\alpha \end{bmatrix} \), the verify that A’A = I
Solution:

Question 7.
(i) Show that the matrix \(A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \) is a symmetric matrix.
(ii) Show that the matrix \(A=\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \) is a skew-symmetric matrix.
Solution:
(i) \(A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \), A’ = \(\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \) = A
A is symmetric matrix, since A’ = A

(ii) \(A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -cos\alpha & \quad sin\alpha \end{bmatrix} \)
A’ = \(\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \) = – 1\(\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \)
A’ = – A
A is a skew-symmetric matrix, since A’ = A

Question 8.
For the matrix, \(A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}\)
(i) (A + A’) is a symmetric matrix.
(ii) (A – A’) is a skew-symmetric matrix.
Solution:

Question 9.
Find \(\\ \frac { 1 }{ 2 } (A+A’)\) and \(\\ \frac { 1 }{ 2 } (A-A’)\), when \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
Solution:

Question 10.
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
(i) \(\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}\)
(ii) \(\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right] \)
(iii) \(\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right] \)
(iv) \(\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}\)
Solution:

Thus A can be represented as a sum of a symmetric and a skew-symmetric matrix.

Thus B can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Thus A can be expressed as the sum of a symmetric and skew-symmetric matrix.

Thus D can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Question 11.
Choose the correct answer in the following questions:
If A, B are symmetric matrices of same order then AB-BA is a
(a) Skew – symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity matrix
Solution:
Now A’ = B, B’ = B
(AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB
= – (AB – BA)
AB – BA is a skew-symmetric matrix Hence, option (a) is correct.

Question 12.
If \(A=\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}\) then A+A’ = I, if the
value of α is
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) π
(d) \(\frac { 3\pi }{ 2 } \)
Solution:
Now

Thus option (b) is correct.

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-2-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Question 1.
cos^-1(cos\(\frac { 13π }{ 6 }\))
Solution:
The principal value brach of cos^-1 is [0, p]

Question 2.
tan^-1(tan\(\frac { 7π }{ 6 }\))
Solution:
The principal value brach of tan^-1 is (- \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))

Question 3.
2sin^-1\(\frac { 3 }{ 5 }\) = tan^-1\(\frac { 24 }{ 7 }\)
Solution:

Question 4.
sin^-1\(\frac { 8 }{ 17 }\) + sin^-1\(\frac { 3 }{ 5 }\) = tan^-1\(\frac { 77 }{ 36 }\)
Solution:

Question 5.
cos^-1\(\frac { 4 }{ 5 }\) + cos^-1\(\frac { 12 }{ 13 }\) = cos^-1\(\frac { 33 }{ 65 }\)
Solution:

Question 6.
cos^-1\(\frac { 12 }{ 13 }\) + sin^-1\(\frac { 3 }{ 5 }\) = sin^-1\(\frac { 56 }{ 65 }\)
Solution:

Question 7.
tan^-1\(\frac { 63 }{ 16 }\) = sin^-1\(\frac { 5 }{ 13 }\) + cos^-1\(\frac { 3 }{ 5 }\)
Solution:

Question 8.
tan^-1\(\frac { 1 }{ 5 }\) + tan^-1\(\frac { 1 }{ 7 }\) + tan^-1\(\frac { 1 }{ 3 }\) + tan^-1\(\frac { 1 }{ 8 }\) = \(\frac { π }{ 4 }\)
Solution:

Question 9.
tan^-1\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos^-1\(\left(\frac{1-x}{1+x}\right), x \in[0,1]\)
Solution:

Question 10.
cot^-1\(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\) = \(\frac { x }{ 2 }\) x ∈ (0, \(\frac { π }{ 4 }\))
Solution:

Question 11.
tan^-1\(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) = \(\frac { π }{ 4 }\) – \(\frac { 1 }{ 2 }\) cos^-1 x, – \(\frac{1}{\sqrt{2}} \leq x \leq 1\)
Solution:

Question 12.
\(\frac { 9p }{ 8 }\) – \(\frac { 9 }{ 4 }\)sin^-1\(\frac { 1 }{ 3 }\) = \(\frac { 9 }{ 4 }\)sin^-1\(\frac{2 \sqrt{2}}{3}\)
Solution:

Question 13.
2tan^-1(cosx) = tan^-1(2cosecx)
Solution:

Question 14.
tan^-1\(\frac{1-x}{1+x}\) = \(\frac { 1 }{ 2 }\) tan^-1x, (x > 0)
Solution:

Question 15.
sin(tan^-1x), |x| < 1 is equal to
a. \(\frac{x}{\sqrt{1-x^{2}}}\)
b. \(\frac{1}{\sqrt{1-x^{2}}}\)
c. \(\frac{1}{\sqrt{1+x^{2}}}\)
d. \(\frac{x}{\sqrt{1+x^{2}}}\)
Solution:
d. \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 16.
sin^-1(1 – x) – 2sin^-1x = \(\frac { π }{ 2 }\), then x is equal to
a. 0, \(\frac { 1 }{ 2 }\)
b. 1, \(\frac { 1 }{ 2 }\)
c. 0
d. \(\frac { 1 }{ 2 }\)
Solution:
c. 0

But x ≠ \(\frac { 1 }{ 2 }\), since sin^-1(1 – \(\frac { 1 }{ 2 }\)) ≠ \(\frac { π }{ 2 }\) + 2 sin^-1\(\frac { 1 }{ 2 }\)
∴ x = 0.

Question 17.
tan^-1 \(\frac { x }{ y }\) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:
a. \(\frac { π }{ 2 }\)
b. \(\frac { π }{ 3 }\)
c. \(\frac { π }{ 4 }\)
d. \(\frac { -3π }{ 4 }\)
Solution:
c. \(\frac { π }{ 4 }\)

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-2-ex-2-2/

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2

Ex 2.2 Class 12 NCERT Solutions Question 1.
3sin^-1x = sin^-1(3x – 4x³). x ∈ [-\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\)]
Solution:
Let θ = sin^-1 ⇒ x = sin θ
∴ 3x – 4x³ = 3sinθ – 4sin³θ = sin3 θ
∴ 3θ = sin^-1(3x – 4x³)
⇒ 3sin^-1x = sin^-1(3x – 4x³)

Class 12th Maths Chapter 2 Exercise 2.2 Question 2.
3cos^-1x = cos^-1(4x³ – 3x). x ∈ [\(\frac { 1 }{ 2 }\), 1]
Solution:
Let θ cos^-1 ⇒ x = cos θ
4x³ – 3x = 4cos³θ – 3cosθ = cos 3 θ
⇒ 3θ = cos^-1(4x³ – 3x)
⇒ 3cos^-1x = cos^-1(4x³ – 3x)

Maths Ex 2.2 Class 12 NCERT Solutions Question 3.
tan^-1\(\frac { 2 }{ 11 }\) + tan^-1\(\frac { 7 }{ 24 }\) = tan^-1\(\frac { 1 }{ 2 }\)
Solution:

Ncert Ex 2.2 Class 12  Question 4.
2tan^-1\(\frac { 1 }{ 2 }\) + tan^-1\(\frac { 1 }{ 7 }\) = tan^-1\(\frac { 31 }{ 17 }\)
Solution:

Exercise 2.2 Maths Class 12 Solutions Question 5.
tan^-1\(\frac{\sqrt{1+x^{2}}-1}{x}\), x ≠ 0
Solution:

Question 6.
tan^-1\(\frac{1}{\sqrt{x^{2}-1}}\), |x| > 1
Solution:

Question 7.
tan^-1\(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\), x < π
Solution:

Question 8.
tan^-1\(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\), x < π
Solution:

Question 9.
tan^-1\(\frac{x}{\sqrt{a^{2}-x^{2}}}\), |x| < a
Solution:

Question 10.
tan^-1\(\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a>0 ; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\).
Solution:

Question 11.
tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\))]
Solution:

Question 12.
cot(tan^-1 a + cot^-1 a)
Solution:
Since tan^-1 a + cot^-1 a = \(\frac{π}{2}\),
cot(tan^-1 a + cot^-1 a = cot(\(\frac{π}{2}\)) = 0

Question 13.
tan\(\frac{1}{2}\)[\(\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]\))], |x| < 1, y > 0 and xy < 1
Solution:

Question 14.
If sin(sin^-1\(\frac{1}{5}\) + cos^-1x) = 1, then find the value of x.
Solution:

Question 15.
If tan^-1\(\frac{x-1}{x-2}\) + tan^-1\(\frac{x+1}{x+2}\), then find the value of x.
Solution:

Question 16.
sin^-1\(\left(\sin \frac{2 \pi}{3}\right)\)
Solution:

Question 17.
tan^-1\(\left(\tan \frac{3 \pi}{4}\right)\)
Solution:

Question 18.
tan\(\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Solution:

Question 19.
cos^-1\(\left(\cos \frac{7 \pi}{6}\right)\) is equal to
a. \(\frac{7π}{6}\)
b. \(\frac{5π}{6}\)
c. \(\frac{π}{3}\)
d. \(\frac{π}{6}\)
Solution:
b. \(\frac{5π}{6}\)

Question 20.
sin\(\left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)\) is equal to
a. \(\frac{1}{2}\)
b. \(\frac{1}{3}\)
c. \(\frac{1}{4}\)
d. 1
Solution:
d. 1

Question 21.
tan^-1\(\sqrt{3}\) – cot^-1\(\sqrt{3}\) is equal to
a. π
b. – \(\frac{π}{2}\)
c. 0
d. 2\(\sqrt{3}\)
Solution:
b. – \(\frac{π}{2}\)
tan^-1\(\sqrt{3}\) – cot^-1(-\(\sqrt{3}\))
= \(\sqrt{3}\) – (π – cot^-1\(\sqrt{3}\))
= (tan^-1\(\sqrt{3}\) + cot^-1\(\sqrt{3}\)) – π
= \(\frac{π}{2}\) – π = – \(\frac{π}{2}\)

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-2-ex-2-1/

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1

Ex 2.1 Class 12 Question 1.
sin^-1\(\frac { -1 }{ 2 }\)
Solution:

Question 2.
cos^-1\(\left(\frac{\sqrt{3}}{2}\right)\)
Solution:

Question 3.
cosec^-1(2)
Solution:
The principal values branch of cosec^-1 is

Question 4.
tan^-1(-\(\sqrt{3}\))
Solution:
The principal values branch of tan^-1 is

Question 5.
cos^-1\(\frac { – 1 }{ 2 }\)
Solution:
The principal values branch of cos^-1 is [0, π]

Question 6.
tan^-1(-1)
Solution:
The principal values branch of tan^-1 is

Question 7.
sec^-1\(\left(\frac{2}{\sqrt{3}}\right)\)
Solution:
The principal values branch of sec^-1 is

Question 8.
cot^-1(\(\sqrt{3}\))
Solution:
The principal values branch of cot^-1 is [0, π]
cot\(\frac { π }{ 6 }\) = \(\sqrt{3}\), \(\frac { π }{ 6 }\) ∈ [0, π]
∴ Priciapal value of cot^-1\(\sqrt{3}\) is \(\frac { π }{ 6 }\)

Question 9.
cos^-1\(\left(\frac{-1}{\sqrt{2}}\right)\)
Solution:
The principal values branch of cos^-1 is [0, π]

Question 10.
cosec^-1(\(\sqrt{-2}\))
Solution:
The principal values branch of cosec^-1 is

Question 11.
tan^-1(1) + cos^-1(\(\frac { – 1 }{ 2 }\)) + sin^-1(\(\frac { – 1 }{ 2 }\))
Solution:

Question 12.
cos^-1(\(\frac { 1 }{ 2 }\)) + 2 sin^-1(\(\frac { 1 }{ 2 }\))
Solution:

Question 13.
If sin^-1 x = y, then
a. 0 ≤ y ≤ π
b. – \(\frac { π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\)
c. 0 < y < π
d. – \(\frac { π }{ 2 }\) < y < \(\frac { π }{ 2 }\)
Solution:
b. – \(\frac { π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\)
The principal values branch of sin^-1 is
[\(\frac { – π }{ 2 }\), \(\frac { π }{ 2 }\)] . i.e., \(\frac { – π }{ 2 }\) ≤ y ≤ \(\frac { π }{ 2 }\)

Question 14.
tan^-1\(\sqrt{3}\) – sec^-1 ( – 2) is equal to
a. π
b . \(\frac { – π }{ 3 }\)
c. \(\frac { π }{ 3 }\)
d. \(\frac { 2π }{ 3 }\)
Solution:
b . \(\frac { – π }{ 3 }\)
tan^-1\(\sqrt{3}\) – sec^-1 ( – 2) = \(\frac { π }{ 3 }\) – \(\frac { 2π }{ 3 }\) = \(\frac { – π }{ 3 }\)

These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-1-ex-1-3/

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.3

Question 1.
Let f : {1,3,4} → (1,2, 5} and g: (1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
Solution:

gof = {(1, 3),(3, 1), (4, 3)}

Question 2.
Let f, g and h be functions from R to R. Show that
i. (f + g)oh = foh + goh
ii. (f. g)oh = (foh) . (goh)
Solution:
i. (f + g)oh(x) = (f + g)h(x)) = f(h(x)) + g(h(x)) = (foh)(x) + (goh)(x)
∴ (f + g)oh = foh + goh

ii. (f. g)oh(x) = (f.g)(h(x))
= f(h(x)).g(h(x))
= (foh)(x). (goh)(x)
= [(foh).(goh)](x)
∴ (f. g)oh = (foh) . (goh)

Question 3.
Find gof and fog, if
i. f(x) = |x| and g(x) = |5x – 2|
ii. f(x) = 8x and g(x) = x\(x^{\frac{1}{3}}\)
(March 2014, SAY 2015, March 2016)
Solution:
i. (gof)(x) = g(f(x)) = g(|x|) = |5|x| – 2|
(fog)(x) = f(g(x)) = f(|5x – 2|)
= ||5x – 2|| = |5x – 2|

ii. (gof)(x) = g(f(x)) = g(8x³) = (8x³ )A^1/3 = 2x
(fog)(x) = f(g(x)) = f(x^1/3) = 8(x^{1/3)³ = 8x}

Inverse Function Calculator. Use this free tool to find the inverse of a function

Question 4.
If f(x) = \(\frac{(4 x+3)}{(6 x-4)}\), x ≠ 3, show that fof (x) = x, for all x ≠ \(\frac{2}{3}\). What is the inverse of f?
Solution:

Another method:
Without using the result fof(x) = x, we can directly find the inverse off.

Question 5.
State with reason whether following functions have inverse
i. f : {1,2, 3,4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
ii. g: {5, 6, 7, 8} → {1, 2, 3, 4} with g ={(5, 4), (6, 3), (7, 4), (8, 2)}
iii. h : {2, 3,4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Solution:
i. The elements 1,2,3 and 4 have the same image.
∴ g is not one-one.
Hence g has no inverse.

ii. The image of the elements 5 and 7 is the same
∴ g is not one-one
Hence g has no inverse.

iii.

From the arrow diagram, it is clear that h is one-one and onto.
Hence h has inverse.

Question 6.
Show that f : [-1, 1] → R, given by f(x) = \(\frac{x}{(x+2)}\) is one-one. Find the inverse of the function f : [- 1, 1] → Range f.
Solution:
Let x₁, x₂ ∈ [-1, 1]
f(x₁) = f(x₂) ⇒ \(\frac{x_{1}}{2+x_{1}}\) = \(\frac{x_{2}}{2+x_{2}}\)
⇒ x₁(2 + x₂) = x₂(2 + x₁)
⇒ 2x₁ + x₁x₂ = 2x₂ + x₁x₂
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂
∴ f is one-one
Let y = \(\frac{x}{x+2}\) ⇒ y(x + 2)= x
⇒ xy – x = – 2y
⇒ x (1 – y) = 2y
⇒ x = \(\frac{2y}{1-y}\), y ≠ 1
∴ f^-1(x) = \(\frac{2x}{1-x}\)

Question 7.
Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
f(x) = 4x + 3, x ∈ R
Domain of f = R and range of f = R
One-one
Let x₁, x₂ ∈ R
f(x₁) = f(x₂) ⇒ 4x₁ + 3 = 4x₂ + 3
⇒ 4x₁ = 4x₂
⇒ x₁ = x₂
∴ f is one-one.

Onto
Let y ∈ range of f such that y = f(x)

i.e., corresponding to every y ∈ R, there exists a real number \(\frac{y-3}{4}\) such that f(\(\frac{y-3}{4}\)) = y
∴ f is onto
Hence f is a bijection and f^-1 exists
To find f^-1
Let y = f(x)
Then x = \(\frac{y-3}{4}\) from (1)
∴ Inverse of/is given by f^-1(x) = \(\frac{x-3}{4}\)

Question 8.
Consider f : R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^-1 of f given by f^-1(y) = \(\sqrt{y-4}\), where R₊ is the set of all non-negative real numbers.
Solution:
Let x₁, x₂ ∈ R₊ such that f(x₁) = f(x₂)
⇒ x²₁ + 4 = x²₂ + 4
⇒ x²₁ = x²₂ ⇒ (x²₁ – x²₂) = 0
⇒ (x₁ – x₂) (x₁ + x₂) = 0
⇒ x₁ – x₂ = 0 since x₁, x₂ ∈ R
⇒ x₁ = x₂
∴ f is one-one.
Let y ∈ [4, ∞) such that y = f(x)
⇒ y = x² + 4 ⇒ x² = y – 4
⇒ x = \(\sqrt{y – 4}\) since x ∈ R₊
i.e., x = \(\sqrt{y – 4}\) ∈ R₊+ …. (1)
f(x) = f(\(\sqrt{y – 4}\)) = (\(\sqrt{y – 4}\))² + 4
= y – 4 + 4 = y
∴ f is onto
Since f is one-one and onto, f is invertible.
To find f^-1
Let y = x² + 4
⇒ x = \(\sqrt{y – 4}\) from(1)
∴ f^-1(y) = \(\sqrt{y – 4}\)

Question 9.
Consider f : R₊ → [- 5, ∞) given by f(x) = 9x² + 6x – 5. Show that f is invertible with f^-1(y) = \(\left(\frac{(\sqrt{y+6})-1}{3}\right)\)
Solution:
Let y = 9x² + 6x – 5 = (3x + 1)²
⇒ (3x + 1)² = y + 6
⇒ 3x + 1 = \(\sqrt{y + 6}\) ⇒ 3x = \(\sqrt{y + 6}\) – 1
⇒ x = \(\frac{\sqrt{y+6}-1}{3}\)

Hence gof = fog = Identity function. Hence f is invertible and
f^-1(y) = \(\left(\frac{(\sqrt{y+6})-1}{3}\right)\)

Question 10.
Let f : X → Y be an invertible function. Show that f has unique inverse.
Solution:
Let g₁ and g₂ be two inverses of f.
Then for all y ∈ Y
(fog₁)y = y and (fog₂)y = y
⇒ (fog₁)y = (fog₂)(y)
⇒ f(g₁(y)) = f(g₂(y))
⇒ g₁(y) = g₂(y) since f is one-one
⇒ g₁ = g₂
Hence the inverse of f is unique.

Question 11.
Consider f : {1,2, 3} → {a, b,c} given by f(1) = a, f (2) = b and f(3) = c. Find f^-1 and show that (f^-1)^-1 = f.
Solution:
f = {(1, a), (2, b), (3, c)}
f^-1 = {(a, 1), (b, 2), (c, 3))
Clearly f is bijective
(f^-1)^-1 = {(1, a), (2, b), (3, c)} = f

Question 12.
Let f : X → Y be an invertible function. Show that the inverse of f^-1 is f, i.e., (f^-1)^-1 = f.
Solution:
f : X → Y is invertible. Then f^-1 : Y → X
Let x ∈ X , then y = f(x) ∈ Y such that f^-1 (y) = x
Now (f^-1of)(x) = f^-1(f(x)) = f^-1(y) = x
and (fof^-1)(y) = f(f^-1(y)) = f(x) = y
i.e., f^-1 of and fof^-1 are identity functions.
Hence they are inverse of each other.
i.e., the inverse of f^-1 is f.
i.e., (f^-1)^-1 = f.

Question 13.
If f : R → R be given by f(x) = (3 – x³)^{\(\frac { 1 }{ 3 }\)}, then fof (x) is
a. x^{\(\frac { 1 }{ 3 }\)}
b. x³
c. x
d. (3 – x³)
Solution:
c. x

Question 14.
Let f : R – { – \(\frac { 4 }{ 3 }\) } → be a function defined as f(x) = \(\frac { 4x }{ 3x+4 }\). The inverse of f is the map g : Range f → R – \(\frac { 4 }{ 3 }\) given by
a. g(y) = \(\frac { 3y }{ 3-4y }\)
b. g(y) = \(\frac { 4y }{ 4-3y }\)
a. g(y) = \(\frac { 4y }{ 3-4y }\)
a. g(y) = \(\frac { 3y }{ 4-3y }\)
Solution:
b. g(y) = \(\frac { 4y }{ 4-3y }\)
Let \(\frac { 4x }{ 3x+4 }\) = y ⇒ 4x = 3xy + 4y
⇒ 4x – 3xy = 4y ⇒ x = \(\frac { 4y }{ 4-3y }\)
∴ g(y) = \(\frac { 4y }{ 4-3y }\).