Class 12

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(b)

Question 1.
State which of the following matrices are symmetric, skew-symmetric, both or not either:

Solution:
(i) Symmetric
(ii) Neither Symmetric nor skew-symmetric
(iii) Symmetric
(iv) Skew symmetric
(v) Both
(vi) Neither symmetric nor skew-symmetric
(vii) Skew symmetric

Question 2.
State ‘True’ or ‘False’:
(i) If A and B are symmetric matrices of the same order and AB – BA ≠ 0, then AB is not symmetric.
Solution:
True

(ii) For any square matrix A, AA’ is symmetric.
Solution:
True

(iii) If A is any skew-symmetric matrix, then A² is also skew-symmetric.
Solution:
False

(iv) If A is symmetric, then A², A³, …, Aⁿ are all symmetric.
Solution:
True

(v) If A is symmetric then A – A¹ is both symmetric and skew-symmetric.
Solution:
False

(vi) For any square matrix (A – A¹)² is skew-symmetric.
Solution:
True

(vii) A matrix which is not symmetric is skew-symmetric.
Solution:
False

Question 3.
(i) If A and B are symmetric matrices of the same order with AB ≠ BA, final whether AB – BA is symmetric or skew symmetric.
Solution:
A and B are symmetric matrices;
Thus A’ = A and B’ = B
Now (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB = – (AB – BA)
∴ AB – BA is skew symmetric.

(ii) If a symmetric/skew-symmetric matrix is expressed as a sum of a symmetric and a skew-symmetric matrix then prove that one of the matrices in the sum must be zero matrix.
Solution:
We know that zero matrix is both symmetric as well as skew-symmetric.
Let A is symmetric.
∴ A = A + O where A is symmetric and O is treated as skew-symmetric. If B is skew-symmetric then we can write B = O + B where O is symmetric and B is skew-symmetric.

Question 4.
A and B are square matrices of the same order, prove that
(i) If A, B and AB are all symmetric, then AB – BA = 0
Solution:
Let A, B and AB are all symmetric.
∴A’ = A, B’ = B and (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB
⇒ AB – BA = 0

(ii) If A, B and AB are all skew symmetric then AB + BA = 0
Solution:
Let A, B and AB are all skew symmetric matrices
∴ A’ = -A, B’ = -B and (AB)’ = -AB
Now (AB)’ = -AB
⇒ B’A’ = -AB
⇒ (-B) (-A) = -AB
⇒ BA = -AB
⇒ AB + BA = 0

Question 5.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A’ = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
2 & 1 & 5 \\
0 & 3 & 3
\end{array}\right]\)

(i) A+A’ is symmetric

(ii) A-A’ is skew-symmetric

Question 6.
Prove that a unit matrix is its own inverse. Is the converse true?
IfA = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\) show that A² = I and hence A= A^-1.
Solution:
No the converse is not true for example:

Question 7.
Here A is an involuntary matrix, recall the definition given earlier.
Solution:

Question 8.
Show that \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\) is its own inverse.
Solution:

Question 9.
Express as a sum of a symmetric and a skew symmetric matrix.

Solutions:

Question 10.
What is the inverse of

Question 11.
Find inverse of the following matrices by elementary row/column operation (transformations):
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find the inverse of the following matrices using elementary transformation:
(i) \(\left[\begin{array}{lll}
\mathbf{0} & \mathbf{0} & 2 \\
\mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{2} & \mathbf{0} & \mathbf{0}
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]\)
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(b)

Question 1.
Differentiate from definition
(i) e^3x
Solution:

(ii) 2^x2
Solution:

(iii) In (3x + 1)
Solution:

(iv) logx⁵ (Hint : logx⁵ = \(\frac{\ln 5}{\ln x}\))
Solution:

(v) In sin x
Solution:

(vi) x² a^2x
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 5 Determinants Ex 5(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 5 Determinants Exercise 5(b)

Question 1.
Write the number of solutions of the following system of equations.
(i) x – 2y = 0
Solution:
No solution

(ii) x – y = 0 and 2x – 2y = 1
Solution:
Infinite

(iii) 2x + y = 2 and -x – 1/2y = 3
Solution:
No solution

(iv) 3x + 2y = 1 and x + 5y = 6
Solution:
One

(v) 2x + 3y + 1 = 0 and x – 3y – 4 = 0
Solution:
One

(vi) x + y + z = 1
x + y + z = 2
2x + 3y + z = 0
Solution:
No solution

(vii) x + 4y – z = 0
3x – 4y – z = 0
x – 3y + z = 0
Solution:
One

(viii) x + y – z = 0
3x – y + z = 0
x – 3y + z = 0
Solution:
One

(ix) a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0
and \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
Infinite solutions as Δ = Δ₁ = Δ₂ = Δ₃ = 0

Question 2.
Show that the following system is inconsistent.
(a – b)x + (b – c)y + (c – a)z = 0
(b – c)x + (c – a)y + (a – b)z = 0
(c – a)x + (a – b)y + (b – c)z =1
Solution:

Question 3.
(i) The system of equations
x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6 has
(a) infinitely many solutions
(b) no solution
(c) a unique solution
(d) none of the three
Solution:
(a) infinitely many solutions

(ii) If the system of equations
2x + 5y + 8z = 0
x + 4y + 7z = 0
6x + 9y – z = 0
has a nontrivial solution, then is equal to
(a) 12
(b) -12
(c) 0
(d) none of the three
Solution:
(b) -12

(iii) The system of linear equations
x + y + z = 2
2x + y – z = 3
3x +2y + kz = 4
has a unique solution if
(a) k ≠ 0
(b) -1 < k < 1
(c) -2 < k < 2
(d) k = 0
Solution:
(a) k ≠ 0

(iv) The equations
x + y + z = 6
x + 2y + 3z = 10
x + 2y + mz = n
give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) none of the three
Solution:
(c) m = 3, n = 10

(v) The system of equations
2x – y + z = 0
x – 2y + z = 0
x – y + 2z = 0
has infinite number of nontrivial solutions for
(a) = 1
(b) = 5
(c) = -5
(d) no real value of
Solution:
(c) = -5

(vi) The system of equations
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z =0
with has
(a) more than two solutions
(b) one trivial and one nontrivial solutions
(c) No solution
(d) only trivial solutions
Solution:
(a) more than two solutions

Question 4.
Can the inverses of the following matrices be found?
(i) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
|A| = 0
∴ A^-1 can not be found.

(ii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
∴ |A| = 4 – 6 = -2 ≠ 0
∴ A^-1 exists.

(iii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) = 1 – 1 = 0
∴ A^-1 does not exist.

(iv) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) = 4 – 4 = 0
∴ A^-1 does not exist.

(v) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1 ≠ 0
∴ A^-1 exists.

Question 5.
Find the inverse of the following:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{cc}
i & -i \\
i & i
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
x & -x \\
x & x^2
\end{array}\right]\), x ≠ 0, x ≠ -1
Solution:

Question 6.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & -1 & 2 \\
1 & 3 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
-2 & 2 & 3 \\
1 & 4 & 2 \\
-2 & -3 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
2 & 2 & 1 \\
1 & 2 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 3 & 0 \\
2 & -1 & 6 \\
5 & -3 & 1
\end{array}\right]\)
Solution:

Question 7.
Which of the following matrices are invertible?
(i) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
2 & 1 & -2 \\
1 & 2 & 1 \\
3 & 6 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
2 & 1 & -4 \\
-1 & 0 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -2 & 1 \\
3 & 2 & 4
\end{array}\right]\)
Solution:

Question 8.
Examining consistency and solvability, solve the following equations by matrix method.
(i) x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:

(ii) x + 2y – 3z = 4
2x + 4y – 5z = 12
3x – y + z = 3
Solution:
Let

(iii) 2x – y + z = 4
x + 3y + 2z = 12
3x + 2y + 3z = 16
Solution:

(iv) x + y + z = 4
2x + 5y – 2x = 3
x + 7y – 7z = 5
Solution:

(v) x + y + z = 4
2x – y + 3z = 1
3x + 2y – z = 1
Solution:

(vi) x + y – z = 6
2x – 3y + z = 1
2x – 4y + 2z = 1
Solution:

(vii) x – 2y = 3
3x + 4y – z = -2
5x – 3z = -1
Solution:

(viii) x + 2y + 3z = 14
2x – y + 5z = 15
2y + 4z – 3x = 13
Solution:

(ix) 2x + 3y +z = 11
x + y + z = 6
5x – y + 10z = 34
Solution:

Question 9.
Given the matrices
A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
write down the linear equations given by AX = C and solve it for x, y, z by matrix method.
Solution:

Question 10.
Find X, if \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & -1 \\
2 & 1 & -1
\end{array}\right]\) X = \(\left[\begin{array}{l}
6 \\
0 \\
1
\end{array}\right]\) where X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\)
Solution:

Question 11.
Answer the following:
(i) If every element of a third order matrix is multiplied by 5, then how many times its determinant value becomes?
Solution:
125

(ii) What is the value of x if \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=,\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\) ?
Solution:
x = 6

(iii) What are the values of x and y if \(\left|\begin{array}{ll}
x & y \\
1 & 1
\end{array}\right|=2,\left|\begin{array}{ll}
x & 3 \\
y & 2
\end{array}\right|=1\) ?
Solution:
x = 5, y = 3

(iv) What is the value of x if \(\left|\begin{array}{ccc}
x+1 & 1 & 1 \\
1 & 1 & -1 \\
-1 & 1 & 1
\end{array}\right|\) = 4?
Solution:
x = 0

(v) What is the value of \(\left|\begin{array}{ccc}
\mathbf{o} & -\mathbf{h} & -\mathbf{g} \\
\mathbf{h} & \mathbf{0} & -\mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{0}
\end{array}\right|\)?
Solution:
0

(vi) What is the value of \(\left|\begin{array}{l}
\frac{1}{a} 1 \mathrm{bc} \\
\frac{1}{b} 1 c a \\
\frac{1}{c} 1 a b
\end{array}\right|\)
Solution:
0

(vii) What is the co-factor of 4 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
4 & 5 & 0 \\
2 & 0 & 1
\end{array}\right|\)
Solution:
-2

(viii)In which interval does the determinant \(\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\) lie?
Solution:
[2, 4]

(ix) Ifx + y + z = n, what is the value of Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & -\tan A & 0
\end{array}\right|\) Where A, B, C are the angles of triangle.
Solution:
0

Question 12.
Evaluate the following determinants:
(i) \(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
= 2\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\) = 0
(∵ C₁ = C₃)

(ii) \(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\) = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
14 & 17 & 11
\end{array}\right|\)
( R₁ = R₁ – R₂, R₂ = R₂ – R₃)
= 0 (∵ R₁ = R₂)

(iii) \(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
= 7\(\left|\begin{array}{ccc}
32 & 777 & 32 \\
105 & 888 & 105 \\
116 & 999 & 116
\end{array}\right|\) = 0
(∵ C₁ = C₂)

(iv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 4 \\
3 & 4 & 6
\end{array}\right|\)
Solution:

(v) \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 7 \\
8 & 14 & 20
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:

= 225 – 256 – 4(100 – 144) + 9(64 – 81)
= -31 – 4(-44) + 9(-17)
= -31 + 176 – 153 = -184 + 176
= -8

(vii) \(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
= 2\(\left|\begin{array}{cc}
1 & -5863 \\
1 & 4137
\end{array}\right|\)
(expanding along 2nd column)
= 2(4137 + 5863)
= 2 × 10000 = 20000

(viii) \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
0 & a^2 & b \\
b^2 & 0 & a^2 \\
a & b^2 & 0
\end{array}\right|\)
Solution:

= -a² (0 –  a²) + b (b⁴ –  0) = a⁵ + b⁵

(x) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
\boldsymbol{x}-\boldsymbol{y} & \boldsymbol{y}-\boldsymbol{z} & z-\boldsymbol{x} \\
\boldsymbol{p}-\boldsymbol{q} & \boldsymbol{q}-\boldsymbol{r} & \boldsymbol{r}-\boldsymbol{p}
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-r & r-p
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0 ( ∵ C₁ = 0)

(xi) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0

(xii) \(\left|\begin{array}{ccc}
-\cos ^2 \theta & \sec ^2 \theta & -0.2 \\
\cot ^2 \theta & -\tan ^2 \theta & 1.2 \\
-1 & 1 & 1
\end{array}\right|\)
Solution:

(Expanding along 3rd row)
= (-cos² θ + sec² θ) (-tan² θ – 1.2) – (sec² θ + 0.2) (cot² θ – tan² θ)
= sin² θ – 1.2 cos² θ – sec² θ tan² θ – 1.2 sec² θ – cosec² θ +  sec² θ tan² θ – 0.2 cot² θ + 0.2 tan² θ
= sin² θ – cosec² θ + 1.2 (cos² θ – sec² θ) + 0.2 (tan² θ – cot² θ) ≠ 0
The question seems to be wrong.

Question 13.
If \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\) = 0 what are x and y?
Solution:

or, xy – 0 = 0 ⇒ xy = 0, ⇒ x = 0, or y = 0

Question 14.
For what value of x \(\left|\begin{array}{ccc}
2 x & 0 & 0 \\
0 & 1 & 2 \\
-1 & 2 & 0
\end{array}\right|\) = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 4 \\
0 & 3 & 5
\end{array}\right|\)?
Solution:

Question 15.
Solve \(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
or, (x – a) \(\left|\begin{array}{cc}
x+b & 0 \\
0 & x+c
\end{array}\right|\) = 0
or, (x + a) (x + b) (x + c) = 0
x = -a, x = -b, x = -c

Question 16.
Solve \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Solution:

Question 17.
Solve \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|\) = 0
Solution:

Question 18.
Show that x = 2 is a root of \(\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|\) = 0 Solve this completely.
Solution:
Putting x = 2,

= (x – 1) (-15x + 30 – 5x² + 10x)
= (x – 1) (-5x² – 5x + 30)
= -5(x – 1) (x² + x – 6)
= -5(x – 1) (x + 3) (x – 2) = 0
⇒ x = 1 or, -3 or 2.

Question 19.
Evaluate \(\left|\begin{array}{ccc}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\) – \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

= (a – b) (b – c) [(-a + c) – (b + c – a – b)]
= (a – b) (b – c) (-a + c – c + a) = 0

Question 20.
\(\left|\begin{array}{lll}
a & a^2-b c & 1 \\
b & b^2-a c & 1 \\
c & c^2-a b & 1
\end{array}\right|\)
Solution:

Question21.
For what value of X the system of equations
x + y + z = 6, 4x + λy – λz = 0, 3x + 2y – 4z = -5 does not possess a solution?
Solution:

= 24 – 6λ – 2λ = 24 – 8λ
when Δ = 0
We have 24 – 8λ, = 0 or, λ = 3
The system of equations does not posses solution for λ = 3.

Question 22.
If A is a 3 × 3 matrix and |A| = 2, then which matrix is represented by A × adj A?
Solution:

Question 23.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\)
show that (I + A) (I – A)^-1 = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) where I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

Question 24.
Prove the following:
(i) \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:

(ii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) = (b – c) (c – a) (a – b) (a + b + c)
Solution:

= (a – b) (b – c) (b² + bc + c² – a² – ab – b²)
= (a – b) (b- c) (c² – a² + bc – ab)
= (a – b) (b – c) {(c – a) (c + a) + b(c – a)}
= (a – b) (b – c) (c – a) (a + b + c) = R.H.S.
(Proved)

(iii) \(\left|\begin{array}{lll}
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{b}
\end{array}\right|\) = 3abc – a³ – b³ – c³
Solution:

= (a + b + c) {(b – c) (a – b) – (c – a)²}
= (a + b + c) (a + b + c) (ab – b² – ca + bc – c² – a² + 2ca)
= (a + b + c) (-a² – b² – c² + ab + bc + ca)
= -(a + b + c) (a² + b² + c² – ab – bc – ca)
=- (a³ + b³ + c³ – 3abc)
= 3abc – a³ – b³ – c³

(iv) \(\left|\begin{array}{lll}
b^2-a b & b-c & b c-a c \\
a b-a^2 & a-b & b^2-a b \\
b c-a c & c-a & a b-a^2
\end{array}\right|\) = 0
Solution:

= (b² – a² + bc – ac) (a – b) {(-a + b) (c – a) – (bc – ac – ab + a²)}
= (b² – a² + bc – ac) (a – b) (- ca + a² + bc – ab – bc + ac + ab – a²)
= (b² – a² + bc – ac) (a – b) × 0 = 0
= R.H.S.
(Proved)

(v) \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\) = 4a²b²c²
Solution:

(vi) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c a \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a² + b² + c² ) (a + b + c) (b – c) (c – a) (a – b)
Solution:

= (a – b) (b – c) (a² + b² + c²) (-a² – ab + bc + c²)
= (a – b) (b – c) (a² + b² + c²) {(c² – a²) + b(c – a)}
= (a² + b² + c²) (a – b) (b – c) (c – a) (c + a + b)

(vii) \(\left|\begin{array}{lll}
b+c & a+b & a \\
c+a & b+c & b \\
a+b & c+a & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b +c) {(a – b) (a – c) – (c – b) (b – c)}
= (a + b + c) (a² – ac – ab + bc – bc + c² + b² – bc)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= (a³ + b³ + c³ – 3abc)

(viii) \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2(b + c) (c + a) (a + b)
Solution:

= -2(a + b) (b + c) (-a – b – c + b)
= 2(a + b) (b + c) (c + a)

(ix) \(\left|\begin{array}{ccc}
a x-b y-c z & a y+b x & a z+c x \\
b x+a y & b y-c z-a x & b z+c y \\
c x+a z & a y+b z & c z-a x-b y
\end{array}\right|\) = (a² + b² + c²) (ax + by + cz) (x² + y² + z²)
Solution:

Question 25.
If 2s = a + b + c show that \(\left|\begin{array}{ccc}
a^2 & (s-a)^2 & (s-a)^2 \\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2
\end{array}\right|\) = 2s³ (s – a) (s – b) (s – c)
Solution:
Let s – a = A, s – b = B, s – c = C
A + B + C = 3s – (a + b + c)
= 3s – 2s = s
Also B + C = s – b + s – c = 2s – (b + c)
= (a + b + c) – b + c = a
Similarly C + A = b, A + B = c

= 2 ABC (A + B + C)²
[Refer Q.No.9 (xii) of Exercise 5(a)]
= 2(s – a) (s – b)(s – c) s³

Question 26.
if \(\left|\begin{array}{ccc}
x & x^2 & x^3-1 \\
y & y^2 & y^3-1 \\
z & z^2 & z^3-1
\end{array}\right|\) = 0 then prove that xyz =1 when x, y, z are non zero and unequal.
Solution:

= (x – y) (y – z) (z – x) (xyz – 1)
It is given that
(x – y) (y – z) (z – x) (xyz – 1) = 0
⇒ xyz – 1 (as x ≠ y ≠ z)

Question 27.
Without expanding show that the following determinant is equal to Ax + B where A and B are determinants of order 3 not involving x.
\(\left|\begin{array}{ccc}
x^2+x & x+1 & x-2 \\
2 x^2+3 x-1 & 3 x & 3 x-3 \\
x^2+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right|\)
Solution:

Question 28.
If x, y, z are positive and are the pth, qth and rth terms of a G.P. then prove that \(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|\) = 0
Solution:
Let the G.P. be
a, aR, aR², aR³ …..aR^n-1
p th term = aR^p-1
q th term = aR^q-1
r th term = aR^r-1
x = aR^p-1, y= aR^q-1, z = aR^r-1
log x = log a + (p – 1) log R,
log y = log a + (q – 1) log R,
log z = log a + (r – 1) log R

Question 29.
If D_j = \(\left|\begin{array}{ccc}
j & a & n(n+2) / 2 \\
j^2 & b & n(n+1)(2 n+1) / 6 \\
j^3 & c & n^2(n+1)^2 / 4
\end{array}\right|\) then prove that \(\sum_{j=1}^n\)D_j = 0.
Solution:

Question 30.
Ifa₁, a₂,……a_n are in G.P. and a_i > 0 for every i, then find the value of
\(\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\
\log a_{n+2} & \log a_{n+3} & \log a_{n+4}
\end{array}\right|\)
Solution:

Question 31.
If f(x)= \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin ^2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin ^2 x
\end{array}\right|\) what is the least value of f(x)?
Solution:

As minimum value of sin 2x is 0. So the minimum value of above function f(x) is 2.

Question 32.
If f_r(x), g_r(x), h_r(x), r = 1, 2, 3 are polynomials in x such that f_r(a) = g_r(a) = h_r(a) and
F(x) = \(\left[\begin{array}{lll}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{array}\right]\) find F'(x) at x = a.
Solution:
We have

[Since f₁a) = g₁(a) = h₁(a), f₂(a) = g₂(a) = h₂(a) and f₃(a) = g₃(a) = h₃(a) So that each determinant is zero due to presence of two identical rows.]

Question 33.
If f(x) = \(\left[\begin{array}{ccc}
\cos x & \sin x & \cos x \\
\cos 2 x & \sin 2 x & 2 \cos 2 x \\
\cos 3 x & \sin 3 x & 3 \cos 3 x
\end{array}\right]\) find f'(\(\frac{\pi}{2}\)).
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(a)

Question 1.
State the order of the following matrices.
(i) [abc]
(ii) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
x & y \\
y & z \\
z & x
\end{array}\right]\)
(iv) \(\left[\begin{array}{cccc}
1 & 0 & 1 & 4 \\
2 & 1 & 3 & 0 \\
-3 & 2 & 1 & 3
\end{array}\right]\)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)

Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a sqare matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p²

Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i) \(\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)\)
(ii) \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\)
(iii) \(\left(\begin{array}{ll}
a & b \\
c & d \\
e & f \\
g & h
\end{array}\right)\)
(iv) (1, 2, 3)

Question 4.
Let A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) What is the order of A?
(ii) Write down the entries a₃₁, a₂₅, a₂₃
(iii) Write down A^T.
(iv) What is the order of A^T?
Solution:
A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) Order of A is (3 x 5)
(ii) a₃₁ = 3, a₂₅= 2, a₂₃ = 6
(iii) A^T = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of A^T is (5 x 3).

Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\), B = \(\left[\begin{array}{c}
-6 \\
9
\end{array}\right]\)
Solution:

(ii) A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Solution:

(iii) A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{5}
\end{array}\right]\), B = \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{2} \\
\frac{1}{2} & \frac{4}{5}
\end{array}\right]\)
Solution:

(iv) A = \(\left[\begin{array}{cc}
1 & a-b \\
a+b & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & b \\
-a & 5
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
-1 & 4 & 3 \\
1 & 2 & -3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & -5 \\
1 & -3 & -3 \\
1 & 2 & 4
\end{array}\right]\)
Solution:

Question 6.
(i) Find the 2×2 matrix X
if X + \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
Solution:

(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] derermine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1

(iii) If \(\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]\) – \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) determine x₁, x₂, y₁, y₂.
Solution:

(iv) Find a matrix which when added to \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:

Question 7.
Calculate whenever possible, the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\) is impossible because number of columns of 1st ≠ number of rows of second.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
1 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\)
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\), C = \(\left[\begin{array}{ll}
2 & 2 \\
1 & 3
\end{array}\right]\)
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:

Question 9.
Find the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:

(vi) \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(x) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False

(ii) k[0] = 0, k ∈ R
Solution:
False

(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False

(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True

(v) \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) + \(\left[\begin{array}{cc}
-1 & 0 \\
2 & 0
\end{array}\right]\) = 0
Solution:
True

(vi) \(\left[\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right]\) = 3 \(\left[\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right]\)
Solution:
False

(vii) With five elements a matrix can not be constructed.
Solution:
False

(viii)The unit matrix is its own transpose.
Solution:
True

Question 11.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 13
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find A – α I, α ∈ R.
Solution:

Question 12.
Find x and y in the following.
(i) \(\left[\begin{array}{cc}
x & -2 y \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -8 \\
0 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{l}
x \\
y
\end{array}\right]+\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1
\end{array}\right]\)
Solution:

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.

Question 14.
Write down the matrix

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) a_ij = i + j
(ii) a_ij = i – j
(iii) a_ij = i × j
(iv) a_ij = i / j
Solution:

Question 16.
If \(\left[\begin{array}{cc}
2 x & y \\
1 & 3
\end{array}\right]+\left[\begin{array}{cc}
4 & 2 \\
0 & -1
\end{array}\right]=\left[\begin{array}{ll}
8 & 3 \\
1 & 2
\end{array}\right]\)
Solution:

Question 17.
Find A such that
\(\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 0 & -2 \\
3 & 1 & -1
\end{array}\right]+A=\left[\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 0 \\
1 & 3 & 2
\end{array}\right]\)
Solution:

Question 18.
If

Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)

Question 20.
Find A if \(\left[\begin{array}{l}
4 \\
1 \\
3
\end{array}\right]\) A = \(\left[\begin{array}{rrr}
-4 & 8 & 4 \\
-1 & 2 & 1 \\
-3 & 6 & 3
\end{array}\right]\)
Solution:

Question 21.
Find B if B² = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:

∴ a² + bc = 17, ab + bd= 8
ca + cd = 8, bc + d² = 17
∴ a² + bc = bc + d²
or, a² + d² or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a² + bc = 17
or, a² + b . b = 17 (∵ b = c)
or, a² + b² = 17
Also (a + b)² = a² + b² + 2ab
∴ (a + b)² = 17 + 8 = 25
or, a + b = 5
And (a – b)² = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

Question 22.
Find x and y when

Question 23.
Find AB and BA given that:

Question 24.
Evaluate

Question 25.
If

Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:

Question 26.
Find A and B where

Question 27.
If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) and I be the 2 × 2 unit matrix find (A – 2I) (A – 3I)
Solution:

Question 28.
Verify that [AB]^T = B^TA^T where

Question 29.
Verify that A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) satisfies the equation x² – (a + d)x + (ad – bc)I = 0 where I is the 2 x 2 matrix.
Solution:

Question 30.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), show that A³ – 23 A – 40 I = 0
Solution:

Question 31.

Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A² – B² = (A – B) (A + B)
(ii) A² + 2AB + B² = (A + B)²
(iii) A² – 2AB + B² = (A – B)²
Solution:
(i) (A – B) (A + B)
= A² + AB – BA – B²
= A² + AB – AB- B²(∵ AB = BA)
= A² – B²
(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² (∵ AB = BA)
= A² + 2AB + B²
(iii) (A – B)² = (A – B) (A – B)
= A² – AB – BA + B²
= A² – AB – AB + B² (∵ AB = BA)
= A² – 2AB + B²

Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A² – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A² – AβI – αIA + αβI²
= A² – βAI – αA + αβI
(∵ IA = A, I² = I)
= A² – βA – αA + αβI) (∵ AI = A)
= A² – (α + β) A + αβI

Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar mltiβlication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Question 42.

Question 43.

	Men	Women	Children
Family A →	4	6	2
Family B →	2	2	4

	Family B
	Calory	Proteins
Men	2400	45
Women	1900	55
Children	1800	33

Solution:
The given informations can be written in matrix form as

∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]

⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(b)

Question 1.
Maximize Z = 5x₁+ 6x₂
Subject to: 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation, we get 2x₁ + 3x₂ = 6
Step – 2 Let us draw the graph

x1	3	0
x2	0	0

Step – 3 Clearly (0,0) statisfies 2x₁ + 3x₂ ≤ 6
The shaded region is the feasible region with vertices 0(0,0), A(3,0), B(0,2).
Step – 4

Corner point	Z = 5x1+ 6x2
0(0.0)	0
A(3,0)	15 → maximum
B(0,2)	12

Z is maximum at A (3,0)
∴ The solution of LPP is x₁ = 3, x₂ = 0
Z_max = 15

Question 2.
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation we get x₁ + 2x₂ = 0
Step – 2 Let us draw the graph of x₁ + 2x₂ = 4

x1	0	4
x2	2	0

Step – 3 Clearly 0(0,0) does not satisfy
x₁ + 2x₂ > 4, x₁ > 0, x₂ > 0 is the first quadrant.
The feasible region is the shaded region with vertices A(4, 0), B(0, 2).
Step – 4 Z (4, 0) = 24
Z (0, 2) = 14 → minimum
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B (0, 2).
Let us draw the half-plane 6x₁ + 7x₂ < 14

x1	0	3.5
x2	2	-1

As this half-plane has no point common with the feasible region, we have Z is minimum for x₁= 0, x₂ = 2 and the minimum value of Z = 14.

Question 3.
Maximize Z = 20x₁+ 40x₂
Subject to: x₁ + x₂ ≤ 1
6x₁ + 2x₂ ≤ 3
x₁, x₂ ≥ 0.
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ = 1    …. (1)
6x₁ + 2x₂ = 3   …. (2)
x₁, x₂ ≥ 0
Step – 2 Let us draw the graph:
Table – 1

x1	0	1
x2	1	0

Table – 2

x1	0	0.5
x2	1.5	0

Step – 3 As (0, 0) satisfies both the inequations the shaded region is the feasible region.
Step – 4 Solving
x₁ + x₂ = 1
6x₁ + 2x₂ = 3
we have x₁ = ¼ x₂ = ¾
The vertices are O(0, 0), A(0.5, 0), B(0,1) and C(¼, ¾)
Now Z(O) = 0
Z(A) = 10
Z(B) = 40
Z(C) = 20 × ¼ + 40 × ¾ = 35
∴ Z attains maximum at B for x₁= 0, x₂ = 1
Z_max = 40

Question 4.
Minimize: Z = 30x₁ + 45x₂
Subject to: 2x₁ + 6x₂ ≥ 4
5x₁ + 2x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Consider the constraints as equations
2x₁ + 6x₂ = 4
5x₁ + 2x₂ = 5
Step – 2
Table – 1

x1	2	-1
x2	0	1

Table – 2

x1	1	0
x2	0	2.5

Step – 3 Clearly 0(0,0) does not satisfy 2x₁ + 6x₂ ≥ 4 and 5x₁ + 2x₂ ≥ 5.
Thus the shaded region is the feasible region.
Solving the equations we get
x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\).
∴ The vertices are A(2, 0)
B(\(\frac{11}{13}\), \(\frac{5}{13}\)) and C(0, \(\frac{5}{2}\)).
Step – 4 Z(A) = 60
Z(B) = \(\frac{555}{13}\) → minimum
Z(C) = \(\frac{225}{2}\)
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B(\(\frac{11}{13}\), \(\frac{5}{13}\))
Let us draw the half plane
30x₁ + 45x₂ < \(\frac{555}{13}\)

x1	\(\frac{11}{13}\)	0
x2	\(\frac{5}{13}\)	\(\frac{27}{39}\)

As this half plane and the feasible region has no point in common we have Z is minimum for x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\), and Z_min = \(\frac{555}{13}\)

Question 5.
Maximize: Z = 3x₁+ 2x₂
Subject to: -2x₁ + x₂ ≤ 1
x₁ ≤ 2
x₁+ x₂ ≤ 3
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-2x₁ + x₂ = 1        …..(1)
x₁ = 2                   …..(2)
x₁+ x₂ = 3            …..(3)
Step – 2 Let us draw the lines.
Table – 1

x1	0	-1
x2	1	-1

Table – 2

x1	2	2
x2	0	1

Table – 3

x1	0	3
x2	3	0

Step – 3 (0, 0) satisfies all the constraints and x₁, x₂ > 0 is the 1st quadrant the shaded region is the feasible region.

Step – 4 Solving -2x₁ + x₂ = 1
x₁+ x₂ = 3
we have 3x₁ = 2
⇒ x₁ = \(\frac{2}{3}\), x₂ = 3 – \(\frac{2}{3}\) = \(\frac{7}{3}\)
From x₁+ x₂ = 3 and x₁ = 2 we have x₁ = 2, x₂ = 1
∴ The vertices are 0(0, 0), A(2, 0), B(2, 1), C(\(\frac{2}{3}\), \(\frac{7}{3}\)), D(0, 1)
Z(0) = 0, Z(A) = 6, Z(B) = 8, Z(C) = 3.\(\frac{2}{3}\) + 2.\(\frac{7}{3}\) = \(\frac{20}{3}\), Z(D) = 2
Z is maximum at B.
∴ The solution of given LPP is x₁ = 2, x₂ = 1, Z_(max) = 8.

Question 6.
Maximize: Z = 50x₁+ 60x₂
Subject to: x₁ + x₂ ≤ 5
x₁+ 2x₂ ≤ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + x₂ = 5     ….(1)
x₁+ 2x₂ = 4    ….(2)
Step – 2 Let us draw the graph
Table – 1

x1	5	5
x2	0	0

Table – 2

x1	4	0
x2	0	2

Step – 3 0(0,0) satisfies x₁ + x₂ ≤ 5 and does not satisfy x₁+ 2x₂ ≤ 4
Thus the shaded region is the feasible region.
Step – 4 The corner points are A(4,0), B(5,0), C(0,5) , D(0,2)
∴

Corner point	z = 50x1+ 60x2
A(4,0)	200
B (5,0)	250 → maximum
C(0,5)	300
D(0,2)	120

Z is maximum for x₁ = 0, x₂ = 5, Z_(max) = 300.

Question 7.
Maximize: Z = 5x₁+ 7x₂
Subject to: x₁ + x₂ ≤ 4
5x₁+ 8x₂ ≤ 30
10x₁+ 7x₂ ≤ 35
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get,
x₁ + x₂ = 4           …. (1)
5x₁+ 8x₂ = 30      …. (2)
10x₁+ 7x₂ = 35    …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	4	0
x2	0	4

Table – 2

x1	6	2
x2	0	2.5

Table – 3

x1	0	3.5
x2	5	0

Step – 3 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get (\(\frac{2}{3}\), \(\frac{10}{3}\))
From (1) and (3) we get
x₁ = \(\frac{7}{3}\), x₁ = \(\frac{5}{3}\)
∴ The corner points are 0(0,0), A(\(\frac{7}{2}\), 0), B(\(\frac{7}{3}\), \(\frac{5}{3}\)), C(\(\frac{2}{3}\), \(\frac{10}{3}\)), D(0, \(\frac{15}{4}\))
Step – 4

Corner point	z = 5x1+ 7x2
0(0,0)	0
A(\(\frac{7}{2}\), 0)	\(\frac{35}{2}\)
B(\(\frac{7}{3}\), \(\frac{5}{3}\))	\(\frac{70}{3}\)
C(\(\frac{2}{3}\), \(\frac{10}{3}\))	\(\frac{80}{3}\)
D(0, \(\frac{15}{4}\))	\(\frac{105}{4}\)

Z attains its maximum value \(\frac{80}{3}\) for x₁ = \(\frac{2}{3}\) and x₂ = \(\frac{10}{3}\).

Question 8.
Maximize: Z = 14x₁ – 4x₂
Subject to: x₁ + 12x₂ ≤ 65
7x₁ – 2x₂ ≤ 25
2x₁+ 3x₂ ≤ 10
x₁, x₂ ≥ 0
Also find two other points which maximize Z.
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + 12x₂ = 65   …. (1)
7x₁ – 2x₂ = 25    …. (2)
2x₁ + 3x₂ = 10   …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	65	5
x2	0	5

Table – 2

x1	5	10
x2	5	22.5

Table – 3

x1	5	2
x2	0	2

Step – 3 Clearly 0(0,0) satisfies x₁ + 12x₂ ≤ 65 and 7x₁ – 2x₂ ≤ 25 but does not satisfy 2x₁+ 3x₂ ≤ 10. Thus shaded region is the feasible region.
Equation (1) and (2) meet at (5, 5).
From (2) and (3)

∴ The corner points of the feasible region are A(0, \(\frac{10}{3}\)), B(\(\frac{19}{5}\), \(\frac{4}{5}\)), C(5, 5), D(0, \(\frac{65}{12}\)).
Step – 4

Corner point	z = 14x1 – 4x2
A(0, \(\frac{10}{3}\))	\(\frac{-40}{3}\)
B(\(\frac{19}{5}\), \(\frac{4}{5}\))	50 → maximum
C(5, 5)	50 → maximum
D(0, \(\frac{65}{12}\))	\(\frac{65}{3}\)

Z is maximum for x₁ = \(\frac{19}{5}\), x₂ = \(\frac{4}{5}\) or x₁ = 5, x₂ = 5 and Z_max = 50
There is no other point that maximizes Z.

Question 9.
Maximize: Z = 10x₁ + 12x₂ + 8x₃
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁ + x₂ + x₃ = 20
x₁, x₂ ≥ 0
[Hints: Eliminate x₃ from all expressions using the given equation in the set of constraints, so that it becomes an LPP in two variables]
Solution:
Eliminating x₃ this LPP can be written as Maximize Z = 2x₁ + 4x₂ + 160
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁, x₂ ≥ 0
Step – 1 Treating the consraints as equations we get
x₁ + 2x₂ = 30    …..(1)
5x₁ – 7x₃ = 12   …..(2)
Step – 2 Let us draw the graph
Table – 1

x1	30	0
x2	0	15

Table – 2

x1	8	1
x2	8	20

Step – 3 Clearly 0(0,0) satisfies x₁ + 2x₂ ≤ 30 and does not satisfy 12x₁ + 7x₂ ≤ 152
∴ The shaded region is the feasible region.

Step – 4

Z is maximum for x₁ = 30, x₂ = 0 and Z_max = 220

Question 10.
Maximize: Z = 20x₁ + 10x₂
Subject to: x₁ + 2x₂ ≤ 40
3x₁ + x₂ ≥ 30
4x₁+ 3x₂ ≥ 60
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equalities we have:
x₁ + 2x₂ = 40   ….(1)
3x₁ + x₂ = 30   ….(2)
4x₁+ 3x₂ = 60  ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + 2x₂ ≤ 40 and does not satisfy 3x₁ + x₂ ≥ 30 and 4x₁+ 3x₂ ≥ 60, x₁, x₂ ≥ 0 is the first quadrant.
∴ The shaded region is the feasible region.
Step – 4 x₁ + 2x₂ = 40 and 3x₁ + x₂ = 30

∴ The vetices are A(15, 0), B(10, 0), C(4, 18) and D(6, 12)
Z(A) = 300, Z(B) = 800
Z (C) = 20 x 4 + 10 x 18 = 260
Z (D) = 120 + 120 = 240
Z attains minimum at D(6 ,12).
∴ The required solution x₁ = 6, x₂ =12 and Z_min = 240

Question 11.
Maximize: Z = 4x₁ + 3x₂
Subject to: x₁ + x₂ ≤ 50
x₁ + 2x₂ ≥ 80
2x₁+ x₂ ≥ 20
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ ≤ 50    ….(1)
x₁ + 2x₂ ≥ 80  ….(2)
2x₁+ x₂ ≥ 20   ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + x₂ < 50, x₁ + 2x₂ < 80 but does not satisfy
2x₁ + x₂ > 20, x₁ > 0, x₂ > 0 is the 1st quadrant.
Hence the shaded region is the feasible region.
Step – 4 x₁ + x₂ = 50
x₁ + 2x₂ = 80
=> x₂ = 30, x₁ = 20
The vertices of feasible region are
A(10, 0), B(50, 0), C(20, 30), D (0, 40) and E (0, 20)

Point	Z = 4x1 + 3x2
A(10,0)	40
5(50,0)	200
C(20,30)	170
D(0,40)	120
E(0,120)	60

Question 12.
Optimize: Z = 5x₁ + 25x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 (0, 0) satisfies -0.5x₁ + x₂ ≤ 2, but does not satisfy x₁ + x₂ ≥ 2 and -x₁+ 5x₂ ≥ 5, x₁ > 0, x₂ > 0 is the 1st quadrant.
The shaded region is the feasible region with vertices A(\(\frac{5}{6}\), \(\frac{7}{6}\)) and B(0, 2).
Step – 4 Z can be made arbitrarily large.
∴ Problem has no maximum.
But Z(A) = \(\frac{100}{3}\), Z(B) = 50
Z is minimum at A(\(\frac{5}{6}\), \(\frac{7}{6}\)).
But the feasible region is unbounded.
Hence we cannot immediately decide, Z is minimum at A.
Let us draw the half plane
5x₁ + 25x₂ < \(\frac{100}{3}\)
⇒ 3x₁ + 15x₂ < 20
As there is no point common to this half plane and the feasible region.
we have Z is minimum for x₁ = \(\frac{5}{6}\), x₂ = \(\frac{7}{6}\) and the minimum value = \(\frac{100}{3}\)

Question 13.
Optimize: Z = 5x₁ + 2x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.

As Z can be made arbitrarily large, the given LPP has no maximum.
Z is minimum at B (0, 2). But we cannot immediately decide, Z is minimum at B.
Let us draw the half plane 5x₁ + 2x₂ < 4

x1	0	4/5
x2	2	0

As there is no point common to this half plane and the feasible region,
we have Z is minimum for x₁ = 0, x₂ = 2 and the minimum value of Z = 4.

Question 14.
Optimize: Z = -10x₁ + 2x₂
Subject to: -x₁ + x₂ ≥ -1
x₁ + x₂ ≤ 6
x₂ ≤ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-x₁ + x₂ = -1     ….(1)
x₁ + x₂ = 6        ….(2)
x₂ = 5                ….(3)
Step – 2 Let us draw the graph
Table – 1

x1	1	0
x2	0	-1

Table – 2

x1	6	0
x2	0	1

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
The vertices are 0(0,0) , A(1,0), B(\(\frac{7}{2}\), \(\frac{5}{2}\)) ,C(1, 5) and D (0, 5)
Step – 4 Z(O) = 0
Z(A) = -10
Z(B) = – 30
Z(C) = 0
Z(D) = 10
∴ Z is maximum for x₁= 0, x, = 5 and Z_(max) = 10
Z is minimum for x₁ = \(\frac{7}{2}\)  x₂ = \(\frac{5}{2}\) and Z_(min) = -30

Question 15.
Solve the L.P.P.s obtained in Exercise 3(a) Q.1 to Q. 9 by graphical method.
(1) Maximise: Z = 1500x + 2000y
Subject to: x + y < 20
x + 2y < 24
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 20
x + 2y = 24
Step – 2 Let us draw of graph.

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get
y = 14
x = 16
With vertices 0(0, 0), A(20, 0), B(16, 4), C(0, 12).
Step – 4 Z(0) = 0
Z(A) = 30,000
Z(B) = 32,000 → Maximum
Z(C) = 24000
Z is maximum for x = 16, y = 4 with Z = 32000
To get maximum profit he must keep 16 sets of model X and 4 sets of model Y.
Maximum profit = 1500 × 16 + 2000 × 4 = ₹32,000

(2) Maximize: 15x + 10y
Subject: x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
2x +3y = 600
2a + y = 480
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
The corner point are 0(0, 0), A (240, 0) B(210, 60),C(0, 200)
Step – 4 Z(0) = 6
Z(A) = 3600
Z(B) = 3150 + 600
= 3750 → maximum
Z(C) = 2000
Thus Z is maximum for x = 210 and y = 60
and Z_(max) = 3750

(3) Maximize: Z = 20x + 30y
Subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10       …(1)
x + y = 6           …(2)
x = 4
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
Solving (1) and (2) we get x = 2, y = 4.
The vertices and 0(0, 0) , A(4, 0), B(4, 2), C(2, 4), D (0, 5).
Step – 4 Z(0) =0
Z(A) = 80
Z (B) =140
Z(C) = 1 60 → maximum
Z (D) = 150
∴ Z is Maximum when x = 2, y = 4 and Z_(max) = 160

(4) Maximize: Z = 15x + 17y
Subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y > 300
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
4x + 7y = 150      ….(1)
x + y = 30            ….(2)
15x + 17y = 300  ….(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
4x + 7y ≤ 150, x + y ≤ 30, but does not satisfy 15x + 17y ≥ 300.
∴ The shaded region is the feasible region.
From (1) and (2) we get

∴ Z is maximum for x = 20. y = 10 and Z_(max) = 470

(5) Maximize: Z = 2x + 4y
Subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
3x + 2y = 10  …(1)
2x + 5y = 15  …(2)
5x + 6y = 21  …(3)
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
From (1) and (3) we get

From (2) and (3) we get

Step-4 Z(O) = 0

(6) Maximize: Z = 1000x + 800y
Subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 5    ….(1)
2x + y = 9  ….(2)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
∴ Thus the shaded region is the feasible region.
From (1) and (2) we get x = 4, y = 1.
∴ The vertices are A(0, 0), A(4.5, 0), B(4, 1) and C(0, 5).
Step – 4 Z(0) =0
Z (A) = 4500
Z (B) = 4800 → Maximum
Z (C) = 4000
Z is maximum for x = 4 and y = 1, Z_(max) = 4800

(7) Minimize: Z = 4960 – 70x – 130y
Subject to: x + y ≤ 12
x + y ≥ 6
x ≤ 8
y ≤ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 12   ….(1)
x + y = 6     ….(2)
x = 8           ….(3)
y = 4           ….(4)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints except x + y > 6.
The shaded region is the feasible region.
The vertices are A(6, 0), B(8, 0), C(8, 4), D(4, 8), E(0, 8) and F(0, 6).
Step – 4 Z (A) = 4540
Z (B) = 4400
Z (C) = 3880
Z (D) = 3640 → Minimum
Z (E) = 3920
Z (F) = 4180
∴ Z is maximum for x = 4 and y = 8 and Z_(min) = 3640.

(8) Minimize: Z = 16x + 20y
Subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10  ….(1)
x + y = 6      …(2)
3x + y = 8    …(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints. Thus the shaded region is the feasible region.
From (1) and (2) we get y = 4, x = 2.
From (2) and (3) we get x = 1, y = 5.
The vertices are A(10, 0), B(2, 4), C(1, 5), D(0, 8).
Step – 4 Z (A) = 160
Z (B) = 112 → Minimum
Z (C) =116
Z (D) = 160
As the region is unbounded, let us draw the half plane Z < Z_(min)
⇒ 16x + 20y < 112
⇒ 4x + 5y < 28

x1	7	0
x2	0	5.6

There is no point common to the shaded region and the half plane 4x + 5y ≤ 28 other than B(2, 4).
∴ Z is minimum for x = 2, y = 4 and Z_(min) = 112.

(9) Minimize: Z = (512.5)x + 800y
Subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0
Solution:
Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3

x1	8	0
x2	0	10

Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3
Step – 2 The line segment AB is the feasible region.
Step – 3 Z (A) = 3587.5 + 1000 = 4587.5
Z (B) = 2870 + 2400 = 5270
Clearly Z is minimum for
x = 7, y = \(\frac{5}{4}\) and Z_(min) = 4587.5

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 2 Inverse Trigonometric Functions Ex 2 Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

Question 1.
Fill in the blanks choosing correct answer from the brackets:
(i) If A = tan^-1 x, then the value of sin 2A = ________. (\(\frac{2 x}{1-x^2}\), \(\frac{2 x}{\sqrt{1-x^2}}\), \(\frac{2 x}{1+x^2}\))
Solution:
\(\frac{2 x}{1+x^2}\)

(ii) If the value of sin^-1 x = \(\frac{\pi}{5}\) for some x ∈ (-1, 1) then the value of cos^-1 x is ________. (\(\frac{3 \pi}{10}\), \(\frac{5 \pi}{10}\),\(\frac{3 \pi}{10}\))
Solution:
\(\frac{3 \pi}{10}\)

(iii) The value of tan^-1 x (2cos\(\frac{\pi}{3}\)) is ________. (1, \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(iv) If x + y = 4, xy = 1, then tan^-1 x + tan^-1 y = ________. (\(\frac{3 \pi}{4}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{2}\)

(v) The value of cot^-1 2 + tan^-1 \(\frac{1}{3}\) = ________. (\(\frac{\pi}{4}\), 1, \(\frac{\pi}{2}\))
Solution:
\(\frac{\pi}{4}\)

(vi) The principal value of sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________. (\(\frac{2 \pi}{3}\), \(\frac{\pi}{3}\), \(\frac{4 \pi}{3}\))
Solution:
\(\frac{\pi}{3}\)

(vii) If sin^-1 \(\frac{x}{5}\) + cosec^-1 \(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x = ________. (2, 3, 4)
Solution:
x = 3

(viii) The value of sin (tan^-1 x + tan^-1 \(\frac{1}{x}\)), x > 0 = ________. (0, 1, 1/2)
Solution:
1

(ix) cot^-1 \(\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\) = ________. (2π – \(\frac{x}{2}\), \(\frac{x}{2}\), π – \(\frac{x}{2}\))
Solution:
π – \(\frac{x}{2}\)

(x) 2sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{24}{25}\) = ________. (π, -π, 0)
Solution:
π

(xi) if Θ = cos^-1 x + sin^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which Θ lies is ________. [(\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), [0, \(\frac{\pi}{2}\)), (0, \(\frac{\pi}{2}\)])
Solution:
(0, \(\frac{\pi}{2}\)]

(xii) sec² (tan^-1 2) + cosec² (cot^-1 3) = ________. (16, 14, 15)
Solution:
15

Question 2.
Write whether the following statements are true or false.
(i) sin^-1 \(\frac{1}{x}\) cosec^-1 x = 1
Solution:
False

(ii) cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{2}{3}\) = tan^-1 \(\frac{17}{6}\)
Solution:
True

(iii) tan^-1 \(\frac{4}{3}\) + cot^-1 (\(\frac{-3}{4}\)) = π
Solution:
True

(iv) sec^-1 \(\frac{1}{2}\) + cosec^-1 \(\frac{1}{2}\) = \(\frac{\pi}{2}\)
Solution:
False

(v) sec^-1 (-\(\frac{7}{5}\)) = π – cos^-1 \(\frac{5}{7}\)
Solution:
True

(vi) tan^-1 (tan 3) = 3
Solution:
False

(vii) The principal value of tan^-1 (tan \(\frac{3 \pi}{4}\)) is \(\frac{3 \pi}{4}\)
Solution:
False

(viii) cot^-1 (-√3) is in the second quadrant.
Solution:
True

(ix) 3 tan^-1 3 = tan^-1 \(\frac{9}{13}\)
Solution:
False

(x) tan^-1 2 + tan^-1 3 = – \(\frac{\pi}{4}\)
Solution:
False

(xi) 2 sin^-1 \(\frac{4}{5}\) = sin^-1 \(\frac{24}{25}\)
Solution:
False

(xii) The equation tan^-1 (cotx) = 2x has exactly two real solutions.
Solution:
True

Question 3.
Express the value of the foilowing in simplest form.
(i) sin (2 sin^-1 0.6)
Solution:
sin (2 sin^-1 0.6)

(ii) tan (\(\frac{\pi}{4}\) + 2 cot^-1 3)
Solution:

(iii) cos (2 sin^-1 x)
Solution:

(iv) tan (cos^-1 x)
Solution:

(v) tan^-1 (\(\frac{x}{y}\)) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:

(vi) cosec (cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\))
Solution:

(vii) sin^-1 \(\frac{1}{\sqrt{5}}\) + cos^-1 \(\frac{3}{\sqrt{10}}\)
Solution:

(viii) sin cos^-1 tan sec √2
Solution:
sin cos^-1 tan sec √2
= sin cos^-1 tan sec \(\frac{\pi}{4}\)
= sin cos^-1 1 = sin 0 = 0

(ix) sin (2 tan^-1 \(\sqrt{\frac{1-x}{1+x}}\))
Solution:

(x) tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
Solution:

(xi) sin cot^-1 cos tan^-1 x.
Solution:

(xii) tan^-1 \(\left(x+\sqrt{1+x^2}\right)\)
Solution:

Question 4.
Prove the following statements:
(i) sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\frac{36}{85}\)
Solution:

(ii) sin^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)
Solution:

(iii) tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
Solution:
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

(iv) tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) tan ( 2tan^-1 \(\frac{1}{5}\) – \(\frac{\pi}{4}\) ) + \(\frac{7}{17}\) = 0
Solution:

Question 5.
Prove the following statements:
(i) cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:

(ii) sin^-1 \(\frac{4}{5}\) + 2 tan^-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:

(iii) 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{70}\) + tan^-1 \(\frac{1}{99}\) = \(\frac{\pi}{4}\)
Solution:

(iv) 2 tan^-1 \(\frac{1}{5}\) + sec^-1 \(\frac{5 \sqrt{2}}{7}\) + 2 tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) cos^-1 \(\frac{12}{13}\) + 2 cos^-1 \(\sqrt{\frac{64}{65}}\) + cos^-1 \(\sqrt{\frac{49}{50}}\) = cos^-1 \(\frac{1}{\sqrt{2}}\)
Solution:


(vi) tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\) = 6
Solution:
tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\)
= tan² tan^-1 √2 + cot² cot^-1 (2)
= 2 + 4 = 6

(vii) cos tan^-1 cot sin^-1 x = x.
Solution.

Question 6.
Prove the following statements:
(i) cot^-1 (tan 2x) + cot^-1 (- tan 2x) = π
Solution:

(ii) tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)
Solution:

(iii) tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\)) = tan^-1 a – tan^-1 c.
Solution:
tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\))
= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c
= tan^-1 a – tan^-1 c.

(iv) cot^-1 \(\frac{p q+1}{p-q}\) + cot^-1 \(\frac{q r+1}{q-r}\) + cot^-1 \(\frac{r p+1}{r-p}\) = 0
Solution:

(v)

Solution:

Question 7.
Prove the following statements:
(i) tan^-1 \(\frac{2 a-b}{b \sqrt{3}}\) + tan^-1 \(\frac{2 b-a}{a \sqrt{3}}\) = \(\frac{\pi}{3}\)
Solution:

(ii) tan^-1 \(\frac{1}{x+y}\) + tan^-1 \(\frac{y}{x^2+x y+1}\) = tan^-1 \(\frac{1}{x}\)
Solution:

(iii) sin^-1 \(\sqrt{\frac{x-q}{p-q}}\) = cos^-1 \(\sqrt{\frac{p-x}{p-q}}\) = cot^-1 \(\sqrt{\frac{p-x}{x-q}}\)
Solution:

(iv) sin² (sin^-1 x + sin^-1 y + sin^-1 z) = cos² (cos^-1 x + cos^-1 y + cos^-1 z)
Solution:

(v) tan (tan^-1 x + tan^-1 y + tan^-1 z) = cot (cot^-1 x + cot^-1 y + cot^-1 z)
Solution:

Question 8.
(i) If sin^-1 x + sin^-1 y + sin^-1 z = π, show that x\(\sqrt{1-x^2}\) + x\(\sqrt{1-y^2}\) + x\(\sqrt{1-z^2}\) = 2xyz
Solution:
Let sin^-1 x = α, sin^-1 y = β, sin^-1 z = γ
∴ α + β + γ = π
∴ x = sin α, y = sin β, z = sin γ
or, α + β = π – γ
or, sin(α + β) = sin(π – γ) = sin γ
and cos(α + β) = cos(π – γ) = – cos γ

(ii) tan^-1 x + tan^-1 y + tan^-1 z = π show that x + y + z = xyz.
Solution:

(iii) tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\). Show that xy + yz + zx = 1
Solution:

or, 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 \(\frac{y z}{x r}\) + tan^-1 \(\frac{z x}{y r}\) + tan^-1 \(\frac{x y}{z r}\) = \(\frac{\pi}{2}\)
Solution:

(v) In a triangle ABC if m∠A = 90°, prove that tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\) = \(\frac{\pi}{4}\). where a, b, and c are sides of the triangle.
Solution:
L.H.S. tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\)

Question 9.
Solve
(i) cos (2 sin^-1 x) = 1/9
Solution:

(ii) sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
Solution:
sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
or, sin^-1 (1 – x) = \(\frac{\pi}{2}\) – sin^-1 x = cos^-1 x
or, sin^-1 (1 – x) = sin^-1 \(\sqrt{1-x^2}\)
or, 1 – x = \(\sqrt{1-x^2}\)
or, 1 + x² – 2x = 1 – x²
or, 2x² – 2x  = 0
or, 2x (x – 1) = 0
∴ x = 0 or, 1

(iii) sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ – 2 sin^-1 x = \(\frac{\pi}{2}\) – sin^-1 (1 – x)
⇒ cos^-1 (1 – x)
⇒ cos (– 2 sin^-1 x) = 1 – x      ….. (1)
Let sin^-1 Θ ⇒ sin Θ
Now cos (– 2 sin^-1 x) = cos (-2Θ)
= cos 2Θ = 1 – 2 sin² Θ = 1 – 2x²
Using in (1) we get
1 – 2x² = 1 – x
⇒ 2x² – x = 0 ⇒ x (2x – 1) = 0
⇒ x = 0, ½, But x = ½ does not
Satisfy the given equation, Thus x = 0.

(iv) cos^-1 x + sin^-1 \(\frac{x}{2}\) = \(\frac{\pi}{6}\)
Solution:

(v) tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}\) = \(\frac{\pi}{4}\)
Solution:

(vi) tan^-1 \(\frac{1}{2 x+1}\) + tan^-1 \(\frac{1}{4 x+1}\) = tan^-1 \(\frac{2}{x^2}\)
Solution:

(vii) 3 sin^-1 \(\frac{2 x}{1+x^2}\) – 4 cos^-1 \(\frac{1-x^2}{1+x^2}\) + 2 tan^-1 \(\frac{2 x}{1-x^2}\) = \(\frac{\pi}{3}\)
Solution:

(viii) cot^-1 \(\frac{1}{x-1}\) + cot^-1 \(\frac{1}{x}\) + cot^-1 \(\frac{1}{x+1}\) = cot^-1 \(\frac{1}{3x}\)
Solution:

(ix) cot^-1 \(\frac{1-x^2}{2 x}\) =  cosec^-1 \(\frac{1+a^2}{2 a}\) – sec^-1 \(\frac{1+b^2}{1-b^2}\)
Solution:

(x) sin^-1 \(\left(\frac{2 a}{1+a^2}\right)\) + sin^-1 \(\left(\frac{2 b}{1+b^2}\right)\) = 2 tan^-1 x
Solution:

(xi) sin^-1 y – cos^-1 x = cos^-1 \(\frac{\sqrt{3}}{2}\)
Solution:

(xii) sin^-1 2x + sin^-1 x = \(\frac{\pi}{3}\)
Solution:

Question 10.
Rectify the error ifany in the following:
sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{33}{65}\)
Solution:

Question 11.
Prove that:
(i) cos^-1 \(\left(\frac{b+a \cos x}{a+b \cos x}\right)\) = 2 tan^-1 \(\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)\)
Solution:

(ii) tan \(\left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\) + tan \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\)
Solution:

(iii) tan^-1 \(\sqrt{\frac{x r}{y z}}\) + tan^-1 \(\sqrt{\frac{y r}{y x}}\) + tan^-1 \(\sqrt{\frac{z r}{x y}}\) = π where r = x + y +z.
Solution:

Question 12.
(i) If cos^-1 (\(\frac{x}{a}\)) + cos^-1 (\(\frac{y}{b}\)) = Θ, prove that \(\frac{x^2}{a^2}\) – \(\frac{2 x}{a b}\) cos Θ + \(\frac{y^2}{b^2}\) = sin² Θ.
Solution:

(ii) If cos^-1 (\(\frac{x}{y}\)) + cos^-1 (\(\frac{y}{3}\)) = Θ, prove that 9x² – 12xy cos Θ + 4y² = 36 sin² Θ.
Solution:

(iii) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = sin^-1 (\(\frac{c^2}{a b}\)) prove that b²x² + 2xy \(\sqrt{a^2 b^2-c^4}\) a²y² = c²
Solution:

(iv) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = α prove that \(\frac{x^2}{a^2}\) + \(\frac{2 x y}{a b}\) cos α + \(\frac{y^2}{b^2}\) = sin² α
Solution:

(v) If sin^-1 x + sin^-1 y + sin^-1 z = π prove that x² + y² + z² + 4x²y²z² = 2 ( x²y² + y²z² + z²x² )
Solution:

Question 13.
Solve the following equations:
(i) tan^-1 \(\frac{x-1}{x+1}\) + tan^-1 \(\frac{2 x-1}{2 x+1}\) = tan^-1 \(\frac{23}{36}\)
Solution:

(ii) tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 x = \(\frac{\pi}{4}\)
Solution:

(iii) cos^-1 \(\left(x+\frac{1}{2}\right)\) + cos^-1 x+ cos^-1 \(\left(x-\frac{1}{2}\right)\) = \(\frac{3 \pi}{2}\)
Solution:

(iv) 3tan^-1 \(\frac{1}{2+\sqrt{3}}\) – tan^-1 \(\frac{1}{x}\) = tan^-1 \(\frac{1}{3}\)
Solution:

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Fundamental Rules for Correcting Grammatical Errors Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Fundamental Rules for Correcting Grammatical Errors

Question 1.
In English Grammar verbs always agree with their subject in persons and numbers.
Example:
(a) I am a teacher.
subject – I
verb – am
(b) He is a student.
subject – He
verb – is
(c) They are farmers.
subject – They
verb – are
So while correcting the errors first underline the subject and verb of each sentence and see whether the verb agrees with its subject.

Question 2.
Prepositions always follow either a noun or an adjective or a verb. So find out preposition in each sentence and see their appropriate use.
Example:
(a) He is jealous of my success.
Adjective – jealous
Preposition – of
(b) The boy always depends on his father’s help.
Adjective – depends
Preposition – on
(c) He has confidence in me.
Noun- Confidence
Preposition – in

Question 3.
To find out errors in sentences articles play important role. In some sentences: there is wrong use of articles i.e., wrong use of ‘a’, ‘an’, ‘the’. So go through the use of articles meticulously and get it corrected.
Example:
A hermit live in a small cottage in a remote village. There is an orchard behind the cottage. The orchard has many rare trees.

Question 4.
Sometimes the use of quantifiers like ‘much’, ‘many’, ‘a lot of, ‘little’, ‘a little’, ‘few’ and ‘a few’ is wrongly used in sentences. It should be corrected carefully.
Example:
(a) They don’t find much time to work in the garden.
Uncountable – time
(b) Is there much water in the pond?
Uncountable – water

From the above examples it is found that ‘much’ is used before uncountable nouns in negative and interrogative sentences.
Example:
(c) He doesn’t have many shirts.
PI. Count. – shirts
(d) Does he have many books in his library.
PI. Count. – books

From the above examples it is found that ‘many’ is used before plural countable noun in negative and interrogative sentences.
Example:
(e) The mechanic repairs a lot of bikes everyday.
PI. Count. – bikes
(f) He has got a lot of money.
Uncountable – money

From the above examples it is found that ‘a lot of can be used with plural countable nouns as well as uncountable nouns in affirmative sentences.

Question 5.
Some verbal expressions like would rather and had better always go with the Bare Infinitive form of a non-finite verbs.
Example:
(a) You would rather go than stay here.
(going, go, gone, to go)
(Choose the correct alternative)
(b) They had better do their duty in time.
(doing, done, do, to do)
(Choose the correct alternative)

Question 6.
An imperative sentence beginning with “Let” can be followed by objective case of a pronoun.
Example:
(a) (i) Let he do whatever he wants to do. (Incorrect)
(ii) Let him do whatever he wants to do. (Correct)
(b) (i) Let you and I solve the problem. (Incorrect)
(ii) Let you and me solve the problem. (Correct).
(c) (i) Let there be no secret between you and we. (Incorrect)
(ii) Let there be no secret between you and us. (Correct)

Pronouns
Subjective case	Objective Case
I	Me
We	Us
You	You
He	Him
She	Her
It	It

Question 7.
The expression “I wish” always goes with “I were” i.e., past simple or “had + past participle” i.e., past perfect.
Example:
(a) (i) I wish I am the king. (Incorrect).
(ii) I wish I were the king. (Correct)
(b) (i) I wish I know your name. (Incorrect)
(ii) I wish I knew /had known your name. (Correct)

Question 8.
When the singular pronouns of first, second and third person go together their order should be “2nd person, third person and first person” i.e., “you + he and I”.
Example: (a) (i) I, you and he are friends. (Incorrect)
(ii) You, he and I are friends. (Correct)
(b) (i) You, they and we must work together. (Incorrect)
(ii) We, you and they must work together. (Correct)

Question 9.
Verbs like absent, apply, acquit, enjoy, resign pride, avail always go with reflexive pronouns (self/selves forms).
I – myself
he – himself
she – herself
you – yourself (singular) yourselves (plural)
we – ourselves
they – themselves
child/animal/ object (singular) – itself.

Example:
(a) (i) He prides on his money. (Incorrect)
(ii) He prides himself on his money. (Correct)

(b) (i) He availed of the chance. (Incorrect)
(ii) He availed himself of the chance. (Correct)

(c) (i) The children are expected to behave in the class. (Incorrect)
(ii) The children are expected to behave themselves in the class. (Correct)

(d) (i) The man resigned to the will of God. (Incorrect)
(ii) The man resigned himself to the will of God. (Correct)

(e) (i) My sister absented from her chemistry class yesterday. (Incorrect)
(ii) My sister absented herself from her chemistry class yesterday. (Correct)

Question 10.
The idiomatic expressions like “with a view to”, “look forward to”, “used to”, “habituated to”, and “objected to” always go with the “V+ing” form of a non-finite verb.
Example:
(a) (i) Sulipta is working hard with a view to win the match. (Incorrect)
(ii) Sulipta is working hard with a view to i. (Correct)
(b) (i) I look forward to see the doctor next month. (Incorrect).
(ii) I look forward to seeing the doctor next month. (Correct)
(c) (i) He is used to get up late. (Incorrect)
(ii) He is used to getting up late. (Correct)

Question 11.
(a) Some nouns look singular but they are originally plural. These nouns are people, police, public, cattle, swine, geese, teeth, oxen, mice and they can take plural verbs after them.
(b) The + Adjective and ‘The + Nationality word’ used as the subjects can take plural verbs.
Example:
(i) The police are on duty.
(ii) The people are shouting.
(iii) The cattle are grazing.
(iv) The poor are helpful.
(v) The wounded have been admitted in the hospital.
(vi) The English are a nation of traders.
(vii) The Japanese are industrious.
(viii) The geese look hungry.
(ix) The swine are the lovers of filth.
(x) When the cat is away, mice dance in joy.
(xi) The blind are the most unfortunate.

Question 12.
Grammatical expressions like “It is no use”, “There is no use” and “There is no point” always go with “-ing” form of non-finite verbs.
Example:
(i) It is no use attending classes today.
(ii) There is no point talking with her.

Question 13.
Some verbs tell us about our feelings, emotions, opinions, relations or about a permanent state. Such verbs are called stative verbs. These verbs are usually used in the simple present form.
(a) Verbs of possession: have, own, possess, belong to, contain, consist.
(b) Verbs of liking/disliking: like, dislike, love, hate, prefer, admire and want.
(c) Verbs of perception: see, hear, smell, taste, feel.
(d) Verbs of thinking: think, believe, understand, know.
(e) Verbs of mental activity: hope, forget, remember.
(f) Verbs of appearance: appear, seem, look (like) resemble.
(g) Some other verbs: depend, weight, cost, measure, sound.

Example:
(i) He belongs to Cuttack.
(ii) A year consists of six seasons.
(iii) Lipsa hates telling a lie.
(iv) Honey tastes sweet.
(v) Rose smells sweet.
(vi) Batu loves this girl.
(vii) Mama resembles her mother.
(viii) He knows me.
(ix) Empty vessel sounds much.
(x) I never believe in ghosts.

Question 14.
Adverb like No sooner, hardly/seldom, little, under no circumstances, nowhere and conjunctions like neither, not only etc. are used with inversion if they are used at the beginning of the sentences, i.e., Adv. + Aux. Verb + Sub. + Main verb.
Example:
(a) (i) No sooner the bell rang than the students came out of the class. (Incorrect)
(ii) No sooner did the bell rang than the students came out of the class. (Correct)

(b) (i) Hardly father had come out of the house when it began to rain. (Incorrect)
(ii) Hardly had father come out of the house when it began to rain. (Correct)

(c) (i) Little he understands my problem. (Incorrect)
(ii) Little does he understand my problem. (Correct)

(d) (i) Under no circumstances I shall compel you to do this work. (Incorrect)
(ii) Under no circumstances shall I compel you to do this work. (Correct)

(e) (i) Not only the robbers captured the city but also destroyed it. (Incorrect)
(ii) Not only did the robbers captured the city but also destroyed it. (Correct)

(f) (i) Neither the girl can dance nor sing. (Incorrect)
(ii) Neither can the girl dance nor sing. (Correct)

Question 15.
Some nouns look like plural but originally they are singular and take singular verbs. News, physics, politics, economics, measles, mathematics, diabetes, gymnastics, mumps, rabies, itches, scabies, electronics, athletics, sports, billiards, hurdles, cards.
Example:
(i) Mathematics is my favourite subjects.
(ii) Diabetes is a disease.
(iii) Gymnastics is good for health.
(iv) Billiards is my favourite indoor game.
(v) Physics is a tough subject.
(vi) The new has been greeted with cheers.
(viii)Politics is dirty game.

Question 16.
When the pronoun is the object of a verb or a preposition, it should be in the objective case.
Example:
(a) (i) These books are for you and I. (Incorrect)
(ii) These books are for you and me. (Correct)
(b) (i) Between he and I there is an understanding. (Incorrect)
(ii) Between he and me, there is an understanding. (Correct)

Question 17.
Subject form of the pronouns like (I/he/she/ you/they/we, etc.) is used with the sentences taking than.
Example:
(a) (i) He is taller than me. (Incorrect)
(ii) He is taller than I. (Correct)
(b) (i) I love you more than him. (Incorrect)
(ii) I love you more than he (Correct)

Question 18.
To avoid repetition of a noun, we use that for singular noun and those for the plural noun.
Example:
(a) (i) The climate of Simla is better than Darjeeling. (Incorrect)
(ii) The climate of Simla is better than that of Darjeeling. (Correct)
(b) (i) The roads of Bhubaneswar are wider than Cuttack. (Incorrect)
(ii) The roads of Bhubaneswar are wider than those of Cuttack. (Correct)

Question 19.
‘Each other’, the reciprocal pronoun is used in speaking of two persons/things, ‘one another’ for more than two.
Example:
(a) (i) Jack and Jill love one another. (Incorrect)
(ii) Jack and Jill love each other. (Correct)
(b) (i) The street dogs barked at each other. (Incorrect)
(ii) The street dogs barked at one another. (Correct)

Question 20.
The expressions like each (of), everyone (of), neither of (of the two), either of (of the two), the number of, one of etc. always take singular verb after them.
Example:
(a) (i) Each boy were given a prize. (Incorrect)
(ii) Each boy was given a prize. (Correct)

(b) (i) Each man and each woman have been sent to the workfield. (Incorrect)
(ii) Each man and each woman has been sent to the workfield. (Correct)

(c) (i) Everyone of the cars look attractive. (Incorrect)
(ii) Everyone of the cars looks attractive. (Correct)

(d) (i) Either of the books have a lot of pictures. (Incorrect)
(ii) Either of the books has a lot of pictures. (Correct)

(e) (i) Neither of the girls were called to appear the test. (Incorrect)
(ii) Neither of the girls was called to appear the test. (Correct)

(f) (i) The number of books stolen are forty. (Incorrect)
(ii) The number of books stolen is forty. (Correct)

(g) (i) One of the chairs have a broken leg. (Incorrect)
(ii) One of the chairs has a broken leg. (Correct)

(h) (i) One of my brothers are in the Indian army. (Incorrect)
(ii) One of my brothers is in the Indian army. (Correct)

Question 21.
The name of the shops (books/ opticals) always takes singular verb after them.
Example:
(a) (i) The Books and Books stand on the main road, Bhubaneswar. (Incorrect)
(ii) The Books and Books sstands on the main road, Bhubaneswar. (Correct)
(b) (i) The Giri Opticals have a good name in the whole town. (Incorrect)
(ii) The Giri Opticals has a good name in the whole town. (Correct)
(c) (i) ‘Gulliver’s Travels’ were written by Jonathan Swift. (Incorrect)
(ii) ‘Gulliver’s Travels’ was written by Jonathan Swift. (Correct)

Question 22.
After same or such, relative pronoun as or that is used.
Example:
(a) (i) This isn’t such a good book with I expected. (Incorrect)
(ii) This isn’t such a good book as I expected. (Correct)
(b) (i) This is the same beggar who came to our house last week. (Incorrect)
(ii) this is the same beggar that came to our house yesterday. (Correct)
(c) (i) My problem is the same which yours. (Incorrect)
(ii) My problem is the same as yours. (Correct)

Question 23.
We use relative pronoun that in the noun phrase on antecedent have the + superlative degree, ordinal (like first/second/ tenth /last, etc.) something/ anything/ all/nothing/ somebody etc. (Indefinite pronouns)
Example:
(a) (i) This is the best film which I have ever seen. (Incorrect)
(ii) This is the best film that I have ever seen. (Correct)
(b) (i) Love is something which money can’t buy. (Incorrect)
(ii) Love is something that money can’t buy. (Correct)
(c) (i) She is the most beautiful girl which has ever lived. (Incorrect)
(ii) She is the most beautiful girl that has ever lived. (Correct)
(d) (i) This is all which is yours. (Incorrect)
(ii) This is all that is yours. (Correct)
(e) (i) The second train which left for Puri just now, has faced an accident. (Incorrect)
(ii) The second train that left for Puri just now, has faced an accident. (Correct)

Question 24.
Comparative degree is used for two persons things/ animals and superlative degree for more than two.
Example:
(a) (i) It is the best of the two books. (Incorrect)
(ii) It is better of the two books. (Correct)
(b) (i) He is the better of the three boys. (Incorrect)
(ii) He is the best of the three boys. (Correct)
(c) (i) Which is the best: bread or butter? (Incorrect)
(ii) Which is better: bread or butter? (Correct)
(d) (i) Out of these two watches, which is the best? (Incorrect)
(ii) Out of these two watches, which is better? (Correct)

Question 25.
We use Fewer/Few/ A Few to denote number and less for quantity.
Example:
(a) (i) There are no less than twenty boys in this class. (Incorrect)
(ii) There are no fewer than twenty boys in this class. (Correct)
(b) (i) He takes no fewer than one litre of milk. (Incorrect)
(ii) He takes no jess than one litre of milk, (milk – u.n.) (Correct)

Question 26.
When comparative degree is used in the superlative sense, it is followed by ‘any other’.
Example:
(a) (i) Akram is better than any bowler. (Incorrect)
(ii) Akram is better than any other bowler. (Correct)
(b) (i) Raman is better than any student in the class. (Incorrect)
(ii) Raman is better than any other student in the class. (Correct)

Question 27.
Comparative Adjectives like senior, junior, superior, inferior take to after them instead of than’.
Example:
(a) (i) My father is senior than you in service. (Incorrect)
(ii) My father is senior to you in service. (Correct)
(b) (i) I am junior than her. (Incorrect)
(ii) I am junior to her. (Correct)
(c) (i) This book is inferior than that book. (Incorrect)
(ii) This book is inferior to that book. (Correct)

Question 28.
When expression of measurement, amount, and quantity are used as adjectives, the nouns occuring after the hyphen (-) are always singular.
Example:
(a) (i) The Prime Minister went on a three day state visit to America. (Incorrect)
(ii) The Prime Minister went on a three day state visit to America. (Correct)
(b) (i) A seven-members jury decided the case. (Incorrect)
(ii) A seven-members jury decided the case. (correct)
(c) (i) I came upon a hundred-rupees note. (Incorrect)
(ii) I came upon a hundred-rupee note. (Correct)

Question 29.
We usually use ‘but’ (not ‘than’) after ‘else’.
Example:
(a) (i) It is nothing else than pride. (Incorrect)
(ii) It is nothing else but pride. (Correct)
(b) (i) Call me anything else than a thief. (Incorrect)
(ii) Call me anything else but a thief. (Correct)

Question 30.
‘Very’ is used with adjectives and adverbs in the positive degree with precedent and much is used with Adjectives and adverbs in the comparative degree with past participle.
Example:
(a) (i) She is very slower than Reena. (Incorrect)
(ii) She is much slower (c.d) than Reena. (Correct)
(b) (i) You are very older than me. (Incorrect)
(ii) You are much older (c.d.) than me. (Correct)
(c) (i) The policeman was walking much slowly. (Incorrect)
(ii) The policeman, was walking very slowly. (Correct)

Question 31.
The verb ‘know’ is followed by ‘how to + infinitive’.
Example:
(a) (i) The professor knows to teach the topic. (Incorrect)
(ii) The professor knows how to teach the topic. (Correct)
(b) (i) An expert pilot knows to land the flight in hostile weather. (Incorrect)
(ii) An expert pilot knows how to land the flight in hostile weather. (Correct)

Question 32.
When a noun or pronoun is placed before a gerund (verb + ing working like a noun is called ‘gerund’), it (that noun/pronoun) should be put in the ‘possessive case’, (my/his/her/our/their/teacher’s/one’s etc)
Example:
(a) (i) Please excuse me being late. (Incorrect)
(ii) Please excuse my being late. (Correct)
(b) (i) I would like you stopping smoking. (Incorrect)
(ii) I would like your stopping smoking. (Correct)
(c) (i) The professor disliked us coming late. (Incorrect)
(ii) The professor disliked our coming late. (Correct)

Question 33.
Certain verbs looking alike have different meanings. They are as follows :
Example:
(a) (i) He hanged the lamp on the wall. (Incorrect)
(ii) He hung the lamp on the wall. (Correct)
(b) (i) The criminal was hung for murder. (Incorrect)
(ii) The criminal was hanged for murder. (Correct)
(c) (i) The hens have laid no eggs today. (Incorrect)
(ii) The hens have lain no eggs today. (Correct)
(d) (i) Let me lay on the bed. (Incorrect)
(ii) Let me lay on the bed. (Correct)
Look at the list of following verbs with their past and perfect or past participle forms.

Present	Past	P.P. / Perfect
lie (rest/lie down)	lay	lain
lay (place, arrange/deposit)	laid	laid
lie (to tell a lie)	lied	lied
hang (put up)	hung	hung
hang (execute the order of death sentence)	hanged	hanged
flow (water)	flowed	flowed
fly (bird)	flew	flown
flee (run away).	fled	fled
bear (put up with/tolerate)	bore	borne
bore (make a hole/make tied)	bored	bored
find (to discover)	found	found
found (establish)	founded	founded
fall (drop in the ground)	fell	fallen
fell (cut down a tree)	felled	felled
feel (sensitize)	felt	felt
fill (to pour till the end)	filled	filled
awake (intransitive verb)	awoke	awoke
awake (transitive verb)	awaked	awaked

Question 34.
Indirect questions have the usual Wh-word + subject + verb order.
Example:
(a) (i) Tell me where are you going. (Incorrect)
(ii) Tell me where you are going. (Correct)
(b) (i) Father asked the servant where had he gone. (Incorrect)
(ii) Father asked the servant where he had gone. (Correct)
(c) (i) Can you tell me why does the girl cry bitterly. (Incorrect)
(ii) Can you tell me why the girl cries bitterly. (Correct)
(d) (i) The gentleman asked the station master when was the next train. (Incorrect)
(ii) The gentleman asked the station master when next train was. (Correct)

Question 35.
The past tense in the Principal Clause or Main Clause is used with the same past tense in the Subordinating or Dependent Clause.
Example:
(a) (i) She knew that I am coming. (Incorrect)
(ii) She knew that I was coming. (Correct)
(b) (i) Father said that he won’t go to office that day. (Incorrect)
(ii) Father said that he wouldn’t go to office that day. (Correct)
(c) (i) The doctor asked the patient if he (the patient) has taken medicine at regular
intervals. (Incorrect)
(ii) The doctor asked the patient if he had taken medicine at regular intervals.(Correct)

Question 36.
It is time + subject + past tense verb.
Example:
(a) (i) It is time you go to bed. (Incorrect)
(ii) It is time you went to bed. (Correct)
(b) (i) It is time the doctor operates the patient. (Incorrect)
(ii) It is time the doctor operated the patient. (Correct)

Question 37.
After imperatives (order, advice, request) we use won’t you? (to invite people to do things) and will you/would you/could you? (to tell people to do things).
Example:
(a) (i) Do sit down, will you? (Incorrect)
(ii) Do sit down, won’t you? (Correct)
(b) (i) Give me sufficient time, won’t you? (Incorrect)
(ii) Give me sufficient time, will you? (Correct)
(c) (i) Shut up, can you? (Incorrect)
(ii) Shut up, can’t you? (Correct)

Question 38.
Certain verbs in English are not used in their progressive forms. They are :
Example:
(a) (i) I amn’t seeing you these days. (Incorrect)
(ii) I don’t see you these days. (Correct)

(b) (i) This bag is containing a lot of story books. (Incorrect)
(ii) This bag is contains a lot of story books. (Correct)

(c) (i) Are you appearing disappointed? (Incorrect)
(ii) Do you appear disappointed? (Correct)

(d) (i) Why is the girl hating me so much? (Incorrect)
(ii) Why does the girl hate me so much? (Correct)

(e) (i) Honey is tasting sweet. (Incorrect)
(ii) Honey tastes sweet. (Correct)

(f) (i) Rose is smelling sweet. (Incorrect)
(ii) Rose smells sweet. (Correct)

(g) (i) We are not believing in ghost. (Incorrect)
(ii) We don’t believe in ghost. (Correct)

(h) (i) I am loving this girl. (Incorrect)
(ii) I love this girl. (Correct)

Question 39.
Would rather + subject takes past simple tense of the verb.
Example:
(a) (i) I would rather you resign the job. (Incorrect)
(ii) I would rather you resigned the job. (Correct)
(b) (i) I would rather he leaves this place. (Incorrect)
(ii) I would rather he left this place. (Correct)

Some incorrect words/expressions and their correct use.

Incorrect – Correct
arm (weapon) – arms
blotting – blotting paper
boarding – boarding school
bowel – bowels
foundation – foundations
arrear – arrears
furnitures – furniture
luggages – luggage
earning – earnings
breads – pieces/slices/loaves of bread
equipments – equipment
informations – information
gentries – gentry
machineries – machinery/machines
poetries – poetry
sceneries – scenery
traffics – traffic
scissor – scissors
trouser – trousers
a coward man – a coward/ a cowardly man
a miser man – a miser/ a miserly man
a man of letter – a man of letters(literate)
arrear bill – arrears bill
birth date – date of birth
cousin brother/sister – cousin
custom duty – customs duty
famous criminal – notorious criminal
in the campus – on the campus
in the committee – on the committee
in leave – on leave
in holiday – on holiday
tennis field – tennis court
Cheque of Rs. 200/- – Cheque for Rs. 200/-
no place (bus/train etc.) – no room (bus/train etc.)
today morning/ – this morning/afternoon/
afternoon/evening – evening
this night – tonight
two dozens pens – two dozen pens
saving bank – savings bank
make noise – make a noise
tell lie – tell a lie/tell lies
white hair – grey hair
miles after miles – mile after mile
in hurry – in a hurry
bad in studies – bad at studies
strong/weak of/ good in – strong/weak in/ good at
what to speak of – not to speak of
with bag and baggage – bag and baggage
with heart and soul – heart and soul
with tooth and nail (Completely) – tooth and nail
with black and blue (mercilessly/beat) – black and blue
clever in figure works – clever at figure works
build a home – build a house
cut jokes – crack jokes
cut the pencil – mend the pencil
cook bread – bake bread
describe about – describe
discuss about – discuss
attack on – attack
give a speech – deliver a speech
goodbye – bid goodbye
eat the poor – feed the poor
give order – give orders
make a goal – score a goal
rise the lid – raise the lid
see the pulse – feel the pulse
to have headache – to have a headache
to have temperature – to have a temperature
speak a lie – tell a lie
make prayers – say prayers
ladies bicycle – ladies’ bicycle
women’s college – womens’ college
family members – members of the family
passing marks – pass marks
decrease fear – allay fear
in class tenth – in class ten/in the tenth class
lecture (person) – lecturer
out of orders – out of order (technically defect)
deny request/invitation – refuse request/invitation
refuse stealing/lying – deny stealing/lying
refuse – reject/not to accept
deny – not acknowledge/tell that something is untrue.
a furniture – a piece of furniture
a luggage – a piece of luggage
an information – a piece of information.
a wood – a piece of wood
a grass – a blade of grass
a chocolate – a bar of chocolate
a paper – a sheet of paper
a news – a piece of news/some news
a work – a piece of work / some work
a bread – a slice of bread
a meat – a piece of meat
a toothbrush – a stick of toothbrush
a toothpaste – a tube of toothpaste

Rewrite the following passage, correcting all the grammatical errors in it:

Question 1.
The streets crowd by traffic even before the daybreak. Taxis have been bringing tired people from the airport and the railway station to hotels. They hope sleeping for a few hours before their busy day in the big city is beginning. Trucks are bringing fresh fruits and vegetables in the city. Ships loaded with food and fuel tie up in the dock. Towards morning the streets are quieter, and they are never deserted in the big city.
Answer:
The streets are crowded by traffic even before the daybreak. Taxis bring tired people from the airport and the railway station to the hotels. They hope to sleep for a few hours before their busy day in the big city begins. Trucks bring fresh fruits and vegetables to the city. Ships loaded with food and fuel tied up at the dock. Towards the morning the streets are quieter, and they are never deserted in the big city.

Question 2.
Every country in the world want rapid economic development today. Some economists tells us that it is possible to remove poverty and make everyone prosperous, provide we adopt the right economic policies. The key to prosperity, we are also told, lies in rapid and large-scale industrialization: setting up more factories which will churn out an endless stream of consumer good – products designed to make life more pleasant; motor cars to carried us in comfort and at high speed along smooth super highways; air conditioners to keeps us cool in summer, television sets which will keep us informed as well as entertain and so on. It is believed that as more and more consumers buy the goods that these factories will produce, more and more workers find employment in them; and as their levels of income rise, they will, in their turn create a farther demand for yet more goods. In this way, everyone becomes rich. There are no limits to economical growth and prosperity.
Answer:
Every country in the world wants rapid economic development today. Some economists tell us that it is possible to remove poverty and make everyone prosperous, provided we adopt the right economic policies. The key to prosperity, we are also told, lies in rapid and large-scale industrialization: setting up more factories which will churn out an endless stream of consumer goods – products designed to make life more pleasant; motor cars to carry us in comfort and at high speed along smooth super highways.

air conditioners to keep us cool in summer, television sets which will keep us informed as well as entertained and so on. It is believed that as more and more consumers buy the goods that these factories will produce, more and more workers find employment in them; and as their level of income rises, they will, in their turn create a further demand for yet more goods. In this way, everyone becomes rich. There are no limits to economical growth and prosperity.

Question 3.
Thrift means regulating expenses by such a way that there might be some saving in income. There could be no hard and fast standards for what shall be one’s expenditure and saving. It varied according to one’s circumstances. The rich man may neglect the duty of saving in special occasions because he has power to make up for this neglect. And people with limited income need thrift the most. It gives them strength by relieving them of anxieties for the future. Good housewives in poor families are found to lay off something from daily expenses.
Answer:
Thrift means regulating expenses in such a way that there might be some saving in income. There should be no hard and fast standard for what could be one’s expenditure and saving. It varies according to one’s circumstances. The rich man may neglect the duty of saving on special occasions because he has the power to make up for this neglect. And people with limited income need thrift the most. It gives them strength by relieving them from anxieties for future. Good housewives in poor families are found laying off something from daily expenses.

Question 4.
Preschools have offered good basic education as well as help the child in becoming much independent but confident. Parents may rely in preschools for all-round development of the children. The pre-primary education of the child generally began in home by parents and grandparents. But the picture changes rapidly.
Answer:
Preschools offer good basic education as well as help the child in becoming much independent and confident. Parents rely on preschools for an all-round development of their children. The pre-primary education of the child begins at home by parents and grandparents. But the picture has changed rapidly.

Question 5.
Once a lion was enjoy a nap in his den. A mouse came out in its hole in the den. It start frisking about. In so doing it leap upon the lion’s face. The lion’s sleep disturbs. He wake up furious. He caught the mouse and had been killed it, but the mouse entreated “Your Majesty, I humbly beg your pardon, I’m a poor and little subject of you. But a tiny creature as 1 am, I shall be of some help to you in time. So, please let me go.” The lion laughed aloud for this, but he released the mouse all the same.
Answer:
Once a lion was enjoying a nap in his den. A mouse came out of its hole in the den. It started frisking about. In so doing it leapt upon the lion’s face. The lion’s sleep was disturbed. He woke up furiously. He caught the mouse and was about to kill it. But the mouse entreated “Your Majesty, I humbly beg your pardon, I’m a poor and little subject of yours. But a tiny creature as I am, I shall be of some help to you in time. So, please let me go.” The lion laughed aloud for this and he released the mouse all the same.

Question 6.
Man is at first like an animal. His power then rested only in physical strength. But in this respect he was no match to many beasts. So he has to live in constant fear of them. In course of time came knowledge and it gave him the power to get mastery the entire animal kingdom. He invented weapons with which he not only scared off beasts but even killed them. He learnt how to hunt beasts like a deer which may run more faster than he did Now man could achieve things which were considered impossible then.
Answer:
Man was at first like an animal. His power then rested only on physical strength. But in this respect he was no match for many beasts. So he had to live in constant fear of them. In course of time came knowledge and it gave him a power to get mastery upon the entire animal kingdom. He invented weapons with which he not only scared off beasts but also killed them. He learnt how to hunt beasts like a deer who could run faster than he. Now man has achieved things which were considered impossible then.

Question 7.
The foodbazar took the entire responsibility in sending a farmer’s produce to the customer, without depending on a chain of middlemen. The private company with its vast resources could set in cold storages, acquire refrigerated trucks to transport the produce to cities. In the end of the food chain there are airconditioned supermarkets where consumers can buy the produce at good condition. A kilo of tomato which a customer will buy for 10 rupees may be available for only 7 rupees in a supermarket. Don’t imagine that any private company will do all this out from charity or love towards the farmers. In time food chains might come on away from the cities and closer to the farms.
Answer:
The foodbazar is taking the entire responsibility of sending a farmer’s produce to the customer, without depending on a chain of middlemen. The private company with its vast resources can set up cold storages and acquire refrigerated trucks to transport the produce to cities. At the end of the food chain there are airconditioned supermarkets where consumers can buy the produce in good condition. A kilo of tomato which a customer will buy for 10 rupees can be available for only 7 rupees in a supermarket. We can’t imagine that any private company will do all this out of charity or love for the farmers. With time food chains may come away from the cities and closer to the farms.

Question 8.
A big fat hen lived of a farmyard in a village. A dove also lived in a large tree besides the same farmyard. Soon both of them became good friend. They use to meet every evening to share their thoughts. Oneday, the hen see by a fox. He enters the farmyard secretly and was able to catch the hen. The clever fox put the hen in a sack, carried at his back and the hen started crying aloud in protest. The dove heard the crying of the hen from the top of the tree. She at once realised that the hen fell in danger. The dove thought off an idea to save her friend. She went ahead and lay on the path motionless as of she was dead. The fox put his sack down at the wayside and went near the dove. In the meantime, the hen managed to escape, and hide behind a bush. When the fox was about to catch the ‘dead’ dove, he surprised to see that the dove quickly flew away. In the evening the dove met the hen at the farmyard and the two began to laugh for the foolishness of the fox.
Answer:
A big fat hen lived in a farmyard in a village. A dove also lived in a large tree beside the same farmyard. Soon both of them became good friends. They used to meet every evening to share their thoughts. Oneday, the hen was seen by a fox. He entered the farmyard secretly and was able to catch the hen. The clever fox put the hen in a sack, carried on his back and the hen started crying aloud in protest. The dove heard the cry of the hen from the top of the tree.

She at once realised that the hen had fallen into danger. The dove thought of an idea to save her friend. She went ahead and lay on the path motionless as if she was dead. The fox put his sack down on the wayside and went near the dove. In the meantime, the hen managed to escape, and hid behind a bush. When the fox was about to catch the ‘dead’ dove, he was surprised to see that the dove quickly flew away. In the evening the dove met the hen at the farmyard and the two began to laugh for the foolishness of the fox.

Question 9.
The food bazaar is taking entire responsibility in sending the farmer’s produce to the consumer. The private company with its vast resource, may set up cold storages, acquire fleets of refrigerated trucks to transport the produce into cities and even construct roads for speedy transportation. In the end of the food chain, there have been air-conditioned supermarkets where consumers could buy produce of high quality, in good condition, at comparatively reasonable prices, in clean and hygienic surrounding. A kilo of tomatoes which a customer could buy from a vegetable-vendor for ten rupees must be available, weighed but neatly packed for only t 7.50 in a super-market. Out of that amount, the farmer is likely to have got at least ? 3.50 a much higher price than he would get if he would sell his produce to a middleman.
Answer:
The food bazaar is taking an responsibility in sending a farmer’s produce to the consumer. The private company with its vast resources, can set up cold storages, acquire a fleet of refrigerated trucks to transport the produce to the cities and even construct roads for speedy transportation. At the end of the food chain, there are air-conditioned supermarkets where consumers can buy produce of high quality, in good condition, in a comparatively reasonable prices, in a clean and hygienic surrounding.

A kilo of tomato which a customer can buy from a vegetable-vendor for ten rupees may be available, weighed but neatly packed for only t 7.50 in a super-market. Out of that amount, the farmer is likely get at least ? 3.50 a much higher price than he would get if he would sell his produce to a middleman.

Question 10.
Few years ago I was spending a week at Port Blair. The week had been over but I was in the airport ready for leaving when I discovered, in my dismay, that I forgot one of my suitcase in my hotel. Quickly, I jumped into a taxi and had explained my situation to the taxi driver. We sped away in the direction of my hotel. Suddenly the taxi driver slowed down so that he would talk with the driver of a truck moving through a road next to us. The truck contained live chickens.
Answer:
A few years ago I was spending a week in Port Blair. As the week was over but 1 was in the airport ready to leave when I discovered, to my dismay, that I had forgotten one of my suitcases in my hotel. Quickly, I jumped into a taxi and explained my situation to the taxi driver. We sped away in the direction of my hotel. Suddenly the taxi driver slowed down so that he might talk to the driver of a truck moving through the road next to us. The truck carried live chichens.

Question 11.
Rahul and Ramesh were best friend studying in school. Rahul was good in sports but poor in studies. Ramesh is good at both. At any competition, Ramesh always managed to win. As a result, Rahul became jealous with Ramesh. The sports day was near. Rahul and Ramesh were practising for it. Rahul was not sure if he score a win over Ramesh. Slowly they talked less among themselves. Ramesh asked him the reason many time but Rahul always put him of with a excuse or other. One day before the sports event, Rahul hit at a plan to defeat Ramesh. He went to the ground before anyone has arrived there. He dug a small pit on the path where Ramesh was supposed to run. Then he covered it with leaves and went back to his classroom. When the race started. Ramesh was ahead of Rahul but after sometimes he stepped on the pit and fell down. Rahul took over him. He was very happy as his plan worked and he had own the first prize. Soonafter, Rahul realized his fault and begged forgiveness.
Answer:
Rahul and Ramesh were best friends when studying in school. Rahul was good at sports but poor in studies. Ramesh was good at both. In any competition, Ramesh always managed to win. As a result, Rahul became jealous of Ramesh. The sports day was near. So Rahul and Ramesh were practising for it. Rahul was not sure if he would score a win over Ramesh. Gradually they talked less between themselves. Ramesh asked him the reason many a time but Rahul always put him off with an excuse or other.

One day before the sports event, Rahul hit upon a plan to defeat Ramesh. He went to the ground before anyone had arrived there. He dug a small pit on the path where Ramesh was supposed to run. Then he covered it with leaves and went back to his classroom. When the race started. Ramesh was ahead of Rahul but after sometime he stepped on the pit and fell down. Rahul took over him. He was very happy as his plan worked and he had won the first prize. Soonafter, Rahul realized his fault and begged forgiveness.

Question 12.
Generally the word ‘superstition’ was used as the term of disgrace or reproach to senseless belief based in ignorant fear. We look down to the uncivilized people because they worshipped the different aspect of nature and are filled in awe by things that are simple enough for us. Many Hindus considered a sneeze, the cry of a lizard as ominous. For the Westerners, number 13 is so unlucky that it can be avoided by all means. If the cat crosses their path they call up their tour. Indeed, every community is subject to some superstition and we could not conceive for any community being completely free of it.
Answer:
Generally the word ‘superstition’ is used as a term of disgrace or reproach to senseless belief based on ignorant fear. We look down upon the uncivilized people because they worshipped different aspects of nature and were fdled with awe by the things that are simple enough for us.

Many Hindus consider a sneeze, the cry of a lizard as ominous. For the Westerners, number 13 is so unlucky that it should be avoided by all means. If the cat crosses their path they call off their tour. Indeed, every community is subject to some superstition and we can not conceive of any community completely free from it.

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(a)

Question 1.
(i) ∫2 dx
Solution:
∫2 dx = 2x + C

(ii) ∫3x² dx
Solution:
∫3x² dx = 3 \(\frac{x^3}{3}\) + C = x³ + C

(iii) ∫4x³ dx
Solution:
∫4x³ dx = 4 \(\frac{x^4}{4}\) + C = x⁴ + C

(iv) ∫x⁵ dx
Solution:
∫x⁵ dx = \(\frac{x^6}{6}\) + C

(v) ∫x³¹ dx
Solution:
∫x³¹ dx = \(\frac{x^{32}}{32}\) + C

(vi) ∫\(\left(2 \sqrt{x}+\frac{3}{\sqrt{x}}\right)\) dx
Solution:

(vii) ∫\(\frac{1}{x \sqrt{x}}\) dx
Solution:

(viii) ∫\(\left(x^{\frac{4}{7}}+\frac{1}{x^{\frac{1}{3}}}\right)\) dx
Solution:

(ix) ∫\(\frac{4}{x}\) dx
Solution:
∫\(\frac{4}{x}\) dx = 4 ln |x| + C

(x) ∫\(\left(\frac{3}{x^2}+\frac{1}{x^{12}}\right)\) dx
Solution:

(xi) ∫(x² + √x)² dx
Solution:

(xii) ∫(x + 3) (2 – x) dx
Solution:
∫(x + 3) (2 – x) dx
= ∫(2x + 6 – x² – 3x) dx
= ∫(-x² – x + 6) dx
= –\(\frac{1}{3}\)x³ – \(\frac{1}{2}\)x² + 6x + C

(xiii) ∫\(\frac{(\sqrt{x}+2)^2}{x^4}\) dx
Solution:

(xiv) ∫\(\frac{(x+\sqrt{x})(2 x+1)}{x^2}\) dx
Solution:

Question 2.
(i) ∫cos dx
Solution:
∫cos dx = sin x + C

(ii) ∫\(\frac{d x}{\cos ^2 x}\)
Solution:
∫\(\frac{d x}{\cos ^2 x}\) = ∫sec² x dx = tan x + C

(iii) ∫\(\frac{d x}{1-\cos ^2 x}\)
Solution:
∫\(\frac{d x}{1-\cos ^2 x}\) = ∫\(\frac{d x}{\sin ^2 x}\)
= ∫cosec² x dx
= -cot x + C

(iv) ∫\(\frac{\sin x}{\cos ^2 x}\) dx
Solution:
∫\(\frac{\sin x}{\cos ^2 x}\) dx = ∫\(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}\) dx
= ∫sec x . tan x dx = sec x + C

(v) ∫\(\frac{2 \cos x}{1-\cos ^2 x}\) dx
Solution:

(vi) ∫\(\frac{1-\sin ^3 x}{\sin ^2 x}\) dx
Solution:
∫\(\frac{1-\sin ^3 x}{\sin ^2 x}\) dx = ∫(cosec² x – sin x) dx
= -cot x + cos x + C

(vii) ∫\(\frac{\sin ^2 x}{1+\cos x}\) dx
Solution:

(viii) ∫\(\frac{\cos 2 x}{\cos x+\sin x}\) dx
Solution:

(ix) ∫\(\frac{\cos ^4 x-\sin ^4 x}{\cos x-\sin x}\) dx
Solution:

(x) ∫\(\frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x}\) dx
Solution:

(xi) ∫\(\frac{a^2 \sin ^2 x+b^2 \cos ^2 x}{\sin ^2 2 x}\) dx
Solution:

(xii) ∫(tan x + cot x)² dx
Solution:

(xiii) ∫\(\frac{1-\cos 2 x}{1+\cos 2 x}\) dx
Solution:

(xiv) ∫sec² x . cosec² x dx
Solution:

(xv) ∫\(\frac{\sin ^3 x+b \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:

(xvi) ∫\(\sqrt{1+\sin 2 x}\) dx
Solution:

(xvii) ∫\(\sqrt{1-\cos 2 x}\) dx
Solution:

(xviii) ∫\(\sqrt{1+\cos 2 x}\) dx
Solution:
∫\(\sqrt{1+\cos 2 x}\) dx = ∫√2 cos x dx
= √2 sin x + C

(xix) ∫\(\frac{\cos 3 x \cos 2 x+\sin 3 x \sin 2 x}{1-\cos ^2 x}\) dx
Solution:

(xx) ∫(a cot x + b tan x)² dx
Solution:
∫(a cot x + b tan x)² dx
= ∫{a² cot² x + b² tan² x + 2ab cot x – tanx} xdx
= ∫(a² cot² x + b² tan² x + 2ab) dx
= a² ∫cot² x dx + b² ∫tan² x dx + 2ab ∫dx
= a² ∫(cosec² x – 1) dx + b² ∫(sec² x – 1) dx + 2ab x + C
= a² (-cot x – x) + b² (tan x – x) + 2ab x + C
= b² tan x – a² cot x – x (a² + b² – 2ab) + C
= b² tan x – a² cot x – (a – b)² x + C
= b² tan x – a² cot x + C

Question 3.
(i) ∫(e^x + 2) dx
Solution:
∫(e^x +2) dx = ∫e^x dx + 2 ∫dx
= e^x + 2x + C

(ii) ∫3^x dx
Solution:
∫3^x dx = ∫\(\frac{3^x}{\ln 3}\) + C

(iii) ∫a^x+2 dx
Solution:
∫a^x+2 dx = ∫a^x a² dx
= a² . \(\frac{a^x}{\ln a}\) + C
= \(\frac{a^{x+2}}{\ln a}\) + C

(iv) ∫a^3x dx
Solution:
∫a^3x dx = \(\frac{1}{3} \frac{a^{3 x}}{\ln a}\) + C

(v) ∫\(\frac{e^{2 x}+1}{e^x}\) dx
Solution:
∫\(\frac{e^{2 x}+1}{e^x}\) dx = ∫(e^x + e^-x) dx
= e^x – e^-x + C

Question 4.
(i) ∫\(\left(\frac{5}{\sqrt{1-x^2}}+\frac{7}{1+x^2}\right)\) dx
Solution:

(ii) ∫\(\frac{3 x^2}{x^2+1}\) dx
Solution:

(iii) ∫\(\frac{x^6}{x^2+1}\) dx
Solution:

(iv) ∫\(\frac{x^4+x^3+x^2+x+2}{x^2+1}\) dx
Solution:

(v) ∫\(\left(\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}\right)\) dx
Solution:

(vi) ∫\(\frac{x^2+\sqrt{x^2+1}}{x^3 \sqrt{x^2-1}}\) dx
Solution:

Question 5.
Find the unique antiderivative F(x) of f(x) = 2x² + 1, whese F(o) = -2.
Solution:
F(x) = ∫f(x) dx = ∫(2x² + 1) dx
= \(\frac{2}{3}\)x³ + x + C
But F(0) = -2
∴ -2 = C
Thus F(x) = \(\frac{2}{3}\)x³ + x – 2

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
Write the maximum value of the function y = x⁵ in the interval [1, 5].
(a) 125
(b) 3125
(c) 625
(d) 225
Solution:
(b) 3125

Question 2.
Differentiate sin^-1 (cos x) with respect to x.
(a) -1
(b) 0
(c) 1
(d) None of the above
Solution:
(a) -1

Question 3.
Write the equation of the tangent to the curve y = |x| at the point (-2, 2).
(a) 2x + 2y = 0
(b) 2x + y = 0
(c) x + y = 1
(d) x + y = 0
Solution:
(d) x + y = 0

Question 4.
Write the value of \(\frac{d}{d x}\)(sin x)
(a) tan x
(b) sin x
(c) cos x
(d) cot x
Solution:
(c) cos x

Question 5.
Write the value of \(\frac{d}{d x}\)(cos x)
(a) cos x
(b) -sin x
(c) sin x
(d) cot x
Solution:
(b) -sin x

Question 6.
Write the value of \(\frac{d}{d x}\)(tan x)
(a) tan x
(b) sec² x
(c) sin x
(d) cot x
Solution:
(c) sin x

Question 7.
Write the value of \(\frac{d}{d x}\)(sec x)
(a) sin x cos x
(b) sin x tan x
(c) cot x tan x
(d) sec x tan x
Solution:
(d) sec x tan x

Question 8.
Write the value of \(\frac{d}{d x}\)(cot x) = -cosec²x
(a) -sin² x
(b) -cos² x
(c) -cosec² x
(d) -tan² x
Solution:
(c) -cosec² x

Question 9.
Write the value of \(\frac{d}{d x}\)(cosec x)
(a) -cosec x cot x
(b) -sin x cot x
(c) -tan x cot x
(d) -cos x cot x
Solution:
(a) -cosec x cot x

Question 10.
Write the value of \(\frac{d}{d x}\)(e^x) = ex
(a) e
(b) x
(c) e^x
(d) e^2x
Solution:
(c) e^x

Question 11.
Write the value of \(\frac{d}{d x}\)(a^x)
(a) a^2x log a
(b) a^x log a
(c) log a
(d) log a^x
Solution:
(b) a^x log a

Question 12.
Write the value of \(\frac{d}{d x}\)(log x)
(a) x
(b) \(\frac{1}{2 x}\)
(c) \(\frac{1}{x^2}\)
(d) \(\frac{1}{x}\)
Solution:
(d) \(\frac{1}{x}\)

Question 13.
Write the value of \(\frac{d}{d x}\)(log_a x)
(a) \(\frac{x}{\log a}\)
(b) \(\frac{1}{x \log a}\)
(c) \(\frac{1}{x^2 \log a}\)
(d) \(\frac{1}{2 x^2 \log a}\)
Solution:
(b) \(\frac{1}{x \log a}\)

Question 14.
Write the value of \(\frac{d}{d x}\)(cot^-1 x) = \(\frac{-1}{1+x^2}\)
(a) \(\frac{1}{1-x^2}\)
(b) \(\frac{1}{1+x^2}\)
(c) \(\frac{-1}{1-x^2}\)
(d) \(\frac{-1}{1+x^2}\)
Solution:
(d) \(\frac{-1}{1+x^2}\)

Question 15.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) for √x + √y = √c.
(a) \(\frac{\sqrt{y}}{\sqrt{x}}\)
(b) \(\frac{\sqrt{x}}{\sqrt{y}}\)
(c) –\(\frac{\sqrt{y}}{\sqrt{x}}\)
(d) –\(\frac{\sqrt{x}}{\sqrt{y}}\)
Solution:
(c) –\(\frac{\sqrt{y}}{\sqrt{x}}\)

Question 16.
If f(x) = \(\log _{x^2}\) (log x), then find f'(e).
(a) \(\frac{1}{e}\)
(b) \(\frac{1}{\mathrm{e}^2}\)
(c) \(\frac{1}{2 \mathrm{e}}\)
(d) \(\frac{2}{\mathrm{e}}\)
Solution:
(c) \(\frac{1}{2 \mathrm{e}}\)

Question 17.
If y = \(\sqrt{\sin x+y}\), then what is \(\frac{\mathrm{dy}}{\mathrm{dx}}\)?
(a) \(\frac{\cos x}{2 y+1}\)
(b) \(\frac{\cos x}{2 y-1}\)
(c) \(\frac{\sin x}{2 y-1}\)
(d) \(\frac{\sin x}{2 y+1}\)
Solution:
(b) \(\frac{\cos x}{2 y-1}\)

Question 18.
If y = 10^x and z = 100^x/2 then find \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}^2}{\mathrm{z}}\right)\)
(a) 10^x log 10
(b) 10^2x log 10
(c) x log 10
(d) x² log 10
Solution:
(a) 10^x log 10

Question 19.
Find the differential of y = sin² x.
(a) 2sinx cot x dx
(b) 2sinx tan x dx
(c) 2sinx cosx dx
(d) 2cosx sinx dx
Solution:
(c) 2sinx cosx dx

Question 20.
Differentiate \(\sqrt{\sin \sqrt{x}}\)
(a) \(\frac{\cos \sqrt{x}}{4 \sqrt{x \sin \sqrt{x}}}\)
(b) \(\frac{\sin x}{4 \sqrt{x \cos \sqrt{x}}}\)
(c) \(\frac{\sin \sqrt{x}}{4 \sqrt{x \cos \sqrt{x}}}\)
(d) \(\frac{\sin x}{4 \sqrt{x \cos x}}\)
Solution:
(a) \(\frac{\cos \sqrt{x}}{4 \sqrt{x \sin \sqrt{x}}}\)

Question 21.
If f(x) = [tan² x], what is f'(0)?
(a) 0
(b) 1
(c) -1
(d) None of the above
Solution:
(a) 0

Question 22.
If xy = e^x-y then find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
(a) \(\frac{\log x}{(1-\log x)^2}\)
(b) \(\frac{\log x}{(1+\log x)^2}\)
(c) \(\frac{\log x}{(1+\log x)}\)
(d) \(\frac{\log x}{(1-\log x)}\)
Solution:
(b) \(\frac{\log x}{(1+\log x)^2}\)

Question 23.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) if y = cot^-1\(\left(\frac{1+\cos x}{1-\cos x}\right)^{\frac{1}{2}}\)
(a) 0
(b) \(\frac{1}{2}\)
(c) 1
(d) \(\frac{1}{4}\)
Solution:
(b) \(\frac{1}{2}\)

Question 24.
Find the derivative of sin x° w.r.t. x.
(a) \(\frac{\pi}{90}\) cos x°
(b) π cos x°
(c) \(\frac{\pi}{180}\) cos x°
(d) π sin x°
Solution:
(c) \(\frac{\pi}{180}\) cos x°

Question 25.
The least value of a such that f(x) = x² + ax + 1 is strictly increasing on (1, 2) is
(a) -2
(b) -4
(c) 2
(d) 4
Solution:
(a) -2

Question 26.
If y = sec^-1\(\left(\frac{x+1}{x-1}\right)\) + sin^-1\(\left(\frac{x-1}{x+1}\right)\) then find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
(a) 0
(b) -1
(c) 1
(d) None of the above
Solution:
(a) 0

Question 27.
Let f(x) = e^x g(x), g(0) = 2 and g'(0) = 1, then find f'(0).
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(c) 3

Question 28.
If x ∈ (0, π/2) then find \(\frac{\mathrm{d}}{\mathrm{dx}}\)[sin x]
(a) 0
(b) 1
(c) 7C
(d) \(\frac{\pi}{2}\)
Solution:
(a) 0

Question 29.
If f(x) = [x²] then f’\(\left(\frac{3}{2}\right)\) = ________.
(a) 1
(b) \(\frac{3}{2}\)
(c) 0
(d) –\(\frac{3}{2}\)
Solution:
(c) 0

Question 30.
A differentiable function f defined for all x > 0 satisfies f(x²) = x³ for all x > 0. What is f'(b)?
(a) 3
(b) 4
(c) 6
(d) 12
Solution:
(c) 6

Question 31.
What is the derivative of f'(In x) w.r.t x where f(x) = ln x.
(a) \(\frac{1}{x^2 \ln x}\)
(b) \(\frac{1}{2 x \ln x}\)
(c) \(\frac{1}{x \ln x^2}\)
(d) \(\frac{1}{x \ln x}\)
Solution:
(d) \(\frac{1}{x \ln x}\)

Question 32.
For what value of a, f(x) = log_a (x) is increasing on R?
(a) a > 1
(c) a < 1
(b) a = 1
(d) a > 1
Solution:
(a) a ≥ 1

Question 33.
Write the points at which tangent to the curve y = x² – 3x is parallel to x – axis.
(a) \(\left(\frac{3}{2}, \frac{9}{4}\right)\)
(b) \(\left(\frac{3}{2}, \frac{-9}{4}\right)\)
(c) \(\left(\frac{-3}{2}, \frac{9}{4}\right)\)
(d) \(\left(\frac{-3}{2}, \frac{-9}{4}\right)\)
Solution:
(b) \(\left(\frac{3}{2}, \frac{-9}{4}\right)\)

Question 34.
If y = e^x + e^-x + 2 has a tangent parallel to x – axis at (α, β) then find the value of α.
(a) 0
(b) 2
(c) 1
(d) -1
Solution:
(a) 0

Question 35.
Write slope of the tangent to the curve y = √3 sin x + cos x at (\(\frac{\pi}{3}\), 2)
(a) 1
(b) -1
(c) 0
(d) None of the above
Solution:
(c) 0

Question 36.
Find the value of x for which f(x) is either a local maximum or a local minimum when
f(x) = x³ – 3x² – 9x + 6.
(a) (3, -1)
(b) (3, 1)
(c) (-1, 3)
(d) (1, -3)
Solution:
(a) (3, -1)

Question 37.
Find the x – coordinates of the extreme points of the function.
y = cos x + sin x , x ∈ [0, \(\frac{\pi}{2}\)]
(a) \(\frac{\pi}{4}\)
(b) \(\frac{2 \pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) π
Solution:
(a) \(\frac{\pi}{4}\)

Question 38.
Find the equation of tangent to the curve x = y² – 2 at the points where slope of the normal equal to (-2).
(a) x + y – 1 = 0
(b) 2x + y – 1 = 0
(c) 2x – y + 1 = 0
(d) x – y + 1 = 0
Solution:
(b) 2x + y – 1 = 0

Question 39.
f(x) = x⁴ – 62x² + ax + 9 attains its maximum value at x = 1 in the interval [0, 2]. Find the value of ‘a’.
(a) 12
(b) 120
(c) 1
(d) 1200
Solution:
(b) 120

Question 40.
If sin (x + y) = log (x + y) then \(\frac{d y}{d x}\) is:
(a) 2
(b) -2
(c) 1
(d) -1
Solution:
(d) -1

Question 41.
If f(x) = 2x and g(x) = \(\frac{x^2+1}{2}\) then which of the following can be a discontinuous function?
(a) F(x) + g(x)
(b) f(x) – g(x)
(c) f(x) . g(x)
(d) \(\frac{f(x)}{g(x)}\)
Solution:
(d) \(\frac{f(x)}{g(x)}\)

Question 42.
If f(x) = x² sin (‘) where x ≠ 0 then the value of the function f at x = 0, so that the function is continuous at x = 0 is
(a) 0
(b) -1
(c) 1
(d) none of these
Solution:
(a) 0

Question 43.
The derivative of cos^-1 (2x² – 1) with respect to cos^-1 x
(a) 2
(b) \(\frac{-1}{2 \sqrt{1-x^2}}\)
(c) \(\frac{2}{x}\)
(d) 1 – x²
Solution:
(a) 2

Question 44.
If y = \(\sqrt{\sin x+y}\) then \(\frac{d y}{d x}\) is equal to:
(a) \(\frac{\cos x}{2 y-1}\)
(b) \(\frac{\cos x}{1-2 y}\)
(c) \(\frac{\sin x}{1-2 y}\)
(d) \(\frac{\sin x}{2 y-1}\)
Solution:
(a) \(\frac{\cos x}{2 y-1}\)

Question 45.
Find the value of \(\frac{d y}{d x}\) at θ = \(\frac{\pi}{3}\) if x = asec³ θ and x = atan³ θ is:
(a) \(\frac{\sqrt{3}}{2}\)
(b) –\(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) 1
Solution:
(a) \(\frac{\sqrt{3}}{2}\)

Question 46.
If x²y = e^x-y then \(\frac{d y}{d x}\) is:
(a) \(\frac{1+x}{1+\log x}\)
(b) \(\frac{1-x}{1+\log x}\)
(c) \(\frac{x}{1+\log x}\)
(d) \(\frac{x}{(1+\log x)^2}\)
Solution:
(d) \(\frac{x}{(1+\log x)^2}\)

Question 47.
Differential coefficient of sec (tan^-1 x) is:
(a) \(\frac{x}{1+x^2}\)
(b) \(x \sqrt{1+x^2}\)
(c) \(\frac{x}{\sqrt{1+x^2}}\)
(d) \(\frac{1}{\sqrt{1+x^2}}\)
Solution:
(c) \(\frac{x}{\sqrt{1+x^2}}\)

Question 48.
The function f(x) = x^x is decreasing in the interval:
(a) (0, e)
(b) (0, \(\frac{1}{e}\))
(c) (0, 1)
(d) none of these
Solution:
(b) (0, \(\frac{1}{e}\))

(B) Very Short Type Questions With Answers

Question 1.
If f'(2⁺) = 0 and f'(2^–) = 0, then is f(x) continuous at x = 2?
Solution:
f'(2⁺) = 0 and f'(2^–) = 0
⇒ f is differentiable at x = 2
⇒ f is continuous at x = 2.

Question 2.
If Φ(x) = f(x) + f(1 – x), f'(x) = 0 for 0 < x < 1, then is x = a point of maxima or minima of Φ(x)?
Solution:
Let Φ(x) = f(x) + f(1 – x)
⇒ Φ’ = f'(x) – f'(1 – x)
Φ”(x) = f”(x) + f”(1 – x)
Now Φ'(\(\frac{1}{2}\)) = f'(\(\frac{1}{2}\)) – f'(\(\frac{1}{2}\)) = 0
⇒ \(\frac{1}{2}\) is a critical point.
Φ”(\(\frac{1}{2}\)) = f”(\(\frac{1}{2}\)) – f”(\(\frac{1}{2}\)) = 0
⇒ x = \(\frac{1}{2}\) is neither a point of local maxima nor a point of local minima.

Question 3.
Write the interval in which the function f(x) = sin^-1 (2 – x) is differentiable.
Solution:
f(x) = sin^-1 (2 – x) is differentiable for
1 – (2 – x)² > 0.
⇒ (2 – x)² < 1
⇒ -1 < 2 – x < 1
⇒ -3 < -x < -1
⇒ 3 > x < 1
⇒ 1 < x < 3
⇒ x ∈ (1, 3)
∴ f is differentiable on (1, 3)

Question 4.
Write the set of values of x for which the function f(x) = sin x – x is increasing.
Solution:
f(x) = sin x – x
⇒ f'(x) = cos x – 1 ≤ 0 for all x ∈ R
∴ f(x) is increasing for all x ∈ R

Question 5.
Write the differential coefficient of e^{[x] ln (x + 1)} where 3 ≤ x < 4 with respect to x.
Solution:
Let y = e
= e^3ln(x+1) (∵ 3 ≤ x < 4)
= \(e^{\ln (x+1)^3}\) = (x + 1)³
\(\frac{d y}{d x}\) = 3(x + 1)²

Question 6.
Write a logarithmic function which is differentiable only in the open interval (-1, 1).
Solution:
f(x) = \(\left\{\begin{array}{l}
\ln \left|\sin ^{-1} x\right|, x \neq 0 \\
1, x=0
\end{array}\right.\) is differentiable only on (-1, 1).

Question 7.
Write a function which has both relative and absolute maximum at the point (1, 2).
Solution:
f(x) = -x² + 2x + 1 on [0, 1] has relative and absolute maximum at (1, 2).

Question 8.
Write the derivation of e^3logx with respect to x².
Solution:
Let u = e^3lnx = e^lnx = \(e^{\ln x^3}\)= x³ and v = x²
∴ \(\frac{d u}{d x}\) = 3x² and \(\frac{d v}{d x}\) = 2x
∴ Derivative of u w.r.t. v is
\(\frac{d u}{d v}\) = \(\frac{3 x^2}{2 x}\) = \(\frac{3 x}{2}\)

Question 9.
Write the maximum value of the function y = x⁵ in the interval [1, 5]:
Solution:
y = x⁵
⇒ \(\frac{d y}{d x}\) = 5x⁴ > 0 ∀ x ∈ [1, 5]
i.e y is strictly increasing in the given interval.
Thus the maximum value of
y = x⁵ = 5⁵ = 3125

Question 10.
Differentiate a^lnx with respect to x
Solution:
\(\frac{d}{d x}\)(a^{ln x}) = a^{ln x} (ln a)\(\frac{d}{d x}\)ln x
= \(\frac{a^{\ln x} \ln a}{x}\)

Question 11.
Mention the values of x for which x the function f(x) = x³ – 12x is increasing.
Solution:
Given f(x) = x³ – 12x
f is increasing if f'(x) > 0
⇒ 3x² – 12 > 0
⇒ x² > 4
⇒ -2 > x > 2
⇒ x ∈ (-∞, -2) ∪ (2, ∞)

Question 12.
Differentiate sin^-1 (cos x) with respect to x.
Solution:
\(\frac{d}{d x}\)sin^-1 (cos x) = \(\frac{d}{d x}\)sin^-1 sin (\(\frac{\pi}{2}\) – x)
= \(\frac{d}{d x}\) (\(\frac{\pi}{2}\) – x) = 1

Question 13.
Write the equation of the tangent to the curve y = |x| at the point (-2, 2).
Solution:
y = |x| = \(\left\{\begin{aligned}
x, & x \geq 0 \\
-x, & x<0
\end{aligned}\right.\)
When x < 0, \(\frac{d y}{d x}\) = -1
which is the slope of the tangent at (-2, 2).
Equation of the required tangent is
y – 2 = (-1) (x + 2)
⇒ y = -x
⇒ x + y = 0.

Question 14.
What is the derivative of sec^-1 x with respect to x if x < -1?
Solution:
Let y = sec^-1 x
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{|x| \sqrt{x^2-1}}\)
when x < (-1) we have |x| = -x
∴ \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}<(-1)}\) = \(\frac{(-1)}{x \sqrt{x^2-1}}\)

Question 15.
Write the set of points where the function f(x) = x³ has relative (local) extrema.
Solution:
f(x) = x³
⇒ f'(x) = 3x²
f'(x) = 0
⇒ x = 0
But f”(x) = 6x = 0 when x = 0
∴ x = 0 is not a point of local extrema.
∴ Hence the set of points where f has relative extrema = Φ.

Question 16.
In which interval of x the function \(\frac{\ln \mathbf{x}}{x}\) is decreasing?
Solution:
Let f(x) = \(\frac{\ln x}{x}\)
⇒ f'(x) = \(\frac{x \cdot \frac{1}{x}-\ln x}{x^2}\) = \(\frac{1-\ln x}{x^2}\)
f is decreasing for
f'(x) < 0 ⇒ \(\frac{1-\ln x}{x^2}\) < 0
⇒ 1 – ln x < 0 (∵ x² > 0)
⇒ 1 < ln x
⇒ ln x > 1
⇒ x e (e, ∞).

(C) Short Type Questions With Answers

Question 1.
If y = \(e^{x^{e^{e^{e^{x^x…..}}}}}\), then find \(\frac{d \mathbf{y}}{\mathbf{d x}}\).
Solution:

Question 2.
If \(\frac{\mathbf{d}^2 \mathbf{y}}{\mathbf{d x}^2}\), if x = a cosΦ and y = b sinΦ.
Solution:
x = a cosΦ, y = b sinΦ

Question 3.
Find the point on the curve x² + y² – 4xy + 2 = 0, where the normal to the curve is parallel to the x – axis.
Solution:
Given equation of the curve is
x² + y² – 4xy + 2 = 0 … (1)
Differentiang we get

Question 4.
Find the intervals in which the function y = \(\frac{\ln x}{x}\) is increasing and decreasing.
Solution:

Question 5.
Differentiate y = tan^-1\(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\)
Solution:

Question 6.
Differentiate y = (sin y)^{sin 2x}
Solution:
y = (sin y)^{sin 2x}
⇒ In y = sin 2x ln (sin y)

Question 7.
Test the differentiability and continuity of the following function at x = 0:
f(x) = \(\left\{\begin{array}{cc}
\frac{1-e^{-x}}{x} & x \neq 0 \\
1 & x=0
\end{array}\right.\)
Solution:
Given function is


As LHD = RHD, we have the given function is differentiable at x = 0 and f'(0) = –\(\frac{1}{2}\).

Question 8.
Show that the sum of intercepts on the co-ordinate axes of any tangent to the curve
√x + √y = √a is constant.
Solution:
Given curve √x + √y = √a … (1)

Question 9.
Find the equation of the normal to the curve y = (log x)² at x = \(\frac{1}{e}\).
Solution:
Given equation of the curve is
y = (log x)²

Question 10.
If y = x + \(\frac{1}{x+\frac{1}{x+\cdots \cdots \infty}}\), find \(\frac{\mathbf{d y}}{\mathbf{d x}}\), the rhs being a valid expression.
Solution:

Question 11.
Differentiate sec^-1 (\(\frac{1}{2 x^2-1}\)) with respect to \(\sqrt{1-x^2}\).
Solution:

Question 12.
Find \(\frac{d \mathbf{y}}{\mathbf{d t}}\) , when y = sin^-1 \(\left(\frac{2 \sqrt{t}-1}{t^2}\right)\)
Solution:

Question 13.
Find \(\frac{\mathbf{d y}}{\mathbf{d x}}\) , if x^myⁿ = \(\left(\frac{\mathbf{x}}{\mathbf{y}}\right)^{\mathrm{m}+\mathrm{n}}\)
Solution:

Question 14.
If x = a sec θ, y = b tan θ, prove that:
\(\frac{d^2 y}{d x^2}\) = –\(\frac{b^4}{a^2 y^3}\)
Solution:

Question 15.
Find the interval where the following function is increasing:
y = sin x + cos x, x ∈ [0, 2π].
Solution:
y = sin x + cos x, x ∈ [0, 2π].

Question 16.
Find \(\frac{d y}{d x}\), when y^x = x^{sin y}.
Solution:
y^x = x^{sin y}
Taking log of both sides we get,
x ln y = (sin y) ln x
Differentiating with respect to x we get,

Question 17.
Show that \(\frac{d y}{d x}\) is independent of t,
if cos x = \(\sqrt{\frac{1}{1+t^2}}\), sin y = \(\frac{2 t}{1+t^2}\)
Solution:
cos x = \(\sqrt{\frac{1}{1+t^2}}\), sin y = \(\frac{2 t}{1+t^2}\)
Let t = tan θ
∴ cos x = cos θ, sin y = sin 2θ
⇒ y = 2θ, x = θ
⇒ y = 2x
⇒ \(\frac{d y}{d x}\) = 2 which is independent of t.

Question 18.
Show that 2sin x + tan x > 3x for all x ∈ (0, \(\frac{\pi}{2}\))
Solution:
Let f(x) = 2sin x + tan x – 3x
⇒ f'(x) = 2cos x + sec² x – 3
Let g(x) = f'(x) = 2cos x + sec² x – 3
⇒ g'(x) = -2sin x + 2sec² x tan x
= 2sin x (sec³ x – 1) ≥ 0
for all x ∈ (0, \(\frac{\pi}{2}\))
∴ g is an increasing function on (0, \(\frac{\pi}{2}\))
But g(0) = 0
∴ g(x) ≥ 0 for all x ∈ (0, \(\frac{\pi}{2}\))
⇒ f(x) ≥ 0 for all x ∈ (0, \(\frac{\pi}{2}\))
⇒ f is an increasing function on (0, \(\frac{\pi}{2}\))
But f(0) = 0
∴ f(x) ≥ 0 for all x ∈ (0, \(\frac{\pi}{2}\))
⇒ 2sin x + tan x ≥ 3x for x ∈ (0, \(\frac{\pi}{2}\))

Question 19.
Show that no two normals to a parabola are parallel.
Solution:
Let us consider the parabola y² = 4ax – (1) and A(at₁², 2at₁) and B (at₂², 2at₂) are any two points on its from (1)

As t₁ ≠ t₂ we have no two normals of a parabola are parallel.

Question 20.
Examine the differentiability of In x² for all real values of x.
Solution:
Let f(x) = In (x²) = 2 ln |x|
Clearly Dom f = R – {0}
Let any a ∈ Dom f.

Question 21.
Find the derivative of x^{sin x} with respect to x.
Solution:
Let y = x^{sin x} = e^{(sin x)/nx}

Question 22.
Differentiate sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) with respect to cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\).
Solution:

Question 23.
Find the slope of the tangent to the curve x = 2(t – sin t), y = 2(1 – cos t) at t = \(\frac{\pi}{4}\).
Solution:

Question 24.
If cos y = x cos(a + y), then show that \(\frac{d y}{d x}\) = \(\frac{\cos ^2(a+y)}{\sin a}\).
Solution:

Question 25.
Find the extreme values of the function y = x + \(\frac{y}{x}\).
Solution:

Question 26.
Find the intervals where the following function is (i) increasing, (ii) decreasing:
f(x) = \(\begin{cases}\mathbf{x}^2+1, & x \leq-3 \\ x^3-8 x+13, & x>-3\end{cases}\)
Solution:
Given f(x) = \(\begin{cases}\mathbf{x}^2+1, & x \leq-3 \\ x^3-8 x+13, & x>-3\end{cases}\)
Clearly f is not differentiable at x = -3.
and f ‘(x) = \(\begin{cases}2 \mathrm{x}, & \mathrm{x}>-3 \\ 3 \mathrm{x}^2-8 & \mathrm{x}<-3 \\ \text { does not exist } & \text { for } \mathrm{x}=-3\end{cases}\)
Case – 1: x > -3
Clearly f'(x) > 0 for x > 0 and f'(x) < 0 for -3 < x < 0
∴ f is increasing in (0, ∞) and decreasing in (-3, 0).
Case – 2: x < -3
Clearly for x < -3, f(x) = 3x² – 8 > 0
i.e., f is increasing in (-∞, -3)
Thus f is increasing in (-∞, -3) ∪ (0, ∞) and decreasing in (-3, 0).

Question 27.
Prove that:
y = In tan \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\) ⇒ \(\frac{d y}{d x}\) = sec x
Solution:

Question 28.
Differentiate with respect to x:
y = \(2^{x^2}\) + tan^-1 \(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\)
Solution:

Question 29.
Find the equation of the tangent to the curve x = y² – 1 at the point where the slope of the normal to the curve is (-2).
Solution:
Given equation of the curve is x = y² – 1.

Question 30.
Find \(\frac{d y}{d x}\) if y = \(\log _{\left(x^2\right)}\)3.
Solution:

Question 31.
Write why the function sin^-1 \(\frac{1}{\sqrt{1-x^2}}\) cannot be differentiated anywhere.
Solution:

⇒ 1 – x² ≥ 1
⇒ x = 0
∴ DOM of {0 3}
But f is not defined in a deleted interval of x = 0.
Hence f is not differentiable at x = 0 and hence not differentiable anywhere.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(c)

Find derivatives of the following functions.
Question 1.
(x² +5)⁸
Solution:
y = (x² +5)⁸
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(x² +5)⁸
= 8(x² +5)⁷ × \(\frac{d}{d x}\)(x² +5) by chain rule
= 8(x² +5)⁷ . 2x
= 16x (x² +5)⁷

Question 2.
\(\frac{1}{\left(x^3+\sin x\right)^2}\)
Solution:

Question 3.
In (√x+1)
Solution:

Question 4.
sin 5x + cos 7x
Solution:
sin 5x + cos 7x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(sin 5x) + \(\frac{d}{d x}\)(cos 7x)
= cos 5x . \(\frac{d}{d x}\)(5x) – sin 7x . \(\frac{d}{d x}\)(7x)
= 5 cos 5x – 7 sin 7x

Question 5.
e^{sin t}
Solution:
y = e^{sin t}
\(\frac{d y}{d x}\) = \(\frac{d}{d t}\)(e^{sin t}) = e^{sin t} . \(\frac{d}{d t}\)(sin t)
= e^{sin t} . cos t

Question 6.
\(\sqrt{a x^2+b x+c}\)
Solution:
y = \(\sqrt{a x^2+b x+c}\)

Question 7.
\(\left(\frac{x+1}{x^2+3}\right)^{-3}\)
Solution:

Question 8.
sec (tan θ)
Solution:
y = sec (tan θ)
\(\frac{d y}{d θ}\) = \(\frac{d}{d θ}\) {sec (tan θ)}
= sec (tan θ) . tan (tan θ) . \(\frac{d}{d θ}\)(tan θ)
= sec (tan θ) . tan (tan θ) . sec² θ

Question 9.
sin \(\left(\frac{1-x^2}{1+x^2}\right)\)
Solution:

Question 10.
\(\sqrt{\tan (3 z)}\)
Solution:

Question 11.
tan³ x
Solution:
y = tan³x = (tan x)³
\(\frac{d y}{d x}\) = 3(tan x)². \(\frac{d}{d x}\)(tan x)
= 3tan² x . sec² x

Question 12.
sin⁴ x
Solution:
y = sin⁴x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(sin⁴ x)
= 4 (sin x)³ . \(\frac{d}{d x}\)(sin x)
= 4 sin³ x . cos x

Question 13.
sin² x cos² x
Solution:
y = sin² x cos² x = \(\frac{1}{4}\)sin² 2x
\(\frac{d y}{d x}\) = \(\frac{1}{4}\) \(\frac{d}{d x}\)(sin 2x)²
= \(\frac{1}{4}\). 2sin 2x . \(\frac{d}{d x}\)(sin 2x)
= \(\frac{1}{2}\) sin 2x . cos 2x . \(\frac{d}{d x}\)(2x)
= \(\frac{1}{2}\) sin 2x cos 2x . 2 = sin 2x . cos 2x

Question 14.
sin 5x cos 7x
Solution:
y = sin 5x cos 7x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(sin 5x) . cos 7x + \(\frac{d}{d x}\)sin 5x . (cos 7x)
= cos 5x . \(\frac{d}{d x}\)(5x) . cos 7x + sin 5x . (-sin 7x) . \(\frac{d}{d x}\)(7x)
= 5 cos 5x . cos 7x – 7 sin 5x . sin 7x

Question 15.
tan x cot 2x
Solution:
y = tan x. cot 2x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(tan x) . cot 2x + tan x . \(\frac{d}{d x}\)(cot 2x)
= sec² x / cot 2x + tan x . (-cosec² x) \(\frac{d}{d x}\)(2x)
= sec² x . cot 2x – 2 tan x . cosec² x

Question 16.
\(\sqrt{\sin \sqrt{x}}\)
Solution:

Question 17.
\(\sqrt{\sec (2 x+1)}\)
Solution:

Question 18.
cosec (ax + b)²
Solution:
y = cosec (ax + b)²
\(\frac{d y}{d x}\) = – cosec (ax + b)² cot (ax + b)² . \(\frac{d}{d x}\)(ax + b)²
[ ∵ \(\frac{d}{d x}\)(cosec u) = -cosec u . cot u . \(\frac{d u}{d x}\)
= – cosec (ax + b)² . cot (ax + b)² . 2(ax + b) . \(\frac{d}{d x}\)(ax + b)
[ ∵ \(\frac{d}{d x}\)(u²) = 2u \(\frac{d u}{d x}\)
= – cosec (ax + b)² . cot (ax + b)² . 2(ax + b) . a
= – 2a (ax + b) cosec (ax + b)² cot (ax + b)²

Question 19.
a^{In x}
Solution:
y = a^{In x} . In a . \(\frac{d}{d x}\)(In x)
[ ∵ \(\frac{d}{d x}\)(a^u) = a^u . In a . \(\frac{d u}{d x}\)
= a^{In x} . In a . \(\frac{1}{x}\) = \(\frac{a^{\ln x} \ln a}{x}\)

Question 20.
\(a^{x^2} b^{x^3}\)
Solution:

Question 21.
In tan x
Solution:

Question 22.
\(5^{\sin x^2}\)
Solution:

Question 23.
In tan\(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)
Solution:

Question 24.
\(\sqrt{\left(a^{\sqrt{x}}\right)}\)
Solution:

Question 25.
In (e^nx + e^-nx)
Solution:
y = In (e^nx + e^-nx)
\(\frac{d y}{d x}\) = \(\frac{1}{e^{n x}+e^{-n x}}\) . \(\frac{d}{d x}\) (e^nx + e^-nx)
= \(\frac{n\left(e^{n x}-e^{-n x}\right)}{e^{n x}+e^{-n x}}\)

Question 26.
\(e^{\sqrt{a x}}\)
Solution:

Question 27.
\(\sqrt{\log x}\)
Solution:
y = \(\sqrt{\log x}\)
\(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{\log x}}\) . \(\frac{d}{d x}\)(log x)
= \(\frac{1}{2 \sqrt{\log x}}\) . \(\frac{1}{x}\)

Question 28.
e^{sin x} – a^{cos x}
Solution:
y = e^{sin x} – a^{cos x}
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)(e^{sin x}) – \(\frac{d}{d x}\)(a^{cos x})
= e^{sin x} . \(\frac{d}{d x}\)(sin x) – a^{cos x} . In a . \(\frac{d}{d x}\)(cos x)
= e^{sin x} . cos x + a^{cos x} . In a . sin x

Question 29.
\(\frac{e^{3 x^2}}{\ln \sin x}\)
Solution:

Question 30.
Prove that
\(\frac{d}{d x}\left[\frac{1-\tan x}{1+\tan x}\right]^{\frac{1}{2}}\) = 1 / \(\sqrt{\cos 2 x}\) (cos x + sin x)
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Ex 8(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 8 Application of Derivatives Exercise 8(c)

Question 1.
Find the intervals where the following functions are (a) increasing and (b) decreasing.
(i) y = sin x, x ∈ [0, 2π]
Solution:

(ii) y = In x, x ∈ R₊
Solution:
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{x}\) > 0
for all x ∈ R₊.
Thus y = In x is increasing in (0, ∞) and decreasing nowhere.

(iii) y = a^x, a > 0, x ∈ R
Solution:
⇒ \(\frac{d y}{d x}\) = a^x In a > 0 for all x ∈ R provided a > 1.
The function is increasing in (-∞, ∞) is a > 1.
Again \(\frac{d y}{d x}\) = a^x In a < 0 if 0 < a < 1.
Thus the function is decreasing in (-∞, ∞) if (0 < a < 1)

(iv) y = sin x + cos x, x ∈ [0, 2π]
Solution:

(v) y = 2x³ + 3x² – 36x – 7
Solution:
\(\frac{d y}{d x}\) = 6x² – 6x – 36
y is increasing if \(\frac{d y}{d x}\) > 0
⇒ x² + x – 6>0
⇒ x² + 3x – 2x – 6>0
⇒ x (x + 3) -2 (x + 3) > 0
⇒ (x + 3) (x – 2) > 0
for x > 2 or x < -3 and (x + 3 ) (x – 2) < 0 for -3 < x < 2
∴ The function is increasing in (-∞, -3) ∪ (2, ∞) and is decreasing in (-3, 2).

(vi) y = \(\frac{1}{x-1^{\prime}}\) x ≠ 1
Solution:
\(\frac{d y}{d x}\) = \(\frac{-1}{(x-1)^2}\)
The function is increasing if \(\frac{d y}{d x}\) > 0
⇒ \(\frac{-1}{(x-1)^2}\) > 0 which is impossible because \(\frac{-1}{(x-1)^2}\) is always -ve for all x ≠ 1. So the function is increasing nowhere. It is decreasing in R – {1}

(vii) y = \(\left\{\begin{array}{cc}
x^2+1, & x \leq-3 \\
x^3-8 x+13, & x>-3
\end{array}\right.\)
Solution:

(viii) y = 4x² + \(\frac{1}{x}\)
Solution:

(ix) y = (x – 1)² (x + 2)
Solution:
\(\frac{d y}{d x}\) = 2 (x – 1) (x + 2) + (x – 1)²
= (x – 1) (2x + 4 + x – 1)
= (x – 1 ) (3x + 3)
= 3 (x – 1) (x + 1 )
The function is increasing if \(\frac{d y}{d x}\) > 0.
⇒ (x + 1) (x – 1) > 0
⇒ x < -1 or x > 1
∴ The function is increasing in (-∞, -1) ∪ (1, ∞)
It is decreasing in (-1, 1).

(x) y = \(\frac{\ln x}{x}\), x > 0
Solution:

(xi) y = tan x – 4 (x – 2), x ∈ (-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Solution:

(xii) y = sin 2x – cos 2x, x ∈ [0, 2π]
Solution:

Question 2.
Give a rough sketch of the functions given in question 1.
Solution:
Do yourself.

Question 3.
Show that the function \(\frac{e^x}{x^p}\) is strictly increasing for x > p > 0.
Solution:

Question 4.
Show that 2 sin x + tan x ≥ 3x for all x ∈ (0, \(\frac{\pi}{2}\)).
Solution:
Let f(x) = 2 sin x + tan x – 3x
Then f(x) = 2 cos x + sec² x – 3