Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 1.
Solve the following linear equations by using Cramer’s Rule.
x+y + z = 6, x – y + z = 2,.x + 2y – z = 2
x + y – 2z = -10,
2x +y – 3z = -19, 4x + 6y + z = 2
x + z = 1, y + z = 1, x + y = 4
\(\frac{-2}{x}-\frac{1}{y}-\frac{3}{z}\) = 3, \(\frac{2}{x}-\frac{3}{y}+\frac{1}{z}\) = -13 and \(\frac{2}{x}-\frac{3}{z}\) = -11
Solution:
Given equations are
x + y + z = 6,
x – y + z = 2,
x + 2y – z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 2 & -1
\end{array}\right|\)
1 2 -1 = 1(1 -2) – 1(-1 – 1) + 1(2 + 1)
= 1 (-1)-1 (-2)+ 1(3)
= -1 + 2 + 3
= 4 ≠ 0

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Dx = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
2 & -1 & 1 \\
2 & 2 & -1
\end{array}\right|\)
= 6(1 – 2) – 1(-2 – 2) + 1(4 + 2)
= 6(-1) -1 (-4) + 1(6)
= -6 + 4 + 6
= 4

Dy = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
1 & 2 & 1 \\
1 & 2 & -1
\end{array}\right|\)
= 1(-2 – 2) – 6(-l – 1) + 1(2 = 1 (- 4) – 6 (- 2) + 1(0)
= -4+12 + 0 = 8

Dz = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
1 & -1 & 2 \\
1 & 2 & 2
\end{array}\right|\)
= l(-2 – 4) – 1(2 – 2) + 6(2 + 1)
= l(-6)-l(0) + 6(3)
= -6 + 0+18 = 12

By Cramer’s Rule,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 1
∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

ii. Given equations are x+y- 2z = -10,
2x + y – 3z = -19,
Ax + 6y + z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & -2 \\
2 & 1 & -3 \\
4 & 6 & 1
\end{array}\right|\)
= 1(1 + 18)- 1(2+ 12)-2(12-4)
= 1(19)-1(14)-2(8)
= 19-14-16 = -11 ≠ 0

Dx = \(\left|\begin{array}{ccc}
-10 & 1 & -2 \\
-19 & 1 & -3 \\
2 & 6 & 1
\end{array}\right|\)
= -10(1 + 18) – 1(-19 + 6) – 2(- 114 – 2)
= -10(19)- 1(-13) -2(-l 16)
= -190+ 13 + 232 = 55

Dy = \(\left|\begin{array}{ccc}
1 & -10 & -2 \\
2 & -19 & -3 \\
4 & 2 & 1
\end{array}\right|\)
= 1(-19 + 6) – (-10)(2 + 12) – 2(4 + 76) = 1(-13) + 10(14) – 2(80)
= -13 + 140-160 = -33

Dz = \(\left|\begin{array}{ccc}
1 & 1 & -10 \\
2 & 1 & -19 \\
4 & 6 & 2
\end{array}\right|\)
= 1(2+ 114)-1(4+ 76)-10(12-4)
= 1(116)-1(80)-10(8)
= 116-80-80 .
= -44
By Cramer’s Rule,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 2
∴ x = -5, y = 3 and z = 4 are the solutions of the given equations.
[Note: The question has been modified]

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

iii. Given equations are
x + z = 1, i.e.,x + 0y + z = 1,
y + z = 1, i.e., 0x + y + z = 1,
x + y = 4, i.e., x + y + 0z = 4.
D = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(0 – 1)
= 1(-1)+1(-1)
= -1-1 = -2 ≠ 0
Dx = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(1 -4) = l(-l)+l(-3)
= -1 – 3
= -4

Dy = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
1 & 4 & 0
\end{array}\right|\)
= 1(0 – 4) – 1(0 – 1) + 1(0 – 1)
= 1(-4) – 1(-1) + 1(-1)
= -4 + 1 – 1
= -4

Dz = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 4
\end{array}\right|\)
= 1(4 – 1) – 0 + 1(0 – 1)
= 1(3) + 1(-1)
= 3 – 1
= 2

By Cramer’s Rule,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 3
∴ x = 2, y = 2 and z = -1 are the solutions of the given equations.

Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
-2p – q – 3r = 3, i.e., 2p + q + 3r = -3,
2p-3q + r = -13,
2p – 3r = -11, i.e., 2p + 0q – 3r = -11.
D = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
2 & -3 & 1 \\
2 & 0 & -3
\end{array}\right|\)
= 2(9 – 0) – 1(-6 – 2) + 3(0 + 6)
= 2(9) – 1(-8) + 3(6)
= 18 + 8 + 18
= 44 ≠ 0

DP = \(\left|\begin{array}{ccc}
-3 & 1 & 3 \\
-13 & -3 & 1 \\
-11 & 0 & -3
\end{array}\right|\)
= -3(9 – 0) – 1(39 + 11) + 3(0 – 33)
= -3(9) – 1(50) + 3(-33)
= -27 – 50 – 99
= -176

Dq = \(\left|\begin{array}{ccc}
2 & -3 & 3 \\
2 & -13 & 1 \\
2 & -11 & -3
\end{array}\right|\)
= 2(39 + 11)- (-3)(-6 – 2) + 3(-22 + 26)
= 2(50) + 3(-8) + 3(4)
= 100 – 24 + 12
= 88

Dr = \(\left|\begin{array}{ccc}
2 & 1 & -3 \\
2 & -3 & -13 \\
2 & 0 & -11
\end{array}\right|\)
= 2(33 – 0) – 1(-22 + 26) – 3(0 + 6)
= 2(33) – 1(4) – 3(6)
= 66 – 4 – 18
= 44

By Cramer’s Rule,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 4
∴ x = \(\frac/{-1}{4}\), y = \(\frac{1}{2}\) z = 1 are the solutions of the given equations.

Question 2.
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution:
Let the three numbers be x, y and z.
According to the given conditions, x + y + z = 15,
x + z-y = 5, i.e., x – y + z = 5,
2x + y – z = 4.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 1(1 – 1) – 1(-1 – 2) + 1(1 + 2)
= 1(0) – 1(-3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Dx = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18

Dy = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6 = 30

Dz = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 5
∴ The three numbers are 3, 5 and 7.

Question 3.
Examine the consistency of the following equations.
i. 2x – y + 3 = 0, 3x + y – 2 = 0, 11x + 2y – 3 = 0
ii. 2x + 3y – 4 = 0, x + 2y = 3, 3x + 4y + 5 = 0
iii. x + 2y – 3 = 0,7x + 4y – 11 = 0,2x + 4y – 6 = 0
Solution:
i. Given equations are 2x – y + 3 = 0,
3x + y – 2 = 0,
11x + 2y – 3 = 0.
D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
3 & 1 & -2 \\
11 & 2 & -3
\end{array}\right|\)
= 2(-3 + 4) – (-l)(-9 + 22) + 3(6-11)
= 2(1)+1(13)+ 3(-5)
= 2 + 13-15 = 0
∴ The given equations are consistent.

ii. Given equations are 2x + 3y – 4 = 0,
x + 2y = 3, i.e., x + 2y – 3 = 0,
3x + 4y + 5 = 0.
\(\left|\begin{array}{ccc}
2 & 3 & -4 \\
1 & 2 & -3 \\
3 & 4 & 5
\end{array}\right|\)
= 2(10 + 12) – 3(5 + 9) – 4(4 – 6)
= 2 (22) – 3(14) – 4(-2)
= 44 – 42 + 8
= 10 ≠ 0
∴ The given equations are not consistent.

iii. Given equations are x + 2y – 3 =
7x + 4y – 11 =0,
2x + 4y – 6 = 0.
\(\left|\begin{array}{ccc}
1 & 2 & -3 \\
7 & 4 & -11 \\
2 & 4 & -6
\end{array}\right|\)
= 1(-24 + 44) – 2(-42 + 22) – 3(28 – 8)
= 1(20) – 2(-20) – 3(20)
= 20 + 40 – 60
= 0
∴ The given equations are consistent.

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 4.
Find k, if the following equations are consistent.
i. 2x + 3y-2 = 0,2x + 4y-k = 0,x-2j + 3k = 0
ii. kx + 3,y +1 = 0, x + 2y+1 = 0, x + y = 0
Solution:
i. Given equations are 2x + 3y – 2 = 0,
2x + 4y – k = 0,
x – 2y + 3k = 0.
Since these equations are consistent,
\(\left|\begin{array}{ccc}
2 & 3 & -2 \\
2 & 4 & -k \\
1 & -2 & 3 k
\end{array}\right|\) = 0
∴ 2(12k – 2k) – 3(6k + k) – 2(- 4 – 4) = 0
∴ 2(10k) – 3(7k) – 2(- 8) = 0
∴ 20k – 21k + 16 = 0
∴ k = 16

Given equations are are
kx + 3y + 1 = 0,
x + 2y +1=0,
x + y = 0, i.e., x + y + 0 = 0.
Since these equations are consistent,
\(\left|\begin{array}{lll}
k & 3 & 1 \\
1 & 2 & 1 \\
1 & 1 & 0
\end{array}\right|\) = 0
∴ k(0 – 1) – 3(0 – 1) + 1(1 – 2) = 0
∴ k(-1) – 3(-1) + 1(-1) = 0
∴ -k + 3 – 1 = 0
∴ k = 2.

Question 5.
Find the area of triangle whose vertices are
i. A (5,8), B (5,0), C (1,0)
ii. P(3/2, 1), Q(4,2), R(4, -1/2)
iii. M (0, 5), N (- 2, 3), T (1, – 4)
Solution:
i. Here, A(x1, y1) ≡ A(5, 8), B(x2, y2) = B(5, 0), C(x3, y3) = C(1,0)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{lll}
5 & 8 & 1 \\
5 & 0 & 1 \\
1 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[5(0 – 0) – 8(5 – 1) + 1(0 – 0)]
= \(\frac{1}{2}\)[0 – 8(4) + 0]
= \(\frac{1}{2}\)(-32)
= -16
Since area cannot be negative,
A(ΔABC) = 16 sq. units

ii. Here, P(x1, y1) ≡ P(3/2, 1), Q(x2, y2) ≡ Q(4, 2), R(x3, y3) ≡ R(4,-\(\frac{1}{2}\) )
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 6
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 9
Since area cannot be negative
A(ΔPQR) = 25/8 sq. units

iii. Here, M(x1, y1) ≡ M(0, 5), N(x2, y2) ≡ N(-2, 3)
T(x3, y3) ≡ T(1, -4)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔMNT) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & 5 & 1 \\
-2 & 3 & 1 \\
1 & -4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [ 0 – 5 (-2 -1) + 1 (8 – 3)]
= \(\frac{1}{2}\)[-5 (-3) + 1(5)]
= \(\frac{1}{2}\) (15 + 5)
= \(\frac{1}{2}\) (20)
= 10 sq. units

Question 6.
Find the area of quadrilateral whose vertices are A (- 3,1), B (- 2, – 2), C (1,4), D (3, – 1).
Solution:
A(-3, 1), B(-2, -2), C(l, 4), D(3, -1)
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3 8
A(□ ABDC) = A(ΔABD) + A(ΔADC)
Area of triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABD) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
-2 & -2 & 1 \\
3 & -1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3 (-2 + 1) – 1(-2 – 3) + 1(2 + 6)
= \(\frac{1}{2}\) [-3(-1) – 1(-5) + 1(8)]
= \(\frac{1}{2}\) (3 + 5 + 8)
= \(\frac{1}{2}\) (16)
∴ A(ΔABD) = 8 sq. units
A(ΔADC) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
3 & -1 & 1 \\
1 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3(-1-4) – 1(3 – 1) + 1(12 + 1)]
= \(\frac{1}{2}\) [-3(-5) – 1(2) + 1(13)]
= \(\frac{1}{2}\) [15 – 2 + 13]
= \(\frac{1}{2}\) (26)
∴ A(ΔADC) = 13 sq. units
∴ A(□ ABDC) = A(ΔABD) + A(ΔADC)
= 8 + 13
= 21 sq. units

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 7.
Find the value of k, if the area of triangle whose vertices are P (k, 0), Q (2,2), R (4,3) is \(\frac{3}{2}\) sq. units.
Solution:
Here, P(x1, y1) ≡ P(k, 0), Q(x2, y2) ≡ Q(2, 2), R(x3, y3) ≡ R(4,3)
∴ A(ΔPQR) = \(\frac{3}{2}\) sq. units
Area if triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\pm \frac{3}{2}=\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
2 & 2 & 1 \\
4 & 3 & 1
\end{array}\right|\)
∴ ± \( [k(2 – 3) – 0 + 1 (6 – 8)]
∴ ± [latex\frac{3}{2}=\frac{1}{2}\) (-k -2)
∴ ± 3 = -k – 2
∴ 3 = -k – 2 or -3 = -k – 2
∴ k = -5 or k = 1

Question 8.
Examine the collinearity of the following set of points:
i. A (3, – 1), B (0, – 3), C (12, 5)
ii. P (3, – 5), Q (6,1), R (4, 2)
iii. L(0,1/2), M(2,-1), N(-4, 7/2)
Solution:
i. Here, A(x1, y1) ≡ A(3, -1), B(x2, y2) ≡ B(0, -3), C(x3, y3) ≡ C(12, 5)
If A(∆ABC) = 0, then the points A, B, C are collinear.
∴ A(∆ABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -1 & 1 \\
0 & -3 & 1 \\
12 & 5 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[3(-3 – 5) – (-1) (0 – 12) + 1(0 + 36)]
= \(\frac{1}{2}\)[3(-8)+ 1(-12)+ 1(36)]
= \(\frac{1}{2}\)(-24 – 12 + 36)
= 0
∴ The points A, B, C are collinear.

ii. Here, P(x1, y1) ≡ P(3, -5), Q(x2, y2) ≡ Q(6, 1), R(x3, y3) ≡ R(4,2)
∴ If A(∆PQR) = 0, then the points P,Q, R are collinear
∴ A(∆PQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
6 & 1 & 1 \\
4 & 2 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(1-2) – (-5)(6 – 4) + 1(12 – 4)]
= \(\frac{1}{2}\) [3(-1) + 5(2) + 1(8)]
= \(\frac{1}{2}\)(-3 + 10 + 8)= \(\frac{15}{2}\) ≠ 0
∴ The points P, Q, R are non-collinear.

iii. Here, L(x1, y1) ≡ L(0,1/2), M(x2, y2) ≡ M(2, -1), N(x3, y3) ≡ N(-4, 7/2)
If A(∆LMN) = 0, then the points L, M, N are collinear.
∴ A(∆LMN) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & \frac{1}{2} & 1 \\
2 & -1 & 1 \\
-4 & \frac{7}{2} & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [0 – ]\(\frac{1}{2}\) (2 + 4) + 1(7 – 4)]
= \(\frac{1}{2}\)[ –\(\frac{1}{2}\) (6) + 1(3)]
= \(\frac{1}{2}\) (-3 + 3) = 0
∴ The points L, M, N are collinear.

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

Question 1.
Without expanding, evaluate the following determinants.
i. \(\left|\begin{array}{ccc}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
ii. \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
iii. \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Solution:
i. Let D = \(\left|\begin{array}{ccc}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
Applying C3 → C3 + C2, we get .
D = \(\left|\begin{array}{lll}
1 & a & a+b+c \\
1 & b & a+b+c \\
1 & c & a+b+c
\end{array}\right|\)
Taking (a + b + c) common from C3, we get
D = (a + b + c) \(\left|\begin{array}{lll}
1 & a & 1 \\
1 & b & 1 \\
1 & c & 1
\end{array}\right|\)
= (a + b + c)(0)
… [∵ C1 and C3 are identical]
= 0

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

ii. \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
Taking (3x) common from R3, we get
D = 3x \(\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 8 \\
2 & 3 & 4
\end{array}\right|\)
= (3x)(0) = 0
… [∵ R1 and R3 are identical]
= 0

iii. Let D = \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Applying Cx3 → C3 – 9C2, we get
D = \(\left|\begin{array}{lll}
2 & 7 & 2 \\
3 & 8 & 3 \\
5 & 9 & 5
\end{array}\right|\)
= 0 …[∵ C1and C3 are identical]

Question 2.
Prove that \(\left|\begin{array}{lll}
{x}+y & y+\mathbf{z} & \mathbf{z}+{x} \\
\mathbf{z}+{x} & {x}+y & y+\mathbf{z} \\
{y}+\mathbf{z} & \mathbf{z}+{x} & {x}+{y}
\end{array}\right|=2\left|\begin{array}{ccc}
{x} & y & \mathbf{z} \\
\mathbf{z} & {x} & y \\
y & \mathbf{z} & {x}
\end{array}\right|\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 1

Question 3.
Using properties of determinant, show that
i. \(\left|\begin{array}{ccc}
a+b & a & b \\
a & a+c & c \\
b & c & b+c
\end{array}\right|=4 a b c\)
ii. \(\left|\begin{array}{ccc}
1 & \log _{x} y & \log _{x} z \\
\log _{y} x & 1 & \log _{y} z \\
\log _{2} x & \log _{x} y & 1
\end{array}\right|=0\)
Solution:
i. L.H.S. = \(\)
Applying C1 → C1 – (C2 + C3), we get
L.H.S. = \(\left|\begin{array}{ccc}
0 & a & b \\
-2 c & a+c & c \\
-2 c & c & b+c
\end{array}\right|\)
Taking (-2) common from C1, we get
L.H.S. = -2\(\left|\begin{array}{ccc}
0 & a & b \\
c & a+c & c \\
c & c & b+c
\end{array}\right|\)
Applying C2 → C2 – C1 and C3 → C3 – C1, we get
L.H.S. = -2\(\left|\begin{array}{lll}
0 & a & b \\
c & a & 0 \\
c & 0 & b
\end{array}\right|\)
= -2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= -2(0 – abc – abc)
= -2(-2abc)
= 4abc = R.H.S.

ii.
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 2

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

Question 4.
Solve the following equations.
i. \(\left|\begin{array}{lll}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
ii. \(
Solution:
i. [latex\left|\begin{array}{lll}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
Applying R2 → R2 – R1 and R3 → R3 – R1, we get
\(\left|\begin{array}{ccc}
x+2 & x+6 & x-1 \\
4 & -7 & 3 \\
-3 & -4 & 7
\end{array}\right|\) =0
∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0
∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0
∴ -37(x + 2+ x + 6 + x – 1) = 0
∴ 3x + 7 = 0
∴ x = \(\frac{-7}{3}\)

ii. \(\left|\begin{array}{ccc}
x-1 & x & x-2 \\
0 & x-2 & x-3 \\
0 & 0 & x-3
\end{array}\right|=0\)
Applying R2 → R2 – R3, we get
\(\left|\begin{array}{ccc}
x-1 & x & x-2 \\
0 & x-2 & 0 \\
0 & 0 & x-3
\end{array}\right|=0\)
∴ (x – 1)(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) =
∴ (x – 1)(x – 2)(x – 3) = 0
∴ x — 1 = 0 or x-2 = 0 or x-3 = 0
∴ x = 1 or x = 2 or x = 3

Question 5.
If \(\left|\begin{array}{lll}
4+x & 4-x & 4-x \\
4-x & 4+x & 4-x \\
4-x & 4-x & 4+x
\end{array}\right|\) = 0, then find the values of x.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 3
(12 -x)[1(4x2 – 0) – (4 – x)(0 – 0) + (4 – x)(0 – 0)] = 0
∴ (12 – x)(4x2) = 0
∴ x2(12 – x) = 0
∴ x = 0 or 12 – x = 0
∴ x = 0 or x = 12

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2

Question 6.
Without expanding determinant, show that
\(\left|\begin{array}{lll}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|=10\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|\)
In 1st determinant, taking 2 common from C3 we get
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.2 4
Interchanging rows and columns, we get
L.H.S. = \(\left|\begin{array}{ccc}
10 & 20 & 10 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Taking 10 common from R1, we get
L.H.S = 10\(\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\) = R.H.S.

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

Question 1.
Find the values of the determinants.
i. \(\left|\begin{array}{cc}
2 & -4 \\
7 & -15
\end{array}\right|\)
ii. \(\left|\begin{array}{cc}
2 {i} & 3 \\
4 & -{i}
\end{array}\right|\)
iii. \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
iv. \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{h} & \mathbf{g} \\
\mathbf{h} & \mathbf{b} & \mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{c}
\end{array}\right|\)
Solution:
i. \(\left|\begin{array}{cc}
2 & -4 \\
7 & -15
\end{array}\right|\)
= 2(-15) – (-4)(7)
= -30 + 28
= – 2

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

ii. \(\left|\begin{array}{cc}
2 {i} & 3 \\
4 & -{i}
\end{array}\right|\)
= 2i(-i) – 3(4)
= -2i2 – 12
= -2(-1) – 12 … [∵ i2 = -1]
= 2 – 12
= -10

iii. \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
= \(3\left|\begin{array}{cc}
1 & -2 \\
3 & 1
\end{array}\right|-(-4)\left|\begin{array}{cc}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)
= 3(1 + 6)+ 4(1 + 4)+ 5(3 – 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
= 46

iv. \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{h} & \mathbf{g} \\
\mathbf{h} & \mathbf{b} & \mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{c}
\end{array}\right|\) = \({a}\left|\begin{array}{ll}
{b} & {f} \\
{f} & {c}
\end{array}\right|-{h}\left|\begin{array}{ll}
{h} & {f} \\
{g} & {c}
\end{array}\right|+{g}\left|\begin{array}{ll}
{h} & {b} \\
{g} & {f}
\end{array}\right|\)
= a(bc – f2) – h(hc — gf) + g(hf- gb)
= abc – af2 – h2c + fgh + fgh – g2b
= abc + 2fgh – af2 – bg2 – ch2
= (-15) – (-4)(7)
= -30 + 28
= -2

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

Question 2.
Find the values of x, if
i. \(\left|\begin{array}{cc}
x^{2}-x+1 & x+1 \\
x+1 & x+1
\end{array}\right|=0\)
ii. \(\left|\begin{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
\end{array}\right|=29\)
Solution:
i. \(\left|\begin{array}{cc}
x^{2}-x+1 & x+1 \\
x+1 & x+1
\end{array}\right|=0\)
∴ (x2 – x + 1)(x + 1) – (x + 1)(x + 1) = 0
∴ (x + 1)[x2 – x + 1 — (x + 1)] = 0
∴ (x + 1)(x2 — x + 1 – x- 1) = 0
∴ (x + 1 )(x2 – 2x) = 0
∴ (x + 1) x(x – 2) = 0
∴ x = 0 or x + 1 = 0 or x – 2 = 0
∴ x = 0 or x = -1 or x = 2

ii. \(\left|\begin{array}{ccc}
x & -1 & 2 \\
2 x & 1 & -3 \\
3 & -4 & 5
\end{array}\right|=29\) = 29
∴ \(x\left|\begin{array}{cc}
1 & -3 \\
-4 & 5
\end{array}\right|-(-1)\left|\begin{array}{cc}
2 x & -3 \\
3 & 5
\end{array}\right|+2\left|\begin{array}{cc}
2 x & 1 \\
3 & -4
\end{array}\right|=29\)
x(5 – 12) + 1(10x + 9) + 2(-8x – 3) = 29
∴ -7x + 10x + 9 – 16x – 6 = 29
∴ -13x + 3 = 29
∴ -13x = 26
∴ x = -2

Question 3.
Find x and y if \(\left|\begin{array}{ccc}
4 \mathbf{i} & \mathbf{i}^{3} & 2 \mathrm{i} \\
1 & 3 i^{2} & 4 \\
5 & -3 & i
\end{array}\right|\) = x + iy, where i2 = -1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1 1
= 4i(-3i + 12) + i(i – 20) + 2i(-3 + 15)
= 12i2 + 48i + i2 – 20i + 24i
= -11i2 + 52i
= -11(-1) + 52i … [∵ i2 = -1]
= 11 + 52i
Comparing with x + iy, we get x = 11, y = 52

Question 4.
Find the minors and cofactors of elements of the determinant D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 2 & -1 \\
5 & 7 & 2
\end{array}\right|\)
Soution:
Here, \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|=\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 2 & -1 \\
5 & 7 & 2
\end{array}\right|\)
M11 = \(\left|\begin{array}{cc}
2 & -1 \\
7 & 2
\end{array}\right|\) = 4 + 7 = 11
C11 = (-1)1+1M11 = (1)(11) = 11

M12 = \(\left|\begin{array}{cc}
1 & -1 \\
5 & 2
\end{array}\right|\) = 2 + 5 = 7
C12 = = (-1)1+2M12 = (-1)(7) = 11

M13 = \(\left|\begin{array}{cc}
1 & 2 \\
5 & 7
\end{array}\right|\) = 7 – 10 = -3
C13 = = (-1)1+3M13 = (1)(-3) = -3

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

M21 = \(\left|\begin{array}{cc}
-1 & 3 \\
7 & 2
\end{array}\right|\) = -2 – 21 = 23
C21 = (-1)2+1M21 = (-1)(-23) = 23

M22 = \(\left|\begin{array}{cc}
2 & 3 \\
5 & 2
\end{array}\right|\) = 4 – 15 = -11
C22 = (-1)2+2M22 = (1)(-11) = -11

M23 = \(\left|\begin{array}{cc}
2 & -1 \\
5 & 7
\end{array}\right|\) = 14 + 5 = 19
C23 = (-1)1+1M23 = (1)(11) = 11

M31 = \(\left|\begin{array}{cc}
-1 & 3 \\
1 & -1
\end{array}\right|\) = 1 – 6 = -5
C31 = (-1)3+1M31 = (1)(-5) = -5

M32 = \(\left|\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right|\) = -2 – 3 = -5
C32 = (-1)3+2M32 = (-1)(-5) = 5

M33 = \(\left|\begin{array}{cc}
2 & -1 \\
1 & 2
\end{array}\right|\) = 4 + 1 = 5
C33 = (-1)3+3M33 = (1)(5) = 5

Question 5.
Evaluate \(\left|\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\) and cofactors of elements in the 2nd determinant and verify:
i. – a21.M21 + a22.M22 – a23.M23 = value of A a21.C21 + a22.C22 + a23.C23 — value of A where M21, M22, M23 are minors of a21, a22, a23 and C21, C22, C23 are cofactors of a21, a22, a23.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1 2
= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30)
= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) + 5(30)
= -40-138+ 150 = -28
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1 3
– a21.M21 + a22.M22 – a23.M23
= – (6)(- 4) + (0)(-19) – (4)(13)
= 24 + 0 – 52
= -28
– a21.M21 + a22.M22 – a23.M23 = value of A

ii. a21.C21 + a22.C22 + a23.C23
= (6)(4) +(0)(-19)+ (4)(-13)
= 24 + 0-52 .
= -28
a21.C21 + a22.C22 + a23.C23 = value of A

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.1

Question 6.
Find the value of determinant expanding along third column \(\left|\begin{array}{ccc}
-1 & 1 & 2 \\
-2 & 3 & -4 \\
-3 & 4 & 0
\end{array}\right|\)
Solution:
Here, \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|=\left|\begin{array}{ccc}
-1 & 1 & 2 \\
-2 & 3 & -4 \\
-3 & 4 & 0
\end{array}\right|\)
Expantion along the third column
= a13C13 + a23C23 + a33C33
= 2 x (-1)1+3 \(\left|\begin{array}{ll}
-2 & 3 \\
-3 & 4
\end{array}\right|\)-4 x (-1)2+3 \(\left|\begin{array}{ll}
-1 & 1 \\
-3 & 4
\end{array}\right|\) + 0 x (-1)3+3 \(\left|\begin{array}{ll}
-1 & 1 \\
-2 & 3
\end{array}\right|\)
= 2 (-8 + 9) +4 (-4 + 3) + 0
= 2 – 4
= -2

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

I. Select the correct option from the given alternatives.

Question 1.
The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to
(a) sin A
(b) cos A
(c) -cos A
(d) sin 2A
Answer:
(b) cos A
Hint:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let (n + 2)A = a and (n + 1)A = b … (i)
∴ L.H.S. = cos a . cos b + sin a . sin b
= cos (a – b)
= cos [(n + 2)A – (n + 1)A] ……..[From (i)]
= cos [(n + 2 – n – 1)A]
= cos A
= R.H.S.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 2.
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) = ________
(a) \(\frac{1}{y}-\frac{1}{x}\)
(b) \(\frac{1}{x}-\frac{1}{y}\)
(c) \(\frac{1}{x}+\frac{1}{y}\)
(d) \(\frac{x y}{x-y}\)
Answer:
(c) \(\frac{1}{x}+\frac{1}{y}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q2

Question 3.
If sin θ = n sin(θ + 2α), then tan(θ + α) is equal to
(a) \(\frac{1+n}{2-n}\) tan α
(b) \(\frac{1-n}{1+n}\) tan α
(c) tan α
(d) \(\frac{1+n}{1-n}\) tan α
Answer:
(d) \(\frac{1+n}{1-n}\) tan α
Hint:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q3

Question 4.
The value of \(\frac{\cos \theta}{1+\sin \theta}\) is equal to ________
(a) \(\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)\)
(b) \(\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
(d) \(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)
Answer:
(c) \(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q4
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q4.1

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 5.
The value of cos A cos (60° – A) cos (60° + A) is equal to ________
(a) \(\frac{1}{2}\) cos 3A
(b) cos 3A
(c) \(\frac{1}{4}\) cos 3A
(d) 4cos 3A
Answer:
(c) \(\frac{1}{4}\) cos 3A
Hint:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q5

Question 6.
The value of \(\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}\) is ________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{64}\)
(c) \(\frac{1}{128}\)
(d) \(\frac{1}{256}\)
Answer:
(b) \(\frac{1}{64}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q6
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q6.1
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q6.2

Question 7.
If α + β + γ = π, then the value of sin2 α + sin2 β – sin2 γ is equal to ________
(a) 2 sin α
(b) 2 sin α cos β sin γ
(c) 2 sin α sin β cos γ
(d) 2 sin α sin β sin γ
Answer:
(c) 2 sin α sin β cos γ
Hint:
sin2 α + sin2 β – sin2 γ
= \(\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^{2} \gamma\)
= 1 – \(\frac{1}{2}\) (cos 2α + cos 2β) – 1 + cos2 γ
= \(\frac{-1}{2}\) × 2 cos(α + β) cos(α – β) + cos2 γ
= cos γ cos (α – β) + cos2 γ …..[∵ α + β + γ = π]
= cos γ [cos (α – β) + cos γ]
= cos γ [cos (α – β) – cos (α + β)]
= 2 sin α sin β cos γ

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 8.
Let 0 < A, B < \(\frac{\pi}{2}\) satisfying the equation 3sin2 A + 2sin2 B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) 2π
Answer:
(b) \(\frac{\pi}{2}\)
Hint:
3 sin 2A – 2sin 2B = 0
sin 2B = \(\frac{3}{2}\) sin 2A …….(i)
3 sin2 A + 2 sin2 B = 1
3 sin2 A = 1 – 2 sin2 B
3 sin2 A = cos 2B ……(ii)
cos(A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin2 A) – sin A (\(\frac{3}{2}\) sin 2A) …..[From (i) and (ii)]
= 3 cos A sin2 A – \(\frac{3}{2}\) (sin A) (2 sin A cos A)
= 3 cos A sin2 A – 3 sin2 A cos A
= 0
= cos \(\frac{\pi}{2}\)
∴ A + 2B = \(\frac{\pi}{2}\) ……..[∵ 0 < A + 2B < \(\frac{3 \pi}{2}\)]

Question 9.
In ∆ABC if cot A cot B cot C > 0, then the triangle is ________
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles right-angled
Answer:
(a) acute angled
Hint:
cot A cot B cot C > 0
Case I:
cot A, cot B, cot C > 0
∴ cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\), 0 < C < \(\frac{\pi}{2}\)
∴ ∆ABC is an acute angled triangle.
Case II:
Two of cot A, cot B, cot C < 0
0 < A, B, C < π and two of cot A, cot B, cot C < 0
∴ Two angles A, B, C are in the 2nd quadrant which is not possible.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 10.
The numerical value of tan 20° tan 80° cot 50° is equal to ________
(a) √3
(b) \(\frac{1}{\sqrt{3}}\)
(c) 2√3
(d) \(\frac{1}{2 \sqrt{3}}\)
Answer:
(a) √3
Hint:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 I Q10
= tan 3(20°)
= tan 60°
= √3
= R.H.S.

II. Prove the following.

Question 1.
tan 20° tan 80° cot 50° = √3
Solution:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q1
= tan 3(20°)
= tan 60°
= √3
= R.H.S.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 2.
If sin α sin β – cos α cos β + 1 = 0, then prove that cot α tan β = -1.
Solution:
sin α sin β – cos α cos β + 1 = 0
∴ cos α cos β – sin α sin β = 1
∴ cos (α + β) = 1
∴ α + β = 0 ……[∵ cos 0 = 1]
∴ β = -α
L.H.S. = cot α tan β
= cot α tan(-α)
= -cot α tan α
= -1
= R.H.S.

Question 3.
\(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q3
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q3.1

Question 4.
\(\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q4
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q4.1

Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = \(-\frac{1}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q5

Question 6.
\(\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q6
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q6.1

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 7.
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q7

Question 8.
sin2 6x – sin2 4x = sin 2x sin 10x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q8

Question 9.
cos2 2x – cos2 6x = sin 4x sin 8x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q9

Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q10

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q11

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 12.
If sin 2A = λ sin 2B, then prove that \(\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q12

Question 13.
\(\frac{2 \cos 2 A+1}{2 \cos 2 A-1}\) = tan (60° + A) tan (60° – A)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q13

Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q14
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q14.1

Question 15.
3 tan6 10° – 27 tan4 10° + 33 tan2 10° = 1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q15

Question 16.
cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384°
= cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°)
= cosec 48° + cosec 84° + cosec (-12°) + cosec 24°
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q16
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q16.1

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 17.
3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) = 13
Solution:
(sin x – cos x)4
= [(sin x – cos x)2]2
= (sin2 x + cos2 x – 2 sin x cos x)2
= (1 – 2 sin x cosx)2
= 1 – 4 sin x cos x + 4 sin2 x cos2 x
(sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x
sin6 x + cos6 x
= (sin2 x)3 + (cos2 x)3
= (sin2 x + cos2 x)3 – 3 sin2 x cos2 x (sin2 x + cos2 x) …..[∵ a3 + b3 = (a + b)3 – 3ab(a + b)]
= 13 – 3 sin2 x cos2 x (1)
= 1 – 3 sin2 x cos2 x
L.H.S. = 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x)
= 3(1 – 4 sin x cos x + 4 sin2 x cos2 x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin2 x cos2 x)
= 3 – 12 sin x cos x + 12 sin2 x cos2 x + 6 + 12 sin x cos x + 4 – 12 sin2 x cos2 x
= 13
= R.H.S.

Question 18.
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution:
We have to prove that,
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
i.e., to prove,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q18
∴ cot θ – tan θ = 2 cot 2θ …..(i)
L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A
= 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)]
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A
= 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)]
= 4(cot 4A – tan 4A) – 8 cot 8A
= 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)]
= 8 cot 8A – 8 cot 8A = 0
= R.H.S.
Alternate Method:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q18.1
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q18.2
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q18.3

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 19.
If A + B + C = \(\frac{3 \pi}{2}\), then cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q19

Question 20.
In any triangle ABC, sin A – cos B = cos C. Show that ∠B = \(\frac{\pi}{2}\).
Solution:
sin A – cos B = cos C
∴ sin A = cos B + cos C
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q20
A = B – C ………(i)
In ∆ABC,
A + B + C = π
∴ B – C + B + C = π
∴ 2B = π
∴ B = \(\frac{\pi}{2}\)

Question 21.
\(\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}\) = sec x cosec x – 2 sin x cos x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q21

Question 22.
sin 20° sin 40° sin 80° = \(\frac{\sqrt{3}}{8}\)
Solution:
L.H.S. = sin 20°. sin 40°. sin 80°
= sin 20°. sin 40°. sin 80°
= \(\frac{1}{2}\) (2 . sin 40°. sin 20°) . sin 80°
= \(\frac{1}{2}\) [cos(40° – 20°) – cos (40° + 20°)] . sin 80°
= \(\frac{1}{2}\) (cos 20° – cos 60°) sin 80°
= \(\frac{1}{2}\) . cos 20° . sin 80° – \(\frac{1}{2}\) . cos 60° . sin 80°
= \(\frac{1}{2 \times 2}\) (2 sin 80° . cos 20°) – \(\frac{1}{2 \times 2}\) . sin 80°
= \(\frac{1}{4}\) [sin(80° + 20°) + sin (80° – 20°)] – \(\frac{1}{2}\) . sin 80°
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q22

Question 23.
sin 18° = \(\frac{\sqrt{5}-1}{4}\)
Solution:
Let θ = 18°
∴ 5θ = 90°
∴ 2θ + 3θ = 90°
∴ 2θ = 90° – 3θ
∴ sin 2θ = sin (90° – 3θ)
∴ sin 2θ = cos 3θ
∴ 2 sin θ cos θ = 4 cos3 θ – 3 cos θ
∴ 2 sin θ = 4 cos2 θ – 3 …..[∵ cos θ ≠ 0]
∴ 2 sin θ = 4 (1 – sin2 θ) – 3
∴ 2 sin θ = 1 – 4 sin2 θ
∴ 4 sin2 θ + 2 sin θ – 1 = 0
∴ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{8}\)
= \(\frac{-2 \pm 2 \sqrt{5}}{8}\)
= \(\frac{-1 \pm \sqrt{5}}{4}\)
Since, sin 18° > 0
∴ sin 18°= \(\frac{\sqrt{5}-1}{4}\)

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 24.
cos 36° = \(\frac{\sqrt{5}+1}{4}\)
Solution:
We know that,
cos 2θ = 1 – 2 sin2 θ
cos 36° = cos 2(18°)
= 1 – 2 sin2 18°
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q24
∴ cos 36° = \(\frac{\sqrt{5}+1}{4}\)

Question 25.
sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\)
Solution:
We know that, sin2 θ = 1 – cos2 θ
sin2 36° = 1 – cos2 36°
= 1 – \(\left(\frac{\sqrt{5}+1}{4}\right)^{2}\)
= \(\frac{16-(5+1+2 \sqrt{5})}{16}\)
= \(\frac{10-2 \sqrt{5}}{16}\)
∴ sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\) ……[∵ sin 36° is positive]

Question 26.
\(\sin \frac{\pi^{c}}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q26
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q26.1

Question 27.
tan \(\frac{\pi}{8}\) = √2 – 1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q27

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q28
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q28.1

Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q29

Question 30.
√3 cosec 20° – sec 20° = 4
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q30
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q30.1

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3

Question 31.
In ∆ABC, ∠C = \(\frac{2 \pi}{3}\), then prove that cos2 A + cos2 B – cos A cos B = \(\frac{3}{4}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q31
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Miscellaneous Exercise 3 II Q31.1

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 1.
Evaluate:
i. \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
{[2} & -4 & 3
\end{array}\right]\)
ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
Solution:
i. \(\begin{aligned}
\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right] &=\left[\begin{array}{lll}
3(2) & 3(-4) & 3(3) \\
2(2) & 2(-4) & 2(3) \\
1(2) & 1(-4) & 1(3)
\end{array}\right] \\
&=\left[\begin{array}{ccc}
6 & -12 & 9 \\
4 & -8 & 6 \\
2 & -4 & 3
\end{array}\right]
\end{aligned}\)

ii. \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
= [2(4)-1(3)+ 3(1)]
= [8 – 3 + 3] = [8]

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -3 \\
4 & 2
\end{array}\right]\) B = \(\left[\begin{array}{cc}
4 & 1 \\
3 & -2
\end{array}\right]\), = show that AB ≠ BA.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 1
From (i) and (ii), we get
AB ≠ BA

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right]\) ,B = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right]\) state whether AB = BA? Justify your answer.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 2
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 3
From (i) and (ii), we get
AB ≠ BA

Question 4.
Show that AB = BA, where
i. A = \(\left[\begin{array}{rrr}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\) , B = \(\left[\begin{array}{rrr}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 4
From (i) and (ii), we get
AB = BA

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 5
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 6
From (i) and (ii), we get
AB = BA
[Note: The question has been modified.]

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 5.
If A = \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\), prove that A2 = 0.
Solution:
A2 = A.A
= \(\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\left[\begin{array}{cc}
4 & 8 \\
-2 & -4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
16-16 & 32-32 \\
-8+8 & -16+16
\end{array}\right] \)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) = 0

Question 6.
Verify A(BC) = (AB)C in each of the following cases:
i. A = \(=\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 i
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 7
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 8
From (i) and (ii), we get
A(BC) = (AB)C.

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 9

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 7.
Verify that A(B + C) = AB + AC in each of the following matrices:
i. A = \(\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(=\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 3 \\
4 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-2 & 0 \\
4 & -3
\end{array}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 10
From (i) and (ii), we get
A(B + C) = AB + AC.
[Note: The question has been modified.]
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 11
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 12

Question 8.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 6
\end{array}\right]\), B = \(\left[\begin{array}{cc}
3 & -1 \\
3 & 7
\end{array}\right]\), find AB – 2I, where I is unit matrix of order 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 13

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 9.
If A = \(\left[\begin{array}{ccc}
4 & 3 & 2 \\
-1 & 2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 0 \\
1 & -2
\end{array}\right]\), show that matrix AB is non singular.
Solution:
im
∴ AB is non-singular matrix.

Question 10.
If A = \(\), find the product (A + I)(A – I).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 14

[Note : Answer given in the textbook is \(\left[\begin{array}{ccc}
9 & 6 & 4 \\
15 & 32 & -2 \\
35 & -7 & 29
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ccc}
10 & 10 & 4 \\
25 & 39 & 2 \\
35 & 7 & 22
\end{array}\right]\). ]

Question 11.
If A = \(\left[\begin{array}{ll}
\alpha & 0 \\
1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]\), find α, if A2 = B.
Solution:
A2 = B
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 15
∴ By equality of matrices, we get
α2 = 1 and α + 1 = 2
∴ α = ± 1 and α = 1
∴ α = 1

Question 12.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), show that A2 – 4A is scalar matrix.
Solution:
A2 – 4A = A.A – 4A
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 16

Question 13.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that A2 – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.
Solution:
A2 – 8A – kI = O
∴ A.A – 8A – kI = O
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 17
∴ by equality of matrices, we get
1 – 8 – k = 0
∴ k = -7

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 14.
If A = \(\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\), show that (A+B)2 = A2 + AB + B2.
Solution:
We have to prove that (A + B)2 = A2 + AB + B2,
i.e., to prove A2 + AB + BA + B2 = A2 + AB + B2,
i.e., to prove BA = 0.
BA = \(\left[\begin{array}{cc}
5 & -4 \\
10 & -8
\end{array}\right]\left[\begin{array}{cc}
8 & 4 \\
10 & 5
\end{array}\right]\)
\(\left[\begin{array}{cc}
40-40 & 20-20 \\
80-80 & 40-40
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 15.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), prove that A2 – 5A + 7I = 0, where I is unit matrix of order 2.
Solution:
A2 – 5A + 7I = 0 = A.A – 5A + 7I = 0
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 18

Question 16.
If A = \(\left[\begin{array}{cc}
3 & 4 \\
-4 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-1 & 2
\end{array}\right]\), show that (A + B)(A – B) = A2 – B2.
Solution:
We have to prove that (A + B)(A – B) = A2 – B2,
i.e., to prove A2 – AB + BA – B2 = A2 – B2,
i.e., to prove – AB + BA = 0,
i.e., to prove AB – BA.
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 19
From (i) and (ii), we get AB = BA

Question 17.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & a \\
-1 & b
\end{array}\right]\) and (A + B)2 = A2 + B2, find the values of a and b.
Solution:
Given, (A + B)2 = A2 + B2
∴ A2 + AB + BA + B2 = A2 + B2
∴ AB + BA = 0
∴ AB = -BA
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 20
∴ by equality of matrices, we get
– 2 + a = 0 and 1 + b = 0
a = 2 and b = -1
[Note: The question has been modified.]

Question 18.
Find matrix X such that AX = B,
where A = \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{c}
-3 \\
-1
\end{array}\right]\)
Solution:
Let X = \(\left[\begin{array}{c}
a \\
b
\end{array}\right]\)
But AX = B
∴ \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{a} \\
\mathrm{b}
\end{array}\right]=\left[\begin{array}{r}
-3 \\
-1
\end{array}\right]\)
∴ \(\left[\begin{array}{c}
a-2 b \\
-2 a+b
\end{array}\right]=\left[\begin{array}{l}
-3 \\
-1
\end{array}\right]\)
By equality of matrices, we get
a – 2b = -3 …(i)
-2a + b = -l …(ii)
By (i) x 2 + (ii), we get
-3b =-7
∴ b = \(\frac{7}{3}\)
Substituting b = \(\frac{7}{3}\) in (i), we get
a – 2 (\(\frac{7}{3}\)) = -3
∴ a = -3 + \(\frac{14}{3}=\frac{5}{3}\)
∴ X = \(\left[\begin{array}{l}
\frac{5}{3} \\
\frac{7}{3}
\end{array}\right]\)

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 19.
Find k, if A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and A2 = KA – 2I
Solution:
A2 = kA – 2I
∴ AA + 2I = kA
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 21
∴ \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]\)
∴ By equality of matrices, we get
3k = 3
∴ k = 1

Question 20.
Find x, if \(\left[\begin{array}{lll}
1 & x & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
3 & 2 & 5
\end{array}\right]\left[\begin{array}{c}
1 \\
-2 \\
3
\end{array}\right]\) = 0
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 22
∴ [6 + 12x + 14] =[0]
∴ By equality of matrices, we get
∴ 12x + 20 = 0
∴ 12x =-20
∴ x = \(\frac{-5}{3}\)

Question 21.
Find x and y, if \(\left\{4\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 0 & 2
\end{array}\right]-\left[\begin{array}{ccc}
3 & -3 & 4 \\
2 & 1 & 1
\end{array}\right]\right\}\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 23
∴ By equality of matrices, we get
x = 19 andy = 12

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 22.
Find x, y, z if
\(\left\{3\left[\begin{array}{ll}
2 & 0 \\
0 & 2 \\
2 & 2
\end{array}\right]-4\left[\begin{array}{cc}
1 & 1 \\
-1 & 2 \\
3 & 1
\end{array}\right]\right\}\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
x-3 \\
y-1 \\
2 z
\end{array}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 24
∴ By equality of matrices, we get
x – 3 = -6,y – 1 = 0, 2z = -2
∴ x = – 3, y = 1, z = – 1

Question 23.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) show that A2 = \(=\left[\begin{array}{cc}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 25

Question 24.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)
show that AB ≠ BA, but |AB| = |A| . |B|.
Solution:
AB = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\left[\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right]\)
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants an6d Matrices Ex 4.6 2
Now, |AB| = \(\left|\begin{array}{cc}
4 & 2 \\
10 & 7
\end{array}\right|\) = 28 – 20 = 8
|A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right|\) = 5 – 6 = -1
|B| = \(\left|\begin{array}{cc}
0 & 4 \\
2 & -1
\end{array}\right|\) = 0 – 8 = -8
∴ |A| . |B| = (-1).(-8) = 8 = |AB|
∴ AB ≠ BA, but |AB| = |A|.|B|

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 25.
Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was 6 and ₹ 10. Using matrix multiplication, find the amount required from each one of them.
Solution:
Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook.
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 27
The total amount required for each one of them is obtained by matrix AB.
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6 28
∴ Jay needs ₹ 104 and Ram needs ₹ 150.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.5

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.5

Question 1.
In Δ ABC, A + B + C = π, show that
cos2A + cos2B + cos2C = – 1 – 4 cosA cosB cosC
Solution:
L.H.S. = cos 2A + cos 2B + cos 2C
= \(2 \cdot \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cdot \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)+\cos 2 \mathrm{C}\)
= 2.cos(A + B).cos (A – B) + 2cos2C – 1
In ΔABC, A + B + C = π
∴ A + B = π – C
∴ cos(A + B) = cos(π – C)
∴ cos(A + B) = – cosC ………….(i)

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5

∴ L.H.S. = – 2.cos C.cos (A – B) + 2.cos2C – 1 …[From(i)]
= – 1 – 2.cosC.[cos(A – B) – cosC]
= – 1 – 2.cos C.[cos(A – B) + cos(A + B)]
… [From (i)]
= – 1 – 2.cos C.(2.cos A.cos B)
= – 1 – 4.cos A.cos B.cos C = R.H.S.

Question 2.
sin A + sin B + sin C = 4 cos A/2 cos B/2 cos C/2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 1

Question 3.
cos A + Cos B + Cos C = 4 cos A/2 cos B/2 cos C/2
= \(2 \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\left(1-2 \sin ^{2} \frac{\mathrm{C}}{2}\right)\)
Solution:
L.H.S. = sin A + sin B + sin C
= \(2 \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\left(1-2 \sin ^{2} \frac{\mathrm{C}}{2}\right)\)
In Δ ABC, A + B + C = π ,
∴ A + B = π – C
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 2

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5

Question 4.
sin2 A + sin2 B – sin2 C = 2 sin A sin B cos C
Solution:
We know that, sin2 = \(\frac{1-\cos 2 \theta}{2}\)
L.H.S.
= sin2 + sin2 B + sin2 C
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 3
= 1 – cos(A + B). cos(A – B) – sin2C
= (1 – sin2 C ) – cos (A + B). cos (A – B)
= cos2 C – cos(A + B). cos(A – B)
∴ cos(A + B) = cos(it — C)
∴ cos(A + B) = — cos C …(i)
∴ L.H.S. = cos2C + cos C.cos(A – B)
… [From (i)]
= cos C[cos C + cos(A – B)]
= cos C[- cos(A + B) + cos(A – B)]
… [From (i)]
= cos C[cos (A-B) – cos(A + B)]
= cos C(2 sin A.sin B)
= 2 sin A.sin B. cos C
= R.H.S.
[Note: The question has been modified.]

Question 5.
\(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}-\sin ^{2} \frac{C}{2}\) = \(1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 4
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 5

Question 6.
tan \(\frac{\mathbf{A}}{2}\) tan \(\frac{\mathbf{B}}{2}\) tan \(\frac{\mathbf{B}}{2}\) tan \(\frac{\mathbf{C}}{2}\) tan \(\frac{\mathbf{C}}{2}\) tan \(\frac{\mathbf{A}}{2}\) = 1
Solution:
In Δ ABC,
A + B + C = π
∴ A + B = π – C
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 6

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5

Question 7.
\(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}\)
Solution:
In Δ ABC,
A + B + C = π
∴ A + B = π – C
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 7

Question 8.
tan 2A + tan 2B + tan 2C = tan 2A tan 2B + tan 2C
Solution:
In Δ ABC,
A + B + C = π
∴ 2A + 2B + 2C = 2π
∴ 2A + 2B = 2π – 2C
tan(2A + 2B) = tan(2n — 2C)
\(\frac{\tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}}{1-\tan 2 \mathrm{~A} \cdot \tan 2 \mathrm{~B}}\) = -tan 2C
∴ tan2A+tan2B=—tan2C.(1-tan2A.tan2B)
∴ tan 2A + tan 2B = – tan2C+ tan2A.tan2B.tan2C
∴ tan 2A + tan 2B + tan 2C = tan2A.tan2B.tan2C

Question 9.
cos2 A + cos2 B – cos2 C = 1 – 2 sin A sin B sin C
Solution:
we know that cos2θ = \(\frac{1+\cos 2 \theta}{2}\)
L.H.S.
= cos2 A + cos2 B + cos2 C
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.5 8
= 1 + cos (A + B).cos(A — B) – cos2 C
In ΔABC,
A + B + C = π
A + B = π — C
cos(A + B) = cos(π — C)
cos(A + B) = -cosC ………….. (i)
L.H.S. = 1 — cos C.cos(A — B) — cos2 C
…[From(i)]
= 1 — cos C.[cos(A — B) + cos C]
= 1 — cos C.[cos(A — B) — cos(A + B)]
.. .[From (i)]
= 1 — cos C.(2.sin A.sin B)
= 1 — 2.sinA.sin B.cos C
= R.H.S.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.4

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.4

Question 1.
Express the following as a sum or difference of two trigonometric functions.
i. 2sin 4x cos 2x
ii. 2sin \(\frac{2 \pi}{3}\) cos \(\frac{\pi}{2}\)
iii. 2cos 4θ cos 2θ
iv. 2cos 35° cos 75°
Solution:
i. 2sin 4x cos 2x = sin(4x + 2x ) + sin (4x – 2x)
= sin 6x + sin 2x

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4

ii.
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4 1

[Note: Answer given in the textbook is sin \(\frac{7 \pi}{12}\) + sin \(\frac{\pi}{12}\) However, as per our calculation it is sin \(\frac{7 \pi}{6}\) + sin \(\frac{\pi}{6}\)

iii. 2cos 4θ cos 2θ = cos(4θ + 2θ)+cos (4θ – 2θ)
= cos 6θ + cos 2θ

iv. 2cos 35° cos75°
= cos(35° + 75°) + cos (35° – 75°)
= cos 110° + cos (-40)°
= cos 110° + cos 40° … [∵ cos(-θ) = cos θ]

Question 2.
Prove the following:
i. \(\frac{\sin 2 x+\sin 2 y}{\sin 2 x-\sin 2 y}=\frac{\tan (x+y)}{\tan (x-y)}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4 2

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4

ii. sin 6x + sin 4x – sin 2x = 4 cos x sin 2x cos 3x
Solution:
L.H.S. = sin 6x + sin 4x — sin 2x
= 2sin \(\left(\frac{6 x+4 x}{2}\right)\) cos \(\left(\frac{6 x-4 x}{2}\right)\) – 2 sin x cos x
= 2 sin 5x cos x — 2 sin x cos x
= 2 cos x (sin 5x — sin x)
= 2 cos \(\left[2 \cos \left(\frac{5 x+x}{2}\right) \sin \left(\frac{5 x-x}{2}\right)\right]\)
= 2 cos x (2 cos 3x sin 2x)
= 4 cos x sin 2x cos 3x
= R.H.S.
[Note: The question has been modified.]

iii. \(\frac{\sin x-\sin 3 x+\sin 5 x-\sin 7 x}{\cos x-\cos 3 x-\cos 5 x+\cos 7 x}\) = cot 2x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4 3

iv. sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33°
Solution:
L.H.S. = sin 18°.cos 39° + sin 6°.cos 15°
= \(\frac{1}{2}\) (2 cos 39°sin 18° + 2.cos 15°.sin 6°)
= \(\frac{1}{2}\)[sin(39° + 18°) — sin(39° — 18°) + sin (15° + 6°) — sin (15° — 6°)]
= \(\frac{1}{2}\)(sin57° – sin21° + sin 21°- sin9°)
= \(\frac{1}{2}\)(sin57° – sin9°)
= \(\frac{1}{2}\) x 2. cos \(\left(\frac{57^{\circ}+9^{\circ}}{2}\right) \cdot \sin \left(\frac{57^{\circ}-9^{\circ}}{2}\right)\)
= cos 33° .sin 24°
= sin 24°. cos 33°
= R.H.S.

v. cos 20° cos 40° cos 60°cos 80° = 1/16
Solution:
L.H.S. = cos 20°.cos 40°.cos 60°.cos 80°
= cos 20°.cos 40°.\(\frac{1}{2}\) .cos 80°
= \(\frac{1}{2 \times 2}\)(2 cos 40°.cos 20°).cos 80°
= \(\frac{1}{4}\)[cos(40° + 20°) + cos(40°- 20°)].cos80°
= \(\frac{1}{4}\)(cos 60° + cos 20°) cos 80°
=\(\frac{1}{4}\)cos 60°. cos 80° + \(\frac{1}{4}\) cos 20°. cos 80°
= \(\frac{1}{4}\left(\frac{1}{2}\right) \cos 80^{\circ}+\frac{1}{2 \times 4}\left(2 \cos 80^{\circ} \cos 20^{\circ}\right)\)
= \(\frac{1}{8}\) cos 80° + \(\frac{1}{8}\)[cos (80° + 20°) + cos (80° — 20°)]
= \(\frac{1}{8}\)cos 80° + \(\frac{1}{8}\)(cos 100° + cos 60°)
= \(\frac{1}{8}\) cos 80° + \(\frac{1}{8}\)cos 100° + \(\frac{1}{8}\)cos 60°
= \(\frac{1}{8}\) cos 80° = \(\frac{1}{8}\) cos (180° – 80°) + \(\frac{1}{8} \times \frac{1}{2}\)
= \(\frac{1}{8}\) cos 80° – \(\frac{1}{8}\) cos 80° + \(\frac{1}{16}\) … [∵ cos (180 – θ) = – cos θ]
= \(\frac{1}{16}\) = R.H.S

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4

vi. sin 20° sin 40° sin 60° sin 80° = 3/16
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.4 4

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.3

Question 1.
Find the values of:
i. sin \(\frac{\pi}{8}\)
ii. \(\frac{\pi}{8}\)
Solution:
We know that sin2 θ = \(\frac{1-\cos 2 \theta}{2}[/atex]
Substituting θ = [latex]\frac{\pi}{8}\), we get
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 1

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3

ii. We know that, cos2 θ = \(\frac{1+\cos 2 \theta}{2}\)
Substituting θ = \(\frac{\pi}{8}\), we get
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 2

Question 2.
Find sin 2x, cos 2x, tan 2x if sec x = \(-\frac{13}{5}\), \(\frac{\pi}{2}\) < x < π
Solution:
sec x = \(-\frac{13}{5}\), \(\frac{\pi}{2}\) < x < π
We know that
Sect2 x = 1 + tan2x
tan2x = \(\frac{169}{25}-1=\frac{144}{25}\)
tan x = \(\pm \frac{12}{5}\)
Since \(\frac{\pi}{2}\) < x < π
x lies in the 2nd quadrant.
tan x < 0
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 3
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 4

Question 3.
i. \(\) = tan2 θ
Solution:
L. H. S. = \(\frac{1-\cos 2 \theta}{1+\cos 2 \theta}\)
= \(\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}\)
= 2tan2 θ
= R.H.S.

ii. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Solution:
L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cosx)cosx
= sin 3x sin x + sin2 x + cos 3x cos x – cos2 x
= (cos 3x cos x + sin 3x sin x)
— (cos2x — sin2x)
= cos (3x – x) – cos 2x
= cos 2x – cos 2x
= 0
= R.H.S.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3

iii. (cos x + cos y )2 + (sin x + sin y)2 = 4cos2 \(\left(\frac{x-y}{2}\right)\)
Solution:
L.H.S. = (cos x + cos y)2 + (sin x + sin y)2
= cos2x + cos2y + 2cos x.cos y + sin2 x + sin2y + 2sin x.siny
= (cos2x + sin2x) + (cos2y + sin2y) + 2(cos x.cos y + si x.sin y)
= 1 + 1 +2cos(x – y)
= 2 + 2 cos (x – y)
= 2[1 + cos(x – y)]
= 2[2cos2 [(\(\left(\frac{x-y}{2}\right)\))] … [∵ 1 + cos θ = 2 cos2 \(\frac{\theta}{2}\)]
= 4 cos2 (\(\frac{x-y}{2}\))
= R.H.S.
[ Note: The question has been modified]

iv. (cos x – cos y)2 + (sin x – sin y)2 = 4sin2 \(\left(\frac{x-y}{2}\right)\)
Solution:

L.H.S. = (cos x – cos y)2 + (sin x – sin y)2
= cos2x + cos2y + 2cos x.cos y + sin2 x + sin2y + 2sin x.siny
= (cos2x + sin2x) + (cos2y + sin2y) – 2(cos x.cos y + sin x.sin y)
= 1 + 1 – 2cos(x – y)
= 2 – 2 cos (x – y)
= 2[1 – cos(x – y)]
= 2[2sin2 [(\(\left(\frac{x-y}{2}\right)\))] … [∵ 1 – cos θ = 2 sin2 \(\frac{\theta}{2}\)]
= 4 sin2 (\(\frac{x-y}{2}\))
= R.H.S.

v. tan x + cot x = 2 cosec 2x
Solution:
L.H.S. = tan x + cot x
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 5

vi. \(\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}\) = 2 tan 2x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 6
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 7

 

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3vii. \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 x}}}\) = 2 cos x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 8
= 2 cos x
= R.H.S.
[Note : The question has been modified.]

viii. 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θ
Solution:
L.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ
= 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ
= 8sin 2θ cos 2θ cos 4θ cos 8θ
= 4(2sin 2θ cos 2θ) cos 4θ cos 8θ
= 4sin 4θ cos 4θ cos 8θ
= 2(2sin 4θ cos 4θ) cos 8θ
= 2sin 8θ cos 8θ
= sin 16θ
= R.H.S.

ix. \( = 2 cot 2x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 9

x. [latex]\frac{\cos x}{1+\sin x}=\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 10
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 11

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3

xi. \(\frac{\tan \left(\frac{\theta}{2}\right)+\cot \left(\frac{\theta}{2}\right)}{\cot \left(\frac{\theta}{2}\right)-\tan \left(\frac{\theta}{2}\right)}=\sec \theta\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 12

xii. \(\frac{1}{\tan 3 \mathbf{A}-\tan A}-\frac{1}{\cot 3 A-\cot A}\) = cot 2A
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 13
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 14

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3

xiii. cos 7° cos 14° cos 28° cos 56° \(\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}\)
Solution:
L.H.S. = cos 7° cos 14° cos 28° cos 56°
= \(\frac{1}{2 \sin 7^{\circ}}\)(2sin 7°cos 7°)cos 14°cos 28°cos 56°
= \(\frac{1}{2 \sin 7^{\circ}}\) (sin 14° cos 14° cos 28° cos 56°)
…[∵ 2sinθ cosθ = sin 2θ]
= [\frac{1}{2\left(2 \sin 7^{\circ}\right)}latex][/latex] (2sin 14° cos 14°) cos 28° cos 56°
= \(\frac{1}{4 \sin 7^{\circ}}\)(sin 28° cos 28° cos 56°)
= \(\frac{1}{2\left(4 \sin 7^{\circ}\right)}\)(2 sin 28° cos 28°) cos 56°
= \(\frac{1}{8 \sin 7^{\circ}}\) (sin 56° cos 56°)
= \(\frac{1}{8 \sin 7^{\circ}}\) (2 sin 56° cos 56°)
= \(\frac{1}{16 \sin 7^{\circ}}\)(sin 112°)
= \(\frac{\sin \left(180^{\circ}-68^{\circ}\right)}{16 \sin \left(90^{\circ}-83^{\circ}\right)}\)
= \(\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}\)
= R.H.S.

xiv. = \(\frac{\sin ^{2}\left(-160^{\circ}\right)}{\sin ^{2} 70^{\circ}}+\frac{\sin \left(180^{\circ}-\theta\right)}{\sin \theta}\) = sec2 20°
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 15
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 16

xv. \(\frac{2 \cos 4 x+1}{2 \cos x+1}\) = (2 cos x – 1)(2 cos 2x – 1)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 17

xvi. = cos2 x + cos2 (x + 120°) + cos2(x – 120°) = \(\frac{3}{2}\)
Solution:
L.H.S = cos2 x + cos2 (x + 120°) + cos2(x – 120°) =
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 18
\(\frac{3}{2}+\frac{1}{2}\) [cos 2x + cos(2x + 240°) + cos(2x 240°)]
= \(\frac{3}{2}+\frac{1}{2}\)(cos 2x + cos 2x cos 240°— sin 2x sin 240° + cos 2x cos 240° + sin 2x sin 240°)
= \(\frac{3}{2}+\frac{1}{2}\)(cos 2x + 2 cos 2x cos 240°)
= \(\frac{3}{2}+\frac{1}{2}\) [cos 2x + 2 cos 2x cos( 180° + 60°)]
= \(\frac{3}{2}+\frac{1}{2}\) [cos 2x + 2cos 2x(-cos 600)]
= \(\frac{3}{2}+\frac{1}{2}\) [cos 2x —2 cos 2x(\(\frac{1}{2}\))]
= \(\frac{3}{2}+\frac{1}{2}\) ( cos 2x – cos 2x)
= \(\frac{3}{2}+\frac{1}{2}\) (0)
= \(\frac{3}{2}\) = R.H.S.

Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3

xvii. 2 cosec 2x + cosec x = sec cot \(\frac{x}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 19

xviii. 4 cos x cos (\(\frac{\pi}{3}\) + x) cos (\(\frac{\pi}{3}\) – x) = cos 3x
\(\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry - II Ex 3.3 20
= cos3x — 3cos x.sin2x
= cos3 x — 3cos x (1— cos2 x)
= cos3x — 3cos x + 3 cos3x
=4 cos3x — 3cos x
= cos 3x = R.H.S.
INote: The question has been modijìed.I

xix. sin x tan \(\frac{x}{2}\) + 2cos x = \(\frac{2}{1+\tan ^{2}\left(\frac{x}{2}\right)}\)
Solution:
L.H.S. = sin x tan (x/2)+ 2cos x
= \(\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)\) + 2cos x
= \(\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)\) + 2 cos x
= 2sin2 x/2 + 2cosx
= 1 – cosx + 2cosx
= 1 + cos x
=2cos2 x/2
= \(\frac{2}{\sec ^{2} \frac{x}{2}}=\frac{2}{1+\tan ^{2} \frac{x}{2}}\) =R.H.S.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 12 Chemical Equilibrium Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 1.
Explain irreversible reaction.
Answer:
Irreversible reaction:
i. Reactions which occur only in one direction, namely, from reactant to products are called irreversible reactions.
ii. They proceed in only a single direction until one of the reactants is exhausted.
iii. The direction in which an irreversible reaction occurs is indicated by an arrow (→) pointing towards the products in the chemical equation.
e.g. a. \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\text { Burn }}{\longrightarrow} \mathrm{CO}_{2(\mathrm{~g})}\)
b. \(2 \mathrm{KClO}_{3(\mathrm{~s})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})}\)

Question 2.
What is a closed system?
Answer:
A system in which there is no exchange of matter with the surroundings is called a closed system.

Question 3.
What is an open system?
Answer:
A system in which exchange of both matter and heat occurs with the surroundings is called an open system.

Question 4.
Why was calcium oxide used in theatre lighting?
Answer:
Calcium oxide (CaO) on strong heating glows with a bright white light. Hence, CaO was used in theatre lighting, which gave rise to the phrase ‘in the limelight’.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 5.
Explain liquid-vapour equilibrium with an example.
Answer:
Liquid-vapour equilibrium:
i. Consider reversible physical process of evaporation of liquid water into water vapour in a closed vessel. Initially, there is practically no water vapour in the vessel.

ii. When the liquid evaporates in the closed container, the liquid molecules escape from the liquid surface into vapour phase building up vapour pressure. They also condense back into liquid state because the container is closed.

iii. In the beginning the rate of evaporation is high and the rate of condensation is low. But with time, as more and more vapour is formed, the rate of evaporation goes down and the rate of condensation increases. Eventually the two rates become equal. This gives rise to a constant vapour pressure. This state is known as an ‘equilibrium state’.
In this state, the rate of evaporation is equal to the rate of condensation.
It may be represented as: H2O(l) ⇌ H2O(Vapour)

iv. At equilibrium, the pressure exerted by the gaseous water molecules at a given temperature remains constant, known as the equilibrium vapour pressure of water (or saturated vapour pressure of water or aqueous tension). The saturated vapour pressure increases with increase of temperature.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 1
[Note: The saturated vapour pressure of water at 100 °C is 1 atm (1.013 bar). Hence, water boils at 100 °C when pressure is 1 atm.]

Question 6.
What is meant by the term ‘normal boiling point’ of a liquid?
Answer:
For any pure liquid at 1 atm pressure, the temperature at which its saturated vapour pressure equals to atmospheric pressure is called the normal boiling point of that liquid.
e.g. The boiling point of ethyl alcohol is 78 °C i.e., the saturated vapour pressure of ethyl alcohol at 78 °C is 1 atm (1.013 bar).

Question 7.
Give an example of solid-liquid equilibrium.
Answer:
A mixture of ice and water in a perfectly insulated thermos flask at 273 K is an example of solid-liquid equilibrium.
H2O(s) ⇌ H2O(l)

Question 8.
Identify the type of equilibrium in the following physical processes:
i. Camphor(s) ⇌ Camphor(g)
ii. Ammonium chloride(s) ⇌ Ammonium chloride(g)
iii. Carbon dioxide gas ⇌ Dry ice
iv. Water ⇌ Ice
Answer:
i. Solid – vapour equilibrium
ii. Solid – vapour equilibrium Solid
iii. Solid – vapour equilibrium
iv. Solid – liquid equilibrium

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 9.
Name two substances that undergoes sublimation.
Answer:
Camphor, ammonium chloride.

Question 10.
Write a short note on chemical equilibrium.
Answer:
Chemical equilibrium:

  • If a reaction takes place in a closed system so that the products and reactants cannot escape, we often find that reaction does not give a 100% yield of products. Instead some reactants remain after the concentrations stop changing.
  • When there is no further change in concentration of reactant and product, the chemical reaction has attained equilibrium, with the rates of forward and reverse reactions being equal.
  • Chemical equilibrium at a given temperature is characterized by constancy of measurable properties such as pressure, concentration, density, etc.
  • Chemical equilibrium can be approached from either side of the chemical reaction.

Question 11.
Explain the law of mass action and give its mathematical representation.
Answer:
Statement: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.
Explanation: A rate equation can be written for a reaction by applying the law of mass action as follows: Consider a reaction, A + B → C
Here A and B are the reactants and C is the product. The concentrations of chemical species are expressed in mol L-1 and denoted by putting the formula in square brackets. On applying the law of mass action to this
reaction, a proportionality expression can be written as: Rate ∝ [A] [B]
This proportionality expression is transformed into an equation by introducing a proportionality constant, k, as follows:
Rate = k [A] [B]
This equation is called the rate equation and the proportionality constant, k, is called the rate constant of the reaction.

Question 12.
Write the rate equation for the following reactions:
i. C + O2 → CO2
ii. 2KClO3 → 2KCl + 3O2
Answer:
The rate equation is written by applying the law of mass action.
i. The reactants are C and O2
Rate ∝ [C] [O2]
∴ Rate = k [C] [O2]
ii. The reactant is KClO3 and its 2 molecules appear in the balanced equation.
∴ Rate ∝ [KClO3]2
∴ Rate = k [KClO3]2

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 13.
Derive the expression of equilibrium constant, KC for the reaction:
A + B ⇌ C + D
Answer:
Consider a hypothetical reversible reaction A + B ⇌ C + D.
Two reactions, namely, forward and reverse reactions occur simultaneously in a reversible chemical reaction. The rate equations for the forward and reverse reactions are:
Rateforward ∝ [A][B]
∴ Rateforward = kf [A] [B] …… (1)
∴ Ratereverse ∝ [C] [D]
∴ Ratereverse = kr [C] [D] …. (2)
At equilibrium, the rates of forward and reverse reactions are equal. Thus,
Rateforward = Ratereverse
∴ kf [A] [B] = kr [C] [D]
∴ \(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{r}}}=\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\) …….. (3)
KC is called the equilibrium constant.

Question 14.
Show that the equilibrium constant of the reverse chemical reaction (KC) is the reciprocal of the equilibrium constant (KC).
Answer:
Consider a reversible chemical reaction:
aA + bB ⇌ cC + dD
The equilibrium constant, KC = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Consider the reverse reaction:
cC + dD ⇌ aA + bB.
The equilibrium constant, KC is:
KC = \(\frac{[\mathrm{A}]^{a}[\mathrm{~B}]^{\mathrm{b}}}{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}=\frac{1}{\mathrm{~K}_{\mathrm{C}}}\)
Thus, equilibrium constant of the reverse chemical reaction (KC) is the reciprocal of the equilibrium constant KC.

Question 15.
Write equilibrium constant expressions for both forward and reverse reaction for the synthesis of ammonia by the Haber process.
Answer:
Synthesis of ammonia by Haber process:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 2

Question 16.
How are the equilibrium constants of the following pair of equilibrium reactions related?
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 3
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 4
ii. KC = \(\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{N}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{N}_{2} \mathrm{O}\right]}\)

Question 17.
Write KP expression for the reaction:
aA(g) + bB(g) ⇌ cC(g) + dD(g)
Answer:
For the given reaction,
KP = \(\frac{\left(P_{c}\right)^{c}\left(P_{D}\right)^{d}}{\left(P_{A}\right)^{a}\left(P_{B}\right)^{b}}\)

Question 18.
N2(g) + 3H2(g) ⇌ 2NH3(g)
Write expressions for KP and substitute expressions for PN2, PH2 and PNH3 using ideal gas equation.
Answer:
For the given reaction, KP = \(\frac{\left(P_{N H_{3}}\right)^{2}}{\left(P_{N_{2}}\right)\left(P_{H_{2}}\right)^{3}}\)
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 5
[Note: The above question is modified to apply appropriate textual context, i. e., to indicate that students need to use ideal gas equation to derive expressions for PN2, PH2 and PNH3]

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 19.
For a chemical equilibrium reaction
H2(g) + I2(g) ⇌ 2HI(g),
write an expression for KP (and relate it to KC).
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 6

Question 20.
Write the relationship between KC and KP for the following equilibria:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 7
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 8
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 9

Question 21.
Write the expressions for KC and KP and the relationship between them for the equilibrium reaction,
2A(g) + B(g) ⇌ 3C(g) + 2D(g)
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 10

Question 22.
Explain in short homogeneous equilibrium and heterogeneous equilibrium.
Answer:
i. In a homogeneous equilibrium, the reactants and products are in the same phase.
e.g. Dissociation of HI:
2HI(g) ⇌ H2(g) + I2(g)
ii. In a heterogeneous equilibrium, the reactants and products exist in different phases, e.g. Formation of NH4Cl:
NH3(g) + HCl(g) ⇌ NH4Cl(s)

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 23.
The unit of KC is different for different reactions. Explain this statement with suitable examples.
Answer:
Unit of equilibrium constant:
i. The unit of equilibrium constant depends upon the expression of KC which is different for different equilibria. Therefore, the unit of KC is also different for different reactions.
ii. Consider the following equilibrium reaction:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 11

iii. Consider the following equilibrium reaction:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 12

Question 24.
Write the equilibrium constant expression for the decomposition of baking soda. Deduce the unit of KC from the above expression.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 13

[Note: Considering gaseous reactants and products, in this reaction, Δn = 2 – 0 = 2
∴ Units of KC = (mol dm-3)Δn
= (mol dm-3)2
= mol2 dm-6
Thus, the units of the above reaction is mol2 dm-6.]

Question 25.
What are the characteristics of equilibrium constant?
Answer:
Characteristics of equilibrium constant:

  • The value of equilibrium constant is independent of initial concentrations of either the reactants or products.
  • Equilibrium constant is temperature dependent. Hence, KC and KP change with change in temperature.
  • Equilibrium constant has a characteristic value for a particular reversible reaction represented by a balanced equation at a given temperature.
  • Higher value of KC or KP means more concentration of products is formed and the equilibrium point is more towards right hand side and vice versa.

Question 26.
Explain how equilibrium constant helps in predicting the direction of the reaction.
Answer:
Prediction of the direction of the reaction:
i. For the reaction, aA + bB ⇌ cC + dD,
The equilibrium constant (KC) is given as:
KC = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
where, all the concentrations are equilibrium concentrations.
ii. When the reaction is not necessarily at equilibrium, the concentration ratio is called QC i.e.,
QC = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
iii. By comparing QC with KC for a reaction under given conditions, we can decide whether the forward or the reverse reaction should occur to establish the equilibrium.
a. QC < KC: The reaction will proceed from left to right, in forward direction, generating more product to attain the equilibrium.
b. QC > KC: The reaction will proceed from right to left, removing product to attain the equilibrium.
c. QC = KC: The reaction is at equilibrium and no net reaction occurs.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 14
[Note: The prediction of the direction of the reaction on the basis of QC and KC values makes no comment on the time required for attaining the equilibrium.]

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 27.
Explain how KC can be used to know the extent of the reaction?
Answer:
Extent of the reaction: The equilibrium constant expression indicates that the magnitude of KC is:
i. directly proportional to the concentrations of the products.
ii. inversely proportional to the concentrations of the reactants.
a. Value of KC is very high (KC > 103):
At equilibrium, there is a high proportion of products compared to reactants.
Forward reaction is favoured.
Reaction is in favour of products and nearly goes to completion.

b. Value of KC is very low (KC < 10-3):
At equilibrium, only a small fraction of the reactants is converted into products.
Reverse reaction is favoured.
Reaction hardly proceeds towards the products.

c. Value of KC is in the range of 10-3 to 103:
Appreciable concentrations of both reactants and products are present at equilibrium.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 15

Question 28.
For the following reactions, write KC expressions and predict direction of the reactions based on the magnitude of their equilibrium constants.
i. 2H2(g) + O2(g) ⇌ 2H2O(g), KC = 2.4 × 1047 at 500 K
ii. 2H2O(g) ⇌ 2H2(g) + O2(g), KC = 4.2 × 10-48 at 500 K
Answer:
i. a. KC expression:
KC = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]^{2}}{\left[\left[\mathrm{H}_{2(\mathrm{~g})}\right]\right]^{2}\left[\mathrm{O}_{2(\mathrm{~g})}\right]}\)
b. For the reaction, KC = 2.4 × 1047 at 500 K
If the value of KC >>> 103, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.

ii. a. KC expression:
KC = \(\frac{\left[\mathrm{H}_{2(\mathrm{~g})}\right]^{2}\left[\mathrm{O}_{2(\mathrm{~g})}\right]}{\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]^{2}}\)
b. For the reaction, KC = 4.2 × 10-48 at 500 K
If the value of KC <<< 10-3, reverse reaction is favoured.
Hence, the given reaction will proceed in the backward direction and will nearly go to completion.

Question 29.
Describe how equilibrium constant can be used to calculate the composition of an equilibrium mixture.
Answer:
An equilibrium constant can be used to calculate the composition of an equilibrium mixture.
Consider an equilibrium reaction, A(aq) + B(aq) ⇌ C(aq) + D(aq)
The equilibrium constant is 4.0 at a certain temperature.
Let the initial amount of A and B be 2.0 mol in ‘V’ litres. Let x mol be the equilibrium amount of C.
Hence, we can construct a table as shown below:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 16
The expression for equilibrium constant can be written as:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 17
Substituting the value of equilibrium concentration, we get
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 18
Therefore, equilibrium concentrations are 0.67 mol of A, 0.67 mol of B, 1.33 mol of C and 1.33 mol of D in V litres.

Question 30.
Explain the link between chemical equilibrium and chemical kinetics:
Answer:
Equilibrium constant (KC) is related to rate or velocity constants of forward reaction (kf) and reverse reaction (kr) as:
KC = \(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{r}}}\)
This equation can be used to determine the composition of the reaction mixture
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 19
[Note: The equilibrium refers to the relative amounts of reactants and products and thus a shift in equilibrium in a particular direction will imply the reaction in that direction will be favoured.]

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 31.
Equal concentrations of hydrogen and iodine are mixed together in a closed container at 700 K and allowed to come to equilibrium. If the concentration of HI at equilibrium is 0.85 mol dm-3, what are the equilibrium concentrations of H2 and I2 if KC = 54 at this temperature?
Solution:
Given: [HI(g)] = 0.85 mol dm-3
KC = 54 at 700 K
Equilibrium concentrations of H2 and I2
Formula: KC = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)
Balanced chemical reaction: 2HI(g)
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 20
Equilibrium concentration of I2(g) = Equilibrium concentration of H2(g)
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 21
Ans: Equilibrium concentrations of H2 and I2 are equal to 0.12 mol dm-3.

Question 32.
Calculate Kc at 500 K for the reaction,
2HI(g) ⇌ H2(g) + I2(g) if the equilibrium concentrations are [HI] = 0.5 M, [H2] = 0.08 M and [I2] = 0.062 M.
Solution:
Given: T = 500 K,
At equilibrium, [HI] = 0.5 M, [H2] = 0.08 M, [I2] = 0.062 M.
To find: Equilibrium constant KC
Formula: KC = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Calculation: The above equilibrium reaction is given as 2HI(g) ⇌ H2(g) + I2(g)
The expression of KC is
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 22
Ans: KC at 500 K for the given reaction is 0.0198.

Question 33.
Calculate KC and KP for the reaction at 295 K, N2O4 ⇌ 2NO2(g) if the equilibrium concentrations are [N2O4] = 0.75 M and [NO2] = 0.062 M, R = 0.08206 L atm K-1 mol-1.
Solution:
Given: R = 0.08206 L atm K-1 mol-1, T = 295 K
At equilibrium , [N2O4] = 0.75 M, [NO2] = 0.062 M
To find: Equilibrium constants, KP and KC
Formulae: i. KC = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)
ii. KP = KC (RT)Δn
Calculation : The equilibrium reaction is given as N2O4(g) ⇌ 2NO2(g)
The expression of KC is
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 23
KP is related to KC by expression: KP = KC (RT)Δn
where, Δn = numbers of moles of gaseous products – number of moles of gaseous reactants
= 2 – 1 = 1
∴ KP = KC(RT)1
∴ KP = 5.13 × 10-3 × 0.08206 × 295
∴ KP= 123.9 × 10-3 = 0.124
Ans: KC and KP for the reaction at 295 K are 5.13 × 10-3 and 0.124 respectively.

Question 34.
The equilibrium constant KC for the reaction of hydrogen with iodine is 54.0 at 700 K.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 24
KC = 54.0 at 700 K
If kf is the rate constant for the formation of HI and kr is the rate constant for the decomposition of HI, deduce whether kr is larger or smaller than kr.
ii. If the value of kr at 700 K is 1.16 × 10-3, what is the value of kf ?
Solution:
Given: i. KC = 54.0 at 700 K
ii. kr = 1.16 × 10-3 at 700 K
To find: i. Whether kf is larger or smaller than kr.
ii. Value of kf.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 25

Question 35.
Given the equilibrium reaction, H2O(g) + CH4(g) ⇌ CO(g) + 3H2(g)
Using Le Chatelier’s principle, predict how concentration of CO will change when the equilibrium is disturbed by
i. adding CH4
ii. adding H2
iii. removing H2O
iv. removing H2
Answer:
i. Adding CH4: Adding CH4 will favour the forward reaction and the yield of CO and H2 will increase.
ii. Adding H2: Adding H2 will favour the reverse reaction and the yield of CO and H2 will decrease.
iii. Removing H2O: Removing H2O will favour the reverse reaction and the yield of CO and H2 will decrease.
iv. Removing H2: Removing H2 will favour the forward reaction and the yield of CO and H2 will increase.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 36.
By using Le Chatelier’s principle, explain the effect of change in pressure (due to volume change) on the composition of equilibrium mixture.
Answer:
Change in pressure:
i. The change in pressure has no effect on the concentrations of solids and liquids. However, it appreciably affects the concentrations of gases.
From the ideal gas equation, PV = nRT or P = \(\frac{\mathrm{n}}{\mathrm{V}}\)RT
∴ P ∝ \(\frac{\mathrm{n}}{\mathrm{V}}\)
where, the ratio n/V is an expression for the concentration of the gas in mol dm-3.
ii. According to Le Chatelier’s principle at constant temperature, when pressure is increased, the equilibrium will shift in a direction in which the number of molecules decreases and when the pressure is decreased the equilibrium will shift in a direction in which the number of molecules increases.

[Note: For a reaction in which decrease in volume takes place, the reaction will be favoured by increasing pressure and for a reaction in which increase in volume takes place, the reaction will be favoured with lowering pressure, temperature being constant.]

Question 37.
An equilibrium mixture of dinitrogen tetroxide (colourless gas) and nitrogen dioxide (brown gas) is set up in a sealed flask at a particular temperature. Observe the effect of change of pressure on the gaseous equilibrium and complete the following table:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 26
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 27

Question 38.
By using Le Chatelier’s principle, explain the effect of change in pressure for the following equilibrium:
H2(g) + I2(g) ⇌ 2HI(g)
Answer:
As there is the same number of molecules of gas on both sides, change of pressure has no effect on the equilibrium.

Question 39.
Explain the effect of change in pressure due to volume change of the following equilibria:
i. 2NO(g) + Cl2(g) ⇌ 2NOCl(g)
ii. 2NO(g) ⇌ N2(g) + O2(g)
Answer:
i. 2NO(g) + Cl2(g) ⇌ 2NOCl(g)
In the forward reaction, the number of molecules decreases (3 to 2) and in the reverse reaction the number of molecules increases (2 to 3).
a. Effect of increase in pressure: According to Le Chatelier’s principle, when pressure is increased the forward reaction is favoured as the number of molecules decreases. Thus, when the pressure of the equilibrium system is increased at constant temperature by reducing the volume, the yield of NOCl increases.
b. Effect of decrease in pressure: When the pressure is decreased the equilibrium will shift from right to left. Therefore, the yield of NOCl will decrease.

ii. 2NO(g) ⇌ N2(g) + O2(g)
As both reactants and products have equal numbers of moles (or molecules), there is no effect of change in pressure (due to volume change) on the composition of the equilibrium mixture.

Question 40.
Explain the effect of change in temperature on the value of KC.
Answer:

  • The value of equilibrium constant is unaffected if temperature remains constant.
  • However, a change in temperature alters the value of equilibrium constant.
  • In a reversible reaction, one of the reactions is exothermic (heat is released) and the other is endothermic (heat is absorbed).
  • The value of equilibrium constant for an exothermic reaction decreases with increase in the temperature and that of endothermic reaction increases with the increase in temperature.

Question 41.
Explain the effect of change in temperature on the following equilibria:
CO(g) + 2H2(g) ⇌ CH3OH(g) ; ΔH = – 90 kJ
Answer:
i. The forward reaction is exothermic and reverse reaction is endothermic. According to Le Chatelier’s principle, when the temperature of the equilibrium mixture increases, the equilibrium shifts from right to left in endothermic direction. Therefore, the yield of CH3OH decreases at high temperature.

ii. When the temperature decreases, the forward exothermic reaction is favoured. Therefore, the yield of CH3OH increases at low temperature.
Thus, the decomposition of CH3OH into CO and H2 is favoured with increase in temperature, whereas formation of CH3OH is favoured with decrease in temperature.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 42.
By using Le Chatelier’s principle, explain the effect of addition of a catalyst on the composition of equilibrium mixture.
Answer:

  • When a catalyst is added to the equilibrium mixture, the rates of forward and reverse reactions increases to the same extent. Hence, the position of equilibrium remains unaffected.
  • A catalyst does not change the composition of equilibrium mixture. The equilibrium concentrations of reactants and products remain same and catalyst does not shift the equilibrium in favour of either reactants or products.
  • The value of equilibrium constant is also not affected by the presence of a catalyst.

[Note: A catalyst does not appear in the balanced chemical equation and in the equilibrium constant expression.]

Question 43.
Consider an esterification reaction:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 28
What will happen if H+ ions are added to the reaction mixture?
Answer:
H+ ions act as catalyst in the esterification reaction. Hence, the addition of H+ ions reduces the time for the completion of reaction.

Question 44.
Complete the following table that shows the shifts in the equilibrium position for the reaction:
N2O4(g) + Heat ⇌ 2NO2(g)
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 29
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 30

Question 45.
Summarize effects of following four factors on the position of equilibrium and value of KC:
i. Concentration
ii. Pressure
iii. Temperature
iv. Catalyst
Answer:

Effect of Position of equilibrium Value of KC
Concentration Changes No change
Pressure Changes if reaction involves change in number of gas molecules No change
Temperature Change Change
Catalyst No change No change

Question 46.
State TRUE or FALSE. Correct the false statement.
i. The value of equilibrium constant depends on temperature.
ii. If QC < KCC, the reaction will proceed from right to left consuming more product to attain equilibrium.
iii. Any change in the pressure of a gaseous reaction mixture at equilibrium, changes the value of KC.
iv. In a reversible reaction, the reverse reaction has an energy change that is equal and opposite to that of the forward reaction.
Answer:
i. True
ii. False
If QC > KC the reaction will proceed from right to left consuming more product to attain equilibrium.
iii. False
Any change in the pressure of a gaseous reaction mixture at equilibrium, does not change the value of KC.
iv. True

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 47.
Draw the flowchart showing the manufacture of NH3 by Haber process.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 31

Question 48.
Explain in short: The Haber process.
Answer:
Haber process:

  • The Haber process is the process of synthesis of ammonia gas by reacting together hydrogen gas and nitrogen gas in a particular stoichiometric ratio by volumes and at selected optimum temperature and pressure.
  • The chemical reaction is: \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \stackrel{\text { Catalyst }}{\rightleftharpoons} 2 \mathrm{NH}_{3(\mathrm{~g})}+\text { Heat }\)
    The reaction proceeds with a decrease in number of moles (Δn = -2) and the forward reaction is exothermic.
  • Iron (containing a small quantity of molybdenum) is used as catalyst.
  • The optimum temperature is about 773 K and the optimum pressure is about 250 atm.

Question 49.
Consider the reaction P(g) + Q(g) ⇌ PQ(g). Diagram ‘X’ represents the reaction at equilibrium.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 32
i. If each molecule (sphere) represents a partial pressure of 1 atm, calculate the value of KP.
ii. Predict the change in equilibrium, when the volume is increased by 50 percentage.
Answer:
i. For the given equilibrium mixture:

Chemical species P Q PQ
Partial pressure 4 6 7

KP = \(\frac{\mathrm{p}_{\mathrm{PQ}}}{\mathrm{p}_{\mathrm{p}} \times \mathrm{p}_{\mathrm{Q}}}=\frac{7}{4 \times 6}\) = 0.29
ii. Increasing the volume will shift the equilibrium position to the side with higher number of gaseous moles. In the given reaction, the equilibrium will shift to the left (toward reactant) resulting in an increase in the concentration of P and Q accompanied by a corresponding decrease in concentration of PQ.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

Multiple Choice Questions

1. Which of the following is expression of KC for
2NH3(g) ⇌ N2(g) + 3H2(g)?
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 33
Answer:
(A) \(\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}{\left[\mathrm{NH}_{3}\right]^{2}}\)

2. For the system 3A + 2B ⇌ C, the expression for equilibrium constant is …………..
Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium 34
Answer:
(D) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

3. For the reaction C(s) + CO2(g) ⇌ 2CO(g) the partial pressure of CO2 and CO are 4 and 8 atm, respectively, then KP for the reaction is ……………
(A) 16 atm
(B) 2 atm
(C) 5 atm
(D) 4 atm
Answer:
(A) 16 atm

4. The equilibrium constant value for the reaction:
2H2(g) + O2(g) ⇌ 2H2O(g) is 2.4 × 1047 at 500 K. What is the value of equilibrium constant for the reaction:
2H2O(g) ⇌ 2H2(g) + O2(g) ?
(A) 0.41 × 10-46
(B) 0.41 × 1047
(C) 0.41 × 10-48
(D) 0.41 × 10-47
Answer:
(D) 0.41 × 10-47

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

5. For the reaction CO(g) + Cl2(g) ⇌ COCl2(g), KP/KC is equal to ……………
(A) \(\frac{1}{\mathrm{RT}}\)
(B) RT
(C) \(\sqrt{\mathrm{RT}}\)
(D) 1.0
Answer:
(A) \(\frac{1}{\mathrm{RT}}\)

6. For which of the following reaction, KP = KC?
(A) PCl5(g) ⇌ PCl3(g) + Cl2(g)
(B) N2(g) + 3H2(g) ⇌ 2NH3(g)
(C) H2(g) + I2(g) ⇌ 2HI(g)
(D) 2NO2(g) ⇌ N2O4(g)
Answer:
(C) H2(g) + I2(g) ⇌ 2HI(g)

7. For the equilibrium reaction
2NO2(g) ⇌ N2O4(g) + 60.0 kJ, the increase in temperature ……………..
(A) favours the formation of N2O4
(B) favours the decomposition of N2O4
(C) does not affect the equilibrium
(D) stops the reaction
Answer:
(B) favours the decomposition of N2O4

8. The following reaction occurs in the blast furnace where iron ore is reduced to iron metal:
3Fe2O3(s) + 3CO(g) ⇌ 2Fe(l) + 3CO2(g)
Using the Le Chatelier’s principle, predict which one of the following will NOT disturb the equilibrium?
(A) Removal of CO
(B) Removal of CO2
(C) Addition of CO2
(D) Addition of Fe2O3
Answer:
(D) Addition of Fe2O3

Maharashtra Board Class 11 Chemistry Important Questions Chapter 12 Chemical Equilibrium

9. The reaction A + B ⇌ C + D + heat, has reached equilibrium. The reaction may be made to proceed forward by
(A) adding more C
(B) adding more D
(C) decreasing the temperature
(D) increasing the temperature
Answer:
(C) decreasing the temperature

10. Identify the CORRECT statement.
(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.
(B) The value of equilibrium constant decreases in presence of a catalyst.
(C) Catalyst affect the position of the equilibrium.
(D) Catalyst changes the equilibrium composition of a reaction mixture.
Answer:
(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.

11. The equilibrium constant for the reaction:
N2(g) + O2(g) ⇌ 2NO(g) is 4 × 10-4 at 2000 K. In presence of a catalyst, the equilibrium is attained ten times faster. Therefore, the equilibrium constant in presence of catalyst of 2000 K is …………..
(A) 40 × 10-4
(B) 4 × 10-2
(C) 4 × 10-3
(D) 4 × 10-4
Answer:
(D) 4 × 10-4

12. The rate of formation of NH3 can be increased by using catalyst …………….
(A) Fe + Co
(B) Mo + Fr
(C) Fe + Mo
(D) Fe + Mg
Answer:
(C) Fe + Mo

Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest रचना विज्ञापन लेखन Notes, Questions and Answers.

Maharashtra State Board 11th Hindi रचना विज्ञापन लेखन

विज्ञापन का सामान्य अर्थ है सूचना या विशिष्ट ज्ञापन वास्तव में आज की उपभोक्तावादी संस्कृति में यह विशेष महत्त्वपूर्ण है। इसका प्रभाव उपभोक्ता, विक्रेता तथा समाज के सभी वर्गों पर गहरा पड़ता है।

विज्ञापन का मुख्य उद्देश्य है –

  • उत्पाद की बिक्री बढ़ाना।
  • सामाजिक अथवा राजनीतिक अभियान को गति देना।
  • विद्यालयों / महाविद्यालयों में प्रवेश हेतु आवेदन-पत्र की जानकारी प्राप्त करना।
  • नाटक, संवाद, कहानी, सिनेमा आदि की जानकारी देना।
  • नौकरी देने / लेने हेतु जानकारी देना।

Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन

विज्ञापन के नमूने :

प्रश्न 1.
निम्नलिखित जानकारी के आधार पर विज्ञापन तैयार कीजिए।
Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन 1
उत्तर:
“घर किराए पर देना है”
500 वर्गफीट, वन बी-एच्.के का फ्लैट गोरेगाँव रेल स्थानक से पाँच मिनट की दूरी पर उपलब्ध है। स्कूल और अस्पताल निकट। जॉगर्स पार्क के बगल/पास में। 24 घंटे पानी की सुविधा।
Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन 2
संपर्क : अभय पांडेय।
मोबाईल : 98xxxxxxx
समय : सुबह 11 से शाम 6
पता : 203 / गजानन कॉलनी, गोरेगाँव (प.), मुंबई।

प्रश्न 2.
निम्नलिखित जानकारी के आधार पर विज्ञापन तैयार कीजिए।
Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन 3
उत्तर:
आवश्यकता है

रामानंद विद्यालय, चेंबूर नाका, चेंबूर, मुंबई 71 के लिए खुले प्रवर्ग के लिए एक हिंदी-मराठी विषय के शिक्षक सेवक की आवश्यकता है। प्रार्थी का प्रशिक्षित एवं हिंदी-मराठी विषय में स्नातक होना अनिवार्य है। अपने शैक्षणिक अनुभव एवं प्रमाणपत्रों की प्रतियों के साथ प्रधानाचार्य से मिले।

दिनांक : 7 और 8 अक्टूबर 2017.
समय : सुबह 10.00 से 3.00 बजे तक
भ्रमणध्वनि : 98xxxxx

Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन

प्रश्न 3.
निम्नलिखित जानकारी के आधार पर विज्ञापन तैयार कीजिए।
Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन 4
उत्तर :
आवश्यकता है ….
सोसायटी के बगीचे की देखभाल करने हेतु अनुभवी माली की आवश्यकता है।

  • पेड़- पौधों की जानकारी आवश्यक
  • सोसायटी कंपाऊंड में रहने की व्यवस्था
  • 10000 से 15000 प्रतिमाह तनख्वाह
  • निर्व्यसनी, ईमानदार माली अपने दो फोटो और आधार कार्ड के साथ संपर्क करें।

सेक्रेटरी.
हरगोविंद सोसायटी
रामनगर, वरली।
भ्रमणध्वनि: 90xxxxxx
केवल इतवार के दिन शाम 4 से 7 के बीच ही संपर्क कर सकते हैं।

प्रश्न 4.
स्वास्थ्यवर्धक पेय के विक्री हेतु विज्ञापन तैयार कीजिए।
खुशखबर! खुशखबर!! खुशखबर!!!
रोजाना नाश्ते के साथ सेवन करें
स्वास्थ्य की हर समस्या से निजात पाएँ

  • शुगर फ्री, मोटापा घटाए
  • कोई साईड इफेक्ट नहीं
  • त्वचा रखे सदाबहार
  • दाम भी कम

Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन

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Maharashtra Board Class 11 Hindi रचना विज्ञापन लेखन 5
प्रथम 100 ग्राहकों को एक पर एक मुफ्त
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