Category: Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1

Question 1.
Check whether the following sequences are G.P. If so, write t_n.
(i) 2, 6, 18, 54, ……
Solution:

(ii) 1, -5, 25, -125, ………
Solution:

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \cdots\)
Solution:

(iv) 3, 4, 5, 6, ……
Solution:

(v) 7, 14, 21, 28, ……
Solution:

Question 2.
For the G.P.
(i) If r = \(\frac{1}{3}\), a = 9, find t₇.
Solution:

(ii) If a = \(\frac{7}{243}\), r = 3, find t₆.
Solution:

(iii) If r = -3 and t₆ = 1701, find a.
Solution:

(iv) If a = \(\frac{2}{3}\), t₆ = 162, find r.
Solution:

Question 3.
Which term of the G. P. 5, 25, 125, 625, …… is 5¹⁰?
Solution:

Question 4.
For what values of x, the terms \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?
Solution:

Question 5.
If for a sequence, \(\mathrm{t}_{\mathrm{n}}=\frac{5^{n-3}}{2^{n-3}}\), show that the sequence is a G. P. Find its first term and the common ratio.
Solution:

Question 6.
Find three numbers in G. P. such that their sum is 21 and the sum of their squares is 189.
Solution:
Let the three numbers in G. P. be \(\frac{a}{r}\), a, ar.
According to the given conditions,


When a = 6, r = 2,
\(\frac{a}{r}\) = 3, a = 6, ar = 12
Hence, the three numbers in G.P. are 12, 6, 3 or 3, 6, 12.

Check:
If sum of the three numbers is 21 and sum of their squares is 189, then our answer is correct.
Sum of the numbers = 12 + 6 + 3 = 21
Sum of the squares of the numbers = 12² + 6² + 3²
= 144 + 36 + 9
= 189
Thus, our answer is correct.

Question 7.
Find four numbers in G. P. such that the sum of the middle two numbers is \(\frac{10}{3}\) and their product is 1.
Solution:
Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\)
According to the given conditions,

Question 8.
Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
Solution:
Let the five numbers in G. P. be
\(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\)
According to the given conditions,

Hence, the five numbers in G.P. are
1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.

Question 9.
The fifth term of a G. P. is x, the eighth term of a G.P. is y and the eleventh term of a G.P. is z, verify whether y² = xz.
Solution:

Question 10.
If p, q, r, s are in G.P., show that p + q, q + r, r + s are also in G. P.
Solution:
p, q, r, s are in G.P.
∴ \(\frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}}{\mathrm{q}}=\frac{\mathrm{s}}{\mathrm{r}}\)
Let \(\frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}}{\mathrm{q}}=\frac{\mathrm{s}}{\mathrm{r}}\) = k
∴ q = pk, r = qk, s = rk
We have to prove that p + q, q + r, r + s are in G.P.
i.e., to prove that \(\frac{\mathrm{q}+\mathrm{r}}{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{r}+\mathrm{s}}{\mathrm{q}+\mathrm{r}}\)

∴ p + q, q + r, r + s are in G.P.

Question 11.
The number of bacteria in a culture doubles every hour. If there were 50 bacteria originally in the culture, how many bacteria will be there at the end of the 5th hour?
Solution:
Since the number of bacteria in culture doubles every hour, increase in number of bacteria after every hour is in G.P.
∴ a = 50, r = \(\frac{100}{50}\) = 2
t_n = ar^n-1
To find the number of bacteria at the end of the 5th hour.
(i.e., to find the number of bacteria at the beginning of the 6th hour, i.e., to find t₆.)
∴ t₆ = ar⁵
= 50 × (2⁵)
= 50 × 32
= 1600

Question 12.
A ball is dropped from a height of 80 ft. The ball is such that it rebounds \(\left(\frac{3}{4}\right)^{\text {th }}\) of the height it has fallen. How high does the ball rebound on the 6th bounce? How high does the ball rebound on the nth bounce?
Solution:
Since the ball rebounds \(\left(\frac{3}{4}\right)^{\text {th }}\) of the height it has fallen, the height in successive bounce is in G.P.
1st height in the bounce = 80 × \(\frac{3}{4}\)

Question 13.
The numbers 3, x and x + 6 are in G. P. Find
(i) x
(ii) 20th term
(iii) nth term.
Solution:
(i) 3, x and x + 6 are in G. P.
\(\frac{x}{3}=\frac{x+6}{x}\)
x² = 3x + 18
x² – 3x – 18 = 0
(x – 6) (x + 3) = 0
x = 6, -3

Question 14.
Mosquitoes are growing at a rate of 10% a year. If there were 200 mosquitoes in the beginning, write down the number of mosquitoes after
(i) 3 years
(ii) 10 years
(iii) n years
Solution:
a = 200, r = 1 + \(\frac{10}{100}\) = \(\frac{11}{10}\)
Mosquitoes at the end of 1st year = 200 × \(\frac{11}{10}\)
(i) Number of mosquitoes after 3 years
= 200 × \(\frac{11}{10} \times\left(\frac{11}{10}\right)^{2}\)
= 200 \(\left(\frac{11}{10}\right)^{3}\)
= 200 (1.1)³

(ii) Number of mosquitoes after 10 years = 200 (1.1)¹⁰

(iii) Number of mosquitoes after n years = 200 (1.1)ⁿ

Question 15.
The numbers x – 6, 2x and x² are in G. P. Find
(i) x
(ii) 1st term
(iii) nth term
Solution:
(i) x – 6, 2x and x are in Geometric progression.
∴ \(\frac{2 x}{x-6}=\frac{x^{2}}{2 x}\)
4x² = x²(x – 6)
4 = x – 6
x = 10

(ii) t₁ = x – 6 = 10 – 6 = 4

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1

(I) Select the correct answer from the given alternatives.

Question 1.
If n is an odd positive integer, then the value of 1 + (i)²ⁿ + (i)⁴ⁿ + (i)⁶ⁿ is:
(A) -4i
(B) 0
(C) 4i
(D) 4
Answer:
(B) 0
Hint:
1 + (i²)ⁿ + (i⁴)ⁿ + (i²)³ⁿ
= 1 – 1 + 1 – 1 …..(n odd positive integer)
= 0

Question 2.
The value of \(\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}\) is equal to:
(A) -2
(B) 1
(C) 0
(D) -1
Answer:
(D) -1
Hint:

Question 3.
√-3 √-6 is equal to
(A) -3√2
(B) 3√2
(C) 3√2 i
(D) -3√2 i
Answer:
(A) -3√2
Hint:
√-3 √-6
= (√3 i) (√6 i)
= 3√2 (-1)
= -3√2

Question 4.
If ω is a complex cube root of unity, then the value of ω⁹⁹ + ω¹⁰⁰ + ω¹⁰¹ is:
(A) -1
(B) 1
(C) 0
(D) 3
Answer:
(C) 0
Hint:
ω⁹⁹ + ω¹⁰⁰ + ω¹⁰¹
= ω⁹⁹ (1 + ω + ω²)
= ω⁹⁹ (0)
= 0

Question 5.
If z = r(cos θ + i sin θ), then the value of \(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\) is
(A) cos 2θ
(B) 2cos 2θ
(C) 2cos θ
(D) 2sin θ
Answer:
(B) 2cos 2θ
Hint:

Question 6.
If ω(≠1) is a cube root of unity and (1 + ω)⁷ = A + Bω, then A and B are respectively the numbers
(A) 0, 1
(B) 1, 1
(C) 1, 0
(D) -1, 1
Answer:
(B) 1, 1
Hint:
(1 + ω)⁷
= (-ω²)⁷
= -ω¹⁴
= -ω²(ω³)⁴
= -ω²
= 1 + ω
A = 1, B = 1

Question 7.
The modulus and argument of (1 + i√3)⁸ are respectively
(A) 2 and \(\frac{2 \pi}{3}\)
(B) 256 and \(\frac{8 \pi}{3}\)
(C) 256 and \(\frac{2 \pi}{3}\)
(D) 64 and \(\frac{4 \pi}{3}\)
Answer:
(C) 256 and \(\frac{2 \pi}{3}\)
Hint:

Question 8.
If arg (z) = θ, then arg \(\overline{(\mathrm{z})}\) =
(A) -θ
(B) θ
(C) π – θ
(D) π + θ
Answer:
(A) -θ
Hint:
Let z = \(\mathrm{re}^{\mathrm{i} \theta}\), then \(\overline{\mathrm{z}}=\mathrm{r} \mathrm{e}^{-\mathrm{i} \theta}\)
∴ arg \(\overline{\mathbf{z}}\) = -θ.

Question 9.
If -1 + √3 i = \(\mathrm{re}^{\mathrm{i} \theta}\), then θ =
(A) –\(\frac{2 \pi}{3}\)
(B) \(\frac{\pi}{3}\)
(C) –\(\frac{\pi}{3}\)
(D) \(\frac{2 \pi}{3}\)
Answer:
(D) \(\frac{2 \pi}{3}\)
Hint:

Question 10.
If z = x + iy and |z – zi| = 1, then
(A) z lies on X-axis
(B) z lies on Y-axis
(C) z lies on a rectangle
(D) z lies on a circle
Answer:
(D) z lies on a circle
Hint:
|z – zi | = |z| |1 – i| = 1
∴ |z| = \(\frac{1}{\sqrt{2}}\)
∴ x² + y² = \(\frac{1}{2}\)

(II) Answer the following:

Question 1.
Simplify the following and express in the form a + ib.
(i) 3 + √-64
Solution:
3 + √-64
= 3 + √64 √-1
= 3 + 8i

(ii) (2i³)²
Solution:
(2i³)²
= 4i⁶
= 4(i²)³
= 4(-1)³
= -4 …..[∵ i² = -1]
= -4 + 0i

(iii) (2 + 3i) (1 – 4i)
Solution:
(2 + 3i)(1 – 4i)
= 2 – 8i + 3i – 12i²
= 2 – 5i – 12(-1) …..[∵ i² = -1]
= 14 – 5i

(iv) \(\frac{5}{2}\)i(-4 – 3i)
Solution:

(v) (1 + 3i)² (3 + i)
Solution:
(1 + 3i)² (3 + i)
= (1 + 6i + 9i²)(3 + i)
= (1 + 6i – 9)(3 + i) ……[∵ i² = -1]
= (-8 + 6i)(3 + i)
= -24 – 8i + 18i + 6i²
= -24 + 10i + 6(-1)
= -24 + 10i – 6
= -30 + 10i

(vi) \(\frac{4+3 i}{1-i}\)
Solution:

(vii) \(\left(1+\frac{2}{i}\right)\left(3+\frac{4}{i}\right)(5+i)^{-1}\)
Solution:

(viii) \(\frac{\sqrt{5}+\sqrt{3 i}}{\sqrt{5}-\sqrt{3} i}\)
Solution:

(ix) \(\frac{3 i^{5}+2 i^{7}+i^{9}}{i^{6}+2 i^{8}+3 i^{18}}\)
Solution:

(x) \(\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}\)
Solution:

Question 2.
Solve the following equations for x, y ∈ R
(i) (4 – 5i)x + (2 + 3i)y = 10 – 7i
Solution:
(4 – 5i)x + (2 + 3i)y = 10 – 7i
(4x + 2y) + (3y – 5x) i = 10 – 7i
Equating real and imaginary parts, we get
4x + 2y= 10 i.e., 2x + y = 5 ……(i)
and 3y – 5x = -7 ……(ii)
Equation (i) × 3 – equation (ii) gives
11x = 22
∴ x = 2
Putting x = 2 in (i), we get
2(2) + y = 5
∴ y = 1
∴ x = 2 and y = 1

(ii) \(\frac{x+i y}{2+3 i}\) = 7 – i
Solution:
\(\frac{x+i y}{2+3 i}\) = 7 – i
x + iy = (7 – i)(2 + 3i)
x + iy = 14 + 21i – 2i – 3i²
x + iy = 14 + 19i – 3(-1)
x + iy = 17 + 19i
Equating real and imaginary parts, we get
∴ x = 17 and y = 19

(iii) (x + iy) (5 + 6i) = 2 + 3i
Solution:

(iv) 2x + i⁹ y(2 + i) = x i⁷ + 10 i¹⁶
Solution:
2x + i⁹ y(2 + i) = x i⁷ + 10 i¹⁶
2x + (i⁴)² . i . y(2 + i) = x(i²)³ . i + 10 . (i⁴)⁴
2x + (1)² . iy(2 + i) = x(-1)³ . i + 10(1)⁴ ……..[∵ i² = -1, i⁴ = 1]
2x + 2yi + y i² = -xi + 10
2x + 2yi – y + xi = 10
(2x – y) + (x + 2y)i = 10 + 0 . i
Equating real and imaginary parts, we get
2x – y = 10 ……(i)
and x + 2y = 0 ……..(ii)
Equation (i) × 2 + equation (ii) gives, we get
5x = 20
∴ x = 4
Putting x = 4 in (i), we get
2(4) – y = 10
y = 8 – 10
∴ y = -2
∴ x = 4 and y = -2

Question 3.
Evaluate
(i) (1 – i + i²)^-15
Solution:

(ii) i¹³¹ + i⁴⁹
Solution:
i¹³¹ + i⁴⁹
= (i⁴)³² . i³ + (i⁴)¹² . i
= (1)³² (-i) + (1)¹² . i
= -i + i
= 0

Question 4.
Find the value of
(i) x³ + 2x² – 3x + 21, if x = 1 + 2i
Solution:

(ii) x⁴ + 9x³ + 35x² – x + 164, if x = -5 + 4i
Solution:

Question 5.
Find the square roots of
(i) -16 + 30i
Solution:
Let \(\sqrt{-16+30 \mathrm{i}}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
-16 + 30i = a² + b² i² + 2abi
-16 + 30i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = -16 and 2ab = 30
a² – b² = -16 and b = \(\frac{15}{a}\)

(ii) 15 – 8i
Solution:
Let \(\sqrt{15-8 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
15 – 8i = a² + b² i² + 2abi
15 – 8i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 15 and 2ab = -8
a² – b² = 15 and b = \(\frac{-4}{a}\)

When a = 4, b = \(\frac{-4}{4}\) = -1
When a = -4, b = \(\frac{-4}{-4}\) = 1
∴ \(\sqrt{15-8 i}\) = ±(4 – i)

(iii) 2 + 2√3 i
Solution:
Let \(\sqrt{2+2 \sqrt{3}}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
2 + 2√3 i = a² + b² i² + 2abi
2 + 2√3 i = a² – b² + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 2 and 2ab = 2√3
a² – b² = 2 and b = \(\frac{\sqrt{3}}{a}\)

(iv) 18i
Solution:
Let √18i = a + bi, where a, b ∈ R.
Squaring on both sides, we get
18i = a² + b² i² + 2abi
0 + 18i = a² – b² + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 0 and 2ab = 18
a² – b² = 0 and b = \(\frac{9}{a}\)
\(a^{2}-\left(\frac{9}{a}\right)^{2}=0\)
\(a^{2}-\frac{81}{a^{2}}=0\)
a⁴ – 81 = 0
(a² – 9) (a² + 9) = 0
a² = 9 or a² = -9
But a ∈ R
∴ a² ≠ -9
∴ a² = 9
∴ a = ± 3
When a = 3, b = \(\frac{9}{3}\) = 3
When a = -3, b = \(\frac{9}{-3}\) = -3
∴ √18i = ±(3 + 3i) = ±3(1 + i)

(v) 3 – 4i
Solution:
Let \(\sqrt{3-4 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
3 – 4i = a² + b² i² + 2abi
3 – 4i = a² – b² + 2abi ……[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 3 and 2ab = -4
a² – b² = 3 and b = \(\frac{-2}{a}\)

(vi) 6 + 8i
Solution:
Let \(\sqrt{6+8 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
6 + 8i = a² + b² i² + 2abi
6 + 8i = a² – b² + 2abi ……[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 6 and 2ab = 8

Question 6.
Find the modulus and argument of each complex number and express it in the polar form.
(i) 8 + 15i
Solution:

(ii) 6 – i
Solution:

(iii) \(\frac{1+\sqrt{3} \mathbf{i}}{2}\)
Solution:

(iv) \(\frac{-1-\mathbf{i}}{\sqrt{2}}\)
Solution:

(v) 2i
Solution:

(vi) -3i
Solution:

(vii) \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \mathbf{i}\)
Solution:

Question 7.
Represent 1 + 21, 2 – i, -3 – 2i, -2 + 3i by points in Argand’s diagram.
Solution:
The complex numbers 1 + 2i, 2 – i, -3 – 2i, -2 + 3i will be represented by the points A(1, 2), B(2, -1), C(-3, -2), D(-2, 3) respectively as shown below:

Question 8.
Show that z = \(\frac{5}{(1-i)(2-i)(3-i)}\) is purely imaginary number.
Solution:

Question 9.
Find the real numbers x and y such that \(\frac{x}{1+2 i}+\frac{y}{3+2 i}=\frac{5+6 i}{-1+8 i}\)
Solution:
\(\frac{x}{1+2 i}+\frac{y}{3+2 i}=\frac{5+6 i}{-1+8 i}\)

(3x + y) + 2(x + y)i = 5 + 6i
Equating real and imaginary parts, we get
3x + y = 5 ……(i)
and 2(x + y) = 6
i.e., x + y = 3 …….(ii)
Subtracting (ii) from (i), we get
2x = 2
∴ x = 1
Putting x = 1 in (ii), we get
1 + y = 3
∴ y = 2
∴ x = 1, y = 2

Question 10.
Show that \(\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{10}+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{10}=0\)
Solution:

Question 11.
Show that \(\left(\frac{1+i}{\sqrt{2}}\right)^{8}+\left(\frac{1-i}{\sqrt{2}}\right)^{8}=2\)
Solution:

Question 12.
Convert the complex numbers in polar form and also in exponential form.
(i) z = \(\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}\)
Solution:

(ii) z = -6 + √2 i
Solution:
z = -6 + √2 i
∴ a = -6, b = √2
i.e. a < 0, b > 0

(iii) \(\frac{-3}{2}+\frac{3 \sqrt{3} i}{2}\)
Solution:

Question 13.
If x + iy = \(\frac{a+i b}{a-i b}\), prove that x² + y² = 1.
Solution:

Question 14.
Show that z = \(\left(\frac{-1+\sqrt{-3}}{2}\right)^{3}\) is a rational number.
Solution:

Question 15.
Show that \(\frac{1-2 i}{3-4 i}+\frac{1+2 i}{3+4 i}\) is real.
Solution:

Question 16.
Simplify
(i) \(\frac{\mathrm{i}^{29}+\mathrm{i}^{39}+\mathrm{i}^{49}}{\mathrm{i}^{30}+\mathrm{i}^{40}+\mathrm{i}^{50}}\)
Solution:

(ii) \(\left(\mathrm{i}^{65}+\frac{1}{\mathrm{i}^{145}}\right)\)
Solution:

(iii) \(\frac{\mathrm{i}^{238}+\mathrm{i}^{236}+\mathrm{i}^{234}+\mathrm{i}^{232}+\mathrm{i}^{230}}{\mathrm{i}^{228}+\mathrm{i}^{226}+\mathrm{i}^{224}+\mathrm{i}^{222}+\mathrm{i}^{220}}\)
Solution:

Question 17.
Simplify \(\left[\frac{1}{1-2 i}+\frac{3}{1+i}\right]\left[\frac{3+4 i}{2-4 i}\right]\)
Solution:

Question 18.
If α and β are complex cube roots of unity, prove that (1 – α) (1 – β) (1 – α²) (1 – β²) = 9.
Solution:
α and β are the complex cube roots of unity.

Question 19.
If ω is a complex cube root of unity, prove that (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = 128.
Solution:
ω is the complex cube root of unity.
∴ ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω²
∴ L.H.S. = (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶
= [(1 + ω²) – ω]⁶ + [(1 + ω) – ω²]⁶
= (-ω – ω))⁶ + (-ω² – ω²)6
= (-2ω)⁶ + (-2ω²)⁶
= 64ω⁶ + 64ω¹²
= 64(ω³)² + 64(ω³)⁴
= 64(1)² + 64(1)⁴
= 128
= R.H.S.

Question 20.
If ω is the cube root of unity, then find the value of \(\left(\frac{-1+\mathbf{i} \sqrt{3}}{2}\right)^{18}+\left(\frac{-1-\mathbf{i} \sqrt{3}}{2}\right)^{18}\)
Solution:
If ω is the complex cube root of unity, then

Given Expression = ω¹⁸ + (ω²)¹⁸
= ω¹⁸ + ω³⁶
= (ω³)⁶ + (ω³)¹²
= (1)⁶ + (1)¹²
= 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4

Question 1.
Find the value of
(i) ω¹⁸
(ii) ω²¹
(iii) ω^-30
(iv) ω^-105
Solution:

Question 2.
If ω is the complex cube root of unity, show that
(i) (2 – ω)(2 – ω²) = 7
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
L.H.S. = (2 – ω)(2 – ω²)
= 4 – 2ω² – 2ω + ω³
= 4 – 2(ω² + ω) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
= R.H.S.

(ii) (1 + ω – ω²)⁶ = 64
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(iii) (1 + ω)³ – (1 + ω²)³ = 0
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(iv) (2 + ω + ω²)³ – (1 – 3ω + ω²)³ = 65
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(v) (3 + 3ω + 5ω²)⁶ – (2 + 6ω + 2ω²)³ = 0
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(vi) \(\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\) = ω²
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(vii) (a + b) + (aω + bω²) + (aω² + bω) = 0
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(viii) (a – b)(a – bω)(a – bω²) = a³ – b³
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(ix) (a + b)² + (aω + bω²)² + (aω²+ bω)² = 6ab
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

Question 3.
If ω is the complex cube root of unity, find the value of
(i) ω + \(\frac{1}{\omega}\)
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
\(\omega+\frac{1}{\omega}=\frac{\omega^{2}+1}{\omega}=\frac{-\omega}{\omega}=-1\)

(ii) ω² + ω³ + ω⁴
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
ω² + ω³ + ω⁴
= ω²(1 + ω + ω²)
= ω²(0)
= 0

(iii) (1 + ω²)³
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(1 + ω²)³
= (-ω)³
= -ω³
= -1

(iv) (1 – ω – ω²)³ + (1 – ω + ω²)³
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(1 – ω – ω²)³ + (1 – ω + ω²)³
= [1 – (ω + ω²)]³ + [(1 + ω²) – ω]³
= [1 – (-1)]² + (-ω – ω)³
= 2³ + (-2ω)³
= 8 – 8ω³
= 8 – 8(1)
= 0

(v) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
Solution:
ω is the complex cube root of unity.
ω³ = 1 and 1 + ω + ω² = 0
Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1
(1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
= (1 + ω)(1 + ω²)(1 + ω)(1 + ω²) …..[∵ ω³ = 1, ω⁴ = ω]
= (-ω²)(-ω)(-ω²)(-ω)
= ω⁶
= (ω³)²
= (1)²
= 1

Question 4.
If α and β are the complex cube roots of unity, show that
(i) α² + β² + αβ = 0
(ii) α⁴ + β⁴ + α^-1β^-1 = 0
Solution:
α and β are the complex cube roots of unity.

Question 5.
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are complex cube roots of unity, show that xyz = a³ + b³.
Solution:
x = a + b, y = αa + βb, z = aβ + bα
α and β are the complex cube roots of unity.
∴ α = \(\frac{-1+i \sqrt{3}}{2}\) and β = \(\frac{-1-i \sqrt{3}}{2}\)

Question 6.
Find the equation in cartesian coordinates of the locus of z if
(i) |z| = 10
Solution:
Let z = x + iy
|z| = 10
|x + iy| = 10
\(\sqrt{x^{2}+y^{2}}\) = 10
∴ x² + y² = 100

(ii) |z – 3| = 2
Solution:
Let z = x + iy
|z – 3| = 2
|x + iy – 3| = 2
|(x – 3) + iy| = 2
\(\sqrt{(x-3)^{2}+y^{2}}\) = 2
∴ (x – 3)² + y² = 4

(iii) |z – 5 + 6i| = 5
Solution:
Let z = x + iy
|z – 5 + 6i| = 5
|x + iy – 5 + 6i| = 5
|(x – 5) + i(y + 6)| = 5
\(\sqrt{(x-5)^{2}+(y+6)^{2}}\) = 5
∴ (x – 5)² + (y + 6)² = 25

(iv) |z + 8| = |z – 4|
Solution:
Let z = x + iy
|z + 8| = |z – 4|
|x + iy + 8| = |x + iy – 4|
|(x + 8) + iy | = |(x – 4) + iy|
\(\sqrt{(x+8)^{2}+y^{2}}=\sqrt{(x-4)^{2}+y^{2}}\)
(x + 8)² + y² = (x – 4)² + y²
x² + 16x + 64 + y² = x² – 8x + 16 + y²
16x + 64 = -8x + 16
24x + 48 = 0
∴ x + 2 = 0

(v) |z – 2 – 2i | = |z + 2 + 2i|
Solution:
Let z = x + iy
|z – 2 – 2i| = |z + 2 + 2i|
|x + iy – 2 – 2i | = |x + iy + 2 + 2i |
|(x – 2) + i(y – 2)| = |(x + 2) + i(y + 2)|
\(\sqrt{(x-2)^{2}+(y-2)^{2}}=\sqrt{(x+2)^{2}+(y+2)^{2}}\)
(x – 2)² + (y – 2)² = (x + 2)² + (y + 2)²
x²– 4x + 4 + y² – 4y + 4 = x² + 4x + 4 + y² + 4y + 4
-4x – 4y = 4x + 4y
8x + 8y = 0
x + y = 0
y = -x

(vi) \(\frac{|z+3 i|}{|z-6 i|}=1\)
Solution:
Let z = x + iy

x² + (y + 3)² = x² + (y – 6)²
y² + 6y + 9 = y² – 12y + 36
18y – 27 = 0
2y – 3 = 0

Question 7.
Use De Moivre’s theorem and simplify the following:
(i) \(\frac{(\cos 2 \theta+i \sin 2 \theta)^{7}}{(\cos 4 \theta+i \sin 4 \theta)^{3}}\)
Solution:

(ii) \(\frac{\cos 5 \theta+i \sin 5 \theta}{(\cos 3 \theta-i \sin 3 \theta)^{2}}\)
Solution:

(iii) \(\frac{\left(\cos \frac{7 \pi}{13}+i \sin \frac{7 \pi}{13}\right)^{4}}{\left(\cos \frac{4 \pi}{13}-i \sin \frac{4 \pi}{13}\right)^{6}}\)
Solution:

Question 8.
Express the following in the form a + ib, a, b ∈ R, using De Moivre’s theorem.
(i) (1 – i)⁵
Solution:

(ii) (1 + i)⁶
Solution:

(iii) (1 – √3 i)⁴
Solution:

(iv) (-2√3 – 2i)⁵
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3

Question 1.
Find the modulus and amplitude for each of the following complex numbers:
(i) 7 – 5i
Solution:
Let z = 7 – 5i
a = 7, b = -5
i.e. a > 0, b < 0
|z| = \(\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+(-5)^{2}}=\sqrt{49+25}=\sqrt{74}\)
Here, (7, -5) lies in 4th quadrant.

(ii) √3 + √2 i
Solution:
Let z = √3 + √2 i
a = √3, b = √2,
i.e. a > 0, b > 0

(iii) -8 + 15i
Solution:
Let z = -8 + 15i
a = -8, b = 15 , i.e. a < 0, b > 0

(iv) -3(1 – i)
Solution:
Let z = -3(1 – i) = -3 + 3i
a = -3, b = 3 , i.e. a < 0, b > 0
|z| = \(\sqrt{a^{2}+b^{2}}=\sqrt{(-3)^{2}+3^{2}}=\sqrt{9+9}\) = 3√2
Here, (-3, 3) lies in 2nd quadrant.
amp(z) = π – \(\tan ^{-1}\left|\frac{b}{a}\right|\)

(v) -4 – 4i
Solution:
Let z = -4 – 4i
a = -4, b = -4 , i.e. a < 0, b < 0

(vi) √3 – i
Solution:
Let z = √3 – i
a = √3, b = -1, i.e. a > 0, b < 0

(vii) 3
Solution:
Let z = 3 + 0i
a = 3, b = 0
z is a real number, it lies on the positive real axis.
|z|= \(\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+0^{2}}=\sqrt{9+0}\) = 3
and amp (z) = 0

(viii) 1 + i
Solution:
Let z = 1 + i
a = 1, b = 1, i.e. a > 0, b > 0
|z| = \(\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}\)
Here, (1, 1) lies in 1st quadrant.
amp (z) = \(\tan ^{-1}\left(\frac{b}{a}\right)=\tan ^{-1}(1)=\frac{\pi}{4}\)

(ix) 1 + i√3
Solution:
Let z = 1 + i√3
a = 1, b = √3, i.e. a > 0, b > 0
|z| = \(\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2\)
Here, (1, √3) lies in 1st quadrant.
amp (z) = \(\tan ^{-1}\left(\frac{b}{a}\right)=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}\)

(x) (1 + 2i)² (1 – i)
Solution:
Let z = (1 + 2i)² (1 – i)
= (1 + 4i + 4i²) (1 – i)
= [1 + 4i + 4(-1)] (1 – i) ….[∵ i² = -1]
= (-3 + 4i) (1 – i)
= -3 + 3i + 4i – 4i²
= -3 + 7i – 4(-1)
= -3 + 7i + 4
∴ z = 1 + 7i
∴ a = 1, b = 7, i. e. a > 0, b > 0
∴ |z| = \(\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}=\sqrt{1^{2}+7^{2}}=\sqrt{1+49}=5 \sqrt{2}\)
Here, (1, 7) lies in 1st quadrant.
∴ amp(z) = \(\tan ^{-1}\left(\frac{b}{a}\right)=\tan ^{-1}(7)\)

Question 2.
Find real values of θ for which \(\left(\frac{4+3 i \sin \theta}{1-2 i \sin \theta}\right)\) is purely real.
Solution:

Question 3.
If z = 3 + 5i, then represent the z, \(\overline{\mathbf{z}}\), -z, \(\overline{\mathbf{-z}}\) in Argand’s diagram.
Solution:
z = 3 + 5i
\(\overline{\mathbf{z}}\) = 3 – 5i
-z = – 3 – 5i
\(\overline{\mathbf{-z}}\)= -3 + 5i
The above complex numbers will be represented by the points
A (3, 5), B (3, -5), C (-3, -5) , D (-3, 5) respectively as shown below:

Question 4.
Express the following complex numbers in polar form and exponential form.
(i) -1 + √3 i
Solution:
Let z = – 1 + √3
a = -1, b = √3

(ii) -i
Solution:
Let z = -i = 0 – i
a = 0, b = -1
z lies on negative imaginary Y-axis.
|z| = r = \(\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}=\sqrt{0^{2}+(-1)^{2}}\) = 1 and
θ = amp z = 270° = \(\frac{3 \pi}{2}\)
The polar form of z = r (cos θ + i sin θ)
= 1 (cos 270° + i sin 270°)
= 1 (cos \(\frac{3 \pi}{2}\) + i sin \(\frac{3 \pi}{2}\))
The exponential form of z = \(r e^{i \theta}=e^{\frac{3 \pi}{2} i}\)

(iii) -1
Solution:
Let z = -1 = -1 + 0.i
a = -1, b = 0
z lies on negative real X-axis.
|z| = r = \(\sqrt{a^{2}+b^{2}}=\sqrt{(-1)^{2}+0^{2}}\) = 1 and
θ = amp z = 180° = π
The polar form of z = r (cos θ + i sin θ)
= 1 (cos 180° + i sin 180°)
= 1 (cos π + i sin π)
The exponential form of z = \(r e^{i \theta}=e^{\pi i}\)

(iv) \(\frac{1}{1+i}\)
Solution:

(v) \(\frac{1+2 i}{1-3 i}\)
Solution:

(vi) \(\frac{1+7 \mathbf{i}}{(2-\mathbf{i})^{2}}\)
Solution:

Question 5.
Express the following numbers in the form x + iy:
(i) \(\sqrt{3}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\)
Solution:

(ii) \(\sqrt{2} \cdot\left(\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right)\)
Solution:

(iii) \(7\left(\cos \left(-\frac{5 \pi}{6}\right)+i \sin \left(-\frac{5 \pi}{6}\right)\right)\)
Solution:

(iv) \(e^{\frac{\pi}{3} i}\)
Solution:

(v) \(e^{\frac{-4 \pi}{3} i}\)
Solution:

(vi) \([latex]e^{\frac{5 \pi}{6} i}\)[/latex]
Solution:

Question 6.
Find the modulus and argument of the complex number \(\frac{1+2 i}{1-3 i}\).
Solution:

Question 7.
Convert the complex number \(\mathrm{z}=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\) in the polar form.
Solution:

Question 8.
For z = 2 + 3i, verify the following:
(i) \(\overline{(\bar{z})}=z\)
Solution:
z = 2 + 3i
∴ \(\bar{z}\) = 2 – 3i
∴ \(\overline{\bar{z}}\) = 2 + 3i = z

(ii) \(\overline{z \bar{z}}=|z|^{2}\)
Solution:
z\(\bar{z}\) = (2 + 3i) (2 – 3i)
= 4 – 9i²
= 4 – 9(-1) …..[∵ i² = -1]
= 13
|z|² = \(\left(\sqrt{2^{2}+3^{2}}\right)^{2}\)
= 2² + 3²
= 4 + 9
= 13
∴ \(\overline{z \bar{z}}=|z|^{2}\)

(iii) (z + \(\bar{z}\)) is real
Solution:
(z + \(\bar{z}\)) = (2 + 3i) + (2 – 3i)
= 2 + 3i + 2 – 3i
= 4, which is a real number.
∴ z + \(\bar{z}\) is real.

(iv) z – \(\bar{z}\) = 6i
Solution:
z – \(\bar{z}\) = (2 + 3i) – (2 – 3i)
= 2 + 3i – 2 + 3i
= 6i

Question 9.
z₁ = 1 + i, z₂ = 2 – 3i, verify the following:
(i) \(\overline{Z_{1}+Z_{2}}=\overline{Z_{1}}+\overline{Z_{2}}\)
Solution:

(ii) \(\overline{Z_{1}-Z_{2}}=\overline{Z_{1}}-\overline{Z_{2}}\)
Solution:

(iii) \(\overline{Z_{1} \cdot Z_{2}}=\overline{Z_{1}} \cdot \overline{Z_{2}}\)
Solution:

(iv) \(\overline{\left(\frac{\mathbf{z}_{1}}{\mathbf{z}_{2}}\right)}=\frac{\overline{\mathbf{z}}_{1}}{\overline{\mathbf{z}}_{2}}\)
Solution:

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

ii. We know that, cos² θ = \(\frac{1+\cos 2 \theta}{2}\)
Substituting θ = \(\frac{\pi}{8}\), we get

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

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.

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

vi. \(\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}\) = 2 tan 2x
Solution:

 

vii. \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 x}}}\) = 2 cos x
Solution:

= 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:

x. [latex]\frac{\cos x}{1+\sin x}=\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1}\)
Solution:

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:

xii. \(\frac{1}{\tan 3 \mathbf{A}-\tan A}-\frac{1}{\cot 3 A-\cot A}\) = cot 2A
Solution:

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:

xv. \(\frac{2 \cos 4 x+1}{2 \cos x+1}\) = (2 cos x – 1)(2 cos 2x – 1)
Solution:

xvi. = cos² x + cos² (x + 120°) + cos²(x – 120°) = \(\frac{3}{2}\)
Solution:
L.H.S = cos² x + cos² (x + 120°) + cos²(x – 120°) =

\(\frac{3}{2}+\frac{1}{2}\) [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.

xvii. 2 cosec 2x + cosec x = sec cot \(\frac{x}{2}\)
Solution:

xviii. 4 cos x cos (\(\frac{\pi}{3}\) + x) cos (\(\frac{\pi}{3}\) – x) = cos 3x
\(\)
Solution:

= 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.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.2

Question 1.
Find the centre and radius of each of the following circles:
(i) x² + y² – 2x + 4y – 4 = 0
(ii) x² + y² – 6x – 8y – 24 = 0
(iii) 4x² + 4y² – 24x – 8y – 24 = 0
Solution:
(i) Given equation of the circle is x² + y² – 2x + 4y – 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 4 and c = -4
⇒ g = -1, f = 2 and c = -4
Centre of the circle = (-g, -f) = (1, -2)
and radius of the circle

(ii) Given equation of the circle is x² + y² – 6x – 8y – 24 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -8 and c = -24
⇒ g = -3, f = -4 and c = -24
Centre of the circle = (-g, -f) = (3, 4)
and radius of the circle

(iii) Given equation of the circle is 4x² + 4y² – 24x – 8y – 24 = 0
Dividing throughout by 4, we get x² + y² – 6x – 2y – 6 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -2 and c = -6
⇒ g = -3, f = -1 and c = -6
Centre of the circle = (-g, -f) = (3, 1)
and radius of the circle

Question 2.
Show that the equation 3x² + 3y² + 12x + 18y – 11 = 0 represents a circle.
Solution:
Given equation is 3x² + 3y² + 12x + 18y – 11 = 0
Dividing throughout by 3, we get
x² + y² + 4x + 6y – \(\frac{11}{3}\) = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 4, 2f = 6, c = \(\frac{-11}{3}\)
⇒ g = 2, f = 3, c = \(\frac{-11}{3}\)
Now, g² + f² – c = (2)² + (3)² – (\(\frac{-11}{3}\))
= 4 + 9 + \(\frac{11}{3}\)
= \(\frac{50}{3}\) > 0
∴ The given equation represents a circle.

Question 3.
Find the equation of the circle passing through the points (5, 7), (6, 6), and (2, -2).
Solution:
Let C(h, k) be the centre of the required circle.
Since the required circle passes through points A(5, 7), B(6, 6), and D(2, -2),
CA = CB = CD = radius

Consider, CA = CD
By distance formula,
\(\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-7)^{2}}=\sqrt{(\mathrm{h}-2)^{2}+[\mathrm{k}-(-2)]^{2}}\)
Squaring both the sides, we get
⇒ (h – 5)² + (k – 7)² = (h – 2)² + (k + 2)²
⇒ h² – 10h + 25 + k² – 14k + 49 = h² – 4h + 4 + k² + 4k + 4
⇒ -10h – 14k + 74 = -4h + 4k + 8
⇒ 6h + 18k – 66 = 0
⇒ h + 3k – 11 = 0 …..(i)
Consider, CB = CD
By distance formula,
\(\sqrt{(h-6)^{2}+(k-6)^{2}}=\sqrt{(h-2)^{2}+[k-(-2)]^{2}}\)
Squaring both the sides, we get
⇒ (h – 6)² + (k – 6)² = (h – 2)² + (k + 2)²
⇒ h² – 12h + 36 + k² – 12k + 36 = h² – 4h + 4 + k² + 4k + 4
⇒ -12h – 12k + 72 = -4h + 4k + 8
⇒ 8h + 16k – 64 = 0
⇒ h + 2k – 8 = 0 ……(ii)
By (i) – (ii), we get k = 3
Substituting k = 3 in (i), we get
h + 3(3) – 11 = 0
⇒ h + 9 – 11 = 0
⇒ h = 2
Centre of the circle is C(2, 3).
radius (r) = CD
= \(\sqrt{(2-2)^{2}+(3+2)^{2}}\)
= \(\sqrt{0+5^{2}}\)
= √25
= 5
The equation of a circle with centre at (h, k) and radius r is given by (x – h)²+ (y – k)² = r²
Here, h = 2, k = 3
The required equation of the circle is
(x – 2)² + (y – 3)² = 5²
⇒ x² – 4x + 4 + y² – 6y + 9 = 25
⇒ x² + y² – 4x – 6y + 4 + 9 – 25 = 0
⇒ x² + y² – 4x – 6y – 12 = 0

Question 4.
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.
Solution:
Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be
x² + y² + 2gx + 2fy + c = 0 …..(i)
For point (3, -2),
Substituting x = 3 and y = -2 in (i), we get
9 + 4 + 6g – 4f + c = 0
⇒ 6g – 4f + c = -13 ….(ii)
For point (1, 0),
Substituting x = 1 andy = 0 in (i), we get
1 + 0 + 2g + 0 + c = 0
⇒ 2g + c = -1 ……(iii)
For point (-1, -2),
Substituting x = -1 and y = -2, we get
1 + 4 – 2g – 4f + c = 0
⇒ 2g + 4f – c = 5 …….(iv)
Adding (ii) and (iv), we get
8g = -8
⇒ g = -1
Substituting g = -1 in (iii), we get
-2 + c = -1
⇒ c = 1
Substituting g = -1 and c = 1 in (iv), we get
-2 + 4f – 1 = 5
⇒ 4f = 8
⇒ f = 2
Substituting g = -1, f = 2 and c = 1 in (i), we get
x² + y² – 2x + 4y + 1 = 0 ……….(v)
If (1, -4) satisfies equation (v), the four points are concyclic.
Substituting x = 1, y = -4 in L.H.S of (v), we get
L.H.S. = (1)² + (-4)² – 2(1) + 4(-4) + 1
= 1 + 16 – 2 – 16 + 1
= 0
= R.H.S.
Point (1, -4) satisfies equation (v).
∴ The given points are concyclic.

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest व्याकरण मुहावरे Notes, Questions and Answers.

Maharashtra State Board 11th Hindi व्याकरण मुहावरे

भाषा को स्पष्ट और प्रभावशाली बनाने के लिए मुहावरों का प्रयोग किया जाता है। मुहावरा ऐसा वाक्यांश होता है जो सामान्य अर्थ से भिन्न किसी विशेष अर्थ का बोध कराता है। उसके अंत में प्राय: किसी क्रिया का सामान्य रूप लगा होता है। इनके प्रयोग से भाषा में सरसता, सुंदरता और स्वाभाविकता आती है।

मुहावरों की विशेषताएँ :

1. मुहावरे लोक जीवन की धरोहर हैं।
2. इनके अंत में प्राय: ‘ना’ होता है।
3. मुहावरे पूर्ण वाक्य नहीं होते।
4. मुहावरों के अर्थ प्रकट करने के लिए क्रियापद का विशेष महत्त्व होता है।
5. मुहावरे भाषा में कलात्मक अभिव्यक्ति की एक शैली है।
6. अन्य भाषा में मुहावरों का शाब्दिक अनुवाद नहीं हो सकता।
7. वाक्य में प्रयुक्त होने पर मुहावरों के शब्दों में रूपांतर हो जाता है। क्रिया लिंग, वचन, कारक आदि के अनुसार बदल जाती है। मुहावरे वाक्य में सरसता, विलक्षणता, तीखापन और प्रवाह उत्पन्न करते हैं। इससे हमारी अभिव्यक्ति में निखार आता है।

मुहावरों के प्रयोग में सावधानी :

• मुहावरों का वाक्यों में प्रयोग करते समय इनके लाक्षणिक अर्थ की पूर्ण जानकारी होनी चाहिए अन्यथा अर्थ के अनर्थ होने की संभावना रहती है।
• मुहावरे ज्यों के त्यों वाक्य में प्रयुक्त नहीं होते इसलिए प्रयोग के अनुसार उसके लिंग, वचन, कारक के अनुसार क्रिया में परिवर्तन करना चाहिए।

पाठ में प्रयुक्त मुहावरे तथा उनके वाक्य प्रयोग :

अंकुर जमाना : प्रारंभ करना
वाक्य : भाई के मन में कपट का अंकुर ऐसा जम गया था कि अब वह वृक्ष बन गया था।

अपने पैरों पर खड़ा होना : आत्मनिर्भर होना।
वाक्य : पढ़-लिखकर सीया अपने पैरों पर खड़ा होना चाहती है।

आँच न आने देना : संकट न आने देना।
वाक्य : गरीबी में भी माता-पिता ने अपने बच्चों पर आँच न आने दी।

आँखों में सैलाब उमड़ना : फूट-फूटकर रोना।
वाक्य : पति की मृत्यु पर पत्नी की आँखों में सैलाब उमड़ आया था।

आँखें फटी रहना : आश्चर्यचकित रह जाना।
वाक्य : बालक कृष्ण के मुख में ब्रह्मांड को देखकर यशोदा मैया की आँखें फटी रह गईं।

आईने में मुँह देखना : अपनी योग्यता जाँचना।
वाक्य : आईने में मुँह देखकर काम करना चाहिए ताकि सफलता का फल प्राप्त हो।

आसमान के तारे तोड़ना : असंभव कार्य करना।
वाक्य : यह प्रतियोगिता जीतकर भार्गव ने आसमान के तारे तोड लाए हैं।

ईंट का जवाब पत्थर से देना : कड़ा जवाब देना।
वाक्य : हमारी टीम ने खेल जीतने के लिए ईंट का जवाब पत्थर से दिया।

उधेड़ वुन में लगना : सोच-विचार करना।
वाक्य : पैसों की उधेड-बून में लगे लोग जीवन का मजा नहीं उठा पाते।

एक आँख से देखना : सामान्य रूप से देखना, पक्षपात न करना।
वाक्य : माँ अपने सभी बच्चों को एक आँख से देखती है।

एक और एक ग्यारह होना : एकता में बल होना।
वाक्य : जब दोनों भाई एक और एक ग्यारह हो गए तो उनका बुरा चाहने वाले उनका कुछ नहीं बिगाड़ सके।

कदम बढ़ाना : प्रगति करना।
वाक्य : समस्या को पीछे छोड़कर कदम बढाना जीवन का सही मार्ग है।

कमर कसना : पूरी तरह तैयार होना।
वाक्य : बरसाती समस्याओं से निपटने के लिए हमने बरसात आने से पहले ही कमर कस ली है।

कमर सीधी करना : आराम करना, सुस्ताना।
वाक्य : इतना पसीना बहाने के बाद कमर सीधी करने का मौका मिला तो नई समस्या खड़ी हो गई।

कलई खुलना : भेद प्रकट होना, राज या रहस्य खुलना।
वाक्य : कोई कितना भी धूर्त क्यों न हो एक न एक दिन उसकी कलई खुल जाती है।

कान देना : ध्यान से सुनना।
वाक्य : अध्यापक की बात पर विद्यार्थी कान देंगे तो सफलता अवश्य मिलेगी।

किस्मत खुलना : भाग्य चमकना।
वावय : आज तो मेरी किस्मत खुल गई जो आपके दर्शन हुए।

गले का हार होना : अत्यंत प्रिय होना।
वाक्य : छोटा शेख घर में सभी के गले का हार था।

गागर में सागर भरना : थोड़े में बहुत कहना।
वाक्य : बिहारी जी ने अपने दोहों में गागर में सागर भर दिया है इस बात को सभी हिंदी प्रेमियों ने स्वीकारा है।

घी के दीये जलाना : खुशी मनाना।
वाक्य : जब श्रीराम जी 14 वर्ष के वनवास के बाद अयोध्या लौटे तो अयोध्या वासियों ने घी के दीये जलाए।

चिकना घड़ा होना : निर्लज्ज होना, किसी बात का असर न होना।
वाक्य : रमेश को समझाना बेकार है क्योंकि वह तो चिकना घड़ा है।

चुटकी लेना : व्यंग्य करना।
वाक्य : चुटकी लेने की आदत कभी-कभी भारी पड़ जाती है।

जबान देना : वचन देना।
वाक्य : रमेश ने अगर जबान दी है तो वह जरूर निभाएगा।

झंडे गाड़ना : पूर्ण रूप से प्रभाव जमाना।
वाक्य : छोटी उम्र में ही शिवाजी महाराज ने 12 मावलों के साथ मुगलो के आधे किले पर झंडे गाड़ दिए थे।

डंका पीटना : प्रचार करना।
वाक्य : अपनी छोटी सी सफलता का भी डंका पीटने में सीया पीछे नहीं हटती।

तितर-बितर होना : बिखर जाना।
वाक्य : माँ की मृत्यु के बाद परिवार तितर-बितर हो गया।

हजारों दीप जल उठना : आनंदित हो उठना।
वाक्य : विदेश जाने के लिए वीजा मिल गया तो रमेश के मन में हजारों दीप जल उठे।

रुपये दाँत से पकड़ना : कंजूसी करना।
वाक्य : इस महँगाई के दौर में हर कोई रुपये दाँत से पकडकर जी रहा है।

दूध का दूध, पानी का पानी करना : इंसाफ करना, न्याय करना।
वाक्य : रंगे हाथ पकड़े जाने पर सच्चाई सबके सामने आ गई और दूध का दूध और पानी का पानी हो गया।

नाम कमाना : यश प्राप्त करना।
वाक्य : कड़ी मेहनत करके राज ने नाम कमाया इसलिए सब उसकी इज्जत करते हैं।

पाँचों उँगलियाँ घी में होना : हर तरफ से लाभ होना।
वाक्य : अब बेटा भी बराबरी से काम करने लगा तो लाला जी की पाँचो उँगलियाँ घी में है।

फला न समाना : अत्यधिक प्रसन्न होना।
वाक्य : मनोकामना पूरी होने पर सीया फूली न समाई।

वीडा उठाना : किसी काम को करने की ठान लेना।
वाक्य : देश के नागरिकों को पर्यावरण सुरक्षा का बीड़ा उठाना होगा।

वाँछे खिलना : अत्यधिक प्रसन्न होना।
वाक्य : चुनाव जीतने के बाद नेता की बाँछे खिल उठीं।

मरजीवा होना : कठोर साधना से लक्ष्य तक पहुँचने वाला होना।
वाक्य : अलवर में सात नदियों को जीवित कर श्री राजेंद्र सिंह जी मरजीवा हो गए।

मल्हार गाना : आनंद मनाना।
वाक्य : समय पर बारिश होने से किसान मल्हार गाने लगे।

राई का पहाड़ बनाना : बात को बढ़ा-चढ़ाकर कहना।
ताक्य : रमेश ने बात को इस ढंग से बताया कि राई का पहाड बन गया।

लोहा मानना : श्रेष्ठता स्वीकार करना।
वाक्य : औरंगजेब भी शिवाजी के युद्ध कौशल का लोहा मानता था।

सफेद झूठ बोलना : पूरी तरह से झूठ बोलना।
वाक्य : दुष्ट प्रवृत्ति के लोग सफेद झूठ बोलने से बाज नहीं आते।

सिर खपाना : ऐसे काम में समय लगाना जिसमें कोई लाभ नहीं।
वाक्य : सुबह से शाम तक सिर खपाते रहे लेकिन पिताजी ने दी पहेली हल नहीं कर पाए।

सिर पर सेहरा बाँधना : अधिक यश प्राप्त करना।
वाक्य : काव्य गायन प्रतियोगिता में रमेश केवल सफल ही नहीं हुआ बल्कि उसके सिर पर सेहरा बँधा।

सोना उगलना : बहुत अधिक लाभ होना।
वाक्य : मेरे देश की मिट्टी ऐसी उपजाऊ है कि सोना उगलती है।

सौ वात की एक वात : असली बात, निचोड़।
वाक्य : सौ बात की एक बात कहूँ, मुझे बेटा-बेटी में भेदभाव बिलकुल पसंद नहीं।

हाथ-पैर मारना : बहुत प्रयत्न करना।
वाक्य : इधर-उधर हाथ-पैर मारने के बाद मेरा लोन सेंक्शन हुआ।

हौसले बुलंद होना : उत्साह बने रहना।
वाक्य : शरीर कमजोर हो गया है लेकिन अभी भी राय साहब के हौसले बुलंद हैं।

श्रीगणेश करना : कार्य आरंभ करना।
वाक्य : दो पैसे जमा होते ही रमेश ने अपने व्यवसाय का श्रीगणेश किया।

दाँतों तले उँगली दबाना : आश्चर्यचकित होना।
वाक्य : रणभूमि में अभिमन्यु की वीरता देखकर कौरवों ने दाँतों तले उँगली दबाई।

अंधे की लाठी होना : निराधार का सहारा बनाना।
वाक्य : मदर टेरेसा भारत आकर अंधे की लाठी बनकर अपना कार्य करने लगी।

आग से खेलना : मुसीबत मोल लेना।
वाक्य : आज़ादी की लड़ाई लड़ते समय आग से खेलकर कई देशवासियों ने अपना घर-परिवार दाँव पर लगा दिया था।

मुट्ठी गर्म करना : रिश्वत देना।
वाक्य : भ्रष्टाचार की जड़ें इतनी गहराई तक पहुँच गई हैं कि जब तक मुट्ठी गर्म न करो कोई काम ही नहीं करता।

इतिश्री होना : समाप्त होना।
वाक्य : 15 अगस्त 1947 को देश आज़ाद हुआ और अंग्रेज शासन की इतिश्री हुई।

उड़ती चिड़िया पहचानना : तीक्ष्ण बुद्धि वाला होना।
वाक्य : बीरबल उडती चिडिया पहचान लेते थे और हर समस्या को सुलझाने में अकबर की सहायता करते है।

हथेली पर सरसों जमाना : कठिन कार्य करना।
वाक्य : दुश्मनों की छावनी में जाकर उनके भेद जानना मतलब हथेली पर सरसों जमाना है।

कंचन बरसना : धन-दौलत से परिपूर्ण होना।
वाक्य : कभी हमारे देश में कंचन बरसता था परंतु विदेशी आक्रमण ने इसे खोखला कर दिया।

कानों कान खबर न होना : बिल्कुल पता न चलना।
वाक्य : सेठ जी ने बेटी का विवाह कर दिया लेकिन किसी को कानों कान खबर न हुई।

गाल बजाना : अपनी प्रशंसा आप करना।
वाक्य : मोहन अपनी सफलता पर खूब गाल बजाता था परंतु परिणाम सामने आने पर शर्मिंदा हुआ।

घड़ों पानी पड़ना : बहुत लज्जित होना।
वाक्य : बेटे की करतूतों का भेद खुलते ही पिता पर घडों पानी पड़ गया।

चिकनी-चुपड़ी बातें करना : चापलूसी करना, मीठी-मीठी बातें बोलना।
वाक्य : अब चिकनी-चुपड़ी बातें करने से कोई लाभ नहीं, सच्चाई सब जान गए हैं।

छाती पर साँप लोटना : ईर्ष्या होना।
वाक्य : गीता के कक्षा में प्रथम आने की खबर सुनते ही मीता की छाती पर साँप लोटने लगा।

तूती बोलना : प्रभाव होना।
वाक्य : मंत्री महोदय के खास आदमी होने की वजह से उसकी तूती बोलती है।

दो टुक जवाब देना : स्पष्ट बोलना।
वाक्य : मैंने आपसे दो टुक बात कर ली है, आगे आपकी मर्जी।

नुक्ताचीनी करना : आलोचना करना।
वाक्य : हर बात में नुक्ताचीनी करने की आदत के चलते रमेश के दोस्त कम और दुश्मन ही अधिक है।

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2

Question 1.
Find the square root of the following complex numbers:
(i) -8 – 6i
Solution:
Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
-8 – 6i = (a + bi)²
-8 – 6i = a² + b²i² + 2abi
-8 – 6i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = -8 and 2ab = -6

(ii) 7 + 24i
Solution:
Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
7 + 24i = (a + bi)²
7 + 24i = a² + b²i² + 2abi
7 + 24i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 7 and 2ab = 24

(iii) 1 + 4√3 i
Solution:
Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
1 + 4√3 i = (a + bi)²
1 + 4√3i = a² + b²i² + 2abi
1 + 4√3i = (a² – b²) + 2abi …..[∵ i² = -1]
Equating real arid imaginary parts, we get
a² – b² = 1 and 2ab = 4√3

(iv) 3 + 2√10 i
Solution:
Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
3 + 2√10 i = a² + b²i² + 2abi
3 + 2√10 i = (a² – b²) + 2abi ……[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 3 and 2ab = 2√10

(v) 2(1 – √3 i)
Solution:
Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
2(1 – √3 i) = a² + b²i² + 2abi
2 – 2√3 i = (a² – b²) + 2abi ….[∵ i² = -1]
Equating real and imaginary parts, we get
a² – b² = 2 and 2ab = -2√3
a² – b² = 2 and b = \(-\frac{\sqrt{3}}{a}\)

Question 2.
Solve the following quadratic equations:
(i) 8x² + 2x + 1 = 0
Solution:
Given equation is 8x² + 2x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 8, b = 2, c = 1
Discriminant = b² – 4ac
= (2)² – 4 × 8 × 1
= 4 – 32
= -28 < 0
So, the given equation has complex roots.
These roots are given by

(ii) 2x² – √3 x + 1 = 0
Solution:
Given equation is 2x² – √3 x + 1 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = -√3, c = 1
Discriminant = b² – 4ac
= (-√3)² – 4 × 2 × 1
= 3 – 8
= -5 < 0
So, the given equation has complex roots.
These roots are given by

(iii) 3x² – 7x + 5 = 0
Solution:
Given equation is 3x² – 7x + 5 = 0
Comparing with ax² + bx + c = 0, we get
a = 3, b = -7, c = 5
Discriminant = b² – 4ac
= (-7)² – 4 × 3 × 5
= 49 – 60
= -11 < 0
So, the given equation has complex roots.
These roots are given by

The roots of the given equation are \(\frac{7+\sqrt{11} \mathrm{i}}{6}\) and \(\frac{7-\sqrt{11} \mathrm{i}}{6}\)

(iv) x² – 4x + 13 = 0
Solution:
Given equation is x² – 4x + 13 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -4, c = 13
Discriminant = b² – 4ac
= (-4)² – 4 × 1 × 13
= 16 – 52
= -36 < 0
So, the given equation has complex roots.
These roots are given by

∴ The roots of the given equation are 2 + 3i and 2 – 3i.

Question 3.
Solve the following quadratic equations:
(i) x² + 3ix + 10 = 0
Solution:
Given equation is x² + 3ix + 10 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 3i, c = 10
Discriminant = b² – 4ac
= (3i)² – 4 × 1 × 10
= 9i² – 40
= -9 – 40 ……[∵ i² = -1]
= -49 < 0
So, the given equation has complex roots.
These roots are given by

∴ x = 2i or x = -5i
∴ The roots of the given equation are 2i and -5i.

(ii) 2x² + 3ix + 2 = 0
Solution:
Given equation is 2x² + 3ix + 2 = 0
Comparing with ax + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b² – 4ac
= (3i)² – 4 × 2 × 2
= 9i² – 16
= -9 – 16 …..[∵ i² = -1]
= -25 < 0
So, the given equation has complex roots.
These roots are given by

∴ The roots of the given equation are \(\frac{1}{2}\)i and -2i.

(iii) x² + 4ix – 4 = 0
Solution:
Given equation is x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × (-4)
= 16i² + 16
= -16 + 16 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ x = -2i
∴ The root of the given equation is -2i.

(iv) ix² – 4x – 4i = 0
Solution:
ix² – 4x – 4i = 0
Multiplying throughout by i, we get
i2x² – 4ix – 4i² = 0
-x² – 4ix + 4 = 0 …[∵ i² = -1]
x² + 4ix – 4 = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = 4i, c = -4
Discriminant = b² – 4ac
= (4i)² – 4 × 1 × (-4)
= 16i² + 16
= -16 + 16
= 0
So, the given equation has equal roots.
These roots are given by

∴ The root of the given equation is -2i

Question 4.
Solve the following quadratic equations:
(i) x² – (2 + i) x – (1 – 7i) = 0
Solution:
Given equation is x² – (2 + i)x – (1 – 7i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(2 + i), c = -(1 – 7i)
Discriminant = b² – 4ac
= [-(2 + i)]² – 4 × 1 × -(1 – 7i)
= 4 + 4i + i² + 4 – 28i
= 4 + 4i – 1 + 4 – 28i …..[∵ i² = – 1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by

Let \(\sqrt{7-24 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 – 24i = a² + i²b² + 2abi
7 – 24i = a² – b² + 2abi
Equating real and imaginary parts, we get
a² – b² = 7 and 2ab = -24

(ii) x² – (3√2 + 2i) x + 6√2 i = 0
Solution:
Given equation is x² – (3√2 + 2i) x + 6√2 i = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(3√2 + 2i), c = 6√2i
Discriminant = b² – 4ac
= [-(3√2 + 2i)]² – 4 × 1 × 6√2 i
= 18 + 12√2i + 4i² – 24√2 i
= 18 – 12√2 i – 4 ……[∵ i² = -1]
= 14 – 12√2 i
So, the given equation has complex roots.
These roots are given by

(iii) x² – (5 – i) x + (18 + i) = 0
Solution:
Given equation is x² – (5 – i)x + (18 + i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = -(5 – i), c = 18 + i
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × 1 × (18 + i)
= 25 – 10i + i² – 72 – 4i
= 25 – 10i – 1 – 72 – 4i ……[∵ i² = -1]
= -48 – 14i
So, the given equation has complex roots.
These roots are given by

(iv) (2 + i) x² – (5 – i) x + 2(1 – i) = 0
Solution:
Given equation is (2 + i) x² – (5 – i) x + 2(1 – i) = 0
Comparing with ax² + bx + c = 0, we get
a = 2 + i, b = -(5 – i), c = 2(1 – i)
Discriminant = b² – 4ac
= [-(5 – i)]² – 4 × (2 + i) × 2(1 – i)
= 25 – 10i + i² – 8(2 + i) (1 – i)
= 25 – 10i + i² – 8(2 – 2i + i – i²)
= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i² = -1]
= 25 – 10i – 1 – 16 + 8i – 8
= -2i
So, the given equation has complex roots.
These roots are given by

Let \(\sqrt{-2 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
-2i = a² + b²i² + 2abi
-2i = a² – b² + 2abi
Equating real and imaginary parts, we get
a² – b² = 0 and 2ab = -2
a² – b² = 0 and b = \(-\frac{1}{a}\)

Question 5.
Find the value of
(i) x³ – x² + x + 46, if x = 2 + 3i
Solution:
x = 2 + 3i
x – 2 = 3i
(x – 2)² = 9i²
x² – 4x + 4 = 9(-1) …..[∵ i² = -1]
x² – 4x + 13 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder
∴ x³ – x² + x + 46 = (x² – 4x + 13) (x + 3) + 7
= 0(x + 3) + 7 …..[from(i)]
= 7

Alternate Method:
x = 2 + 3i
α = 2 + 3i, \(\bar{\alpha}\) = 2 – 3i
α\(\bar{\alpha}\) = (2 + 3i)(2 – 3i)
= 4 – 6i + 6i – 9i²
= 4 – 9(-1)
= 4 + 9
= 13
α + \(\bar{\alpha}\) = 2 + 3i + 2 – 3i = 4
∴ Standard form of quadratic equation,
x² – (Sum of roots) x + Product of roots = 0
x² – 4x + 13 = 0

Dividend = Divisor × Quotient + Remainder
∴ x³ – x² + x + 46 = (x² – 4x + 13).(x + 3) + 7
= 0(x + 3) + 7 …..[From (i)]
= 7

(ii) 2x³ – 11x² + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:


Dividend = Divisor × Quotient + Remainder
2x³ – 11x² + 44x + 27 = (x² – 6x + 25)(2x + 1) + 2
= 0.(2x + 1) + 2 …..[From (i)]
= 0 + 2
= 2

(iii) x³ + x² – x + 22, if x = \(\frac{5}{1-2 i}\)
Solution:

Dividend = Divisor × Quotient + Remainder
x³ + x² – x + 22 = (x² – 2x + 5)(x + 3) + 7
= 0.(x + 3) + 7 …..[From (i)]
= 0 + 7
= 7

(iv) x⁴ + 9x³ + 35x² – x + 4, if x = -5 + √-4
Solution:
x = -5 + √-4
x + 5 = √-4
x + 5 = √4 √-1
x + 5 = 2i
(x + 5)² = 4i²
x² + 10x + 25 = 4(-1) ….[∵ i² = -1]
x² + 10x + 29 = 0 …..(i)

Dividend = Divisor × Quotient + Remainder
x⁴ + 9x³ + 35x² – x + 4 = (x² + 10x + 29) (x² – x + 16) – 132x – 460
= 0.(x² – x + 16) – 132x – 460 …..[From (i)]
= -132 (-5 + 2i) – 460
= 660 – 264i – 460
= 200 – 264i

(v) 2x⁴ + 5x³ + 7x² – x + 41, if x = -2 – √3i
Solution:

Dividend = Divisor × Quotient + Remainder
2x⁴ + 5x³ + 7x² – x + 41 = (x² + 4x + 7) (2x² – 3x + 5) + 6
= 0(2x² – 3x + 5) + 6 ……[From (i)]
= 0 + 6
= 6

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 1 Complex Numbers Ex 1.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1

Question 1.
Simplify:
(i) √-16 + 3√-25 + √-36 – √-625
Solution:

= 4i + 3(5i) + 6i – 25i
= 25i – 25i
= 0

(ii) 4√-4 + 5√-9 – 3√-16
Solution:

Question 2.
Write the conjugates of the following complex numbers
(i) 3 + i
Solution:
Conjugate of (3 + i) is (3 – i).

(ii) 3 – i
Solution:
Conjugate of (3 – i) is (3 + i).

(iii) √-5 – √7 i
Solution:
Conjugate of (√-5 – √7 i) is (√-5 + √7 i).

(iv) -√-5
Solution:
-√-5 = -√5 × √-1 = -√5 i
Conjugate of (-√-5) is √5 i

(v) 5i
Solution:
Conjugate of (5i) is (-5i).

(vi) √5 – i
Solution:
Conjugate of (√5 – i) is (√5 + i).

(vii) √2 + √3 i
Solution:
Conjugate of (√2 + √3 i) is (√2 – √3 i)

(viii) cos θ + i sin θ
Solution:
Conjugate of (cos θ + i sin θ) is (cos θ – i sin θ)

Question 3.
Find a and b if
(i) a + 2b + 2ai = 4 + 6i
Solution:
a + 2b + 2ai = 4 + 6i
Equating real and imaginary parts, we get
a + 2b = 4 …..(i)
2a = 6 ……(ii)
∴ a = 3
Substituting, a = 3 in (i), we get
3 + 2b = 4
∴ b = \(\frac{1}{2}\)
∴ a = 3 and b = \(\frac{1}{2}\)

Check:
For a = 3 and b = \(\frac{1}{2}\)
Consider, L.H.S. = a + 2b + 2ai
= 3 + 2(\(\frac{1}{2}\)) + 2(3)i
= 4 + 6i
= R.H.S.

(ii) (a – b) + (a + b)i = a + 5i
Solution:
(a – b) + (a + b)i = a + 5i
Equating real and imaginary parts, we get
a – b = a ……(i)
a + b = 5 ……(ii)
From (i), b = 0
Substituting b = 0 in (ii), we get
a + 0 = 5
∴ a = 5
∴ a = 5 and b = 0

(iii) (a + b) (2 + i) = b + 1 + (10 + 2a)i
Solution:
(a + b) (2 + i) = b + 1 + (10 + 2a)i
2(a + b) + (a + b)i = (b + 1) + (10 + 2a)i
Equating real and imaginary parts, we get
2(a + b) = b + 1
∴ 2a + b = 1 ……(i)
and a + b = 10 + 2a
-a + b = 10 …….(ii)
Subtracting equation (ii) from (i), we get
3a = -9
∴ a = -3
Substituting a = – 3 in (ii), we get
-(-3) + b = 10
∴ b = 7
∴ a = -3 and b = 7

(iv) abi = 3a – b + 12i
Solution:
abi = 3a – b + 12i
∴ 0 + abi = (3a – b) + 12i
Equating real and imaginary parts, we get
3a – b = 0
∴ 3a = b …..(i)
and ab = 12
∴ b = \(\frac{12}{a}\) ……..(ii)
Substituting b = \(\frac{12}{a}\) in (i), we get
3a = \(\frac{12}{a}\)
3a² = 12
a² = 4
a = ±2
When a = 2, b = \(\frac{12}{a}\) = \(\frac{12}{2}\) = 6
When a = -2, b = \(\frac{12}{a}\) = \(\frac{12}{-2}\) = -6
∴ a = 2 and b = 6 or a = -2 and b = -6

(v) \(\frac{1}{a+i b}\) = 3 – 2i
Solution:

(vi) (a + ib) (1 + i) = 2 + i
Solution:
(a + ib)(1 + i) = 2 + i
a + ai + bi + bi² = 2 + i
a + (a + b)i + b(-1) = 2 + i ……(∵ i² = -1)
(a – b) + (a + b)i = 2 + i
Equating real and imaginary parts, we get
a – b = 2 ……(i)
a + b = 1 …….(ii)
Adding equations (i) and (ii), we get
2a = 3
∴ a = \(\frac{3}{2}\)
Substituting a = \(\frac{3}{2}\) in (ii), we get
\(\frac{3}{2}\) + b = 1
∴ b = 1 – \(\frac{3}{2}\) = \(\frac{-1}{2}\)
∴ a = \(\frac{3}{2}\) and b = \(\frac{-1}{2}\)

Question 4.
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:
(i) (1 + 2i)(-2 + i)
Solution:
(1 + 2i)(-2 + i) = -2 + i – 4i + 2i²
= -2 – 3i + 2(-1) ……[∵ i² = -1]
∴ (1 + 2i)(-2 + i) = -4 – 3i
∴ a = -4 and b = -3

(ii) (1 + i)(1 – i)-1
Solution:

(iii) \(\frac{i(4+3 i)}{1-i}\)
Solution:

(iv) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
Solution:

(v) \(\left(\frac{1+i}{1-1}\right)^{2}\)
Solution:

(vi) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
Solution:

(vii) (1 + i)^-3
Solution:

(viii) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
Solution:

(ix) (-√5 + 2√-4 ) + (1 – √-9 ) + (2 + 3i)(2 – 3i)
Solution:
(-√5 + 2√-4) + (1 – √-9) + (2 + 3i)(2 – 3i)
= (-√5 + 2√4.√-1) + (1 – √9.√-1) + 4 – 9i²
= [-√5 + 2(2)i] + (1 – 3i) + 4 – 9i²
= -√5 + 4i + 1 – 3i + 4 – 9(-1) ……[∵ i² = -1]
= (14 – √5) + i
∴ a = 14 – √5 and b = 1

(x) (2 + 3i)(2 – 3i)
Solution:
(2 + 3i)(2 – 3i)
= 4 – 9i²
= 4 – 9(-1) …[∵ i² = -1]
= 4 + 9
= 13
∴ (2 + 3i)(2 – 3i) = 13 + 0i
∴ a = 13 and b = 0

(xi) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)
Solution:

Question 5.
Show that (-1 + √3i)³ is a real number.
Solution:

Question 6.
Find the value of (3 + \(\frac{2}{\mathrm{i}}\)) (i⁶ – i⁷) (1 + i¹¹).
Solution:

Question 7.
Evaluate the following:
(i) i³⁵
(ii) i⁸⁸⁸
(iii) i⁹³
(iv) i¹¹⁶
(v) i⁴⁰³
(vi) \(\frac{1}{i^{58}}\)
(vii) i^-888
(viii) i³⁰ + i⁴⁰ + i⁵⁰ + i⁶⁰
Solution:

Question 8.
Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.
Solution:

= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 9.
Find the value of
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
(ii) i + i² + i³ + i⁴
Solution:

Question 10.
Simplify: \(\frac{\mathbf{i}^{592}+\mathbf{i}^{590}+\mathbf{i}^{588}+\mathbf{i}^{586}+\mathbf{i}^{584}}{\mathbf{i}^{582}+\mathbf{i}^{580}+\mathbf{i}^{578}+\mathbf{i}^{576}+\mathbf{i}^{574}}\)
Solution:

Question 11.
Find the value of 1 + i² + i⁴ + i⁶ + i⁸ + …… + i²⁰.
Solution:

Question 12.
Show that 1 + i¹⁰ + i¹⁰⁰ – i¹⁰⁰⁰ = 0.
Solution:

Question 13.
Is (1 + i¹⁴ + i¹⁸ + i²²) a real number? Justify your answer.
Solution:

Question 14.
Evaluate: \(\left(\mathbf{i}^{37}+\frac{1}{\mathbf{i}^{67}}\right)\)
Solution:

Question 15.
Prove that: (1 + i)⁴ × \(\left(1+\frac{1}{\mathrm{i}}\right)^{4}\) = 16
Solution:

Question 16.
Find the value of \(\frac{\mathbf{i}^{6}+\mathbf{i}^{7}+\mathbf{i}^{8}+\mathbf{i}^{9}}{\mathbf{i}^{2}+\mathbf{i}^{3}}\)
Solution:

Question 17.
If a = \(\frac{-1+\sqrt{3} i}{2}\), b = \(\frac{-1-\sqrt{3} i}{2}\), then show that a² = b and b² = a.
Solution:

Question 18.
If x + iy = (a + ib)³, show that \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)
Solution:
x + iy = (a + ib)³
x + iy = a³ + 3a²bi + 3ab²i² + b³i³
x + iy = a³ + 3a²bi – 3ab² – b³i ……[∵ i² = -1, i³ = -i]
x + iy = (a³ – 3ab²) + (3a²b – b³)i
Equating real and imaginary parts, we get
x = a³ – 3ab² and y = 3a²b – b³
\(\frac{x}{a}\) = a² – 3b² and \(\frac{y}{b}\) = 3a² – b²
\(\frac{x}{a}+\frac{y}{b}\) = a² – 3b + 3a² – b²
\(\frac{x}{a}+\frac{y}{b}\) = 4a² – 4b²
\(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)

Alternate Method:
x + iy = (a + ib)³
x + iy = a³ + 3a²bi + 3ab²i² + b³i³
x + iy = a³ + 3a²bi – 3ab² – b³i …..[∵ i² = -1, i³ = -i]
x + iy = (a³ – 3ab²) + (3a²b – b³)i
Equating real and imaginary parts, we get
x = a³ – 3ab² and y = 3a²b – b³
Consider

Question 19.
If \(\frac{a+3 i}{2+i b}\) = 1 – i, show that (5a – 7b) = 0.
Solution:
\(\frac{a+3 i}{2+i b}\) = 1 – i
a + 3i = (1 – i)(2 + ib)
= 2 + bi – 2i – bi²
= 2 + (b – 2)i – b(-1) ……[∵ i² = -1]
a + 3i = (2 + b) + (b – 2)i
Equating real and imaginary parts, we get
a = 2 + b and 3 = b – 2
a = 2 + b and b = 5
a = 2 + 5 = 7
5a – 7b = 5(7) – 7(5) = 35 – 35 = 0

Question 20.
If x + iy = \(\sqrt{\frac{a+i b}{c+i d}}\), prove that \(\left(x^{2}+y^{2}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}\)
Solution:

Question 21.
If (a + ib) = \(\frac{1+i}{1-i}\), then prove that a² + b² = 1.
Solution:

∴ a + bi = 0 + i
Equating real and imaginary parts, we get
a = 0 and b = 1
a² + b² = 0² + 1² = 1

Question 22.
Show that \(\left(\frac{\sqrt{7}+i \sqrt{3}}{\sqrt{7}-i \sqrt{3}}+\frac{\sqrt{7}-i \sqrt{3}}{\sqrt{7}+i \sqrt{3}}\right)\) is real.
Solution:

Question 23.
If (x + iy)³ = y + vi, then show that \(\frac{y}{x}+\frac{v}{y}\) = 4(x² – y²).
Solution:

Question 24.
Find the values of x and y which satisfy the following equations (x, y ∈ R)
(i) (x + 2y) + (2x – 3y)i + 4i = 5
Solution:
(x + 2y) + (2x – 3y)i + 4i = 5
(x + 2y) + (2x – 3y)i = 5 – 4i
Equating real and imaginary parts, we get
x + 2y = 5 ……(i)
and 2x – 3y = -4 …..(ii)
Equation (i) x 2 – equation (ii) gives
7y = 14
∴ y = 2
Substituting y = 2 in (i), we get
x + 2(2) = 5
x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2

Check:
For x = 1 and y = 2
Consider, L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.

(ii) \(\frac{x+1}{1+i}+\frac{y-1}{1-i}=i\)
Solution:

(x + y) + (y – x – 2)i = 2i
(x + y) + (y – x – 2)i = 0 + 2i
Equating real and imaginary parts, we get
x + y = 0 and y – x – 2 = 2
∴ x + y = 0 …….(i)
and -x + y = 4 …….(ii)
Adding (i) and (ii), we get
2y = 4
∴ y = 2
Substituting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

(iii) \(\frac{x+i y}{2+3 i}+\frac{2+i}{2-3 i}=\frac{9}{13}(1+i)\)
Solution:

(2x + 3y + 1) + (8 – 3x + 2y)i = 9 + 9i
Equating real and imaginary parts, we get
2x + 3y + 1 = 9 and 8 – 3x + 2y = 9
2x + 3y = 8 ……(i)
and 3x – 2y = – 1 ……(ii)
Equation (i) × 2 + equation (ii) × 3 gives
13x = 13
∴ x = 1
Substituting x = 1 in (i), we get
2(1) + 3y = 8
3y = 6
∴ y = 2
∴ x = 1 and y = 2

(iv) If x(1 + 3i) + y(2 – i) – 5 + i³ = 0, find x + y
Solution:
x(1 + 3i) + y(2 – i) – 5 + i³ = 0
x + 3xi + 2y – yi – 5 – i = 0 ……[∵ i³ = -i]
(x + 2y – 5) + (3x – y – 1)i = 0 + 0i
Equating real and imaginary parts, we get
x + 2y – 5 = 0 …..(i)
and 3x – y – 1 = 0 ……(ii)
Equation (i) + equation (ii) × 2 gives
7x – 7 = 0
7x = 1
∴ x = 1
Substituting x = 1 in (i), we get
1 + 2y – 5 = 0
2y = 4
y = 2
∴ x = 1 and y = 2
∴ x + y = 1 + 2 = 3

(v) If x + 2i + 15i⁶y = 7x + i³(y + 4), find x + y
Solution:
x + 2i + 15i⁶y = 7x + i³(y + 4)
x + 2i + 15(i²)³ y = 7x + i³(y + 4)
x + 2i + 15(-1)³ y = 7x – i(y + 4) ……[∵ i² = -1, i³ = -i]
x + 2i – 15y – 7x + iy + 4i = 0
(-6x – 15y) + i(y + 6) = 0 + 0i
Equating real and imaginary parts, we get
-6x – 15y = 0 and y + 6 = 0
-6x – 15y = 0 and y = -6
-6x – 15(-6) = 0
-6x + 90 = 0
∴ x = 15
∴ x + y = 15 – 6 = 9

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.

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:

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:

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:

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:

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:

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

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.

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°)

= 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°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

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:

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:

Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = \(-\frac{1}{2}\)
Solution:

Question 6.
\(\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x\)
Solution:

Question 7.
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Solution:

Question 8.
sin² 6x – sin² 4x = sin 2x sin 10x
Solution:

Question 9.
cos² 2x – cos² 6x = sin 4x sin 8x
Solution:

Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:

Question 12.
If sin 2A = λ sin 2B, then prove that \(\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}\)
Solution:

Question 13.
\(\frac{2 \cos 2 A+1}{2 \cos 2 A-1}\) = tan (60° + A) tan (60° – A)
Solution:

Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:

Question 15.
3 tan⁶ 10° – 27 tan⁴ 10° + 33 tan² 10° = 1
Solution:

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°

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

∴ 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:

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:

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

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:

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°

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°= \(\frac{\sqrt{5}-1}{4}\)

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°

∴ 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:

Question 27.
tan \(\frac{\pi}{8}\) = √2 – 1
Solution:

Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:

Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:

Question 30.
√3 cosec 20° – sec 20° = 4
Solution:

Question 31.
In ∆ABC, ∠C = \(\frac{2 \pi}{3}\), then prove that cos² A + cos² B – cos A cos B = \(\frac{3}{4}\).
Solution: