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Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.6 Questions and Answers.

11th Maths Part 2 Sequences and Series Exercise 2.6 Questions And Answers Maharashtra Board

Question 1.
Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\).
Solution:

Question 2.
Find \(\sum_{r=1}^{n}\left(3 r^{2}-2 r+1\right)\)
Solution:

Question 3.
Find \(\sum_{r=1}^{n}\left(\frac{1+2+3 \ldots .+r}{r}\right)\)
Solution:

= \(\frac{n}{4}\) [(n + 1) + 2]
= \(\frac{n}{4}\) (n + 3)

Question 4.
Find \(\sum_{r=1}^{n}\left(\frac{1^{3}+2^{3}+\ldots . .+r^{3}}{r(r+1)}\right)\)
Solution:

Question 5.
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms.
Solution:
5 × 7 + 9 × 11 + 13 × 15 + ….. upto n terms
Now, 5, 9, 13, … are in A.P. with
rth term = 5 + (r – 1) (4) = 4r + 1
7, 11, 15, ….. are in A.P. with
rth term = 7 + (r – 1) (4) = 4r + 3
∴ 5 × 7 + 9 × 11 + 13 × 15 + …… upto n terms

Question 6.
Find the sum 2² + 4² + 6² + 8² + ….. upto n terms.
Solution:

Question 7.
Find (70² – 69²) + (68² – 67²) + (66² – 65²) + …… + (2² – 1²)
Solution:
Let S = (70² – 69²) + (68² – 67²) + …… + (2² – 1²)
∴ S = (2² – 1²) + (4² – 3²) + ….. + (70² – 69²)
Here, 2, 4, 6,…, 70 are in A.P. with rth term = 2r
and 1, 3, 5, …,69 are in A.P. with rth term = 2r – 1

Question 8.
Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3)
Solution:
Let S = 1 × 3 × 5 + 3 × 5 × 7 + ….. upto n terms
Here, 1, 3, 5, 7 … are in A.P. with rth term = 2r – 1,
3, 5, 7, 9,… are in A.P. with rth term = 2r + 1,
5, 7, 9, 11,… are in A.P. with rth term = 2r + 3

Question 9.
If \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \text { upto } n \text { terms }}{1+2+3+4+\ldots \text { upto } n \text { terms }}\) = \(\frac{100}{3}\), find n.
Solution:

Question 10.
If S₁, S₂ and S₃ are the sums of first n natural numbers, their squares and their cubes respectively, then show that 9\(\mathrm{S}_{2}{ }^{2}\) = S₃(1 + 8S₁).
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 11 Maths 2
• Exercise 2.2 Class 11 Maths 2
• Exercise 2.3 Class 11 Maths 2
• Exercise 2.4 Class 11 Maths 2
• Exercise 2.5 Class 11 Maths 2
• Exercise 2.6 Class 11 Maths 2
• Miscellaneous Exercise 2 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.5 Questions and Answers.

11th Maths Part 2 Sequences and Series Exercise 2.5 Questions And Answers Maharashtra Board

Question 1.
Find Sn of the following arithmetico-geometric sequences.
(i) 2, 4x, 6x², 8x³, 10x⁴, ……
Solution:
2, 4x, 6x², 8x³, 10x⁴, ……
Here, 2, 4, 6, 8, 10,… are in A.P.
∴ a = 2, d = 2
∴ nth term = a + (n – 1)d
= 2 + (n – 1)(2)
= 2n

(ii) 1, 4x, 7x², 10x³, 13x⁴, ……
Solution:
1, 4x, 7x², 10x³, 13x⁴, ……
Here, 1, 4, 7, 10, 13,… are in A.P.
a = 1, d = 3
∴ nth term = a + (n – 1)d
= 1 + (n – 1)(3)
= 3n – 2
Also, 1, x, x², x³,… are in G.P.
∴ a = 1, r = x,
nth term = ar^n-1 = x^n-1
nth term of arithmetico-geometric sequence is

(iii) 1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, ……
Solution:
1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, …..
Here, 1, 2, 3, 4, 5, … are in A.P.
∴ a = 1, d = 1
∴ nth term = a + (n – 1)d
= 1 + (n – 1)1
= n

(iv) 3, 12, 36, 96, 240, ……
Solution:
3, 12, 36, 96, 240, ……
i.e., 1 × 3, 2 × 6, 3 × 12, 4 × 24, 5 × 48, …….
Here, 1, 2, 3, 4, 5, ….. are in A.P.
∴ nth term = n
Also, 3, 6, 12, 24, 48, ….. are in G.P.
∴ a = 3, r = 2
∴ nth term = ar^n-1 = 3 . (2^n-1)
∴ nth term of arithmetico-geometric sequence is

Question 2.
Find the sum to infinity of the following arithmetico-geometric sequence.
(i) \(1, \frac{2}{4}, \frac{3}{16}, \frac{4}{64}, \ldots\)
Solution:

(ii) \(3, \frac{6}{5}, \frac{9}{25}, \frac{12}{125}, \frac{15}{625}, \ldots\)
Solution:

∴ \(\frac{4}{5} S=\frac{15}{4}\)
∴ S = \(\frac{75}{16}\)

(iii) \(1, \frac{-4}{3}, \frac{7}{9}, \frac{-10}{27} \ldots\)
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 11 Maths 2
• Exercise 2.2 Class 11 Maths 2
• Exercise 2.3 Class 11 Maths 2
• Exercise 2.4 Class 11 Maths 2
• Exercise 2.5 Class 11 Maths 2
• Exercise 2.6 Class 11 Maths 2
• Miscellaneous Exercise 2 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.4 Questions and Answers.

11th Maths Part 2 Sequences and Series Exercise 2.4 Questions And Answers Maharashtra Board

Question 1.
Verify whether the following sequences are H.P.
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Solution:
\(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Here, the reciprocal sequence is 3, 5, 7, 9,…
t₁ = 3, t₂ = 5, t₃ = 7, t₄ = 9, …..
t₂ – t₁ = t₃ – t₂ = t₄ – t₃ = 2 = constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)
Solution:
\(\frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots\)
Here, the reciprocal sequence is 3, 6, 12, 24,…
t₁ = 3, t₂ = 6, t₃ = 12, ……
t₂ – t₁ = 3, t₃ – t₂ = 6
t₂ – t₁ ≠ t₃ – t₂
∴ The reciprocal sequence is not an A.P.
∴ The given sequence is not a H.P.

(iii) \(5, \frac{10}{17}, \frac{10}{32}, \frac{10}{47}, \ldots\)
Solution:

∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.

Question 2.
Find the nth term and hence find the 8th term of the following HPs.
(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\)
Solution:
\(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots\) are in H.P.
∴ 2, 5, 8, 11,… are in A.P.
∴ a = 2, d = 3
t_n = a + (n – 1)d
= 2 + (n – 1)(3)
= 3n – 1
∴ nth term of H.P. = \(\frac{1}{3 n-1}\)
∴ 8th term of H.P. = \(\frac{1}{3(8)-1}\) = \(\frac{1}{23}\)

(ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\)
Solution:
\(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots\) are in H.P.
∴ 4, 6, 8, 10, … are in A.P.
∴ a = 4, d = 2
t_n = a + (n – 1)d
= 4 + (n – 1) (2)
= 2n + 2
∴ nth term of H.P. = \(\frac{1}{2 n+2}\)
∴ 8th term of H.P. = \(\frac{1}{2(8)+2}\) = \(\frac{1}{18}\)

(iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\)
Solution:
\(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \ldots\) are in H.P.
∴ 5, 10, 15, 20, … are in A.P.
∴ a = 5, d = 5
t_n = a + (n – 1)d
= 5 + (n – 1) (5)
= 5n
∴ nth term of H.P. = \(\frac{1}{5 n}\)
∴ 8th term of H.P. = \(\frac{1}{5(8)}\) = \(\frac{1}{40}\)

Question 3.
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\) respectively.
Solution:
G.M. = 4, H.M. = \(\frac{16}{5}\)
Now, (G.M.)² = (A.M.) (H.M.)
∴ 4² = A.M. × \(\frac{16}{5}\)
∴ A.M. = 16 × \(\frac{5}{16}\)
∴ A.M. = 5

Question 4.
Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6.
Solution:
A.M. = \(\frac{15}{2}\), G.M. = 6
Now, (G.M.)² = (A.M.) (H.M.)
∴ 6² = \(\frac{15}{2}\) × H.M.
∴ H.M. = 36 × \(\frac{2}{15}\)
∴ H.M. = \(\frac{24}{5}\)

Question 5.
Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Solution:
A.M. = 75, H.M. = 48
Now, (G.M.)² = (A.M.) (H.M.)
∴ (G.M.)² = 75 × 48
∴ (G.M.)² = 25 × 3 × 16 × 3
∴ (G.M.)² = 5² × 4² × 3²
∴ G.M. = 5 × 4 × 3
∴ G.M. = 60

Question 6.
Insert two numbers between \(\frac{1}{4}\) and \(\frac{1}{3}\) so that the resulting sequence is a H.P.
Solution:
Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\).
∴ \(\frac{1}{4}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{3}\) are in H.P.
∴ 4, H₁, H₂, 3 are in A.P.
t₁ = 4, t₂ = H₁, t₃ = H₂, t₄ = 3
∴ t₁ = a = 4, t₄ = 3
t_n = a + (n – 1)d
t₄ = 4 + (4 – 1)d
3 = 4 + 3d
3d = -1
∴ d = \(\frac{-1}{3}\)
H₁ = t₂ = a + d = 4 – \(\frac{1}{3}\) = \(\frac{11}{3}\)
H₂ = t₃ = a + 2d = 4 – \(\frac{2}{3}\) = \(\frac{10}{3}\)
∴ For resulting sequence to be H.P. we need to insert numbers \(\frac{3}{11}\) and \(\frac{3}{10}\).

Question 7.
Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Solution:
Let the required numbers be G₁ and G₂.
∴ 1, G₁, G₂, -27 are in G.P.
t₁ = 1, t₂ = G₁, t₃ = G₂, t₄ = -27
∴ t₁ = a = 1
t_n = ar^n-1
t₄ = (1) r^4-1
-27 = r³
r³ = (-3)³
∴ r = -3
∴ G₁ = t₂ = ar = 1(-3) = -3
G₂ = t₃ = ar² = 1(-3)² = 9
∴ For resulting sequence to be G.P. we need to insert numbers -3 and 9.

Question 8.
If the A.M. of two numbers exceeds their G.M. by 2 and their H.M. by \(\frac{18}{5}\), find the numbers.
Solution:
Let a and b be the two numbers.
A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)
According to the given conditions,

Consider, G = A – 2 = 10 – 2 = 8
\(\sqrt{a b}\) = 8
ab = 64
a(20 – a) = 64 …..[From (i)]
a² – 20a + 64 = 0
(a – 4)(a – 16) = 0
∴ a = 4 or a = 16
When a = 4, b = 20 – 4 = 16
When a = 16, b = 20 – 16 = 4
∴ The two numbers are 4 and 16.

Question 9.
Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\).
Solution:
Let a and b be the two numbers.
A = \(\frac{a+b}{2}\), G = \(\sqrt{a b}\), H = \(\frac{2 a b}{a+b}\)
According to the given conditions,

\(\sqrt{a b}\) = 6
ab = 36
a(13 – a) = 36 ……[From (i)]
a² – 13a + 36 = 0
(a – 4)(a – 9) = 0
∴ a = 4 or a = 9
When a = 4, b = 13 – 4 = 9
When a = 9, b = 13 – 9 = 4
∴ The two numbers are 4 and 9.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 11 Maths 2
• Exercise 2.2 Class 11 Maths 2
• Exercise 2.3 Class 11 Maths 2
• Exercise 2.4 Class 11 Maths 2
• Exercise 2.5 Class 11 Maths 2
• Exercise 2.6 Class 11 Maths 2
• Miscellaneous Exercise 2 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.3 Questions and Answers.

11th Maths Part 2 Sequences and Series Exercise 2.3 Questions And Answers Maharashtra Board

Question 1.
Determine whether the sum to infinity of the following G.P.s exist, if exists find them.
(i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)
Solution:

(ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\)
Solution:

(iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\)
Solution:

(iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\)
Solution:

(v) 9, 8.1, 7.29, ……
Solution:
9, 8.1, 7.29, …..
Here, a = 9, r = \(\frac{8.1}{9}\) = 0.9, |r| < 1
∴ Sum to infinity exists.
∴ Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)
= \(\frac{9}{1-0.9}\)
= \(\frac{9}{0.1}\)
= 90

Question 2.
Express the following recurring decimals as rational numbers.
(i) \(0 . \overline{7}\)
(ii) \(2 . \overline{4}\)
(iii) \(2.3 \overline{5}\)
(iv) \(51.0 \overline{2}\)
Solution:
(i) \(0 . \overline{7}\) = 0.7777… = 0.7 + 0.07 + 0.007 + ….
The terms 0.7, 0.07, 0.007,… are in G.P.
∴ a = 0.7, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(ii) \(2 . \overline{4}\) = 2.444 … = 2 + 0.4 + 0.04 + 0.004 + …
The terms 0.4, 0.04, 0.004,… are in G.P.
∴ a = 0.4, r = \(\frac{0.07}{0.7}\) = 0.1, |r| = 10.11 < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(iii) \(2.3 \overline{5}\) = 2.3555… = 2.3 + 0.05 + 0.005 + 0.0005 + …
The terms 0.05,0.005,0.0005,… are in G.P.
∴ a = 0.05, r = \(\frac{0.005}{0.05}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

(iv) \(51.0 \overline{2}\) = 51.0222 …. = 51 + 0.02 + 0.002 + 0.0002 + …..
The terms 0.02, 0.002, 0.0002,… are in G.P.
∴ a = 0.02, r = \(\frac{0.002}{0.02}\) = 0.1, |r| = |0.1| < 1
∴ Sum to infinity exists.
∴ Sum to infinity

Question 3.
If the common ratio of a G.P. is \(\frac{2}{3}\) and the sum to infinity is 12, find the first term.
Solution:
r = \(\frac{2}{3}\), sum to infinity = 12 ….. [Given]
Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)
12 = \(\frac{a}{1-\frac{2}{3}}\)
a = 12 × \(\frac{1}{3}\)
∴ a = 4

Question 4.
If the first term of the G.P. is 6 and its sum to infinity is \(\frac{96}{17}\), find the common ratio.
Solution:
a = 6, sum to infinity = \(\frac{96}{17}\) …..[Given]
Sum to infinity = \(\frac{\mathrm{a}}{1-\mathrm{r}}\)

Question 5.
The sum of an infinite G.P. is 5 and the sum of the squares of these terms is 15, find the G.P.
Solution:
Let the required G.P. be a, ar, ar², ar³, …..
Sum to infinity of this G.P. = 5

Question 6.
Find
(i) \(\sum_{r=1}^{\infty} 4(0.5)^{r}\)
Solution:

(ii) \(\sum_{r=1}^{\infty}\left(-\frac{1}{3}\right)^{r}\)
Solution:

(iii) \(\sum_{r=0}^{\infty}(-8)\left(-\frac{1}{2}\right)^{r}\)
Solution:

(iv) \(\sum_{n=1}^{\infty} 0.4^{n}\)
Solution:

Question 7.
The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated indefinitely. Find the sum of
(i) the areas of all the squares.
(ii) the perimeters of all the squares.
Solution:

(i) Area of the 1st square = 12
Area of the 2nd square = \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)
Area of the 3rd square = \(\left(\frac{1}{2}\right)^{2}\)
and so on.
∴ Sum of the areas of all the squares

∴ Sum to infinity exists.
∴ Sum of the areas of all the squares = \(\frac{1}{1-\frac{1}{2}}\) = 2

(ii) Perimeter of 1st square = 4
Perimeter of 2nd square = 4\(\left(\frac{1}{\sqrt{2}}\right)\)
Perimeter of 3rd square = 4\(\left(\frac{1}{2}\right)\)
and so on.
Sum of the perimeters of all the squares

Question 8.
A ball is dropped from a height of 10 m. It bounces to a height of 6m, then 3.6 m, and so on. Find the total distance travelled by the ball.
Solution:
Here, on the first bounce, the ball will go 6 m and it will return 6 m.
On the second bounce, the ball will go 3.6 m and it will return 3.6 m, and so on.
Given that, a ball is dropped from a height of 10 m.
∴ Total distance travelled by the ball is = 10 + 2[6 + 3.6 + …]
The terms 6, 3.6 … are in G.P.
a = 6, r = 0.6, |r| = |0.6| < 1
∴ Sum to infinity exists.
∴ Total distance travelled by the ball

Class 11 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 11 Maths 2
• Exercise 2.2 Class 11 Maths 2
• Exercise 2.3 Class 11 Maths 2
• Exercise 2.4 Class 11 Maths 2
• Exercise 2.5 Class 11 Maths 2
• Exercise 2.6 Class 11 Maths 2
• Miscellaneous Exercise 2 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 2 Sequences and Series Ex 2.2 Questions and Answers.

11th Maths Part 2 Sequences and Series Exercise 2.2 Questions And Answers Maharashtra Board

Question 1.
For the following G.P.s, find S_n.
(i) 3, 6, 12, 24, ……..
Solution:

(ii) p, q, \(\frac{\mathbf{q}^{2}}{\mathbf{p}}, \frac{\mathbf{q}^{3}}{\mathbf{p}^{2}}, \ldots\)
Solution:

(iii) 0.7, 0.07, 0.007, …….
Solution:

(iv) √5, -5, 5√5, -25, …….
Solution:

Question 2.
For a G.P.
(i) a = 2, r = \(-\frac{2}{3}\), find S₆.
Solution:

(ii) If S₅ = 1023, r = 4, find a.
Solution:

Question 3.
For a G.P.
(i) If a = 2, r = 3, S_n = 242, find n.
Solution:

(ii) For a G.P. sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.
Solution:

Question 4.
For a G.P.
(i) If t₃ = 20, t₆ = 160, find S₇.
Solution:

(ii) If t₄ = 16, t₉ = 512, find S₁₀.
Solution:

Question 5.
Find the sum to n terms
(i) 3 + 33 + 333 + 3333 + …..
Solution:
S_n = 3 + 33 + 333 +….. upto n terms
= 3(1 + 11 + 111 +….. upto n terms)
= \(\frac{3}{9}\)(9 + 99 + 999 + ….. upto n terms)
= \(\frac{3}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= \(\frac{3}{9}\)[(10 + 100 + 1000 + … upto nterms) – (1 + 1 + 1 + ….. n times)]
But 10, 100, 1000, ….. n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10

(ii) 8 + 88 + 888 + 8888 + …..
Solution:
S_n = 8 + 88 + 888 + … upto n terms
= 8(1 + 11 + 111 + … upto n terms)
= \(\frac{8}{9}\)(9 + 99 + 999 + … upto n terms)
= \(\frac{8}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= \(\frac{8}{9}\)[(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]
But 10, 100, 1000, … n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10

Question 6.
Find the sum to n terms
(i) 0.4 + 0.44 + 0.444 + …..
Solution:
S_n = 0.4 + 0.44 + 0.444 + ….. upto n terms
= 4(0.1 + 0.11 +0.111 + …. upto n terms)
= \(\frac{4}{9}\)(0.9 + 0.99 + 0.999 + … upto n terms)
= \(\frac{4}{9}\)[(1 – 0.1) + (1 – 0.01) + (1 – 0.001) … upto n terms]
= \(\frac{4}{9}\)[(1 + 1 + 1 + …n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]
But 0.1, 0.01, 0.001, … n terms are in G.P. with
a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1

(ii) 0.7 + 0.77 + 0.777 + ……
Solution:
S_n = 0.7 + 0.77 + 0.777 + … upto n terms
= 7(0.1 + 0.11 + 0.111 + … upton terms)
= \(\frac{7}{9}\)(0.9 + 0.99 + 0.999 + … upto n terms)
= \(\frac{7}{9}\)[(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms]
= \(\frac{7}{9}\)[(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms )]
But 0.1, 0.01, 0.001, … n terms are in G.P. with
a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1

Question 7.
Find the sum to n terms of the sequence
(i) 0.5, 0.05, 0.005, …..
Solution:

(ii) 0.2, 0.02, 0.002, ……
Solution:

Question 8.
For a sequence, if S_n = 2(3n – 1), find the nth term, hence showing that the sequence is a G.P.
Solution:

Question 9.
If S, P, R are the sum, product, and sum of the reciprocals of n terms of a G.P, respectively, then verify that \(\left[\frac{S}{R}\right]^{n}\) = P².
Solution:
Let a be the 1st term and r be the common ratio of the G.P.
∴ the G.P. is a, ar, ar², ar³, …, ar^n-1

Question 10.
If S_n, S_2n, S_3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that S_n(S_3n – S_2n) = (S_2n – S_n)².
Solution:
Let a and r be the 1st term and common ratio of the G.P. respectively.

Question 11.
Find
(i) \(\sum_{r=1}^{10}\left(3 \times 2^{r}\right)\)
Solution:

(ii) \(\sum_{r=1}^{10} 5 \times 3^{r}\)
Solution:

Question 12.
The value of a house appreciates 5% per year. How much is the house worth after 6 years if its current worth is Rs. 15 Lac. [Given: (1.05)⁵ = 1.28, (1.05)⁶ = 1.34]
Solution:
The value of a house is Rs. 15 Lac.
Appreciation rate = 5% = \(\frac{5}{100}\) = 0.05
Value of house after 1st year = 15(1 + 0.05) = 15(1.05)
Value of house after 6 years = 15(1.05) (1.05)⁵
= 15(1.05)⁶
= 15(1.34)
= 20.1 lac.

Question 13.
If one invests Rs. 10,000 in a bank at a rate of interest 8% per annum, how long does it take to double the money by compound interest? [(1.08)⁵ = 1.47]
Solution:
Amount invested = Rs. 10000
Interest rate = \(\frac{8}{100}\) = 0.08
amount after 1st year = 10000(1 + 0.08) = 10000(1.08)
Value of the amount after n years
= 10000(1.08) × (1.08)^n-1
= 10000(1.08)ⁿ
= 20000
∴ (1.08)ⁿ = 2
∴ (1.08)⁵ = 1.47 …..[Given]
∴ n = 10 years, (approximately)

Class 11 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 11 Maths 2
• Exercise 2.2 Class 11 Maths 2
• Exercise 2.3 Class 11 Maths 2
• Exercise 2.4 Class 11 Maths 2
• Exercise 2.5 Class 11 Maths 2
• Exercise 2.6 Class 11 Maths 2
• Miscellaneous Exercise 2 Class 11 Maths 2

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

11th Maths Part 2 Sequences and Series Exercise 2.1 Questions And Answers Maharashtra Board

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

Class 11 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 11 Maths 2
• Exercise 2.2 Class 11 Maths 2
• Exercise 2.3 Class 11 Maths 2
• Exercise 2.4 Class 11 Maths 2
• Exercise 2.5 Class 11 Maths 2
• Exercise 2.6 Class 11 Maths 2
• Miscellaneous Exercise 2 Class 11 Maths 2

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

11th Maths Part 2 Complex Numbers Miscellaneous Exercise 1 Questions And Answers Maharashtra Board

(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

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 2
• Exercise 1.2 Class 11 Maths 2
• Exercise 1.3 Class 11 Maths 2
• Exercise 1.4 Class 11 Maths 2
• Miscellaneous Exercise 1 Class 11 Maths 2

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

11th Maths Part 2 Complex Numbers Exercise 1.4 Questions And Answers Maharashtra Board

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:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 2
• Exercise 1.2 Class 11 Maths 2
• Exercise 1.3 Class 11 Maths 2
• Exercise 1.4 Class 11 Maths 2
• Miscellaneous Exercise 1 Class 11 Maths 2

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

11th Maths Part 2 Complex Numbers Exercise 1.3 Questions And Answers Maharashtra Board

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:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 2
• Exercise 1.2 Class 11 Maths 2
• Exercise 1.3 Class 11 Maths 2
• Exercise 1.4 Class 11 Maths 2
• Miscellaneous Exercise 1 Class 11 Maths 2

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

11th Maths Part 2 Complex Numbers Exercise 1.2 Questions And Answers Maharashtra Board

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

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 2
• Exercise 1.2 Class 11 Maths 2
• Exercise 1.3 Class 11 Maths 2
• Exercise 1.4 Class 11 Maths 2
• Miscellaneous Exercise 1 Class 11 Maths 2