Sequences and Series Class 11 Maths 2 Exercise 2.2 Solutions Maharashtra Board

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 Sn.
(i) 3, 6, 12, 24, ……..
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q1 (i)

(ii) p, q, \(\frac{\mathbf{q}^{2}}{\mathbf{p}}, \frac{\mathbf{q}^{3}}{\mathbf{p}^{2}}, \ldots\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q1 (ii)

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

(iii) 0.7, 0.07, 0.007, …….
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q1 (iii)

(iv) √5, -5, 5√5, -25, …….
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q1 (iv)

Question 2.
For a G.P.
(i) a = 2, r = \(-\frac{2}{3}\), find S6.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q2 (i)

(ii) If S5 = 1023, r = 4, find a.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q2 (ii)

Question 3.
For a G.P.
(i) If a = 2, r = 3, Sn = 242, find n.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q3 (i)

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

(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:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q3 (ii)

Question 4.
For a G.P.
(i) If t3 = 20, t6 = 160, find S7.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q4 (i)
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q4 (i).1

(ii) If t4 = 16, t9 = 512, find S10.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q4 (ii)

Question 5.
Find the sum to n terms
(i) 3 + 33 + 333 + 3333 + …..
Solution:
Sn = 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
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q5 (i)

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

(ii) 8 + 88 + 888 + 8888 + …..
Solution:
Sn = 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
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q5 (ii)

Question 6.
Find the sum to n terms
(i) 0.4 + 0.44 + 0.444 + …..
Solution:
Sn = 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
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q6 (i)

(ii) 0.7 + 0.77 + 0.777 + ……
Solution:
Sn = 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
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q6 (ii)

Question 7.
Find the sum to n terms of the sequence
(i) 0.5, 0.05, 0.005, …..
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q7 (i)

(ii) 0.2, 0.02, 0.002, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q7 (ii)
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q7 (ii).1

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

Question 8.
For a sequence, if Sn = 2(3n – 1), find the nth term, hence showing that the sequence is a G.P.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q8

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, …, arn-1
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q9
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q9.1

Question 10.
If Sn, S2n, S3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that Sn(S3n – S2n) = (S2n – Sn)2.
Solution:
Let a and r be the 1st term and common ratio of the G.P. respectively.
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q10
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q10.1

Question 11.
Find
(i) \(\sum_{r=1}^{10}\left(3 \times 2^{r}\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q11 (i)

(ii) \(\sum_{r=1}^{10} 5 \times 3^{r}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2 Q11 (ii)

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.2

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 

Sequences and Series Class 11 Maths 2 Exercise 2.1 Solutions Maharashtra Board

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 tn.
(i) 2, 6, 18, 54, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (i)

(ii) 1, -5, 25, -125, ………
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (ii)

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \cdots\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (iii)

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

(iv) 3, 4, 5, 6, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (iv)

(v) 7, 14, 21, 28, ……
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q1 (v)

Question 2.
For the G.P.
(i) If r = \(\frac{1}{3}\), a = 9, find t7.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (i)

(ii) If a = \(\frac{7}{243}\), r = 3, find t6.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (ii)

(iii) If r = -3 and t6 = 1701, find a.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (iii)

(iv) If a = \(\frac{2}{3}\), t6 = 162, find r.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q2 (iv)

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

Question 3.
Which term of the G. P. 5, 25, 125, 625, …… is 510?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q3

Question 4.
For what values of x, the terms \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?
Solution:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q4

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:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q5
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q5.1

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,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q6
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q6.1
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.

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

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,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q7
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q7.1

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,
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q8
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:
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q9

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 \(\frac{\mathrm{q}+\mathrm{r}}{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{r}+\mathrm{s}}{\mathrm{q}+\mathrm{r}}\)
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q10
∴ p + q, q + r, r + s are in G.P.

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

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
tn = arn-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 × \(\frac{3}{4}\)
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q12

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
Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q13

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

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

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

(ii) t1 = x – 6 = 10 – 6 = 4

Maharashtra Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.1 Q15

Class 11 Maharashtra State Board Maths Solution 

Complex Numbers Class 11 Maths 2 Miscellaneous Exercise 1 Solutions Maharashtra Board

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

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 I Q2

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

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 I Q5

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
= -ω23)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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 I Q7

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.

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 I Q9

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

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

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q1 (iv)

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q1 (vi)

(vii) \(\left(1+\frac{2}{i}\right)\left(3+\frac{4}{i}\right)(5+i)^{-1}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q1 (vii)

(viii) \(\frac{\sqrt{5}+\sqrt{3 i}}{\sqrt{5}-\sqrt{3} i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q1 (viii)

(ix) \(\frac{3 i^{5}+2 i^{7}+i^{9}}{i^{6}+2 i^{8}+3 i^{18}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q1 (ix)

(x) \(\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q1 (x)

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

Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q2 (iii)

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q3 (i)

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

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

Question 4.
Find the value of
(i) x3 + 2x2 – 3x + 21, if x = 1 + 2i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q4 (i)

(ii) x4 + 9x3 + 35x2 – x + 164, if x = -5 + 4i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q4 (ii)

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 = \(\frac{15}{a}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q5 (i)

(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 = \(\frac{-4}{a}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q5 (ii)
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 = \(\frac{\sqrt{3}}{a}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q5 (iii)

(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)

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

(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 = \(\frac{-2}{a}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q5 (v)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q5 (vi)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q5 (vi).1

Question 6.
Find the modulus and argument of each complex number and express it in the polar form.
(i) 8 + 15i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (i)

(ii) 6 – i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (ii)

(iii) \(\frac{1+\sqrt{3} \mathbf{i}}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (iii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (iii).1

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

(iv) \(\frac{-1-\mathbf{i}}{\sqrt{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (iv)

(v) 2i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (v)

(vi) -3i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (vi)

(vii) \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \mathbf{i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q6 (vii)

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q7

Question 8.
Show that z = \(\frac{5}{(1-i)(2-i)(3-i)}\) is purely imaginary number.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q8

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:
\(\frac{x}{1+2 i}+\frac{y}{3+2 i}=\frac{5+6 i}{-1+8 i}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q9
(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

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q10

Question 11.
Show that \(\left(\frac{1+i}{\sqrt{2}}\right)^{8}+\left(\frac{1-i}{\sqrt{2}}\right)^{8}=2\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q11
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q11.1

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q12 (i)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q12 (i).1

(ii) z = -6 + √2 i
Solution:
z = -6 + √2 i
∴ a = -6, b = √2
i.e. a < 0, b > 0
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q12 (ii)

(iii) \(\frac{-3}{2}+\frac{3 \sqrt{3} i}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q12 (iii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q12 (iii).1

Question 13.
If x + iy = \(\frac{a+i b}{a-i b}\), prove that x2 + y2 = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q13

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

Question 14.
Show that z = \(\left(\frac{-1+\sqrt{-3}}{2}\right)^{3}\) is a rational number.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q14

Question 15.
Show that \(\frac{1-2 i}{3-4 i}+\frac{1+2 i}{3+4 i}\) is real.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q15
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q15.1

Question 16.
Simplify
(i) \(\frac{\mathrm{i}^{29}+\mathrm{i}^{39}+\mathrm{i}^{49}}{\mathrm{i}^{30}+\mathrm{i}^{40}+\mathrm{i}^{50}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q16 (i)

(ii) \(\left(\mathrm{i}^{65}+\frac{1}{\mathrm{i}^{145}}\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q16 (ii)

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q16 (iii)

Question 17.
Simplify \(\left[\frac{1}{1-2 i}+\frac{3}{1+i}\right]\left[\frac{3+4 i}{2-4 i}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q17
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q17.1

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

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.
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q18

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.

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

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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Miscellaneous Exercise 1 II Q20
Given Expression = ω18 + (ω2)18
= ω18 + ω36
= (ω3)6 + (ω3)12
= (1)6 + (1)12
= 2

Class 11 Maharashtra State Board Maths Solution 

Complex Numbers Class 11 Maths 2 Exercise 1.4 Solutions Maharashtra Board

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q1

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.

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (ii)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (iii)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (iv)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (v)

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (vi)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (vii)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (viii)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q2 (ix)

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

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

(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

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

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.
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q4 (i)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q4 (ii)

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 = \(\frac{-1-i \sqrt{3}}{2}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q5

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

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

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q6 (vi)
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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q7 (i)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q7 (i).1

(ii) \(\frac{\cos 5 \theta+i \sin 5 \theta}{(\cos 3 \theta-i \sin 3 \theta)^{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q7 (ii)

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q7 (iii)

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

Question 8.
Express the following in the form a + ib, a, b ∈ R, using De Moivre’s theorem.
(i) (1 – i)5
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (i)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (i).1

(ii) (1 + i)6
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (ii)

(iii) (1 – √3 i)4
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (iii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (iii).1

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

(iv) (-2√3 – 2i)5
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (iv)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.4 Q8 (iv).1

Class 11 Maharashtra State Board Maths Solution 

Complex Numbers Class 11 Maths 2 Exercise 1.3 Solutions Maharashtra Board

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.
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q1 (i)

(ii) √3 + √2 i
Solution:
Let z = √3 + √2 i
a = √3, b = √2,
i.e. a > 0, b > 0
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q1 (ii)

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

(iii) -8 + 15i
Solution:
Let z = -8 + 15i
a = -8, b = 15 , i.e. a < 0, b > 0
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q1 (iii)

(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 – \(\tan ^{-1}\left|\frac{b}{a}\right|\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q1 (iv)

(v) -4 – 4i
Solution:
Let z = -4 – 4i
a = -4, b = -4 , i.e. a < 0, b < 0
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q1 (v)

(vi) √3 – i
Solution:
Let z = √3 – i
a = √3, b = -1, i.e. a > 0, b < 0
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q1 (vi)

(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

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

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q2

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q3

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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (i)

(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 = \(r e^{i \theta}=e^{\pi i}\)

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

(iv) \(\frac{1}{1+i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (iv)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (iv).1
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (iv).2

(v) \(\frac{1+2 i}{1-3 i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (v)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (v).1

(vi) \(\frac{1+7 \mathbf{i}}{(2-\mathbf{i})^{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (vi)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q4 (vi).1

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (i)

(ii) \(\sqrt{2} \cdot\left(\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (ii)

(iii) \(7\left(\cos \left(-\frac{5 \pi}{6}\right)+i \sin \left(-\frac{5 \pi}{6}\right)\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (iii)

(iv) \(e^{\frac{\pi}{3} i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (iv)

(v) \(e^{\frac{-4 \pi}{3} i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (v)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (v).1

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

(vi) \([latex]e^{\frac{5 \pi}{6} i}\)[/latex]
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q5 (vi)

Question 6.
Find the modulus and argument of the complex number \(\frac{1+2 i}{1-3 i}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q6
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q6.1
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q6.2
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q6.3

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q7
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q8

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

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

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

Question 9.
z1 = 1 + i, z2 = 2 – 3i, verify the following:
(i) \(\overline{Z_{1}+Z_{2}}=\overline{Z_{1}}+\overline{Z_{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q9 (i)

(ii) \(\overline{Z_{1}-Z_{2}}=\overline{Z_{1}}-\overline{Z_{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q9 (ii)

(iii) \(\overline{Z_{1} \cdot Z_{2}}=\overline{Z_{1}} \cdot \overline{Z_{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q9 (iii)

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

(iv) \(\overline{\left(\frac{\mathbf{z}_{1}}{\mathbf{z}_{2}}\right)}=\frac{\overline{\mathbf{z}}_{1}}{\overline{\mathbf{z}}_{2}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q9 (iv)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3 Q9 (iv).1

Class 11 Maharashtra State Board Maths Solution 

Complex Numbers Class 11 Maths 2 Exercise 1.2 Solutions Maharashtra Board

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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (i)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (i).1

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (ii)

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (iii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (iii).1

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (iv)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (iv).1

(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 = \(-\frac{\sqrt{3}}{a}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q1 (v)

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

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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q2 (i)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q2 (ii)

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q2 (iii)
The roots of the given equation are \(\frac{7+\sqrt{11} \mathrm{i}}{6}\) and \(\frac{7-\sqrt{11} \mathrm{i}}{6}\)

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q2 (iv)
∴ 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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q3 (i)
∴ 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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q3 (ii)
∴ The roots of the given equation are \(\frac{1}{2}\)i and -2i.

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q3 (iii)
∴ 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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q3 (iv)
∴ The root of the given equation is -2i

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

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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (i)
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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (i).1

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (ii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (ii).1
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (ii).2

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

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (iii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (iii).1

(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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (iv)
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 = \(-\frac{1}{a}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (iv).1
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q4 (iv).2

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

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)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (i).1
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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (ii)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (ii).1
Dividend = Divisor × Quotient + Remainder
2x3 – 11x2 + 44x + 27 = (x2 – 6x + 25)(2x + 1) + 2
= 0.(2x + 1) + 2 …..[From (i)]
= 0 + 2
= 2

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

(iii) x3 + x2 – x + 22, if x = \(\frac{5}{1-2 i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (iii)
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)
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (iv)
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

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

(v) 2x4 + 5x3 + 7x2 – x + 41, if x = -2 – √3i
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.2 Q5 (v)
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 

Complex Numbers Class 11 Maths 2 Exercise 1.1 Solutions Maharashtra Board

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

11th Maths Part 2 Complex Numbers Exercise 1.1 Questions And Answers Maharashtra Board

Question 1.
Simplify:
(i) √-16 + 3√-25 + √-36 – √-625
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q1 (i)
= 4i + 3(5i) + 6i – 25i
= 25i – 25i
= 0

(ii) 4√-4 + 5√-9 – 3√-16
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q1 (ii)

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

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

(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

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

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q3 (v)

(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

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

(ii) (1 + i)(1 – i)-1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (ii)

(iii) \(\frac{i(4+3 i)}{1-i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (iii)

(iv) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (iv)

(v) \(\left(\frac{1+i}{1-1}\right)^{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (v)

(vi) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (vi)

(vii) (1 + i)-3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (vii)

(viii) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (viii)

(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

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

(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q4 (xi)

Question 5.
Show that (-1 + √3i)3 is a real number.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q5

Question 6.
Find the value of (3 + \(\frac{2}{\mathrm{i}}\)) (i6 – i7) (1 + i11).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q6

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q7

Question 8.
Show that 1 + i10 + i20 + i30 is a real number.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q8
= 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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q9

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q10

Question 11.
Find the value of 1 + i2 + i4 + i6 + i8 + …… + i20.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q11

Question 12.
Show that 1 + i10 + i100 – i1000 = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q12

Question 13.
Is (1 + i14 + i18 + i22) a real number? Justify your answer.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q13

Question 14.
Evaluate: \(\left(\mathbf{i}^{37}+\frac{1}{\mathbf{i}^{67}}\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q14

Question 15.
Prove that: (1 + i)4 × \(\left(1+\frac{1}{\mathrm{i}}\right)^{4}\) = 16
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q15

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q16

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q17

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
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q18

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

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q20
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q20.1

Question 21.
If (a + ib) = \(\frac{1+i}{1-i}\), then prove that a2 + b2 = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q21
∴ 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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q22

Question 23.
If (x + iy)3 = y + vi, then show that \(\frac{y}{x}+\frac{v}{y}\) = 4(x2 – y2).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q23

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q24 (ii)
(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:
Maharashtra Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.1 Q24 (iii)
(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

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

(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

Class 11 Maharashtra State Board Maths Solution 

Probability Class 11 Maths 1 Miscellaneous Exercise 9 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Miscellaneous Exercise 9 Questions and Answers.

11th Maths Part 1 Probability Miscellaneous Exercise 9 Questions And Answers Maharashtra Board

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

Question 1.
There are 5 girls and 2 boys, then the probability that no two boys are sitting together for a photograph is
(A) \(\frac{1}{21}\)
(B) \(\frac{4}{7}\)
(C) \(\frac{2}{7}\)
(D) \(\frac{5}{7}\)
Answer:
(D) \(\frac{5}{7}\)
Hint:
There are 5 girls and 2 boys.
They can be arranged among themselves in \({ }^{7} \mathrm{P}_{7}\) = 7! ways.
∴ Girls can be arranged among themselves in \({ }^{5} \mathrm{P}_{5}\) = 5! ways.
No two boys should sit together.
Let girls be denoted by the letter G.
– G – G – G – G – G –
There are 6 places, marked by ‘-’ where boys can sit.
∴ Boys can be arranged in
\({ }^{6} \mathrm{P}_{2}=\frac{6 !}{(6-2) !}\)
= \(\frac{6 \times 5 \times 4 !}{4 !}\)
= 30 ways.
∴ Required probability = \(\frac{5 ! \times 30}{7 !}=\frac{5 ! \times 30}{7 \times 6 \times 5 !}=\frac{5}{7}\)

Question 2.
In a jar, there are 5 black marbles and 3 green marbles. Two marbles are picked randomly one after the other without replacement. What is the possibility that both the marbles are black?
(A) \(\frac{5}{14}\)
(B) \(\frac{5}{8}\)
(C) \(\frac{5}{7}\)
(D) \(\frac{5}{16}\)
Answer:
(A) \(\frac{5}{14}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 3.
Two dice are thrown simultaneously. Then the probability of getting two numbers whose product is even is
(A) \(\frac{3}{4}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{5}{7}\)
(D) \(\frac{1}{2}\)
Answer:
(A) \(\frac{3}{4}\)
Hint:
Two dice are thrown.
∴ n(S) = 36
Getting two numbers whose product is even, i.e., one of the two numbers must be even.
Let event A: Getting even number on first dice,
event B: Getting even number on second dice.
n(A) = 18, n(B) = 18, n(A ∩ B) = 9
Required probability = P(A ∩ B)
= \(\frac{n(A)+n(B)-n(A \cap B)}{n(S)}\)
= \(\frac{18+18-9}{36}\)
= \(\frac{3}{4}\)

Question 4.
In a set of 30 shirts, 17 are white and the rest are black. 4 white and 5 black shirts are tagged as ‘PARTY WEAR’. If a shirt is chosen at random from this set, the possibility of choosing a black shirt or a ‘PARTY WEAR’ shirt is
(A) \(\frac{11}{15}\)
(B) \(\frac{13}{30}\)
(C) \(\frac{9}{13}\)
(D) \(\frac{17}{30}\)
Answer:
(D) \(\frac{17}{30}\)
Hint:
17 white + 13 black = 30 shirts
4 white and 5 black are ‘PARTY WEAR’
A: Choosing a black shirt
∴ P(A) = \(\frac{{ }^{13} C_{1}}{{ }^{30} C_{1}}=\frac{13}{30}\)
B: Choosing a ‘PARTY WEAR’ shirt.
∴ P(B) = \(\frac{{ }^{9} \mathrm{C}_{1}}{{ }^{30} \mathrm{C}_{1}}=\frac{9}{30}\)
There are 5 black ‘PARTY WEAR’ shirts.
∴ P(A ∩ B) = \(\frac{{ }^{5} \mathrm{C}_{1}}{{ }^{30} \mathrm{C}_{1}}=\frac{5}{30}\)
∴ Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{30}\) + \(\frac{9}{30}\) – \(\frac{5}{30}\)
= \(\frac{17}{30}\)

Question 5.
There are 2 shelves. One shelf has 5 Physics and 3 Biology books and the other has 4 Physics and 2 Biology books. The probability of drawing a Physics book is
(A) \(\frac{9}{14}\)
(B) \(\frac{31}{48}\)
(C) \(\frac{9}{38}\)
(D) \(\frac{1}{2}\)
Answer:
(B) \(\frac{31}{48}\)
Hint:
Let event S1: First shelve is selected,
event S2: Second shelve is selected,
event P: Drawing a physics book.
∴ P(S1) = \(\frac{1}{2}\) and P(S2) = \(\frac{1}{2}\)
First shelve has 5 physics and 3 biology books, i.e., total 8 books.
∴ P(P/S1) = \(\frac{{ }^{5} C_{1}}{{ }^{8} C_{1}}=\frac{5}{8}\)
Similarly, P(P/S2) = \(\frac{{ }^{4} C_{1}}{{ }^{6} C_{1}}=\frac{4}{6}=\frac{2}{3}\)
∴ P(P) = P(S1) . P(P/S1) + P(S2) . P(P/S2)
= \(\frac{1}{2} \times \frac{5}{8}+\frac{1}{2} \times \frac{2}{3}\)
= \(\frac{31}{48}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 6.
Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. The probability that both of them get selected is
(A) \(\frac{34}{35}\)
(B) \(\frac{1}{35}\)
(C) \(\frac{8}{35}\)
(D) \(\frac{27}{35}\)
Answer:
(C) \(\frac{8}{35}\)

Question 7.
The probability that a student knows the correct answer to a multiple-choice question is \(\frac{2}{3}\). If the student does not know the answer, then the student guesses the answer. The probability of the guessed answer being correct is \(\frac{1}{4}\). Given that the student has answered the question correctly, the probability that the student knows the correct answer is
(A) \(\frac{5}{6}\)
(B) \(\frac{6}{7}\)
(C) \(\frac{7}{8}\)
(D) \(\frac{8}{9}\)
Answer:
(D) \(\frac{8}{9}\)
Hint:
Let event A: Student knows the correct answer,
event A’: Student guesses the answer,
event B: Answer is correct.
∴ P(A) = \(\frac{2}{3}\), P(A’) = \(\frac{1}{3}\), P(B/A’) = \(\frac{1}{4}\)
Clearly, P(B/A) = 1
Required probability = P(A/B)
= \(\frac{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} / \mathrm{A})}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} / \mathrm{A})+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B} / \mathrm{A}^{\prime}\right)}\)
= \(\frac{\frac{2}{3} \times 1}{\frac{2}{3} \times 1+\frac{1}{3} \times \frac{1}{4}}\)
= \(\frac{8}{9}\)

Question 8.
The bag I contain 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. The probability that it was drawn from Bag II is
(A) \(\frac{33}{68}\)
(B) \(\frac{35}{69}\)
(C) \(\frac{34}{67}\)
(D) \(\frac{35}{68}\)
Answer:
(D) \(\frac{35}{68}\)

Question 9.
A fair die is tossed twice. What are the odds in favour of getting 4, 5, or 6 on the first toss and 1, 2, 3, or 4 on the second toss?
(A) 1 : 3
(B) 3 : 1
(C) 1 : 2
(D) 2 : 1
Answer:
(C) 1 : 2
Hint:
A fair dice is tossed twice.
∴ n(S) = 36
A: Getting 4, 5, or 6 on the first toss and Getting 1, 2, 3, or 4 on the second toss.
∴ A = {(4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{12}{36}=\frac{1}{3}\)
∴ Required answer = P(A) : P(A’) = 1 : 2

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 10.
The odds against an event are 5 : 3 and the odds in favour of another independent event are 7 : 5. The probability that at least one of the two events will occur is
(A) \(\frac{52}{96}\)
(B) \(\frac{71}{96}\)
(C) \(\frac{69}{96}\)
(D) \(\frac{13}{96}\)
Answer:
(B) \(\frac{71}{96}\)

(II) Solve the following.

Question 1.
The letters of the word ‘EQUATION’ are arranged in a row. Find the probability that
(i) all the vowels are together
(ii) arrangement starts with a vowel and ends with a consonant.
Solution:
The letters of the word EQUATION can be arranged in 8! ways.
∴ n(S) = 8!
There are 5 vowels and 3 consonants.
(i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N
Let us consider all vowels as one unit.
So, there are 4 units, which can be arranged in 4! ways.
Also, 5 vowels can be arranged among themselves in 5! ways.
∴ n(A) = 4! × 5!
Required probability = P(A)
= \(\frac{n(A)}{n(S)}\)
= \(\frac{4 ! \times 5 !}{8 !}\)
= \(\frac{1}{14}\)

(ii) B: arrangement start with a vowel and ends with a consonant.
First and last places can be filled in 5 and 3 ways respectively.
Remaining 6 letters are arranged in 6! Ways.
∴ n(B) = 5 × 3 × 6!
Required probability = P(B)
= \(\frac{n(B)}{n(S)}\)
= \(\frac{5 \times 3 \times 6 !}{8 !}\)
= \(\frac{15}{56}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 2.
There are 6 positive and 8 negative numbers. Four numbers are chosen at random, without replacement, and multiplied. Find the probability that the product is a positive number.
Solution:
Let event A: Four positive numbers are chosen,
event B: Four negative numbers are chosen,
event C: Two positive and two negative numbers are chosen.
Since four numbers are chosen without replacement,
n(A) = 6 × 5 × 4 × 3 = 360
n(B) = 8 × 7 × 6 × 5 = 1680
In event C, four numbers are to be chosen without replacement such that two numbers are positive and two numbers ate negative. This can be done in following ways:
+ + – – OR + – + – OR + – – + OR – + – + OR – – + + OR – + + –
∴ n(C) = 6 × 5 × 8 × 7 + 6 × 8 × 5 × 7 + 6 × 8 × 7 × 5 + 8 × 6 × 7 × 5 + 6 × 5 × 8 × 7 + 8 × 6 × 5 × 7
= 6 × (8 × 7 × 6 × 5)
=10080
Here, total number of numbers = 14
∴ n(S) = 14 × 13 × 12 × 11 = 24024
Since A, B, C are mutually exclusive events,
Required probability = P(A) + P(B) + P(C)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q2

Question 3.
Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly, and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
Solution:
S = {1, 2,…., 10}
∴ n(S) = 10
A: Number is more than 3.
A = {4, 5, 6, 7, 8, 9, 10}
∴ n(A) = 7
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{7}{10}\)
B: Number is even.
B = {2, 4, 6, 8, 10}
∴ A ∩ B = {4, 6, 8, 10}
∴ n(A ∩ B) = 4
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{4}{10}\)
Required probability = P(B/A)
= \(\frac{P(A \cap B)}{P(A)}\)
= \(\frac{\left(\frac{4}{10}\right)}{\left(\frac{7}{10}\right)}\)
= \(\frac{4}{7}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 4.
If A, B and C are independent events, P(A ∩ B) = \(\frac{1}{2}\), P(B ∩ C) = \(\frac{1}{3}\), P(C ∩ A) = \(\frac{1}{6}\), then find P(A), P(B) and P(C).
Solution:
Since A and B are independent events,
P(A ∩ B) = P(A) . P(B)
∴ P(A) . P(B) = \(\frac{1}{2}\) ……(i)
B and C are independent events.
∴ P(B ∩ C) = P(B) . P(C)
∴ P(B) . P(C) = \(\frac{1}{3}\) ……(ii)
A and C are independent events.
∴ P(A ∩ C) = P(A) . P(C)
∴ P(A) . P(C) = \(\frac{1}{6}\) ……(iii)
Dividing (i) by (ii), we get
\(\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})}=\frac{\frac{1}{2}}{\frac{1}{3}}\)
P(A) = \(\frac{3}{2}\) P(C) ……(iv)
Substituting equation (iv) in (iii), we get
\(\frac{3}{2}\) P(C) . P(C) = \(\frac{1}{6}\)
[P(C)]2 = \(\frac{1}{9}\)
∴ P(C) = \(\frac{1}{3}\)
Substituting P(C) = \(\frac{1}{3}\) in equation (ii), we get P(B) = 1
Substituting P(B) = 1 in equation (i), we get P(A) = \(\frac{1}{2}\)

Question 5.
If the letters of the word ‘REGULATIONS’ be arranged at random, what is the probability that there will be exactly 4 letters between R and E?
Solution:
There are 11 letters in the word ‘REGULATIONS’ which can be arranged among themselves in 11! ways.
∴ n(S) = 11!
Let event A: There will be exactly 4 letters between R and E.
R, E can occur at (1, 6), (2, 7), ….,(6, 11) positions. So, there are 6 possibilities.
Also, R and E can interchange their positions.
So, R, E can be arranged in 2 × 6 = 12 ways.
Remaining 9 letters can be arranged in 9! ways.
∴ n(A) = 12 × 9!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12 \times 9 !}{11 !}=\frac{12 \times 9 !}{11 \times 10 \times 9 !}=\frac{6}{55}\)

Question 6.
In how many ways can the letters of the word ARRANGEMENTS be arranged?
(i) Find the chance that an arrangement chosen at random begins with the letters EE.
(ii) Find the probability that the consonants are together.
Solution:
The word ‘ARRANGEMENTS’ has 12 letters in which 2A, 2E, 2N, 2R, G, M, T, S are there.
n(S) = Total number of arrangements = \(\frac{12 !}{2 ! 2 ! 2 ! 2 !}=\frac{12 !}{(2 !)^{4}}\)
(i) A: Arrangement chosen at random begins with the letters EE.
If the first and second places are filled with EE, there are 10 letters left in which 2A, 2N, 2R, G, M, T, S are there.
∴ n(A) = \(\frac{10 !}{2 ! 2 ! 2 !}=\frac{10 !}{(2 !)^{3}}\)

(ii) B: Consonants (G, M, T, S, 2N, 2R) are together.
2A, 2E, and the group containing consonants form total 5 units. Which can be arranged in \(\frac{5 !}{2 ! 2 !}\) ways.
Also, 8 consonants can be arranged among themselves in \(\frac{8 !}{2 ! 2 !}\) ways.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q6

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 7.
A letter is taken at random from the letters of the word ‘ASSISTANT’ and another letter is taken at random from the letters of the word ‘STATISTICS’. Find the probability that the selected letters are the same.
Solution:
Word ASSISTANT has 2A, I, N, 3S, 2T, and word STATISTICS has A, C, 2I, 3S, 3T.
C and N are uncommon letters.
In the words ASSISTANT, there are 9 letters out of which 2 letters are ‘A’, and in the word STATISTICS, there are 10 letters, out of which 1 letter is A.
∴ Probability of choosing A from both the letters = \(\frac{{ }^{2} C_{1}}{{ }^{9} C_{1}} \times \frac{{ }^{1} C_{1}}{{ }^{10} C_{1}}=\frac{2}{9} \times \frac{1}{10}=\frac{1}{45}\)
Similarly,
Probability of choosing I from both the letters = \(\frac{{ }^{1} \mathrm{C}_{1}}{{ }^{9} \mathrm{C}_{1}} \times \frac{{ }^{2} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{1}}=\frac{1}{9} \times \frac{2}{10}=\frac{1}{45}\)
Probability of choosing S from both the letters = \(\frac{{ }^{3} \mathrm{C}_{1}}{{ }^{9} \mathrm{C}_{1}} \times \frac{{ }^{3} \mathrm{C}_{1}}{{ }^{10} \mathrm{C}_{1}}=\frac{3}{9} \times \frac{3}{10}=\frac{1}{10}\)
Probability of choosing T from both the letters = \(\frac{{ }^{2} C_{1}}{{ }^{9} C_{1}} \times \frac{{ }^{3} C_{1}}{{ }^{10} C_{1}}=\frac{2}{9} \times \frac{3}{10}=\frac{1}{15}\)
Required probability = \(\frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}\) = \(\frac{19}{90}\)

Question 8.
A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j for j = 1, 2,….., 6. What is the probability, in one roil of the die, that an odd number of dots will turn up?
Solution:
According to the given condition, the probability of the face with 1, 2, 3, 4, 5, 6 dots turning up is proportional to 1, 2, 3, 4, 5, 6.
Let k be the common ration of proportionality.
∴ The probabilities of the faces with 1, 2, 3, 4, 5, 6 dots turning up are 1k , 2k, 3k, 4k, 5k, 6k respectively.
Since sum of the probabilities = 1,
k(1 + 2+ ….. + 6) = 1
k(\(\frac{6 \times 7}{2}\)) = 1
k = \(\frac{1}{21}\)
Required probability = P(1) + P(3) + P(5)
= \(\frac{1}{21}+\frac{3}{21}+\frac{5}{21}\)
= \(\frac{9}{21}\)
= \(\frac{3}{7}\)

Question 9.
An urn contains 5 red balls and 2 green balls. A ball is drawn. If it’s green, a red ball is added to the urn, and if it’s red, a green ball is added to the urn. (The original ball is not returned to the urn). Then a second ball is drawn. What is the probability that the second ball is red?
Solution:
A: Event of drawing a red ball and placing a green ball in the urn
B: Event of drawing a green ball and placing a red ball
C: Event of drawing a red ball in the second draw
P(A) = \(\frac{5}{7}\)
P(B) = \(\frac{2}{7}\)
P(C/A) = \(\frac{4}{7}\)
P(C/B) = \(\frac{6}{7}\)
Required probability
P(C) = P(A) P(C/A) + P(B) P(C/B)
= \(\frac{5}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{6}{7}\)
= \(\frac{32}{49}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 10.
The odds against A solving a certain problem are 4 to 3 and the odds in favour of B solving the same problem are 7 to 5, find the probability that the problem will be solved.
Solution:
The odds against A solving the problems are 4 : 3.
Let P(A’) = P(A does not solve the problem) = \(\frac{4}{4+3}=\frac{4}{7}\)
So, the probability that A solves the problem = P(A) = 1 – P(A’)
= 1 – \(\frac{4}{7}\)
= \(\frac{3}{7}\)
Similarly, let P(B) = P(B solves the problem)
Since odds in favour of B solving the problem are 7 : 5.
∴ P(B) = \(\frac{7}{7+5}=\frac{7}{12}\)
Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Since A and B are independent events.
∴ P(A ∩ B) = P(A) . P(B)
∴ Required probability = P(A) + P(B) – P(A) . P(B)
= \(\frac{3}{7}+\frac{7}{12}-\frac{3}{7} \times \frac{7}{12}\)
= \(\frac{16}{21}\)

Question 11.
If P(A) = P(A/B) = \(\frac{1}{5}\), P(B/A) = \(\frac{1}{3}\), then find
(i) P(A’/B)
(ii) P(B’/A’)
Solution:
Since P(A) = P(A/B) = \(\frac{1}{5}\)
P(A) = \(\frac{1}{5}\)
and \(\frac{P(A \cap B)}{P(B)}=\frac{1}{5}\)
∴ P(A) = \(\frac{1}{5}\) ……(i)
P(B) = 5 P(A ∩ B) ……..(ii)
Since P(B/A) = \(\frac{1}{3}\)
\(\frac{P(A \cap B)}{P(A)}=\frac{1}{3}\)
∴ P(A) = 3 P(A ∩ B) ………(iii)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q11
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q11.1

Question 12.
Let A and B be independent events with P(A) = \(\frac{1}{4}\) and P(A ∪ B) = 2P(B) – P(A). Find
(i) P(B)
(ii) P(A/B)
(iii) P(B’/A)
Solution:
A and B are independent events. .
∴ P(A ∩ B) = P(A) × P(B)
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∪ B) = P(A) + P(B) – P(A) × P(B)
∴ 2P(B) – P(A) = P(A) + P(B) – P(A) × P(B) ……[∵ P(A ∪ B) = 2P(B) – P(A)]
∴ 2P(B) – \(\frac{1}{4}\) = \(\frac{1}{4}\) + P(B) – \(\frac{1}{4}\) × P(B)
∴ 2P(B) – P(B) + \(\frac{1}{4}\) P(B) = \(\frac{1}{4}\) + \(\frac{1}{4}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q12

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 13.
Find the probability that a year selected will have 53 Wednesdays.
Solution:
A leap year comes after 3 years.
∴ The probability of a year being a leap year = \(\frac{1}{4}\)
∴ Probability of a year being a non-leap year = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
In a non-leap year, there are 52 weeks and one extra day, whereas a leap year has 52 weeks and 2 extra days.
∴ 53rd Wednesday’s chance in a non-leap year = \(\frac{1}{7}\)
Two extra days of a leap year can be
(Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)
∴ There are 2 possibilities of 53rd Wednesday in a leap year.
∴ 53rd Wednesday’s chance in a leap year = \(\frac{2}{7}\)
Required probability = P(a non-leap year and Wednesday) + P(a leap year and Wednesday)
= \(\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7}\)
= \(\frac{5}{28}\)

Question 14.
The chances of P, Q and R, getting selected as principal of a college are \(\frac{2}{5}\), \(\frac{2}{5}\), \(\frac{1}{5}\) respectively. Their chances of introducing IT in the college are \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) respectively. Find the probability that
(a) IT is introduced in the college after one of them is selected as a principal.
(b) IT is introduced by Q.
Solution:
Let event P: P become principal,
event Q: Q become principal,
event R: R become principal,
event E: Subject IT is introduced.
Given, P(P) = \(\frac{2}{5}\)
P(Q) = \(\frac{2}{5}\)
P(R) = \(\frac{1}{5}\)
P(E/P) = \(\frac{1}{2}\)
P(E/Q) = \(\frac{1}{3}\)
P(E/R) = \(\frac{1}{4}\)
(a) Required probability
P(E) = P(P) P(E/P) + P(Q) P(E/Q) + P(R) P(E/R)
= \(\frac{2}{5} \times \frac{1}{2}+\frac{2}{5} \times \frac{1}{3}+\frac{1}{5} \times \frac{1}{4}\)
= \(\frac{1}{5}+\frac{2}{15}+\frac{1}{20}\)
= \(\frac{12+8+3}{60}\)
= \(\frac{23}{60}\)

(b) Required probability = P(Q/E)
By Bayes’ theorem,
P(Q/E) = \(\frac{P(Q) P(E / Q)}{P(E)}\)
= \(\frac{\frac{2}{5} \times \frac{1}{3}}{\frac{23}{60}}\)
= \(\frac{8}{23}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 15.
Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement What is the probability that we are lucky and find both of the defective fuses in the first two tests?
Solution:
Number of fuses = 5 + 2 = 7
Testing two fuses one-by-one at random, without replacement from 7 can be done in \({ }^{7} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}\) ways.
∴ n(S) = \({ }^{7} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}\) = 7 × 6 = 42
Let event A: Getting defective fuses in the first two tests without replacement.
There are two defective fuses.
∴ n(A) = \({ }^{2} \mathrm{C}_{1} \times{ }^{1} \mathrm{C}_{1}\) = 2 × 1 = 2
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{42}=\frac{1}{21}\)

Question 16.
For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = \(\frac{2}{3}\), P(B ∪ C) = \(\frac{3}{4}\), P(A ∪ B ∪ C) = \(\frac{11}{12}\). Find P(A), P(B) and P(C).
Solution:
Let P(A) = x, P(B) = y, P(C) = z
Since A, B are disjoint,
A ∩ B = Φ and A ∩ B ∩ C = Φ
∴ P(A ∩ B) = 0, P(A ∩ B ∩ C) = 0 ……(i)
Since A and C are independent,
P(A ∩ C) = P(A) P(C) = xz
Since B and C are independent,
P(B ∩ C) = P(B) P(C) = yz
P(A ∪ C) = P(A) + P(C) – P(A ∩ C)
∴ \(\frac{2}{3}\) = x + z – xz ……..(ii)
P(B ∪ C) = P(B) + P(C) – P(B ∩ C)
∴ \(\frac{3}{4}\) = y + z – yz ………(iii)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
\(\frac{11}{12}\) = x + y + z – 0 – yz – zx + 0 …… [From(i)]
= (x + z – xz) + (y + z – yz) – z
= \(\frac{2}{3}+\frac{3}{4}\) – z ……. [From (ii) and (iii)]
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q16

Question 17.
The ratio of boys to girls in a college is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 of that college are good singers. A good singer is chosen. What is the probability that the chosen singer is a girl?
Solution:
Let event S: The student is a good singer,
event B: The student is a boy,
event G: The student is a girl.
Since the ratio of boys to girls is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 are good singers.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q17

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 18.
A and B throw a die alternatively till one of them gets a 3 and wins the game. Find the respective probabilities of winning. (Assuming A begins the game).
Solution:
Since P(getting 3) = \(\frac{1}{6}\),
P(not getting 3) = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
In 1st throw if A gets 3, A wins
∴ P(A win) = \(\frac{1}{6}\)
In 2nd throw by B (i.e., A does not get 3),
∴ P(B wins) = \(\frac{5}{6} \times \frac{1}{6}\)
In 3rd throw by A, P(A wins) = \(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\)
(3rd throw by A shows that B has lost in 2nd throw) and so on.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q18

Question 19.
Consider independent trials consisting of rolling a pair of fair dice, over and over. What is the probability that a sum of 5 appears before a sum of 7?
Solution:
When two dice are thrown, the sample space is
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let event A: The sum is 5 in a trial.
A = {(2, 3), (3, 2), (1, 4), (4, 1)}
∴ P(A) = \(\frac{4}{36}=\frac{1}{9}\)
Let event B: The sum is 7 in a trial.
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}
∴ P(B) = \(\frac{6}{36}=\frac{1}{6}\)
Let event C: Neither sum is 5 nor 7.
P(C) = 1 – P(A) – P(B)
= 1 – \(\frac{1}{9}\) – \(\frac{1}{6}\)
= \(\frac{26}{36}\)
Let the sum of 5 appear in the nth trial for the first time and the sum of 7 has not occurred in the first (n – 1) trials.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q19

Question 20.
A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which is able to detect any part that is obviously defective and discard it. What is the probability that the quality of the parts that make it through the inspection machine and get shipped?
Solution:
Let event G: The event that machine produces a good part,
event S: The event that machine produces a slightly defective part,
event D: The event that machine produces an obviously defective part.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q20

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 21.
Given three identical boxes, I, II, and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
Solution:
Let event B1: Select box I having two gold coins.
event B2: Selecting box II having two silver coins,
event B3: Selecting box III having one silver and one gold coin,
event G: Coin is gold.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q21
To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from the box I.
∴ Required probability = P(B1/G)
By Bayes’ theorem,
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q21.1

Question 22.
In a factory which manufactures bulbs, machines A, B, and C manufacture respectively 25%, 35% and 40% of the bulbs. Of their outputs, 5, 4, and 2 percent are respectively defective bulbs. A bulb is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by machine B?
Solution:
Let event A: Bulb manufactured by machine A
event B: Bulb manufactured by machine B
event C: Bulb manufactured by machine C
event D: Bulb defective
∴ P(A) = \(\frac{25}{100}\)
P(B) = \(\frac{35}{100}\)
P(C) = \(\frac{40}{100}\)
Machines A, B and C manufacture respectively 25%, 35% and 40% of the bulbs.
Of their outputs, 5, 4, and 2 percent are respectively defective bulbs.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q22
Required probability = P(B/D)
By Bayes’ theorem,
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q22.1

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9

Question 23.
A family has two children. One of them is chosen at random and found that the child is a girl. Find the probability that
(i) both the children are girls.
(ii) both the children are girls given that at least one of them is a girl.
Solution:
A family has two children.
∴ Sample space S = {BB, BG, GB, GG}
(i) A: First child is a girl.
∴ A = {GB, GG}
∴ P(A) = \(\frac{2}{4}=\frac{1}{2}\)
B: Second child is a girl.
∴ B = {BG, GG}
∴ A ∩ B = {GG}
∴ P(A ∩ B) = \(\frac{1}{4}\)
Required probability
P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\)

(ii) A: At least one of the children is a girl.
∴ A = {GG, GB, BG}
∴ P(A) = \(\frac{3}{4}\)
B: both children are girls.
B = {GG}
∴ P(B) = \(\frac{1}{4}\)
Also, A ∩ B = B
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Miscellaneous Exercise 9 II Q23

Class 11 Maharashtra State Board Maths Solution  

Probability Class 11 Maths 1 Exercise 9.5 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.5 Questions and Answers.

11th Maths Part 1 Probability Exercise 9.5 Questions And Answers Maharashtra Board

Question 1.
If odds in favour of X solving a problem are 4 : 3 and odds against Y solving the same problem are 2 : 3. Find the probability of:
(i) X solving the problem
(ii) Y solving the problem
Solution:
(i) Odds in favour of X solving a problem are 4 : 3.
∴ The probability of X solving the problem is
P(X) = \(\frac{4}{4+3}=\frac{4}{7}\)

(ii) Odds against Y solving the problem are 2 : 3.
∴ The probability of Y solving the problem is
P(Y) = 1 – P(Y’)
= 1 – \(\frac{2}{2+3}\)
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5

Question 2.
The odds against John solving a problem are 4 to 3 and the odds in favour of Rafi solving the same problem are 7 to 5. What is the chance that the problem is solved when both of them try it?
Solution:
The odds against John solving a problem are 4 to 3.
Let event P(A’) = P (John does not solve the problem)
= \(\frac{4}{4+3}\)
= \(\frac{4}{7}\)
So, the probability that John solves the problem
P(A) = 1 – P(A’) = 1 – \(\frac{4}{7}\) = \(\frac{3}{7}\)
Similarly, Let P(B) = P(Rafi solves the problem)
Since the odds in favour of Rafi solving the problem are 7 to 5,
P(B) = \(\frac{7}{7+5}\) = \(\frac{7}{12}\)
Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Since A, B are independent events,
P(A ∩ B) = P(A) . P(B)
∴ Required probability = P(A) + P(B) – P(A) . P(B)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5 Q2

Question 3.
The odds against student X solving a statistics problem are 8 : 6 and odds in favour of student Y solving the same problem are 14 : 16. Find the chance that
(i) the problem will be solved if they try it independently.
(ii) neither of them solves the problem.
Solution:
The odds against X solving a problem are 8 : 6.
Let P(X’) = P(X does not solve the problem) = \(\frac{8}{8+6}\) = \(\frac{8}{14}\)
So, the probability that X solves the problem
P(X) = 1 – P(X’) = 1 – \(\frac{8}{14}\) = \(\frac{6}{14}\)
Similarly, let P(Y) = P(Y solves the problem)
Since odds in favour of Y solving the problem are 14 : 16,
P(Y) = \(\frac{14}{14+16}=\frac{14}{30}\)
So, the probability that Y does not solve the problem
P(Y’) = 1 – P(Y)
= 1 – \(\frac{14}{30}\)
= \(\frac{16}{30}\)
(i) Required probability
P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)
Since X and Y are independent events,
P(X ∩ Y) = P(X) . P(Y)
∴ Required probability = P(X) + P(Y) – P(X) . P(Y)
= \(\frac{6}{14}+\frac{14}{30}-\frac{6}{14} \times \frac{14}{30}\)
= \(\frac{73}{105}\)

(ii) Required probability = P(X’ ∩ Y’)
Since X and Y are independent events, X’ and Y’ are also independent events.
∴ Required probability = P(X’) . P(Y’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5

Question 4.
The odds against a husband who is 60 years old, living till he is 85 are 7 : 5. The odds against his wife who is now 56, living till she is 81 are 5 : 3. Find the probability that
(i) at least one of them will be alive 25 years hence.
(ii) exactly one of them will be alive 25 years hence.
Solution:
The odds against her husband living till he is 85 are 7 : 5.
Let P(H’) = P(husband dies before he is 85) = \(\frac{7}{7+5}=\frac{7}{12}\)
So, the probability that the husband would be alive till age 85
P(H) = 1 – P(H’) = 1 – \(\frac{7}{12}\) = \(\frac{5}{12}\)
Similarly, P(W’) = P(Wife dies before she is 81)
Since the odds against wife will be alive till she is 81 are 5 : 3.
∴ P(W’) = \(\frac{5}{5+3}=\frac{5}{8}\)
So, the probability that the wife would be alive till age 81
P(W) = 1 – P(W’) = 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\)
(i) Required probability
P(H ∪ W) = P(H) + P(W) – P(H ∩ W)
Since H and W are independent events,
P(H ∩ W) = P(H) . P(W)
∴ Required probability = P(H) + P(W) – P(H) . P(W)
= \(\frac{5}{12}+\frac{3}{8}-\frac{5}{12} \times \frac{3}{8}\)
= \(\frac{40+36-15}{96}\)
= \(\frac{61}{96}\)

(ii) Required probability = P(H ∩ W’) + P(H’ ∩ W)
Since H and W are independent events, H’ and W’ are also independent events.
∴ Required probability = P(H) . P(W’) + P(H’) . P(W)
= \(\frac{5}{12} \times \frac{5}{8}+\frac{7}{12} \times \frac{3}{8}\)
= \(\frac{25+21}{96}\)
= \(\frac{46}{96}\)
= \(\frac{23}{48}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5

Question 5.
There are three events A, B, and C, one of which must, and only one can happen. The odds against event A are 7 : 4 and odds against event B are 5 : 3. Find the odds against event C.
Solution:
Since odds against A are 7 : 4,
P(A) = \(\frac{4}{7+4}=\frac{4}{11}\)
Since odds against B are 5 : 3,
P(B) = \(\frac{3}{5+3}=\frac{3}{8}\)
Since only one of the events A, B and C can happen,
P(A) + P(B) + P(C) = 1
\(\frac{4}{11}\) + \(\frac{3}{8}\) + P(C) = 1
∴ P(C) = 1 – (\(\frac{4}{11}\) + \(\frac{3}{8}\))
= 1 – \(\left(\frac{32+33}{88}\right)\)
= \(\frac{23}{88}\)
∴ P(C’) = 1 – P(C)
= 1 – \(\frac{23}{88}\)
= \(\frac{65}{88}\)
∴ Odds against the event C are P(C’) : P(C)
= \(\frac{65}{88}\) : \(\frac{23}{88}\)
= 65 : 23

Question 6.
In a single toss of a fair die, what are the odds against the event that number 3 or 4 turns up?
Solution:
When a fair die is tossed, the sample space is
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let event A: 3 or 4 turns up.
∴ A = {3, 4}
∴ n(A) = 2
∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{6}=\frac{1}{3}\)
P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
∴ Odds against the event A are P(A’) : P(A)
= \(\frac{2}{3}: \frac{1}{3}\)
= 2 : 1

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5

Question 7.
The odds in favour of A winning a game of chess against B are 3 : 2. If three games are to be played, what are the odds in favour of A’s winning at least two games out of the three?
Solution:
Let event A: A wins the game and event B: B wins the game.
Since the odds in favour of A winning a game against B are 3 : 2,
the probability of occurrence of event A and B is given by
P(A) = \(\frac{3}{3+2}=\frac{3}{5}\) and P(B) = \(\frac{2}{3+2}=\frac{2}{5}\)
Let event E: A wins at least two games out of three games.
∴ P(E) = P(A) . P(A) . P(B) + P(A) . P(B) . P(A) + P(B) . P(A) . P(A) + P(A) . P(A) . P(A)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5 Q7
∴ Odds in favour of A’s winning at least two games out of three are P(E) : P(E’)
= \(\frac{81}{125}: \frac{44}{125}\)
= 81 : 44

Class 11 Maharashtra State Board Maths Solution  

Probability Class 11 Maths 1 Exercise 9.4 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.4 Questions and Answers.

11th Maths Part 1 Probability Exercise 9.4 Questions And Answers Maharashtra Board

Question 1.
There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and marble is picked from the chosen bag. What is the probability that the chosen marble is red?
Solution:
Let event R: Chosen marble is red.
Let event Bi: ith bag is chosen.
∴ P(Bi) = \(\frac{1}{3}\)
If Bag 1 is chosen, it has 75 red and 25 blue marbles.
∴ Probability that the chosen marble is red under the condition that it is from Bag 1 = P(R/B1)
= \(\frac{{ }^{75} \mathrm{C}_{1}}{{ }^{100} \mathrm{C}_{1}}\)
= \(\frac{75}{100}\)
= 0.75
Similarly we get,
P(R/B2) = \(\frac{60}{100}\) = 0.60
P(R/B3) = \(\frac{45}{100}\) = 0.45
∴ Required probability
P(R) = P(B1) P(R/B1) + P(B2) P(R/B2) + P(B3) P(R/B3)
= \(\frac{1}{3}\)(0.75) + \(\frac{1}{3}\)(0.60) + \(\frac{1}{3}\)(0.45)
= \(\frac{1}{3}\)(1.8)
= 0.60

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4

Question 2.
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from
(i) first box
(ii) second box
Solution:
Let event A1: The ball is drawn from 1st box and
event A2: The ball is drawn from the 2nd box.
∴ P(A1) = \(\frac{1}{2}\), P(A2) = \(\frac{1}{2}\)
Let event B: The ball drawn is pink.
There are 5 balls in the 1st box, of which 3 are pink.
∴ P(B/A1) = \(\frac{3}{5}\)
There are 9 balls in the 2nd box, of which 5 are pink.
∴ P(B/A2) = \(\frac{5}{9}\)
(i) By Bayes’ theorem,
the probability that a pink ball is drawn from the first box, is given by
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q2

(ii) By Bayes’ theorem,
the probability that a pink ball is drawn from the second box, is given by
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q2.1

Question 3.
There is a working women’s hostel in a town, where 75% are from neighbouring town. The rest all are from the same town. 48% of women who hail from the same town are graduates and 83% of the women who have come from the neighbouring town are also graduates. Find the probability that a woman selected at random is a graduate from the same town.
Solution:
Let the total number of women be 100.
∴ n(S) = 100
Let event N: Women are from neighbouring town,
event W: Women are from same town and
event G: Women are graduates.
Number of women from neighbouring town,
n(N) = 75
Number of women from same town,
n(W) = 25
∴ P(N) = \(\frac{n(N)}{n(S)}=\frac{75}{100}\) and
P(W) = \(\frac{n(W)}{n(S)}=\frac{25}{100}\)
P(G/N), P(G/W) represent probabilities that woman is graduate given that she is from neighbouring town or same town respectively.
∴ P(G/N) = \(\frac{\mathrm{n}(\mathrm{G} / \mathrm{N})}{\mathrm{n}(\mathrm{S})}=\frac{83}{100}\) and
P(G/W) = \(\frac{\mathrm{n}(\mathrm{G} / \mathrm{W})}{\mathrm{n}(\mathrm{S})}=\frac{48}{100}\)
By Bayes’ theorem, the probability that a women selected at random is a graduate from the same town, is given by
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q3
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q3.1

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4

Question 4.
If E1 and E2 are equally likely, mutually exclusive and exhaustive events and P(A/E1) = 0.2, P(A/E2) = 0.3. Find P(E1/A).
Solution:
E1 and E2 are equally likely, mutually exclusive and exhaustive events.
∴ P(E1) = P(E2) = \(\frac{1}{2}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q4

Question 5.
Jar I contains 5 white and 7 black balls. Jar II contains 3 white and 12 black balls. A fair coin is flipped; if it is Head, a ball is drawn from Jar I, and if it is Tail, a ball is drawn from Jar II. Suppose that this experiment is done and a white ball was drawn. What is the probability that this ball was in fact taken from Jar II?
Solution:
Let event J1: Ball drawn from jar I,
event J2: Ball drawn from jar II.
P(J1) = P(head) = \(\frac{1}{2}\)
P(J2) = P(tail) = \(\frac{1}{2}\)
Let event W: Ball drawn is white.
In Jar I, there are total 12 balls, out of which 5 balls are white.
∴ Probability that the ball drawn is white under the condtion that it is drawn from Jar I.
P(W/J1) = \(\frac{{ }^{5} C_{1}}{{ }^{12} C_{1}}=\frac{5}{12}\)
Similarly, P(W/J2) = \(\frac{{ }^{3} C_{1}}{{ }^{15} C_{1}}=\frac{3}{15}=\frac{1}{5}\)
Required probability = P(J2/W)
By Bayes’ theorem,
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q5

Question 6.
A diagnostic test has a probability 0.95 of giving a positive result when applied to a person suffering from a certain disease, and a probability 0.10 of giving a (false) positive result when applied to a non-sufferer. It is estimated that 0.5% of the population are sufferers. Suppose that the test is now administered to a person about whom we have no relevant information relating to the disease (apart from the fact that he/she comes from this population). Calculate the probability that:
(i) given a positive result, the person is a sufferer.
(ii) given a negative result, the person is a non-sufferer.
Solution:
Let event T: Test positive
event S: Sufferer
P(S) = \(\frac{0.5}{100}\) = 0.005
∴ P(S’) = 1 – P(S) = 1 – 0.005 = 0.995
Since a probability of getting a positive result when applied to a person suffering from a disease is 0.95 and probability of getting positive result when applied to a non sufferer is 0.10.
∴ P(T/S) = 0.95 and P(T/S’) = 0.10
∴ P(T) = P(S) P(T/S) + P(S’) P(T/S’)
= 0.005 × 0.95 + 0.995 × 0.10
= 0.10425
∴ P(T’) = 1 – P(T) = 1 – 0.10425 = 0.8958
(i) Required probability = P(S/T)
By Bayes’ theorem,
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q6

(ii) P(T’/S’) = 1 – 0.1 = 0.9
Required probability = P(S’/T’)
By Bayes’ theorem
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q6.1

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4

Question 7.
A doctor is called to see a sick child. The doctor has prior information that 80% of the sick children in that area have the flu, while the other 20% are sick with measles. Assume that there is no other disease in that area. A well-known symptom of measles is rash. From the past records, it is known that, chances of having rashes given that sick child is suffering from measles is 0.95. However occasionally children with flu also develop rash, whose chance are 0.08. Upon examining the child, the doctor finds a rash. What is the probability that child is suffering from measles?
Solution:
Let the total number of sick children be 100.
∴ n(S) = 100.
Let event A: The child is sick with flu,
event B: The child is sick with measles,
event C: The child is sick with rash.
∴ n(A) = 80 and n(B) = 20
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{80}{100}=\frac{4}{5}\)
P(B) = \(\frac{n(B)}{n(S)}=\frac{20}{100}=\frac{1}{5}\)
Since the chances of having rashes, if the child is suffering from measles is 0.95 and the chances of having rashes if the child has flu is 0.08,
P(C/B) = 0.95 = \(\frac{95}{100}\) and
P(C/A) = 0.08 = \(\frac{8}{100}\)
Required probability = P(B/C)
By Bayes’ theorem,
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q7

Question 8.
2% of the population have a certain blood disease of a serious form: 10% have it in a mild form; and 88% don’t have it at all. A new blood test is developed; the probability of testing positive is \(\frac{9}{10}\) if the subject has the
serious form, \(\frac{6}{10}\) if the subject has the mild form, and \(\frac{1}{10}\) if the subject doesn’t have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?
Solution:
Let event A1: Disease in serious form,
event A2: Disease in mild form,
event A3: Subject does not have disease,
event B: Subject tests positive.
P(A1) = 0.02, P(A2) = 0.1, P(A3) = 0.88
The probability of testing positive is \(\frac{9}{10}\) if the subject has the serious form, \(\frac{6}{10}\) if the subject has the mild form, and \(\frac{1}{10}\) if the subject doesn’t have the disease.
∴ P(B/A1) = 0.9, P(B/A2) = 0.6, P(B/A3) = 0.1
P(B) = P(A1) P(B/A1) + P(A2) P(B/A2) + P(A3) P(B/A3)
= 0.02 × 0.9 + 0.1 × 0.6 + 0.88 × 0.1
= 0.166
Required probability = P(A1/B)
By Baye’s theorem
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q8

Question 9.
A box contains three coins: two fair coins and one fake two-headed coin. A coin is picked randomly from the box and tossed.
(i) What is the probability that it lands head up?
(ii) If happens to be head, what is the probability that it is the two-headed coin?
Solution:
Let event A: Fair coin is tossed,
event B: Fake coin is tossed
and event H: Head occur.
Clearly, a fair coin has one head.
∴ Probability that head occur under the condition that the fair coin is tossed = P(H/A) = \(\frac{1}{2}\)
Fake coin has two heads.
∴ Probability that head occur under the condition that the fake coin is tossed = P(H/B) = 1
n(A) = 2, n(B) = 1, n(S) = 3
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{3}\)
P(B) = \(\frac{n(B)}{n(S)}=\frac{1}{3}\)
(i) Required probability
P(H) = P(A) P(H/A) + P(B) P(H/B)
= \(\frac{2}{3} \times \frac{1}{2}+\frac{1}{3} \times 1\)
= \(\frac{1}{3}+\frac{1}{3}\)
= \(\frac{2}{3}\)

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4

(ii) Required probability = P(B/H)
By Baye’s theorem
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q9

Question 10.
There are three social media groups on a mobile: Group I, Group II and Group III. The probabilities that Group I, Group II and Group III sending the messages on sports are \(\frac{2}{5}\), \(\frac{1}{2}\) and \(\frac{2}{3}\) respectively. The probability of opening the messages by Group I, Group II and Group III are \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{4}\) respectively. Randomly one of the messages is opened and found a message on sports. What is the probability that the message was from Group III.
Solution:
Let event A: Message sent on sports by group I,
event B: Message sent on sports by group II,
event C: Message sent on sports by group III,
event E: Message is opened.
Given that the probabilities that Group I, Group II and Group III sending the messages on sports are \(\frac{2}{5}\), \(\frac{1}{2}\) and \(\frac{2}{3}\) respectively and the probability of opening the messages by Group I, Group II and Group III are \(\frac{1}{2}\), \(\frac{1}{4}\) and \(\frac{1}{4}\) respectively.
∴ P(A) = \(\frac{2}{5}\)
P(B) = \(\frac{1}{2}\)
P(C) = \(\frac{2}{3}\)
P(E/A) = \(\frac{1}{2}\)
P(E/B) = \(\frac{1}{4}\)
P(E/C) = \(\frac{1}{4}\)
Required probability = P(C/E)
By Baye’s theorem
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q10

Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4

Question 11.
Mr. X goes to office by Auto, Car and train. The probabilities of him travelling by these modes are \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{2}{7}\) respectively. The chances of him being late to the office are \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{4}\) respectively by Auto, Car and train. On one particular day he was late to the office. Find the probability that he travelled by car.
Solution:
Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively.
Let L be event that he is late.
Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.4 Q11

Class 11 Maharashtra State Board Maths Solution