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Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2

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

(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

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

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

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

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

∴ the roots of the given equation are \(\frac{-1+\sqrt{7} \mathrm{i}}{8}\) and \(\frac{-1-\sqrt{7} \mathrm{i}}{8}\)

(ii) 2x² – √3x + 1 = 0
Solution:
Given equation is 2x² – √3x + 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

∴ the roots of the given equation are \(\frac{\sqrt{3}+\sqrt{5} i}{4}\) and \(\frac{\sqrt{3}-\sqrt{5} i}{4}\)

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

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

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

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

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

∴ x = 2i or x = -5i
∴ the roots of the given equation are 2i and -5i.
Check:
If x = 2i and x = -5i satisfy the given equation, then our answer is correct.
L.H.S. = x² + 3ix + 10
= (2i)² + 3i(2i) + 10i
= 4i² + 6i² + 10
= 10i² + 10
= -10 + 10 ……[∵ i² = -1]
= 0
= R.H.S.
L.H.S. = x² + 3ix + 10
= (-5i)² + 3i(-5i) + 10
= 25i² – 15i² + 10
= 10i² + 10
= -10 + 10 …..[∵ i² = -1]
= 0
= R.H.S.
Thus, our answer is correct.

(ii) 2x² + 3ix + 2 = 0
Solution:
Given equation is 2x² + 3ix + 2 = 0
Comparing with ax² + bx + c = 0, we get
a = 2, b = 3i, c = 2
Discriminant = b² – 4ac
= (3i)² – 4 × 2 × 2
= 9i² – 16
= -9 – 16
= -25 < 0
So, the given equation has complex roots.
These roots are given by

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

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

∴ the roots of the given equation are -2i and -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 …..[∵ i² = -1]
= 0
So, the given equation has equal roots.
These roots are given by

∴ the roots of the given equation are -2i and -2i.

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

(ii) x² – (3√2 + 2i) x + 6√2i = 0
Solution:
Given equation is x² – (3√2 + 2i) x + 6√2i = 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√2i
= 18 + 12√2i + 4i² – 24√2i
= 18 – 12√2i – 4 …..[∵ i² = -1]
= 14 – 12√2i
So, the given equation has complex roots.
These roots are given by

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

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

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 1.
Write the conjugates of the following complex numbers:
(i) 3 + i
(ii) 3 – i
(iii) -√5 – √7i
(iv) -√-5
(v) 5i
(vi) √5 – i
(vii) √2 + √3i
Solution:
(i) Conjugate of (3 + i) is (3 – i)
(ii) Conjugate of (3 – i) is (3 + i)
(iii) Conjugate of (-√5 – √7i) is (-√5 + √7i)
(iv) -√-5 = -√5 × √-1 = -√5i
Conjugate of -√-5 is √5i
(v) Conjugate of 5i is -5i
(vi) Conjugate of √5 – i is √5 + i
(vii) Conjugate of √2 + √3i is √2 – √3i

Question 2.
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)
(ii) \(\frac{\mathrm{i}(4+3 \mathrm{i})}{(1-\mathrm{i})}\)
(iii) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
(iv) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
(v) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
(vi) (2 + 3i)(2 – 3i)
(vii) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)
Solution:

Question 3.
Show that (-1 + √3i)³ is a real number.
Solution:
(-1 + √3i)³
= (-1)³ + 3(-1)² (√3i) + 3(-1)(√3i)² +(√3i)³ [∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= -1 + 3√3i – 3(3i²) + 3√3 i³
= -1 + 3√3i – 3(-3) – 3√3i [∵ i² = -1, i³ = -1]
= -1 + 9
= 8, which is a real number.

Question 4.
Evaluate the following:
(i) i³⁵
(ii) i⁸⁸⁸
(iii) i⁹³
(iv) i¹¹⁶
(v) i⁴⁰³
(vi) \(\frac{1}{i^{58}}\)
(vii) i³⁰ + i⁴⁰ + i⁵⁰ + i⁶⁰
Solution:
We know that, i² = -1, i³ = -i, i⁴ = 1
(i) i³⁵ = (i⁴)⁸ (i²) i = (1)⁸ (-1) i = -i
(ii) i⁸⁸⁸ = (i⁴)²²² = (1)²²² = 1
(iii) i⁹³ = (i⁴)²³ . i = (1)²³ . i = i
(iv) i¹¹⁶ = (i⁴)²⁹ = (1)²⁹ = 1
(v) i⁴⁰³ = (i⁴)¹⁰⁰ (i²) i = (1)¹⁰⁰ (-1) i = -i
(vi) \(\frac{1}{i^{88}}=\frac{1}{\left(i^{4}\right)^{14} \cdot i^{2}}=\frac{1}{(1)^{14}(-1)}=-1\)
(vii) i³⁰ + i⁴⁰ + i⁵⁰ + i⁶⁰
= (i⁴)⁷ i² + (i⁴)¹⁰ + (i⁴)¹² i² + (i⁴)¹⁵
= (1)⁷ (-1) + (1)¹⁰ + (1)¹² (-1) + (1)¹⁵
= -1 + 1 – 1 + 1
= 0

Question 5.
Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.
Solution:
1 + i¹⁰ + i²⁰ + i³⁰
= 1 + (i⁴)² . i² + (i⁴)⁵ + (i⁴)⁷ . i²
= 1 + (1)² (-1) + (1)⁵ + (1)⁷ (-1) [∵ i⁴ = 1, i² = -1]
= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 6.
Find the value of
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
(ii) i + i² + i³ + i⁴
Solution:
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
= (i⁴)¹² . i + (i⁴)¹⁷ + (i⁴)²² . i + (i⁴)²⁷ . i²
= (1)¹² . i + (1)¹⁷ + (1)²² . i + (1)²⁷(-1) ……[∵ i⁴ = 1, i² = -1]
= i + 1 + i – 1
= 2i

(ii) i + i² + i³ + i⁴
= i + i² + i² . i + i⁴
= i – 1 – i + 1 [∵ i² = -1, i⁴ = 1]
= 0

Question 7.
Find the value of 1 + i² + i⁴ + i⁶ + i⁸ + …… + i²⁰.
Solution:
1 + i² + i⁴ + i⁶ + i⁸ + ….. + i²⁰
= 1 + (i² + i⁴) + (i⁶ + i⁸) + (i¹⁰ + i¹²) + (i¹⁴ + i¹⁶) + (i¹⁸ + i²⁰)
= 1 + [i² + (i²)²] + [(i²)³ + (i²)⁴] + [(i²)⁵ + (i²)⁶] + [(i²)⁷ + (i²)⁸] + [(i²)⁹ + (i²)¹⁰]
= 1 + [-1 + (- 1)²] + [(-1)³ + (-1)⁴] + [(-1)⁵ + (-1)⁶] + [(-1)⁷ + (-1)⁸] + [(-1)⁹ + (-1)¹⁰] [∵ i² = -1]
= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1)
= 1 + 0 + 0 + 0 + 0 + 0
= 1

Question 8.
Find the values of x and y which satisfy the following equations (x, y ∈ R):
(i) (x + 2y) + (2x – 3y)i + 4i = 5
(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)
Solution:
(i) (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) × 2 – equation (ii) gives
7y = 14
∴ y = 2
Putting y- 2 in (i), we get
x + 2(2) = 5
∴ x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2
Check:
If x = 1 and y = 2 satisfy the given condition, then our answer is correct.
L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.
Thus, our answer is correct.

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

(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
Putting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

Question 9.
Find the value of:
(i) x³ – x² + x + 46, if x = 2 + 3i
(ii) 2x³ – 11x² + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:
(i) x = 2 + 3i
∴ x – 2 = 3i
∴ (x – 2)² = 9i²
∴ x² – 4x + 4 = 9(-1) …..[∵ i² = -1]
∴ x² – 4x + 13 = 0 ……(i)

∴ x³ – x² + x + 46 = (x² – 4x + 13)(x + 3) + 7
= 0(x + 3) + 7 ……[From (i)]
= 7

(ii) x = \(\frac{25}{3-4 i}\)

∴ x = 3 + 4i
∴ x – 3 = 4i
∴ (x – 3)² = 16i²
∴ x² – 6x + 9 = 16(-1) …….[∵ i² = -1]
∴ x² – 6x + 25 = 0 …….(i)

∴ 2x³ – 11x² + 44x + 27
= (x² – 6x + 25) (2x + 1) + 2
= 0 . (2x + 1) + 2 ……[From (i)]
= 0 + 2
= 2

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Functions Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2

Question 1.
Which of the following relations are functions? If it is a function determine its domain and range.
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
(iii) {(1, 1), (3, 1), (5, 2)}
Solution:
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}

Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {2, 4, 6, 8, 10, 12, 14},
Range = {1, 2, 3, 4, 5, 6, 7}

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
∴ (1, 1), (1, -1) ∈ the relation
∴ Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.

(iii) {(1, 1), (3, 1), (5, 2)}

Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {1, 3, 5}, Range = {1, 2}

Question 2.
A function f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2, x ∈ R. Show that f is one-one and onto. Hence, find f^-1.
Solution:
f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2
First we have to prove that f is one-one function for that we have to prove if
f(x₁) = f(x₂) then x₁ = x₂
Here f(x) = \(\frac{3 x}{5}\) + 2
Let f(x₁) = f(x₂)
∴ \(\frac{3 x_{1}}{5}+2=\frac{3 x_{2}}{5}+2\)
∴ \(\frac{3 x_{1}}{5}=\frac{3 x_{2}}{5}\)
∴ x₁ = x₂
∴ f is a one-one function.
Now, we have to prove that f is an onto function.
Let y ∈ R be such that
y = f(x)
∴ y = \(\frac{3 x}{5}\) + 2
∴ y – 2 = \(\frac{3 x}{5}\)
∴ x = \(\frac{5(y-2)}{3}\) ∈ R
∴ for any y ∈ co-domain R, there exist an element x = \(\frac{5(y-2)}{3}\) ∈ domain R such that f(x) = y
∴ f is an onto function.
∴ f is one-one onto function.
∴ f^-1 exists.
∴ \(\mathrm{f}^{-1}(y)=\frac{5(y-2)}{3}\)
∴ \(f^{-1}(x)=\frac{5(x-2)}{3}\)

Question 3.
A function f is defined as follows:
f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.
Solution:
f(x) = 4x + 5, -4 ≤ x < 0
f(-1) = 4(-1) + 5 = -4 + 5 = 1
f(-2) = 4(-2) + 5 = -8 + 5 = -3
x = 0 ∉ domain of f
∴ f(0) does not exist.

Question 4.
A function f is defined as follows:
f(x) = 5 – x for 0 ≤ x ≤ 4. Find the value of x such that f(x) = 3.
Solution:
f(x) = 5 – x
f(x) = 3
∴ 5 – x = 3
∴ x = 5 – 3 = 2

Question 5.
If f(x) = 3x² – 5x + 7, find f(x – 1).
Solution:
f(x) = 3x² – 5x + 7
∴ f(x – 1) = 3(x – 1)² – 5(x – 1) + 7
= 3(x² – 2x + 1) – 5(x – 1) + 7
= 3x² – 6x + 3 – 5x + 5 + 7
= 3x² – 11x + 15

Question 6.
If f(x) = 3x + a and f(1) = 7, find a and f(4).
Solution:
f(x) = 3x + a,
f(1) = 7
∴ 3(1) + a = 7
∴ a = 7 – 3 = 4
∴ f(x) = 3x + 4
∴ f(4) = 3(4) + 4
= 12 + 4
= 16

Question 7.
If f(x) = ax² + bx + 2 and f(1) = 3, f(4) = 42, find a and b.
Solution:
f(x) = ax² + bx + 2
f(1) = 3
∴ a(1)² + b(1) + 2 = 3
∴ a + b = 1 …….(i)
f(4) = 42
∴ a(4)² + b(4) + 2 = 42
∴ 16a + 4b = 40
Dividing by 4, we get
4a + b = 10 ……….(ii)
Solving (i) and (ii), we get
a = 3, b = -2

Question 8.
If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), verify whether (fof)(x) = x
Solution:
(fof)(x) = f(f(x))

Question 9.
If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then verify that (fog)(x) = x.
Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Functions Ex 2.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Functions Ex 2.1

Question 1.
Check if the following relations are functions.


Solution:
(a) Yes
Reason: Every element of set A has been assigned a unique element in set B.

(b) No
Reason: An element of set A has been assigned more than one element from set B.

(c) No
Reason: Not every element of set A has been assigned an image from set B.

Question 2.
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}
(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}
(iii) {(1, 3), (4, 1), (2, 2)}
(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}
Solution:
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.
Reason: (2, -1) and (2, 2) show that element 2 ∈ A has been assigned two images -1 and 2 from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason: 3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.

Question 3.
If f(m) = m² – 3m + 1, find
(i) f(0)
(ii) f(-3)
(iii) f(\(\frac{1}{2}\))
(iv) f(x + 1)
(v) f(-x)
Solution:
f(m) = m² – 3m + 1
(i) f(0) = 0² – 3(0) + 1 = 1

(ii) f(-3) = (-3)² – 3(-3) + 1
= 9 + 9 + 1
= 19

(iii) \(f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{4}-\frac{3}{2}+1\)
= \(\frac{1-6+4}{4}\)
= \(-\frac{1}{4}\)

(iv) f(x + 1) = (x + 1)² – 3(x + 1) + 1
= x² + 2x + 1 – 3x – 3 + 1
= x² – x – 1

(v) f(-x) = (-x)² – 3(-x) + 1 = x² + 3x + 1

Question 4.
Find x, if g(x) = 0 where
(i) g(x) = \(\frac{5 x-6}{7}\)
(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
(iii) g(x) = 6x² + x – 2
Solution:
(i) g(x) = \(\frac{5 x-6}{7}\)
g(x) = 0
∴ \(\frac{5 x-6}{7}\) = 0
∴ 5x – 6 = 0
∴ x = \(\frac{6}{5}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
g(x) = 0
∴ \(\frac{18-2 x^{2}}{7}\) = 0
∴ 18 – 2x² = 0
∴ x² = 9
∴ x = ±3

(iii) g(x) = 6x² + x – 2
g(x) = 0
∴ 6x² + x – 2 = 0
∴ 6x² + 4x – 3x – 2 = 0
∴ 2x(3x + 2) – 1(3x + 2) = 0
∴ (2x – 1)(3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

Question 5.
Find x, if f(x) = g(x) where f(x) = x⁴ + 2x², g(x) = 11x².
Solution:
f(x) = x⁴ + 2x², g(x) = 11x²
f(x) = g(x)
∴ x⁴ + 2x² = 11x²
∴ x⁴ – 9x² = 0
∴ x²(x² – 9) = 0
∴ x² = 0 or x² – 9 = 0
∴ x = 0 or x² = 9
∴ x = 0 or x = ±3

Question 6.
If f(x) = \(\begin{cases}x^{2}+3, & x \leq 2 \\ 5 x+7, & x>2\end{cases}\), then find
(i) f(3)
(ii) f(2)
(iii) f(0)
Solution:
f(x) = x² + 3, x ≤ 2
= 5x + 7, x > 2
(i) f(3) = 5(3) + 7 = 15 + 7 = 22
(ii) f(2) = 2² + 3 = 4 + 3 = 7
(iii) f(0) = 0² + 3 = 3

Question 7.
If f(x) = \(\left\{\begin{array}{cl}
4 x-2, & x \leq-3 \\
5, & -3<x<3 \\
x^{2}, & x \geq 3
\end{array}\right.\), then fmd
(i) f(-4)
(ii) f(-3)
(iii) f(1)
(iv) f(5)
Solution:
f(x) = 4x – 2, x ≤ -3
= 5, -3 < x < 3
= x², x ≥ 3
(i) f(-4) = 4(-4) – 2 = -16 – 2 = -18
(ii) f(-3) = 4(-3) – 2 = -12 – 2 = -14
(iii) f(1) = 5
(iv) f(5) = 5² = 25

Question 8.
If f(x) = 3x + 5, g(x) = 6x – 1, then find
(i) (f + g)(x)
(ii) (f – g)(2)
(iii) (fg)(3)
(iv) \(\left(\frac{\mathbf{f}}{\mathbf{g}}\right)(x)\) and its domain
Solution:
f(x) = 3x + 5, g(x) = 6x – 1
(i) (f + g)(x) = f(x) + g(x)
= 3x + 5 + 6x – 1
= 9x + 4

(ii) (f – g) (2) = f(2) – g(2)
= [3(2) + 5] – [6(2) – 1]
= 6 + 5 – 12 + 1
= 0

(iii) (fg)(3) = f(3) g(3)
= [3(3) + 5] [6(3) – 1]
= (14) (17)
= 238

(iv) \(\left(\frac{\mathrm{f}}{\mathrm{g}}\right) x=\frac{\mathrm{f}(x)}{\mathrm{g}(x)}=\frac{3 x+5}{6 x-1}, x \neq \frac{1}{6}\)
Domain = R – {\(\frac{1}{6}\)}

Question 9.
If f(x) = 2x² + 3, g(x) = 5x – 2, then find
(i) fog
(ii) gof
(iii) fof
(iv) gog
Solution:
f(x) = 2x² + 3, g(x) = 5x – 2
(i) (fog)(x) = f(g(x))
= f(5x – 2)
= 2(5x – 2)² + 3
= 2(25x² – 20x + 4) + 3
= 50x² – 40x + 8 + 3
= 50x² – 40x + 11

(ii) (gof)(x) = g(f(x))
= g(2x² + 3)
= 5(2x² + 3) – 2
= 10x² + 15 – 2
= 10x² + 13

(iii) (fof)(x) = f(f(x))
= f(2x² + 3)
= 2(2x² + 3)² + 3
= 2(4x⁴ + 12x² + 9) + 3
= 8x⁴ + 24x² + 18 + 3
= 8x⁴ + 24x² + 21

(iv) (gog)(x) = g(g(x))
= g(5x – 2)
= 5(5x – 2) – 2
= 25x – 10 – 2
= 25x – 12

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Sets and Relations Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Sets and Relations Miscellaneous Exercise 1

Question 1.
Write the following sets in set builder form:
(i) {10, 20, 30, 40, 50}
(ii) {a, e, i, o, u}
(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Solution:
(i) Let A = {10, 20, 30, 40, 50}
∴ A = {x / x = 10n, n ∈ N and n ≤ 5}

(ii) Let B = {a, e, i, o, u}
∴ B = {x / x is a vowel of English alphabets}

(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
∴ C = {x / x represents days of a week}

Question 2.
If U = {x / x ∈ N, 1 ≤ x ≤ 12}, A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}.
Write the sets
(i) A ∪ B
(ii) B ∩ C
(iii) A – B
(iv) B – C
(v) A ∪ B ∪ C
(vi) A ∩ (B ∪ C)
Solution:
U = {x / x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}
(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

(ii) B ∩ C = { }

(iii) A – B = {1, 10}

(iv) B – C = {2, 4, 6, 7, 11}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(vi) B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}
∴ A ∩ (B ∪ C) = {4, 7}

Question 3.
In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?
Solution:
Let A = set of students who drink apple juice
B = set of students who drink orange juice
X = set of all students
∴ n(X) = 425, n(A) = 115, n(B) = 160, n(A ∩ B) = 80

No. of students who neither drink apple juice nor orange juice
n(A’ ∩ B’) = n(A ∪ B)’
= n(X) – n(A ∪ B)
= 425 – [n(A) + n(B) – n(A ∩ B)]
= 425 – (115 + 160 – 80)
= 230

Question 4.
In a school, there are 20 teachers who teach Mathematics or Physics. of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?
Solution:
Let A = set of teachers who teach Mathematics
B = set of teachers who teach Physics
n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4

Since, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 20 = 12 + n(B) – 4
∴ n(B) = 12
∴ Number of teachers who teach physics = 12

Question 5.
(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A × B) ∩ (B × A).
(ii) If A = {-1, 1}, find A × A × A.
Solution:
(i) A = {1, 2, 3} and B = {2, 4}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
(A × B) ∩ (B × A) = {(2, 2)}

(ii) A = {-1, 1}
∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1),(1, 1, 1)}

Question 6.
If A = {1, 2, 3}, B = {4, 5, 6}, which of the following are relations from A to B.
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Solution:
A = {1, 2, 3}, B = {4, 5, 6}
∴ A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
Since, R₁ ⊆ A × B
∴ R₁ is a relation from A to B.

(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
Since, R₂ ⊆ A × B
∴ R₂ is a relation from A to B.

(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since, R₃ ⊆ A × B
∴ R₃ is a relation from A to B.

(iv) R₄ = {(4,2), (2, 6), (5,1), (2, 4)}
Since, (4, 2) ∈ R₄, but (4, 2) ∉ A × B
∴ R₄ ⊄ A × B
∴ R₄ is not a relation from A to B.

Question 7.
Determine the domain and range of the following relation.
R = {(a, b) / a ∈ N, a < 5, b = 4}
Solution:
R = {(a, b) / a ∈ N, a < 5, b = 4}
∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}
Range (R) = {b / b = 4} = {4}

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Sets and Relations Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Sets and Relations Ex 1.2

Question 1.
If (x – 1, y + 4) = (1, 2), find the values of x and y.
Solution:
(x – 1, y + 4) = (1, 2)
By the definition of equality of ordered pairs, we have
x – 1 = 1 and y + 4 = 2
∴ x = 2 and y = -2

Question 2.
If \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{3}, \frac{3}{2}\right)\), find x and y.
Solution:
\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{3}, \frac{3}{2}\right)\)
By the definition of equality of ordered pairs, we have
\(x+\frac{1}{3}=\frac{1}{3}\) and \(\frac{y}{3}-1=\frac{3}{2}\)
\(x=\frac{1}{3}-\frac{1}{3}\) and \(\frac{y}{3}=\frac{3}{2}+1=\frac{5}{2}\)
x = 0 and y = \(\frac{15}{2}\)

Question 3.
If A = {a, b, c}, B = {x, y}, find A × B, B × A, A × A, B × B.
Solution:
A = {a, b, c}, B = {x, y}
A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}
B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
B × B = {(x, x), (x, y), (y, x), (y, y)}

Question 4.
If P = {1, 2, 3} and Q = {6, 4}, find the sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {6, 4}
P × Q = {(1, 6), (1, 4), (2, 6), (2, 4), (3, 6), (3, 4)}
Q × P = {(6, 1), (6, 2), (6, 3), (4, 1), (4, 2), (4, 3)}

Question 5.
Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}. Find
(i) A × (B ∩ C)
(ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C)
(iv) (A × B) ∪ (A × C)
Solution:
A= {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
(i) B ∩ C = {5, 6}
∴ A × (B ∩ C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

(ii) A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

(iii) B ∪ C = {4, 5, 6}
∴ A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

(iv) A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

Question 6.
Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.
Solution:
{(x, y) / x² + y² = 100, where x, y ∈ W}
We have, x² + y² = 100
When x = 0 and y = 10,
x² + y² = 0² + 10² = 100
When x = 6 andy = 8,
x² + y² = 6² + 8² = 100
When x = 8 and y = 6,
x² + y² = 8² + 6² = 100
When x = 10 and y = 0,
x² + y² = 10² + 0² = 100
∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

Question 7.
Write the domain and range of the following relations.
(i) {(a, b) / a ∈ N, a < 6 and b = 4}
(ii) {(a, b) / a, b ∈ N, a + b = 12}
(iii) {(2, 4), (2, 5), (2, 6), (2, 7)}
Solution:
(i) Let R₁ = {(a, b)/ a ∈ N, a < 6 and b = 4}
Set of values of ‘a’ are domain and set of values of ‘b’ are range.
a ∈ N and a < 6
∴ a = 1, 2, 3, 4, 5 and b = 4
Domain (R₁) = {1, 2, 3, 4, 5}
Range (R₁) = {4}

(ii) Let R₂ = {(a, b)/a, b ∈ N and a + b = 12}
Now, a, b ∈ N and a + b = 12
When a = 1, b = 11
When a = 2, b = 10
When a = 3, b = 9
When a = 4, b = 8
When a = 5, b = 7
When a = 6, b = 6
When a = 7, b = 5
When a = 8, b = 4
When a = 9, b = 3
When a = 10, b = 2
When a = 11, b = 1
∴ Domain (R₂) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Range (R₂) = {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1}

(iii) Let R₃ = {(2, 4), (2, 5), (2, 6), (2, 7)}
Domain (R₃) = {2}
Range (R₃) = {4, 5, 6, 7}

Question 8.
Let A = {6, 8} and B = {1, 3, 5}.
Let R = {(a, b) / a ∈ A, b ∈ B, a – b is an even number}.
Show that R is an empty relation from A to B.
Solution:
A= {6, 8}, B = {1, 3, 5}
R = {(a, b)/ a ∈ A, b ∈ B, a – b is an even number}
a ∈ A
∴ a = 6, 8
b ∈ B
∴ b = 1, 3, 5
When a = 6 and b = 1, a – b = 5 which is odd
When a = 6 and b = 3, a – b = 3 which is odd
When a = 6 and b = 5, a – b = 1 which is odd
When a = 8 and b = 1, a – b = 7 which is odd
When a = 8 and b = 3, a – b = 5 which is odd
When a = 8 and b = 5, a – b = 3 which is odd
Thus, no set of values of a and b gives a – b even.
∴ R is an empty relation from A to B.

Question 9.
Write the relation in the Roster form and hence find its domain and range.
(i) R₁ = {(a, a²) / a is a prime number less than 15}
(ii) R₂ = {(a, \(\frac{1}{a}\)) / 0 < a ≤ 5, a ∈ N}
Solution:
(i) R₁ = {(a, a²) / a is a prime number less than 15}
∴ a = 2, 3, 5, 7, 11, 13
∴ a2 = 4, 9, 25, 49, 121, 169
∴ R₁ = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}
∴ Domain (R₁) = {a/a is a prime number less than 15} = {2, 3, 5, 7, 11, 13}
Range (R₁) = {a²/a is a prime number less than 15} = {4, 9, 25, 49, 121, 169}

Question 10.
R = {(a, b) / b = a + 1, a ∈ Z, 0 < a < 5}. Find the range of R.
Solution:
R = {(a, b) / b = a + 1, a ∈ Z, 0 < a < 5}
∴ a = 1, 2, 3, 4
∴ b = 2, 3, 4, 5
∴ Range (R) = {2, 3, 4, 5}

Question 11.
Find the following relations as sets of ordered pairs.
(i) {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
(ii) {(x,y) / y > x + 1, x ∈ {1, 2} and y ∈ {2, 4, 6}}
(iii) {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
Solution:
(i) {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
Here y = 3x
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴ Ordered pairs are {(1, 3), (2, 6), (3, 9)}

(ii) {(x, y) / y > x + 1, x ∈ {1, 2} and y ∈ {2, 4, 6}}
Here, y > x + 1
When x = 1 and y = 2, 2 ≯ 1 + 1
When x = 1 and y = 4, 4 > 1 + 1
When x = 1 and y = 6, 6 > 1 + 1
When x = 2 and y = 2, 2 ≯ 2 + 1
When x = 2 and y = 4, 4 > 2 + 1
When x = 2 and y = 6, 6 > 2 + 1
∴ Ordered pairs are {(1, 4), (1, 6), (2, 4), (2, 6)}

(iii) {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
Here, x + y = 3
When x = 0, y = 3
When x = 1, y = 2
When x = 2, y = 1
When x = 3, y = 0
∴ Ordered pairs are {(0, 3), (1, 2), (2, 1), (3, 0)}

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Sets and Relations Ex 1.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Sets and Relations Ex 1.1

Question 1.
Describe the following sets in Roster form:
(i) {x / x is a letter of the word ‘MARRIAGE’}
(ii) {x / x is an integer, –\(\frac{1}{2}\) < x < \(\frac{9}{2}\)}
(iii) {x / x = 2n, n ∈ N}
Solution:
(i) Let A = {x / x is a letter of the word ‘MARRIAGE’}
∴ A = {M, A, R, I, G, E}

(ii) Let B = {x / x is an integer, –\(\frac{1}{2}\) < x < \(\frac{9}{2}\)}
∴ B = {0, 1, 2, 3, 4}

(iii) Let C = {x / x = 2n, n ∈ N}
∴ C = {2, 4, 6, 8, ….}

Question 2.
Describe the following sets in Set-Builder form:
(i) {0}
(ii) {0, ±1, ±2, ±3}
(iii) \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
Solution:
(i) Let A = {0}
0 is a whole number but it is not a natural number.
∴ A = {x / x ∈ W, x ∉ N}

(ii) Let B = {0, ±1, ±2, ±3}
B is the set of elements which belongs to Z from -3 to 3.
∴ B = {x / x ∈ Z, -3 ≤ x ≤ 3}

(iii) Let C = \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
∴ C = {x / x = \(\frac{n}{n^{2}+1}\), n ∈ N, n ≤ 7}

Question 3.
If A = {x / 6x² + x – 15 = 0}, B = {x / 2x² – 5x – 3 = 0}, C = {x / 2x² – x – 3 = 0}, then find (i) (A ∪ B ∪ C) (ii) (A ∩ B ∩ C)
Solution:
A = {x / 6x² + x – 15 = o}
∴ 6x² + x – 15 = 0
∴ 6x² + 10x – 9x – 15 = 0
∴ 2x(3x + 5) – 3(3x + 5) = 0
∴ (3x + 5) (2x – 3) = 0
∴ 3x + 5 = 0 or 2x – 3 = 0
∴ x = \(\frac{-5}{3}\) or x = \(\frac{3}{2}\)
∴ A = \(\left\{\frac{-5}{3}, \frac{3}{2}\right\}\)

B = {x / 2x² – 5x – 3 = 0}
∴ 2x² – 5x – 3 = 0
∴ 2x² – 6x + x – 3 = 0
∴ 2x(x – 3) + 1(x – 3) = 0
∴ (x – 3)(2x + 1) = 0
∴ x – 3 = 0 or 2x + 1 = 0
∴ x = 3 or x = \(\frac{-1}{2}\)
∴ B = {\(\frac{-1}{2}\), 3}

C = {x / 2x² – x – 3 = 0}
∴ 2x² – x – 3 = 0
∴ 2x² – 3x + 2x – 3 = 0
∴ x(2x – 3) + 1(2x – 3) = 0
∴ (2x – 3) (x + 1) = 0
∴ 2x – 3 = 0 or x + 1 = 0
∴ x = \(\frac{3}{2}\) or x = -1
∴ C = {-1, \(\frac{3}{2}\)}

(i) A ∪ B ∪ C = \(\left\{-\frac{5}{3}, \frac{3}{2}\right\} \cup\left\{\frac{-1}{2}, 3\right\} \cup\left\{-1, \frac{3}{2}\right\}\) = \(\left\{\frac{-5}{3},-1, \frac{-1}{2}, \frac{3}{2}, 3\right\}\)

(ii) A ∩ B ∩ C = { }

Question 4.
If A, B, C are the sets for the letters in the words ‘college’, ‘marriage’ and ‘luggage’ respectively, then verify that [A – (B ∪ C)] = [(A – B) ∩ (A – C)].
Solution:
A = {c, o, l, g, e}
B = {m, a, r, i, g, e}
C = {l, u, g, a, e}
B ∪ C = {m, a, r, i, g, e, l, u}
A – (B ∪ C) = {c, o}
A – B = {c, o, l}
A – C = {c, o}
∴ [(A – B) ∩ (A – C)] = {c, o} = A – (B ∪ C)
∴ [A – (B ∪ C)] = [(A – B) ∩ (A – C)]

Question 5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify the following:
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(iii) (A ∪ B)’ = A’ ∩ B’
(iv) (A ∩ B)’ = A’ ∪ B’
(v) A = (A ∩ B) ∪ (A ∩ B’)
(vi) B = (A ∩ B) ∪ (A’ ∩ B)
(vii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Solution:
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(i) B ∩ C = {4, 5, 6}
∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6} ……(i)
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} ……(ii)
From (i) and (ii), we get
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) B ∪ C = {3, 4, 5, 6, 7, 8}
∴ A ∩ (B ∪ C) = {3, 4} …..(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) ∪ (A ∩ C) = {3, 4} …..(ii)
From (i) and (ii), we get
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) A ∪ B = {1, 2, 3, 4, 5, 6}
∴ (A ∪ B)’ = {7, 8, 9, 10} …….(i)
A’ = {5, 6, 7, 8, 9, 10}, B’ = {1, 2, 7, 8, 9, 10}
∴ A’ ∩ B’ = {7, 8, 9, 10} ……(ii)
From (i) and (ii), we get
(A ∪ B)’ = A’ ∩ B’

(iv) A ∩ B = {3, 4}
∴ (A ∩ B)’ = {1, 2, 5, 6, 7, 8, 9, 10} ……(i)
A’ = {5, 6, 7, 8, 9, 10}
B’ = {1, 2, 7, 8, 9, 10}
∴ A’ ∪ B’ = {1, 2, 5, 6, 7, 8, 9, 10} ……(ii)
From (i) and (ii), we get
(A ∩ B)’ = A’ ∪ B’

(v) A = {1, 2, 3, 4} …..(i)
A ∩ B = {3, 4}
B’ = {1, 2, 7, 8, 9, 10}
A ∩ B’ = {1, 2}
∴ (A ∩ B) ∪ (A ∩ B’) = {1, 2, 3, 4} ……(ii)
From (i) and (ii), we get
A = (A ∩ B) ∪ (A ∩ B’)

(vi) B = {3, 4, 5, 6} …..(i)
A ∩ B = {3, 4}
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ (A ∩ B) ∪ (A’ ∩ B) = {3, 4, 5, 6} …..(ii)
From (i) and (ii), we get
B = (A ∩ B) ∪ (A’ ∩ B)

(vii) A = {1, 2, 3, 4}, B = {3, 4, 5, 6},
A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6}
∴ n(A) = 4, n(B) = 4,
n(A ∩ B) = 2,
n(A ∪ B) = 6 …..(i)
∴ n(A) + n(B) – n(A ∩ B) = 4 + 4 – 2
∴ n(A) + n(B) – n(A ∩ B) = 6 …..(ii)
From (i) and (ii), we get
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Question 6.
If A and B are subsets of the universal set X and n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5, find
(i) n(A ∪ B)
(ii) n(A ∩ B)
(iii) n(A’ ∩ B)
(iv) n(A ∩ B’)
Solution:
n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5
(i) n(A ∪ B) = n(X) – [n(A ∪ B)’]
= n(X) – n(A’ ∩ B’)
= 50 – 5
= 45

(ii) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 35 + 20 – 45
= 10

(iii) n(A’ ∩ B) = n(B) – n(A ∩ B)
= 20 – 10
= 10

(iv) n(A ∩ B’) = n(A) – n(A ∩ B)
= 35 – 10
= 25

Question 7.
Out of 200 students, 35 students failed in MHT-CET, 40 in AIEEE and 40 in IIT entrance, 20 failed in MHT-CET and AIEEE, 17 in AIEEE and IIT entrance, 15 in MHT-CET and IIT entrance, and 5 failed in all three examinations. Find how many students
(i) did not fail in any examination.
(ii) failed in AIEEE or IIT entrance.
Solution:
Let A = set of students who failed in MHT-CET
B = set of students who failed in AIEEE
C = set of students who failed in IIT entrance
X = set of all students
∴ n(X) = 200, n(A) = 35, n(B) = 40, n(C) = 40,
n(A ∩ B) = 20, n(B ∩ C) = 17, n(A ∩ C) = 15, n(A ∩ B ∩ C) = 5

(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 35 + 40 + 40 – 20 – 17 – 15 + 5
= 68
∴ No. of students who did not fail in any exam = n(X) – n(A ∪ B ∪ C)
= 200 – 68
= 132

(ii) No. of students who failed in AIEEE or IIT entrance = n(B ∪ C)
= n(B) + n(C) – n(B ∩ C)
= 40 + 40 – 17
= 63

Question 8.
From amongst 2000 literate individuals of a town, 70% read Marathi newspapers, 50% read English newspapers and 32.5% read both Marathi and English newspapers. Find the number of individuals who read
(i) at least one of the newspapers.
(ii) neither Marathi nor English newspaper.
(iii) only one of the newspapers.
Solution:
Let M = set of individuals who read Marathi newspapers
E = set of individuals who read English newspapers
X = set of all literate individuals
∴ n(X) = 2000,
n(M) = \(\frac{70}{100}\) × 2000 = 1400
n(E) = \(\frac{50}{100}\) × 2000 = 1000
n(M ∩ E) = \(\frac{32.5}{2}\) × 2000 = 650
n(M ∪ E) = n(M) + n(E) – n(M ∩ E)
= 1400 + 1000 – 650
= 1750

(i) No. of individuals who read at least one of the newspapers = n(M ∪ E) = 1750.
(ii) No. of individuals who read neither Marathi nor English newspaper = n(M’ ∩ E’)
= n(M ∪ E)’
= n(X) – n(M ∪ E)
= 2000 – 1750
= 250
(iii) No. of individuals who read only one of the newspapers = n(M ∩ E’) + n(M’ ∩ E)
= n(M ∪ E) – n(M ∩ E)
= 1750 – 650
= 1100

Question 9.
In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take tea and milk both and everyone takes atleast one beverage, find the number of students in the hostel.
Solution:
Let T = set of students who take tea
C = set of students who take coffee
M = set of students who take milk
∴ n(T) = 25, n(C) = 20, n(M) = 15,
n(T ∩ C) = 10, n(M ∩ C) = 8, n(T ∩ M) = 0, n(T ∩ M ∩ C) = 0

∴ Number of students in the hostel = n(T ∪ C ∪ M)
= n(T) + n(C) + n(M) – n(T ∩ C) – n(M ∩ C) – n(T ∩ M) + n(T ∩ M ∩ C)
= 25 + 20 + 15 – 10 – 8 – 0 + 0
= 42

Question 10.
There are 260 persons with skin disorders. If 150 had been exposed to the chemical A, 74 to the chemical B, and 36 to both chemicals A and B, find the number of persons exposed to
(i) Chemical A but not Chemical B
(ii) Chemical B but not Chemical A
(iii) Chemical A or Chemical B.
Solution:
Let A = set of persons exposed to chemical A
B = set of persons exposed to chemical B
X = set of all persons
∴ n(X) = 260, n(A) = 150, n(B) = 74, n(A ∩ B) = 36

(i) No. of persons exposed to chemical A but not to chemical B = n(A ∩ B’)
= n(A) – n(A ∩ B)
= 150 – 36
= 114

(ii) No. of persons exposed to chemical B but not to chemical A = n(A’ ∩ B)
= n(B) – n(A ∩ B)
= 74 – 36
= 38

(iii) No. of persons exposed to chemical A or chemical B = n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 150 + 74 – 36
= 188

Question 11.
If A = {1, 2, 3}, write the set of all possible subsets of A.
Solution:
A = {1, 2, 3}
∴ { }, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} and {1, 2, 3} are all the possible subsets of A.

Question 12.
Write the following intervals in set-builder form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, 12)
(iv) (-23, 5)
Solution:
(i) (-3, 0) = {x / x ∈ R, -3 < x < 0}
(ii) [6, 12] = {x / x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12) = {x / x ∈ R, 6 < x < 12}
(iv) (-23, 5) = {x / x ∈ R, -23 < x < 5}

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(I) Choose the correct alternative.

Question 1.
F(x) is c.d.f. of discreter r.v. X whose p.m.f. is given by P(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\), for x = 0, 1, 2, 3, 4 & P(x) = 0 otherwise then F(5) = __________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{4}\)
(d) 1
Answer:
(d) 1

Question 2.
F(x) is c.d.f. of discrete r.v. X whose distribution is

then F(-3) = __________
(a) 0
(b) 1
(c) 0.2
(d) 0.15
Answer:
(a) 0

Question 3.
X : number obtained on uppermost face when a fair die is thrown then E(X) = __________
(a) 3.0
(b) 3.5
(c) 4.0
(d) 4.5
Answer:
(b) 3.5

Question 4.
If p.m.f. of r.v. X is given below.

then Var(X) = __________
(a) p²
(b) q²
(c) pq
(d) 2pq
Answer:
(d) 2pq

Question 5.
The expected value of the sum of two numbers obtained when two fair dice are rolled is __________
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 6.
Given p.d.f. of a continuous r.v. X as
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0 otherwise then F(1) =
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{3}{9}\)
(d) \(\frac{4}{9}\)
Answer:
(b) \(\frac{2}{9}\)

Question 7.
X is r.v. with p.d.f.
f(x) = \(\frac{k}{\sqrt{x}}\), 0 < x < 4
= 0 otherwise then E(X) = __________
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{2}{3}\)
(d) 1
Answer:
(b) \(\frac{4}{3}\)

Question 8.
If X follows B(20, \(\frac{1}{10}\)) then E(X) = __________
(a) 2
(b) 5
(c) 4
(d) 3
Answer:
(a) 2

Question 9.
If E(X) = m and Var(X) = m then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(b) Possion distribution

Question 10.
If E(X) > Var(X) then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(a) Binomial distribution

(II) Fill in the blanks.

Question 1.
The values of discrete r.v. are generally obtained by __________
Answer:
counting

Question 2.
The values of continuous r.v. are generally obtained by __________
Answer:
measurement

Question 3.
If X is dicrete random variable takes the values x₁, x₂, x₃, …… x_n then \(\sum_{i=1}^{n} p\left(x_{i}\right)\) = __________
Answer:
1

Question 4.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(\frac{x-1}{3}\) for x = 1, 2, 3, and p(x) = 0 otherwise then F(4) = __________
Answer:
1

Question 5.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, and p(x) = 0 otherwise then F(-1) = __________
Answer:
0

Question 6.
E(X) is considered to be __________ of the probability distribution of X.
Answer:
centre of gravity

Question 7.
If X is continuous r.v. and f(x_i) = P(X ≤ x_i) = \(\int_{-\infty}^{x_{i}} f(x) d x\) then f(x) is called __________
Answer:
Cumulative Distribution Function

Question 8.
In Binomial distribution probability of success ________ from trial to trial.
Answer:
remains constant/independent

Question 9.
In Binomial distribution, if n is very large and probability success of p is very small such that np = m (constant) then ________ distribution is applied.
Answer:
Possion

(III) State whether each of the following is True or False.

Question 1.
If P(X = x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, then F(5) = \(\frac{1}{4}\) when f(x) is c.d.f.
Answer:
False

Question 2.

If F(x) is c.d.f. of discrete r.v. X then F(-3) = 0.
Answer:
True

Question 3.
X is the number obtained on the uppermost face when a die is thrown the E(X) = 3.5.
Answer:
True

Question 4.
If p.m.f. of discrete r.v.X is

then E(X) = 2p.
Answer:
True

Question 5.
The p.m.f. of a r.v. X is p(x) = \(\frac{2 x}{n(n+1)}\), x = 1, 2,……n
= 0 otherwise,
Then E(X) = \(\frac{2 n+1}{3}\)
Answer:
True

Question 6.
If f(x) = kx (1 – x) for 0 < x < 1
= 0 otherwise then k = 12
Answer:
False

Question 7.
If X ~ B(n, p) and n = 6 and P(X = 4) = P(X = 2) then p = \(\frac{1}{2}\).
Answer:
True

Question 8.
If r.v. X assumes values 1, 2, 3,………, n with equal probabilities then E(X) = \(\frac{(n+1)}{2}\)
Answer:
True

Question 9.
If r.v. X assumes the values 1, 2, 3,………, 9 with equal probabilities, E(X) = 5.
Answer:
True

(IV) Solve the following problems.

Part – I

Question 1.
Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
(i) An economist is interested in knowing the number of unemployed graduates in the town with a population of 1 lakh.
Solution:
X = No. of unemployed graduates in a town.
∵ The population of the town is 1 lakh
∴ X takes finite values
∴ X is a Discrete Random Variable
∴ Range of = {0, 1, 2, 4, …. 1,00,000}

(ii) Amount of syrup prescribed by a physician.
Solution:
X : Amount of syrup prescribed.
∴ X Takes infinite values
∴ X is a Continuous Random Variable.

(iii) A person on a high protein diet is interested in the weight gained in a week.
Solution:
X : Gain in weight in a week.
X takes infinite values
∴ X is a Continuous Random Variable.

(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them.
Solution:
X : No. of female rats selected
X takes finite values.
∴ X is a Discrete Random Variable.
Range of X = {0, 1, 2, 3, 4, 5}

(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint.
Solution:
X : Speed of car in km/hr
X takes infinite values
∴ X is a Continuous Random Variable.

Question 2.
The probability distribution of a discrete r.v. X is as follows.

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:
(i) Assuming that the given distribution is a p.m.f. of X
∴ Each P(X = x) ≥ 0 for x = 1, 2, 3, 4, 5, 6
k ≥ 0
ΣP(X = x) = 1 and
k + 2k + 3k + 4k + 5k + 6k = 1
∴ 21k = 1 ∴ k = \(\frac{1}{21}\)

(ii) P(X ≤ 4) = 1 – P(X > 4)
= 1 – [P(X = 5) + P(X = 6)]
= 1 – [latex]\frac{5}{21}+\frac{6}{21}[/latex]
= 1 – \(\frac{11}{21}\)
= \(\frac{10}{21}\)
P(2 < X < 6) = p(3) + p(4) + p(5)
= 3k + 4k + 5k
= \(\frac{3}{21}+\frac{4}{21}+\frac{5}{21}\)
= \(\frac{12}{21}\)
= \(\frac{4}{7}\)

(iii) P(X ≥ 3) = p(3) + p(4) + p(5) + p(6)
= 3k + 4k + 5k + 6k

Question 3.
Following is the probability distribution of an r.v. X.

Find the probability that
(i) X is positive.
(ii) X is non-negative.
(iii) X is odd.
(iv) X is even.
Solution:
(i) P(X is positive)
P(X = 0) = p(1) + p(2) + p(3)
= 0.25 + 0.15 + 0.10
= 0.50

(ii) P(X is non-negative)
P(X ≥ 0) = p(0) + p(1) + p(2) + p(3)
= 0.20 + 0.25 + 0.15 + 0.10
= 0.70

(iii) P(X is odd)
P(X = -3, -1, 1, 3)
= p(- 3) +p(-1) + p(1) + p(3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(iv) P(X is even)
= 1 – P(X is odd)
= 1 – 0.55
= 0.45

Question 4.
The p.m.f of a r.v. X is given by
\(P(X=x)= \begin{cases}\left(\begin{array}{l}
5 \\
x
\end{array}\right) \frac{1}{2^{5}}, & x=0,1,2,3,4,5 . \\
0 & \text { otherwise }\end{cases}\)
Show that P(X ≤ 2) = P(X ≥ 3).
Solution:
For x = 0, 1, 2, 3, 4, 5

Question 5.
In the following probability distribution of an r.v. X

Find a and obtain the c.d.f. of X.
Solution:
Given distribution is p.m.f. of r.v. X
ΣP(X = x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) = 1

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p.m.f. of X.
Solution:
A fair coin is tossed 4 times
∴ Sample space contains 16 outcomes
Let X = Number of heads obtained
∴ X takes the values x = 0, 1, 2, 3, 4.
∴ The number of heads obtained in a toss is an even

Question 7.
Find the probability of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appearing in at least one toss.
Solution:
S : A die is tossed two times
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
(i) X : No. is greater than 4
Range of X = {0, 1, 2}

(ii) X : Six appears on aleast one die.
Range of X = {0, 1, 2}

Question 8.
A random variable X has the following probability distribution.

Determine (i) k, (ii) P(X < 3), (iii) P(X > 6), (iv) P(0 < X < 3).
Solution:
(i) It is a p.m.f. of r.v. X
Σp(x) = 1
p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
9k + 10k² = 1
10k² + 9k – 1 = 0
10k² +10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1) (k + 1) = 0
∴ 10k – 1 = 0r k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
k = -1 is not accepted, p(x) ≥ 0, ∀ x ∈ R
∴ k = \(\frac{1}{10}\)

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iii) P(X > 6) = p(7)
= 7k² + k
= \(7\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{7}{100}+\frac{1}{10}\)
= \(\frac{17}{100}\)

(iv) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

Question 9.
The following is the c.d.f. of a r.v. X.

Find the probability distribution of X and P(-1 ≤ X ≤ 2).
Solution:

P(-1 ≤ X ≤ 2) = p(-1) + p(0) + p(1) + p(2)
= 0.2 + 0.15 + 0.10 + 0.10
= 0.55

Question 10.
Find the expected value and variance of the r.v. X if its probability distribution is as follows.
(i)

Solution:

(ii)

Solution:
E(X) = Σx . p(x)

(iii)

Solution:

(iv)

Solution:

= 1.25
S.D. of X = σ_x = √Var(X)
= √1.25
= 1.118

Question 11.
A player tosses two coins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears, and ₹ 2 if no head appears. Find the expected value and variance of the winning amount.
Solution:
S : Two fair coin are tossed
S = {HH, HT, TT, TH}
n(S) = 4
∴ Range of X = {0, 1, 2}
∴ Let Y = amount received corresponds to values of X

Expected winning amount
E(Y) = Σpy = \(\frac{22}{4}\) = ₹ 5.5
V(Y) = Σpy² – (Σpy)²
= \(\frac{154}{4}\) – (5.5)²
= 38.5 – 30.25
= ₹ 8.25

Question 12.
Let the p.m.f. of the r.v. X be
\(p(x)= \begin{cases}\frac{3-x}{10} & \text { for } x=-1,0,1,2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate E(X) and Var(X).
Solution:

Question 13.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
We know that

Question 14.
The p.d.f. of the r.v. X is given by
\(f(x)= \begin{cases}\frac{1}{2 a} & \text { for } 0<x<2 a \\ 0 & \text { otherwise }\end{cases}\)
Show that P(X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
Determine k if
\(f(x)= \begin{cases}k e^{-\theta x} & \text { for } 0 \leq x<\infty, \theta>0 \\ 0 & \text { otherwise }\end{cases}\)
is the p.d.f. of the r.v. X. Also find P(X > \(\frac{1}{\theta}\)). Find M if P(0 < X < M) = \(\frac{1}{2}\)
Solution:
We know that

Question 16.
The p.d.f. of the r.v. X is given by
\(f_{x}(x)=\left\{\begin{array}{l}
\frac{k}{\sqrt{x}}, 0<x<4 \\
0, \text { otherwise }
\end{array}\right.\)
Determine k, c.d.f. of X and hence find P(X ≤ 2) and P(X ≥ 1).
Solution:
We know that

Question 17.
Let X denote the reaction temperature (in °C) of a certain chemical process. Let X be a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{1}{10}, & -5 \leq x \leq 5 \\ 0, & \text { otherwise }\end{cases}\)
Compute P(X < 0).
Solution:
Given p.d.f. is f(x) = \(\frac{1}{10}\), for -5 ≤ x ≤ 5
Let its c.d.f. F(x) be given by

Part – II

Question 1.
Let X ~ B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8)
Solution:
X ~ B(10, 0.2)
n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
(i) P(X = 1) = ¹⁰C₁ (0.2)¹ (0.8)⁹ = 0.2684

(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – ¹⁰C₀ (0.2)⁰ (0.8)¹⁰
= 1 – 0.1074
= 0.8926

(iii) P(X ≤ 8) = 1 – P(x > 1)
= 1 – [p(9) + p(10)]
= 1 – [¹⁰C₉ (0.2)⁹ (0.8)¹ + ¹⁰C₁₀ (0.2)¹⁰]
= 1 – 0.00000041984
= 0.9999

Question 2.
Let X ~ B(n, p) (i) If n = 10 and E(X) = 5, find p and Var(X), (ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) n = 10, E(X) = 5
∴ np = 5
∴ 10p = 5
∴ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
V(X) = npq
= 10 × \(\frac{1}{2}\) × \(\frac{1}{2}\)
= 2.5

(ii) E(X) = 5, V(X) = 2.5
∴ np = 5, ∴ npq = 2.5
∴ 5q = 2.5
∴ q = \(\frac{2.5}{5}\) = 0.5, p = 1 – 0.5 = 0.5
But np = 5
∴ n(0.5) = 5
∴ n = 10

Question 3.
If a fair coin is tossed 4 times, find the probability that it shows (i) 3 heads, (ii) head in the first 2 tosses, and tail in the last 2 tosses.
Solution:
n : No. of times a coin is tossed
∴ n = 4
X : No. of heads
P : Probability of getting heads

Question 4.
The probability that a bomb will hit the target is 0.8. Find the probability that, out of 5 bombs, exactly 2 will miss the target.
Solution:
X : No. of bombs miss the target
p : Probability that bomb miss the target
∴ q = 0.8
∴ p = 1 – q = 1 – 0.8 = 0.2
n = No. of bombs = 5
∴ X ~ B(5, 0.2)
∴ p(x) = ⁿC_x p^x q^n-x
P(X = 2) = ⁵C₂ (0.2)² (0.8)^5-2
= 10 × 0.04 × (0.8)³
= 10 × 0.04 × 0.512
= 0.4 × 0.512
= 0.2048

Question 5.
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom can not be used on a random occasion.
Solution:
X : No. of lamps not burning
p : Probability that the lamp is not burning
∴ q = 0.3
∴ p = 1 – q = 1 – 0.3 = 0.7
n = No. of lamps fitted = 3
∴ X ~ B(3, 0.7)
∴ p(x) = ⁿC_x p^x q^n-x
P(classroom cannot be used)
P(X < 2) = p(0) + p(1)
= ³C₀ (0.7)⁰ (0.3)^3-0 + ³C₁ (0.7)¹ (0.3)^3-1
= 1 × 1 × (0.3)³ + 3 × 0.7 × (0.3)²
= (0.3)2 [0.3 + 3 × 0.7]
= 0.09 [0.3 + 2.1]
= 0.09 [2.4]
= 0.216

Question 6.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
X : No. of defective items
n : No. of items selected = 4
p : Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
P(At most one defective item)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)^4-0 + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)3 [0.9 + 4 × 0.1]
= (0.9)3 × [0.9 + 0.4]
= 0.729 × 1.3
= 0.9477

Question 7.
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Solution:
p = 0.6, q = 1 – 0.6 = 0.4, n = 4
x = 2
∴ p(x) = ⁿC_x p^x q^n-x
P(X = 2) = ⁴C₂ (0.6)² (0.4)² = 0.3456

Question 8.
An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer randomly. Find the probability that this student gets 4 or more correct answers.
Solution:
n : No. of multiple-choice questions
∴ n = 5
X : No. of correct answers
p : Probability of getting correct answer
∵ There are 4 options out of which one is correct
∴ p = \(\frac{1}{4}\)
∴ q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∵ X ~ B(5, \(\frac{1}{4}\))
∴ p(x) = ⁿC_x p^x q^n-x
P(Four or more correct answers)
P(X ≥ 4) = p(4) + p(5)

Question 9.
The probability that a machine will produce all bolts in a production run with in the specification is 0.9. A sample of 3 machines is taken at random. Calculate the probability that all machines will produce all bolts in a production run within the specification.
Solution:
n : No. of samples selected
∴ n = 3
X : No. of bolts produce by machines
p : Probability of getting bolts
∴ p = 0.9
∴ q = 1 – p = 1 – 0.9 = 0.1
∴ X ~ B(3, 0.9)
∴ p(x) = ⁿC_x p^x q^n-x
P(Machine will produce all bolts)
P(X = 3) = ³C₃ (0.9)³ (0.1)^3-3
= 1 × (0.9)³ × (0.1)⁰
= 1 × (0.9)³ × 1
= (0.9)³
= 0.729

Question 10.
A computer installation has 3 terminals. The probability that anyone terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that (i) 0 (ii) 1 terminal requires attention during a week.
Solution:
n : No. of terminals
∴ n = 3
X : No. of terminals need attention
p : Probability of getting terminals need attention
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
∵ X ~ B(3, 0.1)
∴ p(x) = ⁿC_x p^x q^n-x
(i) P(No attention)
∴ P(X = 0) = ³C₀ × (0.1)⁰ (0.9)^3-1
= 1 × 1 × (0.9)³
= 0.729

(ii) P(One terminal need attention)
∴ P(X = 1) = ³C₁ (0.1)¹ (0.9)^3-1
= 3 × 0.1 × (0.9)²
= 0.3 × 0.81
= 0.243

Question 11.
In a large school, 80% of the students like mathematics. A visitor asks each of 4 students, selected at random, whether they like mathematics, (i) Calculate the probabilities of obtaining an answer yes from all of the selected students, (ii) Find the probability that the visitor obtains the answer yes from at least 3 students.
Solution:
X : No. of students like mathematics
p: Probability that students like mathematics
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
n : No. of students selected
∴ n = 4
∵ X ~ B(4, 0.8)
∴ p(x) = ⁿC_x p^x q^n-x
(i) P(All students like mathematics)
∴ P(X = 4) = ⁴C₄ (0.8)⁴ (0.2)^4-4
= 1 × (0.8)⁴ × (0.2)⁰
= 1 × (0.8)⁴ × 1
= 0.4096

(ii) P(Atleast 3 students like mathematics)
∴ P(X ≥ 3) = p(3) + p(4)
= ⁴C₃ (0.8)³ (0.2)^4-3 + 0.4096
= 4 × (0.8)³ (0.2)¹ + 0.4096
= 0.8 × (0.8)³ + 0.4096
= (0.8)⁴ × 0.4096
= 0.4096 + 0.4096
= 0.8192

Question 12.
It is observed that it rains on 10 days out of 30 days. Find the probability that
(i) it rains on exactly 3 days of a week.
(ii) it rains at most 2 days a week.
Solution:
X : No. of days it rains in a week
p : Probability that it rains
∴ p = \(\frac{10}{30}=\frac{1}{3}\)
∴ q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
n : No. of days in a week
∴ n = 7
∴ X ~ B(7, \(\frac{1}{3}\))
(i) P(Rains on Exactly 3 days of a week)

(ii) P(Rains on at most 2 days of a week)
∴ P(X ≤ 2) = p(0) + p(1) + p(2)

Question 13.
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
Solution:
X : Follows Possion Distribution

∴ m = 1
∴ Mean = m = Variance of X = 1

Question 14.
If X has Poisson distribution with parameter m, such that
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
find probabilities P(X = 1) and P(X = 2), when X follows Poisson distribution with m = 2 and P(X = 0) = 0.1353.
Solution:
Given that the random variable X follows the Poisson distribution with parameter m = 2
i.e. X ~ P(2)
Its p.m.f. is satisfying the given equation.
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
When x = 0,
\(\frac{\mathrm{P}(\mathrm{X}=1)}{\mathrm{P}(\mathrm{X}=0)}=\frac{2}{0+1}\)
P(X = 1) = 2P(X = 0)
= 2(0.1353)
= 0.2706
When x = 1,
\(\frac{\mathrm{P}(\mathrm{X}=2)}{\mathrm{P}(\mathrm{X}=1)}=\frac{2}{1+1}\)
P(X = 2) = P(X = 1) = 0.2706

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e^-1 = 0.3678.
Solution:
∵ m = 1
∵ X follows Poisson Distribution

= e^-m × 1 + e^-m × 1
= e^-1 + e^-1
= 2 × e^-1
= 2 × 0.3678
= 0.7356

Question 2.
If X ~ P(\(\frac{1}{2}\)), then find P(X = 3) given e^-0.5 = 0.6065.
Solution:

Question 3.
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e^-3 = 0.0497
Solution:
∵ X follows Poisson Distribution

Question 4.
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e^-4 = 0.0183.
Solution:
∵ m = 1
∵ X ~ P(m = 4)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
X = No. of complaints recieved
(i) P(Only two complaints on a given day)

(ii) P(Atmost two complaints on a given day)
P(X ≤ 2) = p(0) + p(1) + p(2)
= \(\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}\) + 0.1464
= e^-4 + e^-4 × 4 + 0.1464
= e^-4 [1 + 4] + 0.1464
= 0.0183 × 5 + 0.1464
= 0.0915 + 0.1464
= 0.2379

Question 5.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given e^-1.5 = 0.2231.
Solution:
Let X = No. of demands for a car on any day
∴ No. of cars hired
n = 2
m = 1.5
∵ X ~ P(m = 1.5)

Question 6.
Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e^-1 = 0.3678.
Solution:
∵ X = No. of defects on a plywood sheet
∵ m = -1
∵ X ~ P(m = -1)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(No defect)
P(X = 0) = \(\frac{e^{-1} \times 1^{0}}{0 !}\)
= e^-1
= 0.3678

(ii) P(At least one defect)
P(X ≥ 1) = 1 – P(X < 1)
= 1 – p(0)
= 1 – 0.3678
= 0.6322

Question 7.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive. Given e^-5 = 0.0067.
Solution:
X = No. of rats
∵ m = 5
∴ X ~ P(m = 5)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(Exactly five rats)

(ii) P(More than five rats)
P(X > 5) = 1 – P(X ≤ 5)

(iii) P(between 5 and 7 rats, inclusive)
P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)

= 0.0067 × 3125 × 0.02
= 0.0067 × 62.5
= 0.42

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 1.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
∴ n = 4
∵ p = \(\frac{3}{6}=\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∵ X ~ B(3, \(\frac{1}{2}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(Two Successes)

(ii) P(Atleast 3 Successes)

(iii) P(Atmost 2 Successes)

Question 2.
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Question 3.
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
X ~ B(4, 0.1)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} \mathrm{q}^{n-x}\)
P(Not include more than 1 defective)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)⁴ + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)³ [0.9 + 0.4]
= (0.9)³ × 1.3
= 0.977

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution:
X: No. of spade cards
Number of cards drawn
∴ n = 5
p: Probability of getting spade card

(i) P(All five cards are spades)

(ii) P(Only 3 cards are spades)

(iii) P(None is a spade)

Question 5.
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution:
X : No. of bulbs fuse after 200 days of use
p : Probability of getting fuse bulbs
No. of bulbs in a sample
∴ n = 5
∴ p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
∵ X ~ B(5, 0.2)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(X = 0) = ⁵C₀ (0.2)⁰ (0.8)^5-0
= 1 × 1 × (0.8)⁵
= (0.8)⁵

(ii) P(X ≤ 1) = p(0) + p(1)
= ⁵C₀ (0.2)⁰ (0.8)^5-0 + ⁵C₁ (0.2)¹ (0.8)^5-1
= 1 × 1 × (0.8)⁵ + 5 × 0.2 × (0.8)⁴
= (0.8)⁴ [0.8 + 1]
= 1.8 × (0.8)⁴

(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)⁴

(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)⁵

Question 6.
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution:
X : No. of balls drawn marked with the digit 0
n : No. of balls drawn
∴ n = 4
p : Probability of balls marked with 0.
∴ p = \(\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
p(x) = \({ }^{n} C_{x} p^{x} q^{n-x}\)
P(None of the ball is marked with digit 0)

Question 7.
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution:
n: No. of Questions
∴ n = 5
X: No. of correct answers by guessing
p: Probability of getting correct answers

Question 8.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
X : No. of sixes in 6 throws
n : No. of times dice thrown
∴ n = 6
p : Probability of getting six
∴ p = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∵ X ~ B(6, \(\frac{1}{6}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
P(At most 2 sixes)
P(X ≤ 2) = p(0) + p(1) + p(2)

Question 9.
Given that X ~ B(n, p),
(i) if n = 10 and p = 0.4, find E(X) and Var(X).
(ii) if p = 0.6 and E(X) = 6, find n and Var(X).
(iii) if n = 25, E(X) = 10, find p and Var(X).
(iv) if n = 10, E(X) = 8, find Var(X).
Solution:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p
(i) E(X) = np = 10 × 0.4 = 4
∵ q = 1 – p = 1 – 0.4 = 0.6
V(X) = npq = 10 × 0.4 × 0.6 = 2.4

(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4

(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
= \(\sqrt{n p q}\)
= \(\sqrt{25 \times 0.4 \times 0.6}\)
= √6
= 2.4494

(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6