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Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 5 Index Numbers Ex 5.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Index Numbers Ex 5.3

Calculate the cost of living index in problems 1 to 3.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Base year weights (W) and current year price relatives (I) are given in problems 4 to 8. Calculate the cost of living index in each case.

Question 4.

Solution:

ΣW = 20, ΣIW = 1540
CLI = \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)
= \(\frac{1540}{20}\)
= 77

Question 5.

Solution:

ΣW = 17, ΣIW = 3500
CLI = \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)
= \(\frac{3500}{17}\)
= 205.88

Question 6.

Solution:

ΣW = 150, ΣIW = 25400
CLI = \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)
= \(\frac{25400}{150}\)
= 169.33

Question 7.

Solution:

ΣW = x + 18, ΣIW = 100x + 3600
CLI = 150
∴ \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\) = 150
∴ \(\frac{100 x+3600}{x+18}\) = 150
∴ 100x + 3600 = 150x + 2700
∴ 50x = 900
∴ x = 18

Question 8.
Find y if the cost of living index is 200

Solution:

ΣW = y + 18, ΣIW = 300y + 2200
CLI = 200
∴ \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\) = 200
∴ \(\frac{300 y+2200}{y+14}\) = 200
∴ 300y + 2200 = 200y + 2800
∴ 100y = 600
∴ y = 6

Question 9.
The cost of living Index numbers for years 1995 and 1999 are 140 and 200 respectively. A person earns ₹ 11,200 per month in the year 1995. What should be his monthly earning in the year 1999 in order to maintain his standard of living as in the year 1995?
Solution:
CLI (1995) = 140
CLI (1999) = 200
Income (1995) = 11200
Income (1999) = ?
For year 1995
∴ Real Income = \(\frac{\text { Income }}{\text { CLI }} \times 100\)
= \(\frac{11200}{140}\) × 100
= ₹ 8000
For year 1999
∴ Real Income = \(\frac{\text { Income }}{\text { CLI }} \times 100\)
∴ 8000 = \(\frac{\text { Income } \times 100}{200}\)
∴ Income = 16000
∴ Income in 1999 = ₹ 16000

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3

Question 1.
If ω is a complex cube root of unity, show that
(i) (2 – ω)(2 – ω²) = 7
(ii) (2 + ω + ω²)³ – (1 – 3ω + ω²)³ = 65
(iii) \(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\) = 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
(i) L.H.S. = (2 – ω)(2 – ω²)
= 4 – 2ω² – 2ω + ω³
= 4 – 2(ω² + ω) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
= R.H.S.

(ii) L.H.S. = (2 + ω + ω²)³ – (1 – 3ω + ω²)³
= [2 + (ω + ω²)]³ – [-3ω + (1 + ω²)]³
= (2 – 1)³ – (-3ω – ω)³
= 13 – (-4ω)³
= 1 + 64ω³
= 1 + 64(1)
= 65
= R.H.S.

(iii) L.H.S. =\(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\)
= \(\frac{a \omega^{3}+b \omega^{4}+c \omega^{2}}{c+a \omega+b \omega^{2}}\) ……[∵ ω³ = 1, ω⁴ = ω]
= \(\frac{\omega^{2}\left(c+a \omega+b \omega^{2}\right)}{c+a \omega+b \omega^{2}}\)
= ω²
= R.H.S.

Question 2.
If ω is a complex cube root of unity, find the value of
(i) ω + \(\frac{1}{\omega}\)
(ii) ω² + ω³ + ω⁴
(iii) (1 + ω²)³
(iv) (1 – ω – ω²)³ + (1 – ω + ω²)³
(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
(i) ω + \(\frac{1}{\omega}\)
= \(\frac{\omega^{2}+1}{\omega}\)
= \(\frac{-\omega}{\omega}\)
= -1

(ii) ω² + ω³ + ω⁴
= ω² (1 + ω + ω²)
= ω² (0)
= 0

(iii) (1 + ω²)³
= (-ω)³
= -ω³
= -1

(iv) (1 – ω – ω²)³ + (1 – ω + ω²)³
= [1 – (ω + ω²)]³ + [(1 + ω²) – ω]³
= [1 – (-1)]³ + (-ω – ω)³
= 2³ + (-2ω)³
= 8 – 8ω³
= 8 – 8(1)
= 0

(v) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)
= (1 + ω)(1 + ω²)(1 + ω)(1 + ω²) …..[∵ ω³ = 1, ω⁴ = ω]
= (-ω²)(-ω)(-ω²)(-ω)
= ω⁶
= (ω³)²
= (1)²
= 1

Question 3.
If α and β are the complex cube roots of unity, show that α² + β² + αβ = 0.
Solution:
α and β are the complex cube roots of unity.

∴ α – β = -1
L.H.S. = α² + β² + αβ
= α² + 2αβ + β² + αβ – 2αβ ……[Adding and subtracting 2αβ]
= (α² + 2αβ + β²) – αβ
= (α + β)² – αβ
= (-1)² – 1
= 1 – 1
= 0
= R.H.S.

Question 4.
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are the 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.

Question 5.
If ω is a complex cube root of unity, then prove the following:
(i) (ω² + ω – 1)³ = -8
(ii) (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
(i) L.H.S. = (ω² + ω – 1)³
= (-1 – 1)³
= (-2)³
= -8
= R.H.S.

(ii) L.H.S. = (a + b) + (aω + bω²) + (aω² + bω)
= (a + aω + aω²) + (b + bω + bω²)
= a(1 + ω + ω²) + b(1 + ω + ω²)
= a(0) + b(0)
= 0
= R.H.S.

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 5 Index Numbers Ex 5.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Index Numbers Ex 5.2

Calculate Laspeyres, Paasche’s, Dorbish-Bowely’s, and Marshall-Edegworth’s Price Index Numbers in Problems 1 and 2.

Question 1.

Solution:

Question 2.

Solution:

Calculate Walsh’s Price Index Number in Problems 3 and 4.

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If p₀₁(L) = 90, and p₀₁(P) = 40, find p₀₁(D – B) and p₀₁(F)
Solution:

Question 6.
If Σp₀q₀ = 140, Σp₀q₁ = 200, Σp₁q₀ = 350, Σp₁q₁ = 460, find Laspeyre’s Paasche’s Dorbish-Bowley’s and Marshall- Edgeworth’s Price Index Numbers.
Solution:

Question 7.
Given that Laspeyre’s and Dorbish-Bowley’s Price Index Numbers are 160.32 and 164.18 respectively. Find Paasche’s Price Index Number.
Solution:

Question 8.
Given that Σp₀q₀ = 220, Σp₀q₁ = 380, Σp₁q₁ = 350 is Marshall-Edgeworth’s Price Index Number is 150, find Laspeyre’s Price Index Number.
Solution:

Question 9.
Find x in the following table if Laspeyres and Paasche’s Price Index Numbers are equal.

Solution:

Question 10.
If Laspeyre’s Price Index Number is four times Paasche’s Price Index Number, then find the relation between Dorbish-Bowley’s and Fisher’s Price Index Numbers.
Solution:

Question 11.
If Dorbish-Bowley’s and Fisher’s Price Index Numbers are 5 and 4, respectively, then find Laspeyres and Paasche’s Price Index Numbers.
Solution: