Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.3 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.3 Questions And Answers Maharashtra Board

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^{2}}-\sqrt{6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^{2}-9}\right]\)
Solution:

Question 3.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^{2}}-\sqrt{1+y^{2}}}{y^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\right]\)
Solution:

Question 4.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{a+y}-\sqrt{a}}{y \sqrt{a+y}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{x^{2}+9}-\sqrt{2 x^{2}+9}}{\sqrt{3 x^{2}+4}-\sqrt{2 x^{2}+4}}\right)\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 1}\left[\frac{x^{2}+x \sqrt{x}-2}{x-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 4}\left[\frac{x^{2}+x-20}{\sqrt{3 x+4}-4}\right]\)
Solution:

Question 4.
\(\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right)\)
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 11 Maths 2
• Exercise 7.2 Class 11 Maths 2
• Exercise 7.3 Class 11 Maths 2
• Exercise 7.4 Class 11 Maths 2
• Exercise 7.5 Class 11 Maths 2
• Exercise 7.6 Class 11 Maths 2
• Exercise 7.7 Class 11 Maths 2
• Miscellaneous Exercise 7 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.2 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.2 Questions And Answers Maharashtra Board

I. Evaluate the following limits:

Question 1.
\(\lim _{z \rightarrow 2}\left[\frac{z^{2}-5 z+6}{z^{2}-4}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow-3}\left[\frac{x+3}{x^{2}+4 x+3}\right]\)
Solution:

Question 3.
\(\lim _{y \rightarrow 0}\left[\frac{5 y^{3}+8 y^{2}}{3 y^{4}-16 y^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow-2}\left[\frac{-2 x-4}{x^{3}+2 x^{2}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 3}\left[\frac{x^{2}+2 x-15}{x^{2}-5 x+6}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{u \rightarrow 1}\left[\frac{u^{4}-1}{u^{3}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^{3}-27}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-1}\right]\)
Solution:

Question 4.
\(\lim _{\Delta x \rightarrow 0}\left[\frac{(x+\Delta x)^{2}-2(x+\Delta x)+1-\left(x^{2}-2 x+1\right)}{\Delta x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow \sqrt{2}}\left[\frac{x^{2}+x \sqrt{2}-4}{x^{2}-3 x \sqrt{2}+4}\right]\)
Solution:

Question 6.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-7 x+6}{x^{3}-7 x^{2}+16 x-12}\right]\)
Solution:
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-7 x+6}{x^{3}-7 x^{2}+16 x-12}\right]\)
As x → 2, numerator and denominator both tend to zero
∴ x – 2 is a factor of both.
To find the other factor for both of them, by synthetic division
Consider, Numerator = x³ + 0x² – 7x + 6


∴ The limit does not exist

III. Evaluate the following limits:

Question 1.
\(\lim _{y \rightarrow \frac{1}{2}}\left[\frac{1-8 y^{3}}{y-4 y^{3}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 1}\left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 1}\left[\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right]\)
Solution:
\(\lim _{x \rightarrow 1}\left[\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right]\)
As x → 1, numerator and denominator both tend to zero
∴ x – 1 is a factor of both.
To find the factor of numerator and denominator by synthetic division
Consider, numerator = x⁴ + 0x³ – 3x² + 0x + 2

Question 4.
\(\lim _{x \rightarrow 1}\left[\frac{x+2}{x^{2}-5 x+4}+\frac{x-4}{3\left(x^{2}-3 x+2\right)}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow a}\left[\frac{1}{x^{2}-3 a x+2 a^{2}}+\frac{1}{2 x^{2}-3 a x+a^{2}}\right]\)
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 11 Maths 2
• Exercise 7.2 Class 11 Maths 2
• Exercise 7.3 Class 11 Maths 2
• Exercise 7.4 Class 11 Maths 2
• Exercise 7.5 Class 11 Maths 2
• Exercise 7.6 Class 11 Maths 2
• Exercise 7.7 Class 11 Maths 2
• Miscellaneous Exercise 7 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.1 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.1 Questions And Answers Maharashtra Board

I. Evaluate the following limits:

Question 1.
\(\lim _{z \rightarrow-3}\left[\frac{\sqrt{Z+6}}{Z}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow-3}\left[\frac{y^{5}+243}{y^{3}+27}\right]\)
Solution:

Question 3.
\(\lim _{z \rightarrow-5}\left[\frac{\left(\frac{1}{z}+\frac{1}{5}\right)}{z+5}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{5}-3125}\right]\)
Solution:

Question 4.
If \(\lim _{x \rightarrow 1}\left[\frac{x^{4}-1}{x-1}\right]=\lim _{x \rightarrow a}\left[\frac{x^{3}-a^{3}}{x-a}\right]\), find all possible values of a.
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 1}\left[\frac{x+x^{2}+x^{3}+\ldots \ldots \ldots+x^{n}-n}{x-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]\)
Solution:

Question 3.
If \(\lim _{x \rightarrow 5}\left[\frac{x^{k}-5^{k}}{x-5}\right]\) = 500, find all possible values of k.
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{8}-1}{(1-x)^{2}-1}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}\right]\)
Solution:

Question 6.
\(\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]\)
Solution:

Question 7.
\(\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 7}\left[\frac{x^{3}-343}{\sqrt{x}-\sqrt{7}}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 1}\left(\frac{x+x^{3}+x^{5}+\ldots+x^{2 n-1}-n}{x-1}\right)\)
Solution:

IV. In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

Question 1.
\(\lim _{x \rightarrow 2}(2 x+3)=7\)
Solution:
We have to find some δ so that
\(\lim _{x \rightarrow 2}(2 x+3)=7\)
Here a = 2, l = 1 and f(x) = 2x + 3
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |(2x + 3) – 7| < ∈
∴ |2x + 4| < ∈
∴ 2(x – 2)|< ∈
∴ |x – 2| < \(\frac{\epsilon}{2}\)
∴ δ ≤ \(\frac{\epsilon}{2}\) such that
|2x + 4| < δ ⇒ |f(x) – 7| < ∈

Question 2.
\(\lim _{x \rightarrow-3}(3 x+2)=-7\)
Solution:
We have to find some δ so that
\(\lim _{x \rightarrow-3}(3 x+2)=-7\)
Here a = -3, l = -7 and f(x) = 3x + 2
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |3x + 2 – (-7)| < ∈
∴ |3x + 9| < ∈
∴ |3(x + 3)| < ∈
∴ |x + 3| < \(\frac{\epsilon}{3}\)
∴ δ < \(\frac{\epsilon}{3}\) such that
|x + 3| ≤ δ ⇒ |f(x) + 7| < ∈

Question 3.
\(\lim _{x \rightarrow 2}\left(x^{2}-1\right)=3\)
Solution:
We have to find some δ > 0 such that
\(\lim _{x \rightarrow 2}\left(x^{2}-1\right)=3\)
Here, a = 2, l = 3 and f(x) = x² – 1
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |(x² – 1) – 3| < ∈
∴ |x² – 4| < ∈
∴ |(x + 2)(x – 2)| < ∈ …..(i)
We have to get rid of the factor |x + 2|
As |x – 2| < δ
-δ < x – 2 < δ
∴ 2 – δ < x < 2 + δ
Since δ can be assumed as very small, let us choose δ < 1
∴ 1 < x < 3
∴ 3 < x + 2 < 5 …..(Adding 2 throughout)
∴ |x + 2| < 5
∴ |(x + 2)(x – 2)| < 5|x – 2| ……(ii)
From (i) and (ii), we get
5|x – 2|< ∈
∴ x – 2 < \(\frac{\epsilon}{5}\)
If δ = \(\frac{\epsilon}{5}\), |x – 2| < δ ⇒ |x² – 4| < ∈
∴ We choose δ = min{\(\frac{\epsilon}{5}\), 1} then
|x – 2| < δ ⇒ |f(x) – 3| < ∈

Question 4.
\(\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)=3\)
Solution:
We have to find some δ > 0 such that
\(\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)=3\)
Here a = 1, l = 3 and f(x) = x² + x + 1
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |x² + x + 1 – 3| < ∈
∴ |x² + x – 2| < ∈
∴ |(x + 2)(x – 1)| < ∈ …..(i)
We have to get rid of the factor |x + 2|
As |x – 1| < δ
-δ < x – 1 < δ
∴ 1 – δ < x < 1 + δ
Since δ can be assumed as very small, let us choose δ < 1
∴ 0 < x < 2
∴ 2 < x + 2 < 4
∴ |x + 2| < 4
∴ |(x + 2)(x – 1)|< 4 |x – 1| …..(ii)
From (i) and (ii), we get
4|x – 1| < ∈
∴ |x – 1| < \(\frac{\epsilon}{4}\)
If δ = \(\frac{\epsilon}{4}\),
|x – 1| < δ ⇒ x² + x – 2 < ∈
∴ We choose δ = min{\(\frac{\epsilon}{4}\), 1} then
|x – 1| < δ ⇒ |f(x) – 3| < ∈

Class 11 Maharashtra State Board Maths Solution 

• Exercise 7.1 Class 11 Maths 2
• Exercise 7.2 Class 11 Maths 2
• Exercise 7.3 Class 11 Maths 2
• Exercise 7.4 Class 11 Maths 2
• Exercise 7.5 Class 11 Maths 2
• Exercise 7.6 Class 11 Maths 2
• Exercise 7.7 Class 11 Maths 2
• Miscellaneous Exercise 7 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Miscellaneous Exercise 6 Questions and Answers.

11th Maths Part 2 Functions Miscellaneous Exercise 6 Questions And Answers Maharashtra Board

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

Question 1.
If log (5x – 9) – log (x + 3) = log 2, then x = ________
(A) 3
(B) 5
(C) 2
(D) 7
Answer:
(B) 5
Hint:
log (5x – 9) – log (x + 3) = log 2
∴ \(\frac{5 x-9}{x+3}\) = 2
∴ 3x = 9 + 6
∴ x = 5

Question 2.
If log₁₀ (log₁₀ (log₁₀ x)) = 0, then x = ________
(A) 1000
(B) 10¹⁰
(C) 10
(D) 0
Answer:
(B) 10¹⁰
Hint:
log₁₀ log₁₀ log₁₀ x = 0
∴ log₁₀ (log₁₀ (x)) = 10⁰ = 1
∴ log₁₀ x = 10¹ = 10
∴ x = 10¹⁰

Question 3.
Find x, if 2 log₂ x = 4
(A) 4, -4
(B) 4
(C) -4
(D) not defined
Answer:
(B) 4
Hint:
2 log₂ x = 4, x > 0
∴ log₂ (x²) = 4
∴ x² = 16
∴ x = ±4
∴ x = 4

Question 4.
The equation \(\log _{x^{2}} 16+\log _{2 x} 64=3\) has,
(A) one irrational solution
(B) no prime solution
(C) two real solutions
(D) one integral solution
Answer:
(A), (B), (C), (D)
Hint:
\(\log _{x^{2}} 16+\log _{2 x} 64=3\)
∴ \(\frac{\log 16}{\log x^{2}}+\frac{\log 64}{\log 2 x}=3\)
∴ 4 log 2 [log x + log 2] + (6 log 2) (2 log x) = 3 (2 log x) (log 2 + log x)
Let log 2 = a, log x = t. Then
∴ 4at + 4a² + 12at = 6at + 6t²
∴ 6t² – 10at – 4a² = 0
∴ 3t² – 5at – 2a² = 0
∴ (3t + a) (t – 2a) = 0
∴ t = \(-\frac{1}{3}\)a, 2a
∴ log x = \(\log (2)^{-\frac{1}{3}}\), log (2²)
∴ x = \(2^{-\frac{1}{3}}\), 4
∴ x = \(\frac{1}{\sqrt[3]{2}}\), 4

Question 5.
If f(x) = \(\frac{1}{1-x}\), then f(f(f(x))) is
(A) x – 1
(B) 1 – x
(C) x
(D) -x
Answer:
(C) x
Hint:

Question 6.
If f: R → R is defined by f(x) = x³, then f^-1 (8) is equal to:
(A) {2}
(B) {-2.2}
(C) {-2}
(D) (-2.2)
Answer:
(A) {2}
Hint:

Question 7.
Let the function f be defined by f(x) = \(\frac{2 x+1}{1-3 x}\) then f^-1 (x) is:
(A) \(\frac{x-1}{3 x+2}\)
(B) \(\frac{x+1}{3 x-2}\)
(C) \(\frac{2 x+1}{1-3 x}\)
(D) \(\frac{3 x+2}{x-1}\)
Answer:
(A) \(\frac{x-1}{3 x+2}\)
Hint:
f(x) = \(\frac{2 x+1}{1-3 x}\) = y, say. then
2x + 1 = y(1 – 3x)
∴ y – 1 = x(2 + 3y)
∴ x = \(\frac{y-1}{2+3 y}\) = f^-1 (y)
∴ f^-1 (x) = \(\frac{x-1}{2+3 x}\)

Question 8.
If f(x) = 2x² + bx + c and f(0) = 3 and f(2) = 1, then f(1) is equal to
(A) -2
(B) 0
(C) 1
(D) 2
Answer:
(B) 0
Hint:
f(x) = 2x² + bx + c
f(0) = 3
∴ 2(0) + b(0) + c = 3
∴ c = 3 ……..(i)
∴ f(2) = 1
∴ 2(4) + 2b + c = 1
∴ 2b + c = -7
∴ 2b + 3 = -7 …..[From (i)]
∴ b = -5
∴ f(x) = 2x² – 5x + 3
∴ f(1) = 2(1)² – 5(1) + 3 = 0

Question 9.
The domain of \(\frac{1}{[x]-x}\), where [x] is greatest integer function is
(A) R
(B) Z
(C) R – Z
(D) Q – {0}
Answer:
(C) R – Z
Hint:
f(x) = \(\frac{1}{[x]-x}=\frac{1}{-\{x\}}\)
For f to be defined, {x} ≠ 0
∴ x cannot be integer.
∴ Domain = R – Z

Question 10.
The domain and range of f(x) = 2 – |x – 5| are
(A) R+, (-∞, 1]
(B) R, (-∞, 2]
(C) R, (-∞, 2)
(D) R+, (-∞, 2]
Answer:
(B) R, (-∞, 2]
Hint:

f(x) = 2 – |x – 5|
= 2 – (5 – x), x < 5
= 2 – (x – 5), x ≥ 5
∴ f(x) = x – 3, x < 5
= 7 – x, x ≥ 5
Domain = R,
Range (from graph) = (-∞, 2]

(II) Answer the following:

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) {(2, 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) {(2, 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 = {2, 3, 5}, Range = {1, 2}

Question 2.
Find whether the following functions are one-one.
(i) f: R → R defined by f(x) = x² + 5
(ii) f: R – {3} → R defined by f(x) = \(\frac{5 x+7}{x-3}\) for x ∈ R – {3}
Solution:
(i) f: R → R, defined by f(x) = x² + 5
Note that f(-x) = f(x) = x² + 5
∴ f is not one-one (i.e., many-one) function.

(ii) f: R – {3} → R, defined by f(x) = \(\frac{5 x+7}{x-3}\)
Let f(x₁) = f(x₂)
∴ \(\frac{5 x_{1}+7}{x_{1}-3}=\frac{5 x_{2}+7}{x_{2}-3}\)
∴ 5x₁ x₂ – 15x₁ + 7x₂ – 21 = 5x₁ x₂ – 15x₂ + 7x₁ – 21
∴ 22(x₁ – x₂) = 0
∴ x₁ = x₂
∴ f is a one-one function.

Question 3.
Find whether the following functions are onto or not.
(i) f: Z → Z defined by f(x) = 6x – 7 for all x ∈ Z
(ii) f: R → R defined by f(x) = x² + 3 for all x ∈ R
Solution:
(i) f(x) = 6x – 7 = y (say)
(x, y ∈ Z)
∴ x = \(\frac{7+y}{6}\)
Since every integer y does not give integer x, f is not onto.

(ii) f(x) = x² + 3 = y (say)
(x, y ∈ R)
Clearly y ≥ 3 …..[x² ≥ 0]
∴ All the real numbers less than 3 from codomain R, have not been pre-assigned any element from the domain R.
∴ f is not onto.

Question 4.
Let f: R → R be a function defined by f(x) = 5x³ – 8 for all x ∈ R. Show that f is one-one and onto. Hence find f^-1.
Solution:

Question 5.
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:

Question 6.
A function f is defined as 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 7.
A function f is defined as f(x) = 5 – x for 0 ≤ x ≤ 4. Find the values of x such that
(i) f(x) = 3
(ii) f(x) = 5
Solution:
(i) f(x) = 3
∴ 5 – x = 3
∴ x = 5 – 3 = 2

(ii) f(x) = 5
∴ 5 – x = 5
∴ x = 0

Question 8.
If f(x) = 3x⁴ – 5x² + 7, find f(x – 1).
Solution:
f(x) = 3x⁴ – 5x² + 7
∴ f(x – 1) = 3(x – 1)⁴ – 5(x – 1)² + 7
= 3(x⁴ – ⁴C₁ x³ + ⁴C₂ x² – ⁴C₃ x + ⁴C₄) – 5(x² – 2x + 1) + 7
= 3(x⁴ – 4x³ + 6x² – 4x + 1) – 5(x² – 2x + 1) + 7
= 3x⁴ – 12x³ + 18x² – 12x + 3 – 5x² + 10x – 5 + 7
= 3x⁴ – 12x³ + 13x² – 2x + 5

Question 9.
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 10.
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 11.
Find composite of f and g:
(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
(ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}
g = {(1, 1), (3, 27), (4, 64)}
Solution:
(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
∴ f(1) = 3, g(3) = 6
f(2) = 4, g(4) = 8
f(3) = 5, g(5)=10
f(4) = 6, g(6) = 12
(gof) (x) = g (f(x))
(gof)(1) = g(f(1)) = g(3) = 6
(gof)(2) – g(f(2)) = g(4) = 8
(gof)(3) = g(f(3)) = g(5) = 10
(gof)(4) = g(f(4)) = g(6) = 12
∴ gof = {(1, 6), (2, 8), (3, 10), (4, 12)}

(ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}
g = {(1, 1), (3, 27), (4, 64)}
f(1) = 1, g(1) = 1
f(2) = 4, g(3) = 27
f(3) = 4, g(4) = 64
f(4) = 3
(gof) (x) = g(f(x))
(gof) (1) = g(f(1)) = g(1) = 1
(gof) (2) = g(f(2)) = g(4) = 64
(gof) (3) = g(f(3)) = g(4) = 64
(gof) (4) = g(f(4)) = g(3) = 27
∴ gof = {(1, 1), (2, 64), (3, 64), (4, 27)}

Question 12.
Find fog and gof:
(i) f(x) = x² + 5, g(x) = x – 8
(ii) f(x) = 3x – 2, g(x) = x²
(iii) f(x) = 256x⁴, g(x) = √x
Solution:
(i) f(x) = x² + 5, g(x) = x – 8
(fog) (x) = f(g(x))
= f(x – 8)
= (x – 8)² + 5
= x² – 16x + 64 + 5
= x² – 16x + 69
(gof) (x) = g(f(x))
= g(x² + 5)
= x² + 5 – 8
= x – 3

(ii) f(x) = 3x – 2, g(x) = x²
(fog) (x) = f(g(x)) = f(x²) = 3x² – 2
(gof) (x) = g(f(x))
= g(3x – 2)
= (3x – 2)²
= 9x² – 12x + 4

(iii) f(x) = 256x⁴, g(x) = √x
(fog) (x) = f(g(x)) = f (√x) = 256 (√x)⁴ = 256x²
(gof) (x) = g(f(x)) = g(256x⁴) = \(\sqrt{256 x^{4}}\) = 16x²

Question 13.
If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), show that (fof) (x) = x.
Solution:

Question 14.
If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then show that (fog) (x) = x.
Solution:
f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\)
(fog)(x) = f(g(x))
= f(\(\frac{3+5 x}{4 x-1}\))

Question 15.
Let f: R – {2} → R be defined by f(x) = \(\frac{x^{2}-4}{x-2}\) and g: R → R be defined by g(x) = x + 2. Examine whether f = g or not.
Solution:
f(x) = \(\frac{x^{2}-4}{x-2}\), x ≠ 2
∴ f(x) = x + 2, x ≠ 2 and g(x) = x + 2,
The domain of f = R – {2}
The domain of g = R
Here, f and g have different domains.
∴ f ≠ g

Question 16.
Let f: R → R be given by f(x) = x + 5 for all x ∈ R. Draw its graph.
Solution:
f(x) = x + 5

Question 17.
Let f: R → R be given by f(x) = x³ + 1 for all x ∈ R. Draw its graph.
Solution:
Let y = f(x) = x³ + 1

Question 18.
For any base show that log(1 + 2 + 3) = log 1 + log 2 + log 3
Solution:
L.H.S. = log(1 + 2 + 3) = log 6
R.H.S. = log 1 + log 2 + log 3
= 0 + log (2 × 3)
= log 6
∴ L.H.S. = R.H.S.

Question 19.
Find x, if x = \(3^{3 \log _{3} 2}\).
Solution:
x = \(3^{3 \log _{3} 2}\)
= \(3^{\log 3\left(2^{3}\right)}\)
= 2³ ….[\(a^{\log _{a} b}\) = b]
= 8

Question 20.
Show that, log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x| = 0.
Solution:
L.H.S. = log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x|
= \(\log \left|\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)\right|\)
= log|x² + 1 – x²|
= log 1
= 0
= R.H.S.

Question 21.
Show that \(\log \frac{\mathrm{a}^{2}}{\mathrm{bc}}+\log \frac{\mathrm{b}^{2}}{\mathrm{ca}}+\log \frac{\mathrm{c}^{2}}{\mathrm{ab}}=0\).
Solution:

Question 22.
Simplify log (log x⁴) – log(log x).
Solution:
log (log x⁴) – log (log x)
= log (4 log x) – log (log x) …..[log mⁿ = n log m]
= log 4 + log (log x) – log (log x) …..[log (mn) = log m + log n]
= log 4

Question 23.
Simplify \(\log _{10} \frac{28}{45}-\log _{10} \frac{35}{324}+\log _{10} \frac{325}{432}-\log _{10} \frac{13}{15}\)
Solution:

Question 24.
If log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b), then show that a = b.
Solution:
log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b)
∴ 2 log (\(\frac{a+b}{2}\)) = log a + log b
∴ log \(\left(\frac{a+b}{2}\right)^{2}\) = log ab
∴ \(\frac{(a+b)^{2}}{4}\) = ab
∴ a² + 2ab + b² = 4ab
∴ a² + 2ab – 4ab + b² = 0
∴ a² – 2ab + b² = 0
∴ (a – b)² = 0
∴ a – b = 0
∴ a = b

Question 25.
If b² = ac. Prove that, log a + log c = 2 log b.
Solution:
b² = ac
Taking log on both sides, we get
log b² = log ac
∴ 2 log b = log a + log c
∴ log a + log c = 2 log b

Question 26.
Solve for x, log_x (8x – 3) – log_x 4 = 2.
Solution:
log_x (8x – 3) – log_x 4 = 2
∴ \(\log _{x}\left(\frac{8 x-3}{4}\right)\) = 2
∴ x² = \(\frac{8 x-3}{4}\)
∴ 4x² = 8x – 3
∴ 4x² – 8x + 3 = 0
∴ 4x² – 2x – 6x + 3 = 0
∴ 2x(2x – 1) – 3(2x – 1) = 0
∴ (2x – 1)(2x – 3) = 0
∴ 2x – 1 = 0 or 2x – 3 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{3}{2}\)

Question 27.
If a² + b² = 7ab, show that \(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)
Solution:
a² + b² = 7ab
a² + 2ab + b² = 7ab + 2ab
(a + b)² = 9ab
\(\frac{(a+b)^{2}}{9}\) = ab
\(\left(\frac{a+b}{3}\right)^{2}\) = ab
Taking log on both sides, we get
log \(\left(\frac{a+b}{3}\right)^{2}\) = log (ab)
2 log \(\left(\frac{a+b}{3}\right)\) = log a + log b
Dividing throughout by 2, we get
\(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)

Question 28.
If \(\log \left(\frac{x-y}{5}\right)=\frac{1}{2} \log x+\frac{1}{2} \log y\), show that x² + y² = 27xy.
Solution:

Question 29.
If log₃ [log₂ (log₃ x)] = 1, show that x = 6561.
Solution:
log₃ [log₂ (log₃ x)] = 1
∴ log₂ (log₃ x) = 3¹
∴ log₃ x = 2³
∴ log₃ x = 8
∴ x = 3⁸
∴ x = 6561

Question 30.
If f(x) = log(1 – x), 0 ≤ x < 1, show that f(\(\frac{1}{1+x}\)) = f(1 – x) – f(-x).
Solution:

Question 31.
Without using log tables, prove that \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\).
Solution:
We have to prove that, \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\)
i.e., to prove that \(\frac{2}{5}\) < log₁₀ 3 and log₁₀ 3 < \(\frac{1}{2}\)
i.e., to prove that 2 < 5 log₁₀ 3 and 2 log₁₀ 3 < 1
i.e., to prove that 2 log₁₀ 10 < 5 log₁₀ 3 and 2 log₁₀ 3 < log₁₀ 10 ……[∵ log_a a = 1]
i.e., to prove that log₁₀ 10² < log₁₀ 3⁵ and log₁₀ 3² < log₁₀ 10
i.e., to prove that 10² < 3⁵ and 3² < 10
i.e., to prove that 100 < 243 and 9 < 10 which is true
∴ \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\)

Question 32.
Show that \(7 \log \left(\frac{15}{16}\right)+6 \log \left(\frac{8}{3}\right)+5 \log \left(\frac{2}{5}\right)+\log \left(\frac{32}{25}\right)\) = log 3
Solution:

Question 33.
Solve : \(\sqrt{\log _{2} x^{4}}+4 \log _{4} \sqrt{\frac{2}{x}}=2\)
Solution:

Question 34.
Find the value of \(\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10}\left(\frac{49}{4}\right)+\frac{1}{2} \log _{10}\left(\frac{1}{25}\right)}\)
Solution:

Question 35.
If \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\), show that abc = 1.
Solution:
Let \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\) = k
∴ log a = k(x + y – 2z), log b = k(y + z – 2x), log c = k(z + x – 2y)
log a + log b + log c = k(x + y – 2z) + k(y + z – 2x) + k(z + x – 2y)
= k(x + y – 2z + y + z – 2x + z + x – 2y)
= k(0)
= 0
∴ log (abc) = log 1 …….[∵ log 1 = 0]
∴ abc = 1

Question 36.
Show that, log_y x³ . log_z y⁴ . log_x z⁵ = 60.
Solution:

Question 37.
If \(\frac{\log _{2} \mathrm{a}}{4}=\frac{\log _{2} \mathrm{~b}}{6}=\frac{\log _{2} \mathrm{c}}{3 \mathrm{k}}\) and a³b²c = 1, find the value of k.
Solution:

Question 38.
If a² = b³ = c⁴ = d⁵, show that loga bcd = \(\frac{47}{30}\).
Solution:

Question 39.
Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.
(i) 1 < |x – 1| < 4
(ii) |x² – x – 6| = x + 2
(iii) |x² – 9| + |x² – 4| = 5
(iv) -2 < [x] ≤ 7
(v) 2[2x – 5] – 1 = 7
(vi) [x]2 – 5 [x] + 6 = 0
(vii) [x – 2] + [x + 2] + {x} = 0
(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)
Solution:
(i) 1 < |x – 1| < 4
∴ -4 < x – 1 < -1 or 1 < x – 1 < 4
∴ -3 < x < 0 or 2 < x < 5
∴ Solution set = (-3, 0) ∪ (2, 5)

(ii) |x² – x – 6| = x + 2 …..(i)
R.H.S. must be non-negative
∴ x ≥ -2 …..(ii)
|(x – 3) (x + 2)| = x + 2
∴ (x + 2) |x – 3| = x + 2 as x + 2 ≥ 0
∴ |x – 3| = 1 if x ≠ -2
∴ x – 3 = ±1
∴ x = 4 or 2
∴ x = -2 also satisfies the equation
∴ Solution set = {-2, 2, 4}

(iii) |x² – 9| + |x² – 4| = 5
∴ |(x – 3) (x + 3)| + |(x – 2) ( x + 2)| = 5 ………(i)
Case I: x < -3
Also, x < -2, x < 2, x < 3
∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0
Equation (i) reduces to
x² – 9 + x² – 4 = 5
∴ 2x² = 18
∴ x = -3 or 3 (both rejected as x < -3)

Case II: -3 ≤ x < -2
As x < -2, x < 3
∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0
Equation (i) reduces to
-(x² – 9) + x² – 4 = 5
∴ 5 = 5 (true)
-3 ≤ x < -2 is a solution ….(ii)

Case III: -2 ≤ x < 2 As x > -3, x < 3
∴ (x – 3) (x + 3) < 0,
(x – 2) (x + 2) < 0
Equation (i) reduces to
9 – x² + 4 – x2 = 5
∴ 2x² = 13 – 5
∴ x² = 4
∴ x = -2 is a solution …..(iii)

Case IV: 2 ≤ x < 3 As x > -3, x > -2
∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0
Equation (i) reduces to
9 – x² + x² – 4 = 5
∴ 5 = 5 (true)
∴ 2 ≤ x < 3 is a solution ……(iv)

Case V: 3 ≤ x As x > -3, x > -2, x > 2
∴ (x + 3) (x – 3) > 0,
(x – 2) (x + 2) > 0
Equation (i) reduces to
x² – 9 + x² – 4 = 5
∴ 2x² = 18
∴ x² = 9
∴ x = 3 …..(v)
(x = -3 rejected as x ≥ 3)
From (ii), (iii), (iv), (v), we get
∴ Solution set = [-3, -2] ∪ [2, 3]

(iv) -2 < [x] ≤ 7
∴ -2 < x < 8
∴ Solution set = (-2, 8)

(v) 2[2x – 5] – 1 = 7
∴ [2x – 5] = \(\frac{7+1}{2}\) = 4
∴ [2x] – 5 = 4
∴ [2x] = 9
∴ 9 ≤ 2x < 10
∴ \(\frac{9}{2}\) ≤ x < 5
∴ Solution set = [\(\frac{9}{2}\), 5)

(vi) [x]² – 5[x] + 6 = 0
∴ ([x] – 3)([x] – 2) = 0
∴ [x] = 3 or 2
If [x] = 2, then 2 ≤ x < 3
If [x] = 3, then 3 ≤ x < 4
∴ Solution set = [2, 4)

(vii) [x – 2] + [x + 2] + {x} = 0
∴ [x] – 2 + [x] + 2 + {x} = 0
∴ [x] + x = 0 …..[{x} + [x] = x]
∴ x = 0

(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)
L.H.S. = an integer
R.H.S. = an integer
∴ x = 6k, where k is an integer

Question 40.
Find the domain of the following functions.
(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}\)
(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)
(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)
(iv) f(x) = x!
(v) f(x) = \({ }^{5-x} P_{x-1}\)
(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)
(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)
Solution:
(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}=\frac{x^{2}+4 x+4}{(x+3)(x-2)}\)
For f to be defined, x ≠ -3, 2
∴ Domain of f = (-∞, -3) ∪ (-3, 2) ∪ (2, ∞)

(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)
For f to be defined,
x – 3 ≥ 0, 5 – x > 0 and 5 – x ≠ 1
x ≥ 3, x < 5 and x ≠ 4
∴ Domain of f = [3, 4) ∪ (4, 5)

(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)

Equation (i) gives solution set = [-1, 1]
∴ Domain of f = [-1, 1]

(iv) f(x) = x!
∴ Domain of f = set of whole numbers (W)

(v) f(x) = \({ }^{5-x} P_{x-1}\)
5 – x > 0, x – 1 ≥ 0, x – 1 ≤ 5 – x
∴ x < 5, x ≥ 1 and 2x ≤ 6
∴ x ≤ 3
∴ Domain of f = {1, 2, 3}

(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)
x – x² ≥ 0
∴ x² – x ≤ 0
∴ x(x – 1) ≤ 0
∴ 0 ≤ x ≤ 1 …..(i)
5 – x ≥ 0
∴ x ≤ 5 …..(ii)
Intersection of intervals given in (i) and (ii) gives
Solution set = [0, 1]
∴ Domain of f = [0, 1]

(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)
For f to be defined,
log (x² – 6x + 6) ≥ 0
∴ x² – 6x + 6 ≥ 1
∴ x² – 6x + 5 ≥ 0
∴ (x – 5)(x – 1) ≥ 0
∴ x ≤ 1 or x ≥ 5 …..(i)
[∵ The solution of (x – a) (x – b) ≥ 0 is x ≤ a or x ≥ b, for a < b]
and x² – 6x + 6 > 0
∴ (x – 3)² > -6 + 9
∴ (x – 3)² > 3
∴ x < 3 – √3 0r x > 3 + √3 ……..(ii)
From (i) and (ii), we get
x ≤ 1 or x ≥ 5
Solution set = (-∞, 1] ∪ [5, ∞)
∴ Domain of f = (-∞, 1] ∪ [5, ∞)

Question 41.
(i) f(x) = |x – 5|
(ii) f(x) = \(\frac{x}{9+x^{2}}\)
(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\)
(iv) f(x) = [x] – x
(v) f(x) = 1 + 2x + 4x
Solution:
(i) f(x) = |x – 5|

∴ Range of f = [0, ∞)

(ii) f(x) = \(\frac{x}{9+x^{2}}\) = y (say)
∴ x²y – x + 9y = 0
For real x, Discriminant > 0
∴ 1 – 4(y)(9y) ≥ 0
∴ y² ≤ \(\frac{1}{36}\)
∴ \(\frac{-1}{6}\) ≤ y ≤ \(\frac{1}{6}\)
∴ Range of f = [\(\frac{-1}{6}\), \(\frac{1}{6}\)]

(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\) = y, (say)
∴ √x y + y = 1
∴ √x = \(\frac{1-y}{y}\) ≥ 0
∴ \(\frac{y-1}{y}\) ≤ 0
∴ o < y ≤ 1
∴ Range of f = (0, 1]

(iv) f(x) = [x] – x = -{x}
∴ Range of f = (-1, 0] …..[0 ≤ {x} < 1]

(v) f(x) = 1 + 2x + 4x
Since, 2x > 0, 4x > 0
∴ f(x) > 1
∴ Range of f = (1, ∞)

Question 42.
Find (fog) (x) and (gof) (x)
(i) f(x) = e^x, g(x) = log x
(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)
Solution:
(i) f(x) = e^x, g(x) = log x
(fog) (x) = f(g(x))
= f(log x)
= e^{log x}
= x
(gof) (x) = g(f(x))
= g(e^x)
= log (e^x)
= x log e
= x …..[∵ log e = 1]

(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)

Question 43.
Find f(x), if
(i) g(x) = x² + x – 2 and (gof) (x) = 4x² – 10x + 4
(ii) g(x) = 1 + √x and f [g(x)] = 3 + 2√x + x.
Solution:
(i) g(x) = x² + x – 2
(gof) (x) = 4x² – 10x + 4
= (2x – 3)² + (2x – 3) – 2
= g(2x – 3)
= g(f(x))
∴ f(x) = 2x – 3
(gof) (x) = 4x² – 10x + 4
= (-2x + 2)² + (-2x + 2) – 2
= g(-2x + 2)
= g(f(x))
∴ f(x) = -2x + 2

(ii) g(x) = 1 + √x
f(g(x)) = 3 + 2√x + x
= x + 2√x + 1 + 2
= (√x + 1)² + 2
f(√x + 1) = (√x + 1)² + 2
∴ f(x) = x² + 2

Question 44.
Find (fof) (x) if
(i) f(x) = \(\frac{x}{\sqrt{1+x^{2}}}\)
(ii) f(x) = \(\frac{2 x+1}{3 x-2}\)
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 2
• Exercise 5.2 Class 11 Maths 2
• Miscellaneous Exercise 5 Class 11 Maths 2
• Exercise 6.1 Class 11 Maths 2
• Exercise 6.2 Class 11 Maths 2
• Miscellaneous Exercise 6 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.2 Questions and Answers.

11th Maths Part 2 Functions Exercise 6.2 Questions And Answers Maharashtra Board

Question 1.
If f(x) = 3x + 5, g(x) = 6x – 1, then find
(i) (f + g) (x)
(ii) (f – g) (2)
(iii) (fg) (3)
(iv) (f/g) (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 2.
Let f: (2, 4, 5} → {2, 3, 6} and g: {2, 3, 6} → {2, 4} be given by f = {(2, 3), (4, 6), (5, 2)} and g = {(2, 4), (3, 4), (6, 2)}. Write down gof.
Solution:
f = {(2, 3), (4, 6), (5, 2)}
∴ f(2) = 3, f(4) = 6, f(5) = 2
g ={(2, 4), (3, 4), (6, 2)}
∴ g(2) = 4, g(3) = 4, g(6) = 2
gof: {2, 4, 5} → {2, 4}
(gof) (2) = g(f(2)) = g(3) = 4
(gof) (4) = g(f(4)) = g(6) = 2
(gof) (5) = g(f(5)) = g(2) = 4
∴ gof = {(2, 4), (4, 2), (5, 4)}

Question 3.
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 + 11

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

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

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

Question 4.
Verify that f and g are inverse functions of each other, where
(i) f(x) = \(\frac{x-7}{4}\), g(x) = 4x + 7
(ii) f(x) = x³ + 4, g(x) = \(\sqrt[3]{x-4}\)
(iii) f(x) = \(\frac{x+3}{x-2}\), g(x) = \(\frac{2 x+3}{x-1}\)
Solution:
(i) f(x) = \(\frac{x-7}{4}\)
Replacing x by g(x), we get
f[g(x)] = \(\frac{g(x)-7}{4}=\frac{4 x+7-7}{4}\) = x
g(x) = 4x + 7
Replacing x by f(x), we get
g[f(x)] = 4f(x) + 7 = 4(\(\frac{x-7}{4}\)) + 7 = x
Here, f[g(x)] = x and g[f(x)] = x.
∴ f and g are inverse functions of each other.

(ii) f(x) = x³ + 4
Replacing x by g(x), we get
f[g(x)] = [g(x)]³ + 4
= \((\sqrt[3]{x-4})^{3}+4\)
= x – 4 + 4
= x
g(x) = \(\sqrt[3]{x-4}\)
Replacing x by f(x), we get
g[f(x)] = \(\sqrt[3]{f(x)-4}=\sqrt[3]{x^{3}+4-4}=\sqrt[3]{x^{3}}\) = x
Here, f[g(x)] = x and g[f(x)] = x
∴ f and g are inverse functions of each other.

(iii) f(x) = \(\frac{x+3}{x-2}\)
Replacing x by g(x), we get

Here, f[g(x)] = x and g[f(x)] = x.
∴ f and g are inverse functions of each other.

Question 5.
Check if the following functions have an inverse function. If yes, find the inverse function.
(i) f(x) = 5x²
(ii) f(x) = 8
(iii) f(x) = \(\frac{6 x-7}{3}\)
(iv) f(x) = \(\sqrt{4 x+5}\)
(v) f(x) = 9x³ + 8
(vi) f(x) =
Solution:
(i) f(x) = 5x² = y (say)

For two values (x₁ and x₂) of x, values of the function are equal.
∴ f is not one-one.
∴ f does not have an inverse.

(ii) f(x) = 8 = y (say)
For every value of x, the value of the function f is the same.
∴ f is not one-one i.e. (many-one) function.
∴ f does not have the inverse.

(iii) f(x) = \(\frac{6 x-7}{3}\)
Let f(x₁) = f(x₂)
∴ \(\frac{6 x_{1}-7}{3}=\frac{6 x_{2}-7}{3}\)
∴ x₁ = x₂
∴ f is a one-one function.
f(x) = \(\frac{6 x-7}{3}\) = y (say)
∴ x = \(\frac{3y+7}{6}\)
∴ For every y, we can get x
∴ f is an onto function.
∴ x = \(\frac{3y+7}{6}\) = f^-1 (y)
Replacing y by x, we get
f-1 (x) = \(\frac{3x+7}{6}\)

(iv) f(x) = \(\sqrt{4 x+5}, x \geq \frac{-5}{4}\)
Let f(x₁) = f(x₂)
∴ \(\sqrt{4 x_{1}+5}=\sqrt{4 x_{2}+5}\)
∴ x₁ = x₂
∴ f is a one-one function.
f(x) = \(\sqrt{4 x+5}\) = y, (say) y ≥ 0
Squaring on both sides, we get
y² = 4x + 5
∴ x = \(\frac{y^{2}-5}{4}\)
∴ For every y we can get x.
∴ f is an onto function.
∴ x = \(\frac{y^{2}-5}{4}\) = f^-1 (y)
Replacing y by x, we get
f^-1 (x) = \(\frac{x^{2}-5}{4}\)

(v) f(x) 9x³ + 8
Let f(x₁) = f(x₂)
∴ \(9 x_{1}^{3}+8=9 x_{2}^{3}+8\)
∴ x₁ = x₂
∴ f is a one-one function.
∴ f(x) = 9x³ + 8 = y, (say)
∴ x = \(\sqrt[3]{\frac{y-8}{9}}\)
∴ For every y we can get x.
∴ f is an onto function.
∴ x = \(\sqrt[3]{\frac{y-8}{9}}\) = f^-1 (y)
Replacing y by x, we get
f^-1 (x) = \(\sqrt[3]{\frac{x-8}{9}}\)

(vi) f(x) = x + 7, x < 0
= 8 – x, x ≥ 0

We observe from the graph that for two values of x, say x₁, x₂ the values of the function are equal.
i.e. f(x₁) = f(x₂)
∴ f is not one-one (i.e. many-one) function.
∴ f does not have inverse.

Question 6.
If f(x) = , 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) = , then find
(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) = 2 |x| + 3x, then find
(i) f(2)
(ii) f(-5)
Solution:
f(x) = 2 |x| + 3x
(i) f(2) = 2|2| + 3(2)
= 2 (2) + 6 ….. [∵ |x| = x, x > 0]
= 10

(ii) f(-5) = 2 |-5| + 3(-5)
= 2(5) – 15 …..[∵ |x| = -x, x < 0]
= 10 – 15
= -5

Question 9.
If f(x) = 4[x] – 3, where [x] is greatest integer function of x, then find
(i) f(7.2)
(ii) f(0.5)
(iii) \(f\left(-\frac{5}{2}\right)\)
(iv) f(2π), where π = 3.14
Solution:
f(x) = 4[x] – 3
(i) f(7.2) = 4 [7.2] – 3
= 4(7) – 3 ………[∵ 7 ≤ 7.2 < 8, [7.2] = 7]
= 25

(ii) f(0.5) = 4[0.5] – 3
= 4(0) – 3 ………[∵ 0 ≤ 0.5 < 1, [0.5] = 0]
= -3

(iii) \(f\left(-\frac{5}{2}\right)\) = f(-2.5)
= 4[-2.5] – 3
= 4(-3) – 3 …….[∵-3 ≤ -2.5 ≤ -2, [-2.5] = -3]
= -15

(iv) f(2π) = 4[2π] – 3
= 4[6.28] – 3 …..[∵ π = 3.14]
= 4(6) – 3 …….[∵ 6 ≤ 6.28 < 7, [6.28] = 6]
= 21

Question 10.
If f(x) = 2{x} + 5x, where {x} is fractional part function of x, then find
(i) f(-1)
(ii) f(\(\frac{1}{4}\))
(iii) f(-1.2)
(iv) f(-6)
Solution:
f(x) = 2{x} + 5x
(i) {-1} = -1 – [-1] = -1 + 1 = 0
∴ f(-1) = 2 {-1} + 5(-1)
= 2(0) – 5
= -5

(ii) {\(\frac{1}{4}\)} = \(\frac{1}{4}\) – [latex]\frac{1}{4}[/latex] = \(\frac{1}{4}\) – 0 = \(\frac{1}{4}\)
f(\(\frac{1}{4}\)) = 2{\(\frac{1}{4}\)} + 5(\(\frac{1}{4}\))
= 2(\(\frac{1}{4}\)) + \(\frac{5}{4}\)
= \(\frac{7}{4}\)
= 1.75

(iii) {-1.2} = -1.2 – [-1.2] = -1.2 + 2 = 0.8
f(-1.2) = 2{-1.2} + 5(-1.2)
= 2(0.8) + (-6)
= -4.4

(iv) {-6} = -6 – [-6] = -6 + 6 = 0
f(-6) = 2{-6} + 5(-6)
= 2(0) – 30
= -30

Question 11.
Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.
(i) |x + 4| ≥ 5
(ii) |x – 4| + |x – 2| = 3
(iii) x² + 7|x| + 12 = 0
(iv) |x| ≤ 3
(v) 2|x| = 5
(vi) [x + [x + [x]]] = 9
(vii) {x} > 4
(viii) {x} = o
(ix) {x} = 0.5
(x) 2{x} = x + [x]
Solution:
(i) |x + 4| ≥ 5
The solution of |x| ≥ a is x ≤ -a or x ≥ a
∴ |x + 4| ≥ 5 gives
∴ x + 4 ≤ -5 or x + 4 ≥ 5
∴ x ≤ -5 – 4 or x ≥ 5 – 4
∴ x ≤ -9 or x ≥ 1
∴ The solution set = (-∞, – 9] ∪ [1, ∞)

(ii) |x – 4| + |x – 2| = 3 …..(i)
Case I: x < 2
Equation (i) reduces to
4 – x + 2 – x = 3 …….[x < 2 < 4, x – 4 < 0, x – 2 < 0]
∴ 6 – 3 = 2x
∴ x = \(\frac{3}{2}\)

Case II: 2 ≤ x < 4
Equation (i) reduces to
4 – x + x – 2 = 3
∴ 2 = 3 (absurd)
There is no solution in [2, 4)

Case III: x ≥ 4
Equation (i) reduces to
x – 4 + x – 2 = 3
∴ 2x = 6 + 3 = 9
∴ x = \(\frac{9}{2}\)
∴ x = \(\frac{3}{2}\), \(\frac{9}{2}\) are solutions.
The solution set = {\(\frac{3}{2}\), \(\frac{9}{2}\)}

(iii) x² + 7|x| + 12 = 0
∴ (|x|)² + 7|x| + 12 = 0
∴ (|x| + 3) (|x| + 4) = 0
∴ There is no x that satisfies the equation.
The solution set = { } or Φ

(iv) |x| ≤ 3 The solution set of |x| ≤ a is -a ≤ x ≤ a
∴ The required solution is -3 ≤ x ≤ 3
∴ The solution set is [-3, 3]

(v) 2|x| = 5
∴ |x| = \(\frac{5}{2}\)
∴ x = ±\(\frac{5}{2}\)

(vi) [x + [x + [x]]] = 9
∴ [x + [x] + [x] ] = 9 …….[[x + n] = [x] + n, if n is an integer]
∴ [x + 2[x]] = 9
∴ [x] + 2[x] = 9 …..[[2[x] is an integer]]
∴ [x] = 3
∴ x ∈ [3, 4)

(vii) {x} > 4
This is a meaningless statement as 0 ≤ {x} < 1
∴ The solution set = { } or Φ

(viii) {x} = 0
∴ x is an integer
∴ The solution set is Z.

(ix) {x} = 0.5
∴ x = ….., -2.5, -1.5, -0.5, 0.5, 1.5, …..
∴ The solution set = {x : x = n + 0.5, n ∈ Z}

(x) 2{x} = x + [x]
= [x] + {x} + [x] ……[x = [x] + {x}]
∴ {x} = 2[x]
R.H.S. is an integer
∴ L.H.S. is an integer
∴ {x} = 0
∴ [x] = 0
∴ x = 0

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 2
• Exercise 5.2 Class 11 Maths 2
• Miscellaneous Exercise 5 Class 11 Maths 2
• Exercise 6.1 Class 11 Maths 2
• Exercise 6.2 Class 11 Maths 2
• Miscellaneous Exercise 6 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.1 Questions and Answers.

11th Maths Part 2 Functions Exercise 6.1 Questions And Answers Maharashtra Board

Question 1.
Check if the following relations are functions.
(a)

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

(b)

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

(c)

Solution:
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), (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 been assigned 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.
Check if the relation given by the equation represents y as function of x.
(i) 2x + 3y = 12
(ii) x + y² = 9
(iii) x² – y = 25
(iv) 2y + 10 = 0
(v) 3x – 6 = 21
Solution:
(i) 2x + 3y = 12
∴ y = \(\frac{12-2 x}{3}\)
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(ii) x + y² = 9
∴ y² = 9 – x
∴ y = ±\(\sqrt{9-x}\)
∴ For one value of x, there are two values of y.
∴ y is not a function of x.

(iii) x² – y = 25
∴ y = x² – 25
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(iv) 2y + 10 = 0
∴ y = -5
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(v) 3x – 6 = 21
∴ x = 9
∴ x = 9 represents a point on the X-axis.
There is no y involved in the equation.
So the given equation does not represent a function.

Question 4.
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)
(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\), h ≠ 0.
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(\(\frac{1}{2}\)) = \(\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{4}-\frac{3}{2}+1\)
= \(\frac{1-6+4}{4}\)
= \(-\frac{1}{4}\)

(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

(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\)
= \(\frac{(2+h)^{2}-3(2+h)+1-\left(2^{2}-3(2)+1\right)}{h}\)
= \(\frac{\mathrm{h}^{2}+\mathrm{h}}{\mathrm{h}}\)
= h + 1

Question 5.
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
(iv) g(x) = x³ – 2x² – 5x + 6
Solution:
(i) g(x) = \(\frac{5 x-6}{7}\)
g(x) = 0
∴ \(\frac{5 x-6}{7}\) = 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
∴ (2x – 1) (3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

(iv) g(x) = x³ – 2x² – 5x + 6
= ( x- 1) (x² – x – 6)
= (x – 1) (x + 2) (x – 3)
g(x) = 0
∴ (x – 1) (x + 2) (x – 3) = 0
∴ x – 1 = 0 or x + 2 = 0 or x – 3 = 0
∴ x = 1, -2, 3

Question 6.
Find x, if f(x) = g(x) where
(i) f(x) = x⁴ + 2x², g(x) = 11x²
(ii) f(x) = √x – 3, g(x) = 5 – x
Solution:
(i) 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, ±3

(ii) f(x) = √x – 3, g(x) = 5 – x
f(x) = g(x)
∴ √x – 3 = 5 – x
∴ √x = 5 – x + 3
∴ √x = 8 – x
on squaring, we get
x = 64 + x² – 16x
∴ x² – 17x + 64 = 0
∴ x = \(\frac{17 \pm \sqrt{(-17)^{2}-4(64)}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{289-256}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{33}}{2}\)

Question 7.
If f(x) = \(\frac{a-x}{b-x}\), f(2) is undefined, and f(3) = 5, find a and b.
Solution:
f(x) = \(\frac{a-x}{b-x}\)
Given that,
f(2) is undefined
b – 2 = 0
∴ b = 2 …..(i)
f(3) = 5
∴ \(\frac{a-3}{b-3}\) = 5
∴ \(\frac{a-3}{2-3}\) = 5 ….. [From (i)]
∴ a – 3 = -5
∴ a = -2
∴ a = -2, b = 2

Question 8.
Find the domain and range of the following functions.
(i) f(x) = 7x² + 4x – 1
Solution:
f(x) = 7x² + 4x – 1
f is defined for all x.
∴ Domain of f = R (i.e., the set of real numbers)

∴ Range of f = [\(-\frac{11}{7}\), ∞)

(ii) g(x) = \(\frac{x+4}{x-2}\)
Solution:
g(x) = \(\frac{x+4}{x-2}\)
Function g is defined everywhere except at x = 2.
∴ Domain of g = R – {2}
Let y = g(x) = \(\frac{x+4}{x-2}\)
∴ (x – 2) y = x + 4
∴ x(y – 1) = 4 + 2y
∴ For every y, we can find x, except for y = 1.
∴ y = 1 ∉ range of function g
∴ Range of g = R – {1}

(iii) h(x) = \(\frac{\sqrt{x+5}}{5+x}\)
Solution:
h(x) = \(\frac{\sqrt{x+5}}{5+x}=\frac{1}{\sqrt{x+5}}\), x ≠ -5
For x = -5, function h is not defined.
∴ x + 5 > 0 for function h to be well defined.
∴ x > -5
∴ Domain of h = (-5, ∞)
Let y = \(\frac{1}{\sqrt{x+5}}\)
∴ y > 0
Range of h = (0, ∞) or R+

(iv) f(x) = \(\sqrt[3]{x+1}\)
Solution:
f(x) = \(\sqrt[3]{x+1}\)
f is defined for all real x and the values of f(x) ∈ R
∴ Domain of f = R, Range of f = R

(v) f(x) = \(\sqrt{(x-2)(5-x)}\)
Solution:
f(x) = \(\sqrt{(x-2)(5-x)}\)
For f to be defined,
(x – 2)(5 – x) ≥ 0
∴ (x – 2)(x – 5) ≤ 0
∴ 2 ≤ x ≤ 5 ……[∵ The solution of (x – a) (x – b) ≤ 0 is a ≤ x ≤ b, for a < b]
∴ Domain of f = [2, 5]
(x – 2) (5 – x) = -x² + 7x – 10
= \(-\left(x-\frac{7}{2}\right)^{2}+\frac{49}{4}-10\)
= \(\frac{9}{4}-\left(x-\frac{7}{2}\right)^{2} \leq \frac{9}{4}\)
∴ \(\sqrt{(x-2)(5-x)} \leq \sqrt{\frac{9}{4}} \leq \frac{3}{2}\)
Range of f = [0, \(\frac{3}{2}\)]

(vi) f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
Solution:
f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
For f to be defined,
\(\sqrt{\frac{x-3}{7-x}}\) ≥ 0, 7 – x ≠ 0
∴ \(\sqrt{\frac{x-3}{7-x}}\) ≤ 0 and x ≠ 7
∴ 3 ≤ x < 7
Let a < b, \(\frac{x-a}{x-b}\) ≤ 0 ⇒ a ≤ x < b
∴ Domain of f = [3, 7)
f(x) ≥ 0 … [∵ The value of square root function is non-negative]
∴ Range of f = [0, ∞)

(vii) f(x) = \(\sqrt{16-x^{2}}\)
Solution:
f(x) = \(\sqrt{16-x^{2}}\)
For f to be defined,
16 – x² ≥ 0
∴ x² ≤ 16
∴ -4 ≤ x ≤ 4
∴ Domain of f = [-4, 4]
Clearly, f(x) ≥ 0 and the value of f(x) would be maximum when the quantity subtracted from 16 is minimum i.e. x = 0
∴ Maximum value of f(x) = √16 = 4
∴ Range of f = [0, 4]

Question 9.
Express the area A of a square as a function of its
(a) side s
(b) perimeter P
Solution:
(a) area (A) = s²
(b) perimeter (P) = 4s
∴ s = \(\frac{\mathrm{P}}{4}\)
Area (A) = s² = \(\left(\frac{\mathrm{P}}{4}\right)^{2}\)
∴ A = \(\frac{\mathrm{P}^{2}}{16}\)

Question 10.
Express the area A of a circle as a function of its
(i) radius r
(ii) diameter d
(iii) circumference C
Solution:
(i) Area (A) = πr²

(ii) Diameter (d) = 2r
∴ r = \(\frac{\mathrm{d}}{2}\)
∴ Area (A) = πr² = \(\frac{\pi \mathrm{d}^{2}}{4}\)

(iii) Circumference (C) = 2πr
∴ r = \(\frac{C}{2 \pi}\)
Area (A) = πr² = \(\pi\left(\frac{\mathrm{C}}{2 \pi}\right)^{2}\)
∴ A = \(\frac{C^{2}}{4 \pi}\)

Question 11.
An open box is made from a square of cardboard of 30 cms side, by cutting squares of length x centimeters from each corner and folding the sides up. Express the volume of the box as a function of x. Also, find its domain.
Solution:

Length of the box = 30 – 2x
Breadth of the box = 30 – 2x
Height of the box = x
Volume = (30 – 2x)² x, x < 15, x ≠ 15, x > 0
= 4x(15 – x)², x ≠ 15, x > 0
Domain (0, 15)

Question 12.
Let f be a subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify?
Solution:
f = {(ab, a + b): a, b ∈ Z}
Let a = 1, b = 1. Then, ab = 1, a + b = 2
∴ (1, 2) ∈ f
Let a = -1, b = -1. Then, ab = 1, a + b = -2
∴ (1, -2) ∈ f
Since (1, 2) ∈ f and (1, -2) ∈ f,
f is not a function as element 1 does not have a unique image.

Question 13.
Check the injectivity and surjectivity of the following functions.
(i) f : N → N given by f(x) = x²
Solution:
f: N → N given by f(x) = x²

∴ f is injective.
For every y = x² ∈ N, there does not exist x ∈ N.
Example: 7 ∈ N (codomain) for which there is no x in domain N such that x² = 7
∴ f is not surjective.

(ii) f : Z → Z given by f(x) = x²
Solution:
f: Z → Z given by f(x) = x2

∴ f is not injective.
(Example: f(-2) = 4 = f(2). So, -2, 2 have the same image. So, f is not injective.)
Since x² ≥ 0,
f(x) ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iii) f : R → R given by f(x) = x²
Solution:
f : R → R given by f(x) = x²

∴ f is not injective.
f(x) = x² ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iv) f : N → N given by f(x) = x³
Solution:
f: N → N given by f(x) = x³

∴ f is injective.
Numbers from codomain which are not cubes of natural numbers are not images under f.
∴ f is not surjective.

(v) f : R → R given by f(x) = x³
Solution:
f: R → R given by f(x) = x³

∴ For every y ∈ R, there is some x ∈ R.
∴ f is surjective.

Question 14.
Show that if f : A → B and g : B → C are one-one, then gof is also one-one.
Solution:
f is a one-one function.
Let f(x₁) = f(x₂)
Then, x₁ = x₂ for all x₁, x₂ …..(i)
g is a one-one function.
Let g(y₁) = g(y₂)
Then, y₁ = y₂ for all y₁, y₂ …..(ii)
Let (gof) (x₁) = (gof) (x₂)
∴ g(f(x₁)) = g(f(x₂))
∴ g(y₁) = g(y₂),
where y₁ = f(x₁), y₂ = f(x₂) ∈ B
∴ y₁ = y₂ …..[From (ii)]
i.e., f(x₁) = f(x₂)
∴ x₁ = x₂ ….[From (i)]
∴ gof is one-one.

Question 15.
Show that if f : A → B and g : B → C are onto, then gof is also onto.
Solution:
Since g is surjective (onto),
there exists y ∈ B for every z ∈ C such that
g(y) = z …….(i)
Since f is surjective,
there exists x ∈ A for every y ∈ B such that
f(x) = y …….(ii)
(gof) x = g(f(x))
= g(y) ……[From (ii)]
= z …..[From(i)]
i.e., for every z ∈ C, there is x in A such that (gof) x = z
∴ gof is surjective (onto).

Question 16.
If f(x) = 3(4^x+1), find f(-3).
Solution:
f(x) = 3(4^x+1)
∴ f(-3) = 3(4^-3+1)
= 3(4^-2)
= \(\frac{3}{16}\)

Question 17.
Express the following exponential equations in logarithmic form:
(i) 2⁵ = 32
(ii) 54⁰ = 1
(iii) 23¹ = 23
(iv) \(9^{\frac{3}{2}}\) = 27
(v) 3^-4 = \(\frac{1}{81}\)
(vi) 10^-2 = 0.01
(vii) e² = 7.3890
(viii) \(e^{\frac{1}{2}}\) = 1.6487
(ix) e^-x = 6
Solution:

Question 18.
Express the following logarithmic equations in exponential form:
(i) log₂ 64 = 6
(ii) \(\log _{5} \frac{1}{25}\) = -2
(iii) log₁₀ 0.001 = -3
(iv) \(\log _{\frac{1}{2}}\)(8) = -3
(v) ln 1 = 0
(vi) ln e = 1
(vii) ln \(\frac{1}{2}\) = -0.693
Solution:
(i) log₂ 64 = 6
∴ 64 = 2⁶, i.e., 2⁶ = 64

Question 19.
Find the domain of
(i) f(x) = ln (x – 5)
(ii) f(x) = log10 (x² – 5x + 6)
Solution:
(i) f(x) = ln (x – 5)
f is defined, when x – 5 > 0
∴ x > 5
∴ Domain of f = (5, ∞)

(ii) f(x) = log₁₀ (x² – 5x + 6)
x² – 5x + 6 = (x – 2) (x – 3)
f is defined, when (x – 2) (x – 3) > 0
∴ x < 2 or x > 3
Solution of (x – a) (x – b) > 0 is x < a or x > b where a < b
∴ Domain of f = (-∞, 2) ∪ (3, ∞)

Question 20.
Write the following expressions as sum or difference of logarithms:
(a) \(\log \left(\frac{p q}{r s}\right)\)
Solution:

(b) \(\log (\sqrt{x} \sqrt[3]{y})\)
Solution:

(c) \(\ln \left(\frac{a^{3}(a-2)^{2}}{\sqrt{b^{2}+5}}\right)\)
Solution:

(d) \(\ln \left[\frac{\sqrt[3]{x-2}(2 x+1)^{4}}{(x+4) \sqrt{2 x+4}}\right]^{2}\)
Solution:

Question 21.
Write the following expressions as a single logarithm.
(i) 5 log x + 7 log y – log z
Solution:

(ii) \(\frac{1}{3}\) log(x – 1) + \(\frac{1}{2}\) log(x)
Solution:

(iii) ln (x + 2) + ln (x – 2) – 3 ln (x + 5)
Solution:

Question 22.
Given that log 2 = a and log 3 = b, write log √96 terms of a and b.
Solution:
log 2 = a and log 3 = b
log √96 = \(\frac{1}{2}\) log (96)
= \(\frac{1}{2}\) log (2⁵ x 3)
= \(\frac{1}{2}\) (log 2⁵ + log 3) …..[∵ log mn = log m + log n]
= \(\frac{1}{2}\) (5 log 2 + log 3) ……[∵ log mⁿ = n log m]
= \(\frac{5 a+b}{2}\)

Question 23.
Prove that:
(a) \(b^{\log _{b} a}=a\)
Solution:
We have to prove that \(b^{\log _{b} a}=a\)
i.e., to prove that (logb a) (log_b b) = log_b a
(Taking log on both sides with base b)
L.H.S. = (log_b a) (log_b b)
= log_b a …..[∵ log_b b = 1]
= R.H.S.

(b) \(\log _{b^{m}} a=\frac{1}{m} \log _{b} a\)
Solution:

(c) \(a^{\log _{c} b}=b^{\log _{c} a}\)
Solution:

Question 24.
If f(x) = ax² – bx + 6 and f(2) = 3 and f(4) = 30, find a and b.
Solulion:
f(x) = ax² – bx + 6
f(2) = 3
∴ a(2)² – b(2) + 6 = 3
∴ 4a – 2b + 6 = 3
∴ 4a – 2b + 3 = 0 …..(i)
f(4) = 30
∴ a(4)² – b(4) + 6 = 30
∴ 16a – 4b + 6 = 30
∴ 16a – 4b – 24 = 0 …..(ii)
By (ii) – 2 × (i), we get
8a – 30 = 0
∴ a = \(\frac{30}{8}=\frac{15}{4}\)
Substiting a = \(\frac{15}{4}\) in (i), we get
4(\(\frac{15}{4}\)) – 2b + 3 = 0
∴ 2b = 18
∴ b = 9
∴ a = \(\frac{15}{4}\), b = 9

Question 25.
Solve for x:
(i) log 2 + log (x + 3) – log (3x – 5) = log 3
Solution:
log 2 + log (x + 3) – log (3x – 5) = log 3
∴ log 2(x + 3) – log(3x – 5) = log 3 …..[∵ log m + log n = log mn]
∴ log \(\frac{2(x+3)}{3 x-5}\) = log 3 …..[∵ log m – log n = log \(\frac{m}{n}\)]
∴ \(\frac{2(x+3)}{3 x-5}\) = 3
∴ 2x + 6 = 9x – 15
∴ 7x = 21
∴ x = 3

Check:
If x = 3 satisfies the given condition, then our answer is correct.
L.H.S. = log 2 + log (x + 3) – log (3x – 5)
= log 2 + log (3 + 3) – log (9 – 5)
= log 2 + log 6 – log 4
= log (2 × 6) – log 4
= log \(\frac{12}{4}\)
= log 3
= R.H.S.
Thus, our answer is correct.

(ii) 2log₁₀ x = 1 + \(\log _{10}\left(x+\frac{11}{10}\right)\)
Solution:


∴ x² = 10x + 11
∴ x² – 10x – 11 = 0
∴ (x – 11)(x + 1) = 0
∴ x = 11 or x = -1
But log of a negative numbers does not exist
∴ x ≠ -1
∴ x = 11

(iii) log₂ x + log₄ x + log₁₆ x = \(\frac{21}{4}\)
Solution:

(iv) x + log₁₀ (1 + 2^x) = x log₁₀ 5 + log₁₀ 6
Solution:

∴ a + a² = 6
∴ a² + a – 6 = 0
∴ (a + 3)(a – 2) = 0
∴ a + 3 = 0 or a – 2 = 0
∴ a = -3 or a = 2
Since 2x = -3 is not possible,
2x = 2 = 21
∴ x = 1

Question 26.
If log \(\left(\frac{x+y}{3}\right)\) = \(\frac{1}{2}\) log x + \(\frac{1}{2}\) log y, show that \(\frac{x}{y}+\frac{y}{x}\) = 7.
Solution:

Question 27.
If log\(\left(\frac{x-y}{4}\right)\) = log√x + log√y, show that (x + y)² = 20xy.
Solution:

Question 28.
If x = log_abc, y = log_b ca, z = log_c ab, then prove that \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) = 1.
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 2
• Exercise 5.2 Class 11 Maths 2
• Miscellaneous Exercise 5 Class 11 Maths 2
• Exercise 6.1 Class 11 Maths 2
• Exercise 6.2 Class 11 Maths 2
• Miscellaneous Exercise 6 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Miscellaneous Exercise 5 Questions and Answers.

11th Maths Part 2 Sets and Relations Miscellaneous Exercise 5 Questions And Answers Maharashtra Board

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

Question 1.
For the set A = {a, b, c, d, e} the correct statement is
(A) {a, b} ∈ A
(B) {a} ∈ A
(C) a ∈ A
(D) a ∉ A
Answer:
(C) a ∈ A

Question 2.
If aN = {ax : x ∈ N}, then set 6N ∩ 8N =
(A) 8N
(B) 48N
(C) 12N
(D) 24N
Answer:
(D) 24N
Hint:
6N = {6x : x ∈ N} = {6, 12, 18, 24, 30, ……}
8N = {8x : x ∈ N} = {8, 16, 24, 32, ……}
∴ 6N ∩ 8N = {24, 48, 72, …..}
= {24x : x ∈ N}
= 24N

Question 3.
If set A is empty set then n[P[P[P(A)]]] is
(A) 6
(B) 16
(C) 2
(D) 4
Answer:
(D) 4
Hint:
A = Φ
∴ n(A) = 0
∴ n[P(A)] = 2^n(A) = 2⁰ = 1
∴ n[P[P(A)]] = 2^n[P(A)] = 2¹ = 2
∴ n[P[P[P(A)]]] = 2^n[P[P(A)]] = 2² = 4

Question 4.
In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus are
(A) 80%
(B) 40%
(C) 60%
(D) 70%
Answer:
(C) 60%
Hint:
Let C = Population travels by car
B = Population travels by bus
n(C) = 20%, n(B) = 50%, n(C ∩ B) = 10%
n(C ∪ B) = n(C) + n(B) – n(C ∩ B)
= 20% + 50% – 10%
= 60%

Question 5.
If the two sets A and B are having 43 elements in common, then the number of elements common to each of the sets A × B and B × A is
(A) 43²
(B) 2⁴³
(C) 43⁴³
(D) 2⁸⁶
Answer:
(A) 43²

Question 6.
Let R be a relation on the set N be defied by {(x, y) / x, y ∈ N, 2x + y = 41} Then R is
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) None of these
Answer:
(D) None of these

Question 7.
The relation “>” in the set of N (Natural number) is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalent relation
Answer:
(C) Transitive
Hint:
For any a ∈ N, a ≯ a
∴ (a, a) ∉ R
∴ > is not reflexive.
For any a, b ∈ N, if a > b, then b ≯ a.
∴ > is not symmetric.
For any a, b, c ∈ N,
if a > b and b > c, then a > c
∴ > is transitive.

Question 8.
A relation between A and B is
(A) only A × B
(B) An Universal set of A × B
(C) An equivalent set of A × B
(D) A subset of A × B
Answer:
(D) A subset of A × B

Question 9.
If (x, y) ∈ N × N, then xy = x² is a relation that is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalence
Answer:
(D) Equivalence
Hint:
Let x ∈ R, then xx = x²
∴ x is related to x.
∴ Given relation is reflexive.
Letx = 0 and y = 2,
then xy = 0 × 2 = 0 = x²
∴ x is related to y.
Consider, yx = 2 × 0 = 0 ≠ y²
∴ y is not related to x.
∴ Given relation is not symmetric.
Let x be related to y and y be related to z.
∴ xy = x² and yz = y²
∴ x = \(\frac{x^{2}}{y}\) and z = \(\frac{y^{2}}{y}\) = y …..[if y ≠ 0]
Consider, xz = \(\frac{x^{2}}{y}\) × y = x²
∴ x is related to z.
∴ Given relation is transitive.

Question 10.
If A = {a, b, c}, The total no. of distinct relations in A × A is
(A) 3
(B) 9
(C) 8
(D) 29
Answer:
(D) 29

(II) Answer the following.

Question 1.
Write down 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 is a day 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 down 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) C’ = {1, 2, 4, 6, 7, 10, 11}
∴ 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}, check if the following are relations from A to B. Also, write its domain and range.
(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.
Domain (R₁) = Set of first components of R₁ = {1}
Range (R₁) = Set of second components of R₁ = {4, 5, 6}

(ii) R₂ = {(1, 5),(2, 4),(3, 6)}
Since R₂ ⊆ A × B,
R₂ is a relation from A to B.
Domain (R₂) = Set of first components of R₂ = {1, 2, 3}
Range (R₂) = Set of second components of R₂ = {4, 5, 6}

(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since R₃ ⊆ A × B,
R₃ is a relation from A to B.
Domain (R₃) = Set of first components of R₃ = {1, 2, 3}
Range (R₃) = Set of second components of R₃ = {4, 5, 6}

(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 relations.
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Solution:
(i) 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}

(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Since a ∈ Z and |a| < 3,
a < 3 and a > -3
∴ -3 < a < 3
∴ a = -2, -1, 0, 1, 2
b = |a – 1|
When a = -2, b = 3
When a = -1, b = 2
When a = 0, b = 1
When a = 1, b = 0
When a = 2, b = 1
Domain (R) = {-2, -1, 0, 1, 2}
Range (R) = {0, 1, 2, 3}

Question 8.
Find R : A → A when A = {1, 2, 3, 4} such that
(i) R = {(a, b) / a – b = 10}
(ii) R = {(a, b) / |a – b| ≥ 0}
Solution:
R : A → A, A = {1, 2, 3,4}
(i) R = {(a, b)/a – b = 10} = { }

(ii) R = {(a, b) / |a – b| ≥ 0}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ R = A × A

Question 9.
R : {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Check if R is
(i) reflexive
(ii) symmetric
(iii) transitive
Solution:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
(i) Here, (x, x) ∈ R, for x ∈ {1, 2, 3}
∴ R is reflexive.

(ii) Here, (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.

(iii) Here, (1, 2), (2, 3) ∈ R,
But (1, 3) ∉ R.
∴ R is not transitive.

Question 10.
Check if R : Z → Z, R = {(a, b) | 2 divides a – b} is an equivalence relation.
Solution:
(i) Since 2 divides a – a,
(a, a) ∈ R
∴ R is reflexive. .

(ii) Let (a, b) ∈ R
Then 2 divides a – b
∴ 2 divides b – a
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b) ∈ R, (b, c) ∈ R
Then a – b = 2m, b – c = 2n,
∴ a – c = 2(m + n), where m, n are integers.
∴ 2 divides a – c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Question 11.
Show that the relation R in the set A = {1, 2, 3, 4, 5} Given by R = {(a, b) / |a – b| is even} is an equivalence relation.
Solution:
(i) Since |a – a| is even,
∴ (a, a) ∈ R
∴ R is reflexive.

(ii) Let (a, b) ∈ R
Then |a – b| is even
∴ |b – a| is even
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b), (b, c) ∈ R
Then a – b = ±2m, b – c = ±2n
∴ a – c = ±2(m + n), where m, n are integers.
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

Question 12.
Show that the following are equivalence relations:
(i) R in A is set of all books given by R = {(x, y) / x and y have same number of pages}
(ii) R in A = {x ∈ Z | 0 ≤ x ≤ 12} given by R = {(a, b) / |a – b| is a multiple of 4}
(iii) R in A = (x ∈ N/x ≤ 10} given by R = {(a, b) | a = b}
Solution:
(i) a. Clearly (x, x) ∈ R
∴ R is reflexive.

b. If (x, y) ∈ R then (y, x) ∈ R.
∴ R is symmetric.

c. Let (x, y) ∈ R, (y, x) ∈ R.
Then x, y, and z are 3 books having the same number of pages.
∴ (x, z) ∈ R as x, z has the same number of pages.
∴ R is transitive.
Thus, R is an equivalence relation.

(ii) a. Since |a – a| is a multiple of 4,
(a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R
Then a – b = ±4m,
∴ b – a = ±4m, where m is an integer
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
a – b = ± 4m, b – c = ± 4n,
∴ a – c = ±4(m + n), where m, n are integers
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

(iii) a. Since a = a
∴ (a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R Then a = b
∴ b = a
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
Then, a = b, b = c
∴ a = c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 2
• Exercise 5.2 Class 11 Maths 2
• Miscellaneous Exercise 5 Class 11 Maths 2
• Exercise 6.1 Class 11 Maths 2
• Exercise 6.2 Class 11 Maths 2
• Miscellaneous Exercise 6 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.2 Questions and Answers.

11th Maths Part 2 Sets and Relations Exercise 5.2 Questions And Answers Maharashtra Board

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 \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\), find x and y.
Solution:
\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\)
By the definition of equality of ordered pairs, we have
x + \(\frac{1}{3}\) = \(\frac{1}{2}\) and \(\frac{y}{3}\) – 1 = \(\frac{3}{2}\)
∴ x = \(\frac{1}{2}\) – \(\frac{1}{3}\) and \(\frac{y}{3}\) = \(\frac{3}{2}\) + 1
∴ x = \(\frac{1}{6}\) 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 = {1, 4}, find sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {1, 4}
∴ P × Q = {(1, 1), (1, 4), (2, 1), (2, 4), (3, 1), (3, 4)}
and Q × P = {(1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3)}

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

(ii) 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)}
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)}
∴ A × (B ∪ C) = (A × B) ∪ (A × C)

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 and y = 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.
Let A = {6, 8} and B = {1, 3, 5}. Show that R₁ = {(a, b) / a ∈ A, b ∈ B, a – b is an even number} is a null relation, R₂ = {(a, b) / a ∈ A, b ∈ B, a + b is an odd number} is a universal relation.
Solution:
A = {6, 8}, B = {1, 3, 5}
R₁ = {(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 as even.
∴ R₁ has a null relation from A to B.
A × B = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
When a = 6 and b = 1, a + b = 7, which is odd
When a = 6 and b = 3, a + b = 9, which is odd
When a = 6 and b = 5, a + b = 11, which is odd
When a = 8 and b = 1, a + b = 9, which is odd
When a = 8 and b = 3, a + b = 11, which is odd
When a = 8 and b = 5, a + b = 13, which is odd
∴ R₂ = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
Here, R₂ = A × B
∴ R₂ has a universal relation from A to B.

Question 8.
Write the relation in the Roster form. State 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}
(iii) R₃ = {(x, y / y = 3x, y ∈ {3, 6, 9, 12}, x ∈ {1, 2, 3}}
(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
(v) R₅ = {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
(vi) R₆ = {(a, b) / a ∈ N, a < 6 and b = 4}
(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
(viii) R₈ = {(a, b)/ b = a + 2, a ∈ Z, 0 < a < 5}
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}

(iii) R₃ = {(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
∴ R₃ = {(1, 3), (2, 6), (3, 9)}
∴ Domain (R₃) ={1, 2, 3}
∴ Range (R₃) = {3, 6, 9}

(iv) R₄ = {(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
∴ R₄ = {(1, 4), (1, 6), (2, 4), (2, 6)}
Domain (R₄) = {1, 2}
Range (R₄) = {4, 6}

(v) R₅ = {{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
∴ R₅ = {(0, 3), (1, 2), (2, 1), (3, 0)}
Domain (R₅) = {0, 1, 2, 3}
Range (R₅) = {3, 2, 1, 0}

(vi) R₆ = {(a, b)/ a ∈ N, a < 6 and b = 4}
a ∈ N and a < 6
∴ a = 1, 2, 3, 4, 5 and b = 4
R₆ = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}
Domain (R₆) = {1, 2, 3, 4, 5}
Range (R₆) = {4}

(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
Here, a + b = 6
When a = 1, b = 5
When a = 2, b = 4
When a = 3, b = 3
When a = 4, b = 2
When a = 5, b = 1
∴ R₇ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Domain (R₇) = {1, 2, 3, 4, 5}
Range (R₇) = {5, 4, 3, 2, 1}

(viii) R₈ = {(a, b) / b = a + 2, a ∈ Z, 0 < a < 5}
Here, b = a + 2
When a = 1, b = 3
When a = 2, b = 4
When a = 3, b = 5
When a = 4, b = 6
∴ R₈ = {(1, 3), (2, 4), (3, 5), (4, 6)}
Domain (R₈) = {1, 2, 3, 4}
Range (R₈) = {3, 4, 5, 6}

Question 9.
Identify which of the following relations are reflexive, symmetric, and transitive.

Solution:

(i) Given, R = {(a, b): a, b ∈ Z, a – b is an integer}
Let a ∈ Z, then a – a ∈ Z
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a – b ∈ Z
∴ -(a – b) ∈ Z, i.e., b – a ∈ Z
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a – b ∈ Z and b – c ∈ Z
∴ (a – b) + (b – c) ∈ Z
∴ a – c ∈ Z
∴ (a, c) ∈ R
∴ R is transitive.

(ii) Given, R = {(a, b) : a, b ∈ N, a + b is even}
Let a ∈ N, then a + a = 2a, which is even.
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a + b is even
∴ b + a is even
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a + b and b + c is even
Let a + b = 2x and b + c = 2y for x, y ∈ N
∴ (a + b) + (b + c) = 2x + 2y
∴ a + 2b + c = 2(x + y)
∴ a + c = 2(x + y) – 2b = 2(x + y – b)
∴ a + c is even ……..[∵ x, y, b ∈ N, x + y – b ∈ N]
∴ (a, c) ∈ R
∴ R is transitive.

(iii) Given, R = {(a, b) : a, b ∈ N, a divides b}
Let a ∈ N, then a divides a.
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 2 and b = 8, then 2 divides 8
∴ (a, b) ∈ R
But 8 does not divide 2.
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a divides b and b divides c.
Let b = ax and c = by for x, y ∈ N.
∴ c = (ax) y = a(xy)
i.e., a divides c.
∴ (a, c) ∈ R
∴ R is transitive.

(iv) Given, R = {(a, b) : a, b ∈ N, a² – 4ab + 3b² = 0}
Let a ∈ N, then a² – 4aa + 3a² = a² – 4a² + 3a² = 0
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 3 and b = 1,
then a² – 4ab + 3b² = 9 – 12 + 3 = 0
∴ (a, b) ∈ R
Consider, b² – 4ba + 3a² = 1 – 12 + 9 = -2 ≠ 0
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 3, b = 1 and c = \(\frac{1}{3}\),
then a² – 4ab + 3b² = 9 – 12 + 3 = 0 and
b² – 4bc + 3c² = 1 – \(\frac{4}{3}\) + \(\frac{1}{3}\) = 1 – 1 = 0
∴ we get (a, b) and (b, c) ∈ R.
Consider, a² – 4ac + 3c² = 9 – 4 + \(\frac{1}{3}\) = \(\frac{16}{3}\) ≠ 0
∴ (a, c) ∉ R
∴ R is not transitive.

(v) Given, R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}
Let a ∈ G, then ‘a’ cannot be a sister of herself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ ‘a’ is a sister of ‘b’.
∴ ‘b’ is a sister of ‘a’.
∴ (b, c) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ ‘a’ is a sister of ‘b’ and ‘b’ is a sister of ‘c’
∴ ‘a’ is a sister of ‘c’.
∴ (a, c) ∈ R
∴ R is transitive.

(vi) Given, R = {(a, b) : Line a is perpendicular to line b in a plane}
Let a be any line in the plane, then a cannot be perpendicular to itself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ a is perpendicular to b.
∴ b is perpendicular to a.
∴ (b, a) ∈ R.
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R.
∴ a is perpendicular to b and b is perpendicular to c.
∴ a is parallel to c.
∴ (a, c) ∉ R
∴ R is not transitive.

(vii) Given, R = {(a, b) : a, b ∈ R, a < b}
Let a ∈ R, then a ≮ a.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 1 and b = 2, then 1 < 2
∴ (a, b) ∈ R
But 2 ≮ 1
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a < b and b < c
∴ a < c
∴ (a, c) ∈ R
∴ R is transitive.

(viii) Given, R = {(a, b) : a, b ∈ R, a ≤ b³}
Let a = -3, then a³ = -27.
Here, a ≮ a
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 2 and b = 9, then b³ = 729
Here, a < b³
∴ (a, b) ∈ R
Consider, a3 = 8
Here, b ≮ a³
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 10, b = 3, c = 2,
then b³ = 27 and c³ = 8
Here, a < b³ and b < c³.
∴ (a, b) and (b, c) ∈ R
But a ≮ c³
∴ (a, c) ∉ R.
∴ R is not transitive.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 2
• Exercise 5.2 Class 11 Maths 2
• Miscellaneous Exercise 5 Class 11 Maths 2
• Exercise 6.1 Class 11 Maths 2
• Exercise 6.2 Class 11 Maths 2
• Miscellaneous Exercise 6 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.1 Questions and Answers.

11th Maths Part 2 Sets and Relations Exercise 5.1 Questions And Answers Maharashtra Board

Question 1.
Describe the following sets in Roster form:
(i) A = {x/x is a letter of the word ‘MOVEMENT’}
(ii) B = {x/x is an integer, –\(\frac{3}{2}\) < x < \(\frac{9}{2}\)>
(iii) C = {x/x = 2n + 1, n ∈ N}
Solution:
(i) A = {M, O, V, E, N, T}
(ii) B = {-1, 0, 1, 2, 3, 4}
(iii) C = {3, 5, 7, 9, … }

Question 2.
Describe the following sets in Set-Builder form:
(i) {0}
(ii) {0, ±1, ±2, ±3}
(iii) \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
(iv) {0, -1, 2, -3, 4, -5, 6,…}
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}

(iv) Let D = {0, -1, 2, -3, 4, -5, 6, …}
∴ D = {x/x = (-1)n-1 × (n – 1), n ∈ N}

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 = 0)
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 = {\(\frac{-5}{3}\), \(\frac{3}{2}\)}

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

(viii) B – C = {3}
C – B = {7, 8}
B Δ C = (B – C) ∪ (C – B) = {3, 7, 8}
∴ A ∩ (B Δ C) = {3} ……(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) Δ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)] = {3} …..(ii)
From (i) and (ii), we get
A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

(ix) 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)

(x) B = {3, 4, 5, 6}
∴ n(B) = 4 …..(i)
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ n(A’ ∩ B) = 2
A ∩ B = {3, 4}
∴ n(A ∩ B) = 2
∴ n(A’ ∩ B) + n(A ∩ B) = 2 + 2 = 4 …..(ii)
From (i) and (ii), we get
n(B) = n(A’ ∩ 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.
In a class of 200 students who appeared in certain examinations, 35 students faded in CET, 40 in NEET and 40 in JEE, 20 faded in CET and NEET, 17 in NEET and JEE, 15 in CET and JEE and 5 faded in ad three examinations. Find how many students
(i) did not fail in any examination.
(ii) faded in NEET or JEE entrance.
Solution:
Let A = set of students who failed in CET
B = set of students who failed in NEET
C = set of students who failed in JEE
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 NEET or JEE entrance = n(B ∪ C)
= n(B) + n(C) – n(B ∩ C)
= 40 + 40 – 17
= 63

Question 8.
From amongst 2000 Uterate 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) = \(\frac{32.5}{100}\) × 2000 = 650
(i) n(M ∪ E) = n(M) + n(E) – n(M ∩ E)
= 1400 + 1000 – 650
= 1750

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 total 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

∴ Total 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.
Write down the power set of A = {1, 2, 3}.
Solution:
A = {1, 2, 3}
The power set of A is given by
P(A) = {{Φ}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

Question 12.
Write the following intervals in Set-Builder form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, ∞)
(iv) (-∞, 5]
(v) (2, 5]
(vi) [-3, 4)
Solution:
(i) (-3, 0) = {x / x ∈ R, -3 < x < 0}

(ii) [6, 12] = {x / x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, ∞) = {x / x ∈ R, x > 6}

(iv) (-∞, 5] = {x / x ∈ R, x ≤ 5}

(v) (2, 5] = {x / x ∈ R, 2 < x ≤ 5}

(vi) [-3, 4) = {x / x ∈ R, -3 ≤ x < 4}

Question 13.
A college awarded 38 medals in volleyball, 15 in football, and 20 in basketball. The medals were awarded to a total of 58 players and only 3 players got medals in all three sports. How many received medals in exactly two of the three sports?
Solution:
Let A = Set of students who received medals in volleyball
B = Set of students who received medals in football
C = Set of students who received medals in basketball
n(A) = 38, n(B) = 15, n(C) = 20, n(A ∪ B ∪ C) = 58, n(A ∩ B ∩ C) = 3
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
58 = 38 + 15 + 20 – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + 3
∴ n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = 18 ……(i)
Number of players who got exactly two medals = p + q + r
Here, s = n(A ∩ B ∩ C) = 3

n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = 18 …..[From (i)]
∴ p + s + s + r + q + s = 18
∴ p + q + r + 3s = 18
∴ p + q + r + 3(3) = 18
∴ p + q + r = 18 – 9 = 9
∴ Number of players who received exactly two medals = 9.

Question 14.
Solve the following inequalities and write the solution set using interval notation.
(i) -9 < 2x + 7 ≤ 19
(ii) x² – x > 20
(iii) \(\frac{2 x}{x-4}\) ≤ 5
(iv) 6x² + 1 ≤ 5x
Solution:
(i) -9 < 2x + 7 ≤ 19
∴ -16 < 2x ≤ 12
∴ -8< x ≤ 6
∴ x ∈ (-8, 6]

(ii) x² – x > 20
∴ x² – x – 20 > 0
∴ x² – 5x + 4x – 20 > 0
∴ (x – 5) (x + 4) > 0
∴ either x – 5 > 0 and x + 4 > 0 or x – 5 < 0 and x + 4 < 0

Case I: x – 5 > 0 and x + 4 > 0
∴ x > 5 and x > -4
∴ x > 5 ….(i)

Case II:
x – 5 < 0 and x + 4 < 0
∴ x < 5 and x < -4
∴ x < -4 …..(ii)
From (i) and (ii), we get
x ∈ (-∞, – 4) ∪ (5, ∞)

(iii) \(\frac{2 x}{x-4}\) ≤ 5
∴ \(\frac{2 x}{x-4}\) – 5 ≤ 0
∴ \(\frac{2 x-5 x+20}{x-4}\) ≤ 0
∴ \(\frac{20-3 x}{x-4}\) ≤ 0
When \(\frac{a}{b}\) ≤ 0,
a ≥ 0 and b < 0 or a ≤ 0 and b > 0
∴ either 20 – 3x ≥ 0 and x – 4 < 0 or 20 – 3x ≤ 0 and x – 4 > 0
Case I:
20 – 3x ≥ 0 and x – 4 < 0
∴ x ≤ \(\frac{20}{3}\) and x < 4
∴ x < 4 ……(I)

Case II: 20 – 3x ≤ 0 and x – 4 > 0
∴ x ≥ \(\frac{20}{3}\) and x > 4
∴ x ≥ \(\frac{20}{3}\) ……(ii)
From (i) and (ii), we get
x ∈ (-∞, 4) ∪ [\(\frac{20}{3}\), ∞)

(iv) 6x² + 1 ≤ 5x
6x² – 5x + 1 ≤ 0
6x² – 3x – 2x + 1 ≤ 0
(3x – 1) (2x – 1) ≤ 0
either 3x – 1 ≤ 0 and 2x – 1 ≥ 0 or 3x – 1 ≥ 0 and 2x – 1 ≤ 0
Case I:
3x – 1 ≤ 0 and 2x – 1 ≥ 0
∴ x ≤ \(\frac{1}{3}\) and x ≥ \(\frac{1}{2}\), which is not possible.

Case II:
3x – 1 ≥ 0 and 2x – 1 ≤ 0
∴ x ≥ \(\frac{1}{3}\) and x ≤ \(\frac{1}{2}\)
∴ x ∈ [\(\frac{1}{3}\), \(\frac{1}{2}\)]

Question 15.
If A = (-7, 3], B = [2, 6] and C = [4, 9], then find
(i) A ∪ B
(ii) B ∪ C
(iii) A ∪ C
(iv) A ∩ B
(v) B ∩ C
(vi) A ∩ C
(vii) A’ ∩ B
(viii) B’ ∩ C’
(ix) B – C
(x) A – B
Solution:
A = (-7, 3], B = [2, 6], C = [4, 9]
(i) A ∪ B = (-7, 6]

(ii) B ∪ C = [2, 9]

(iii) A ∪ C = (-7, 3] ∪ [4, 9]

(iv) A ∩ B = [2, 3]

(v) B ∩ C = [4, 6]

(vi) A ∩ C = { }

(vii) A’ = (-∞, – 7] ∪ (3, ∞)
∴ A’ ∩ B = (3, 6]

(viii) B’ = (-∞, 2) ∪ (6, ∞)
C’ = (-∞, 4) ∪ (9, ∞)
∴ B’ ∩ C’ = (-∞, 2) ∪ (9, ∞)

(ix) B – C = [2, 4)

(x) A – B = (-7, 2)

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 2
• Exercise 5.2 Class 11 Maths 2
• Miscellaneous Exercise 5 Class 11 Maths 2
• Exercise 6.1 Class 11 Maths 2
• Exercise 6.2 Class 11 Maths 2
• Miscellaneous Exercise 6 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions And Answers Maharashtra Board

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

Question 1.
The total number of terms in the expression of (x + y)¹⁰⁰ + (x – y)¹⁰⁰ after simplification is:
(A) 50
(B) 51
(C) 100
(D) 202
Answer:
(B) 51
Hint:

Question 2.
The middle term in the expansion of (1 + x)²ⁿ will be:
(A) (n – 1)^th
(B) n^th
(C) (n + 1)^th
(D) (n + 2)^th
Answer:
(C) (n + 1)^th
Hint:
(1 + x)²ⁿ has (2n + 1) terms.
∴ (n + 1 )^th term is the middle term.

Question 3.
In the expansion of (x² – 2x)¹⁰, the coefficient of x¹⁶ is
(A) -1680
(B) 1680
(C) 3360
(D) 6720
Answer:
(C) 3360
Hint:
(x² – 2x)¹⁰ = x¹⁰ (x – 2)¹⁰
To get the coefficient of x¹⁶ in (x² – 2x)¹⁰,
we need to check coefficient of x⁶ in (x – 2)¹⁰
∴ Required coefficient = ¹⁰C₆ (-2)⁴
= 210 × 16
= 3360

Question 4.
The term not containing x in expansion of \((1-x)^{2}\left(x+\frac{1}{x}\right)^{10}\) is
(A) ¹¹C₅
(B) ¹⁰C₅
(C) ¹⁰C₄
(D) ¹⁰C₇
Answer:
(A) ¹¹C₅
Hint:

Question 5.
The number of terms in expansion of (4y + x)⁸ – (4y – x)⁸ is
(A) 4
(B) 5
(C) 8
(D) 9
Answer:
(A) 4
Hint:

Question 6.
The value of ¹⁴C₁ + ¹⁴C₃ + ¹⁴C₅ + …. + ¹⁴C₁₁ is
(A) 2¹⁴ – 1
(B) 2¹⁴ – 14
(C) 2¹²
(D) 2¹³ – 14
Answer:
(D) 2¹³ – 14
Hint:

Question 7.
The value of ¹¹C₂ + ¹¹C₄ + ¹¹C₆ + ¹¹C₈ is equal to
(A) 2¹⁰ – 1
(B) 2¹⁰ – 11
(C) 2¹⁰ + 12
(D) 2¹⁰ – 12
Answer:
(D) 2¹⁰ – 12
Hint:

Question 8.
In the expansion of (3x + 2)⁴, the coefficient of the middle term is
(A) 36
(B) 54
(C) 81
(D) 216
Answer:
(D) 216
Hint:
(3x + 2)⁴ has 5 terms.
∴ (3x + 2)⁴ has 3rd term as the middle term.
The coefficient of the middle term

= 6 × 9 × 4
= 216

Question 9.
The coefficient of the 8th term in the expansion of (1 + x)¹⁰ is:
(A) 7
(B) 120
(C) ¹⁰C₈
(D) 210
Answer:
(B) 120
Hint:
r = 7
t₈ = ¹⁰C₇ x⁷ = ¹⁰C₃ x⁷
∴ Coefficient of 8th term = ¹⁰C₃ = 120

Question 10.
If the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are the same, then the value of a is
(A) \(-\frac{7}{9}\)
(B) \(-\frac{9}{7}\)
(C) \(\frac{7}{9}\)
(D) \(\frac{9}{7}\)
Answer:
(D) \(\frac{9}{7}\)
Hint:

(II) Answer the following.

Question 1.
Prove by the method of induction, for all n ∈ N.
(i) 8 + 17 + 26 + ….. + (9n – 1) = \(\frac{n}{2}\) (9n + 7)
Solution:
Let P(n) ≡ 8 + 17 + 26 +…..+(9n – 1) = \(\frac{n}{2}\) (9n + 7), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 8
R.H.S. = \(\frac{1}{2}\) [9(1) + 7] = 8
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 8 + 17 + 26 +…..+ (9k – 1) = \(\frac{k}{2}\) (9k + 7) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., 8 + 17 + 26 + …… + [9(k + 1) – 1]

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 8 + 17 + 26 +…..+ (9n – 1) = \(\frac{n}{2}\) (9n + 7) for all n ∈ N.

(ii) 1² + 4² + 7² + …… + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1)
Solution:
Let P(n) = 1² + 4² + 7² + ….. + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S.= 1² = 1
R.H.S.= \(\frac{1}{2}\) [6(1)² – 3(1) – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 4² + 7² +…..+ (3k – 2)² = \(\frac{k}{2}\) (6k² – 3k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 4² + 7² + … + (3n – 2)² = \(\frac{n}{2}\) (6n² – 3n – 1) for all n ∈ N.

(iii) 2 + 3.2 + 4.2² + …… + (n + 1) 2^n-1 = n . 2ⁿ
Solution:
Let P(n) ≡ 2 + 3.2 + 4.2² +…..+ (n + 1) 2^n-1 = n.2ⁿ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(2¹) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 3.2 + 4.2² + ….. + (k + 1) 2^k-1 = k.2^k …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 3.2 + 4.2² +….+ (k + 2) 2^k = (k + 1) 2^k+1

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 3.2 + 4.2² +……+ (n + 1) 2^n-1 = n.2ⁿ for all n ∈ N.

(iv) \(\frac{1}{3.4 .5}+\frac{2}{4.5 .6}+\frac{3}{5.6 .7}+\ldots+\frac{n}{(n+2)(n+3)(n+4)}\) = \(\frac{n(n+1)}{6(n+3)(n+4)}\)
Solution:

Question 2.
Given that t_n+1 = 5t_n – 8, t₁ = 3, prove by method of induction that t_n = 5^n-1 + 2.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation t_n+1 = 5t_n – 8, t₁ = 3
and R.H.S. a general statement t_n = 5^n-1 + 2.
Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 5^1-1 + 2 = 1 + 2 = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S = t₂ = 5t₁ – 8 = 5(3) – 8 = 7
R.H.S. = t₂ = 5^2-1 + 2 = 5 + 2 = 7
∴ L.H.S. = R.H.S.
∴ P(n) is tme for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1 = 5t_k – 8 and t_k = 5^k-1 + 2

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
t_k+1 = 5^k+1-1 + 2 = 5^k + 2
t_k+1 = 5t_k – 8 and t_k = 5^k-1 + 2 ……[From Step II]
∴ t_k+1 = 5(5^k-1 + 2) – 8 = 5^k + 2
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n = 5^n-1 + 2, for all n ∈ N.

Question 3.
Prove by method of induction
\(\left(\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right)^{n}=\left(\begin{array}{cc}
2 n+1 & -4 n \\
n & -2 n+1
\end{array}\right)\), ∀ n ∈ N.
Solution:


Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\left(\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right)^{n}=\left(\begin{array}{cc}
2 n+1 & -4 n \\
n & -2 n+1
\end{array}\right)\), ∀ n ∈ N.

Question 4.
Expand (3x² + 2y)⁵
Solution:
Here, a = 3x², b = 2y, n = 5.
Using binomial theorem,

Question 5.
Expand \(\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{4}\)
Solution:

Question 6.
Find third term in the expansion of \(\left(9 x^{2}-\frac{y^{3}}{6}\right)^{4}\)
Solution:

Question 7.
Find tenth term in the expansion of \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\)
Solution:

Question 8.
Find the middle term(s) in the expansion of
(i) \(\left(\frac{2 a}{3}-\frac{3}{2 a}\right)^{6}\)
Solution:
Here, a = \(\frac{2 a}{3}\), b = \(\frac{-3}{2 a}\), n = 6.
Now, n is even.
∴ \(\frac{\mathrm{n}+2}{2}=\frac{6+2}{2}=4\)
∴ Middle term is t4, for which r = 3.

∴ The Middle term is -20.

(ii) \(\left(x-\frac{1}{2 y}\right)^{10}\)
Solution:
Here, a = x, b = \(-\frac{1}{2 y}\), n = 10.
Now, n is even.
∴ \(\frac{\mathrm{n}+2}{2}=\frac{10+2}{2}=6\)
∴ Middle term is t6, for which r = 5

(iii) (x² + 2y²)⁷
Solution:
Here, a = x², b = 2y², n = 7.
Now, n is odd.
∴ \(\frac{\mathrm{n}+1}{2}=\frac{7+1}{2}=4, \frac{\mathrm{n}+3}{2}=\frac{7+3}{2}=5\)
∴ Middle terms are t₄ and t₅, for which r = 3 and r = 4 respectively.

∴ Middle terms are 280x⁸y⁶ and 560x⁶y⁸.

(iv) \(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)
Solution:

Question 9.
Find the coefficients of
(i) x⁶ in the expantion of \(\left(3 x^{2}-\frac{1}{3 x}\right)^{9}\)
Solution:

(ii) x⁶⁰ in the expansion of \(\left(\frac{1}{x^{2}}+x^{4}\right)^{18}\)
Solution:

Question 10.
Find the constant term in the expansion of
(i) \(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Solution:

(ii) \(\left(2 x^{2}-\frac{1}{x}\right)^{12}\)
Solution:

Question 11.
Prove by method of induction
(i) log_a xⁿ = n log_a x, x > 0, n ∈ N
Solution:

(ii) 15^2n-1 + 1 is divisible by 16, for all n ∈ N.
Solution:
15^2n-1 + 1 is divisible by 16, if and only if (15^2n-1 + 1) is is a multiple of 16.
Let P(n) ≡ 15^2n-1 + 1 = 16m, where m ∈ N.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 15^2n-1 + 1 is divisible by 16, for all n ∈ N.

(iii) 5²ⁿ – 2²ⁿ is divisible by 3, for all n ∈ N.
Solution:

Question 12.
If the coefficient of x¹⁶ in the expansion of (x² + ax)¹⁰ is 3360, find a.
Solution:

Question 13.
If the middle term in the expansion of \(\left(x+\frac{b}{x}\right)^{6}\) is 160, find b.
Solution:

∴ 160 = \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times b^{3}\)
∴ 160 = 20b³
∴ 8 = b³
∴ b = 2

Question 14.
If the coefficients of x² and x³ in theexpansion of (3 + kx)⁹ are equal, find k.
Solution:

Question 15.
If the constant term in the expansion of \(\left(x^{3}+\frac{\mathrm{k}}{x^{8}}\right)^{11}\) is 1320, find k.
Solution:

Question 16.
Show that there is no term containing x⁶ in the expansion of \(\left(x^{2}-\frac{3}{x}\right)^{11}\).
Solution:

Question 17.
Show that there is no constant term in the expansion of \(\left(2 x-\frac{x^{2}}{4}\right)^{9}\)
Solution:

Question 18.
State, first four terms in the expansion of \(\left(1-\frac{2 x}{3}\right)^{-1 / 2}\)
Solution:

Question 19.
State, first four terms in the expansion of \((1-x)^{-1 / 4}\).
Solution:

Question 20.
State, first three terms in the expansion of \((5+4 x)^{-1 / 2}\)
Solution:

Question 21.
Using the binomial theorem, find the value of \(\sqrt[3]{995}\) upto four places of decimals.
Solution:

Question 22.
Find approximate value of \(\frac{1}{4.08}\) upto four places of decimals.
Solution:

Question 23.
Find the term independent of x in the expansion of (1 – x²) \(\left(x+\frac{2}{x}\right)^{6}\).
Solution:

Question 24.
(a + bx) (1 – x)⁶ = 3 – 20x + cx² + …, then find a, b, c.
Solution:

Question 25.
The 3rd term of (1 + x)ⁿ is 36x². Find 5th term.
Solution:

Question 26.
Suppose (1 + kx)ⁿ = 1 – 12x + 60x² – …… find k and n.
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 4.1 Class 11 Maths 2
• Exercise 4.2 Class 11 Maths 2
• Exercise 4.3 Class 11 Maths 2
• Exercise 4.4 Class 11 Maths 2
• Exercise 4.5 Class 11 Maths 2
• Miscellaneous Exercise 4 Class 11 Maths 2