Bhagya

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.7 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.7 Questions And Answers Maharashtra Board

I. Evaluate the following:

Question 1.
$\lim _{x \rightarrow \infty}\left[\frac{a x^{3}+b x^{2}+c x+d}{e x^{3}+f x^{2}+g x+h}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow \infty}\left[\frac{x^{3}+3 x+2}{(x+4)(x-6)(x-3)}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow \infty}\left[\frac{7 x^{2}+5 x-3}{8 x^{2}-2 x+7}\right]$
Solution:

II. Evaluate the following:

Question 1.
$\lim _{x \rightarrow \infty}\left[\frac{7 x^{2}+2 x-3}{\sqrt{x^{4}+x+2}}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow \infty}\left[\sqrt{x^{2}+4 x+16}-\sqrt{x^{2}+16}\right]$
Solution:

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

III. Evaluate the following:

Question 1.
$\lim _{x \rightarrow \infty}\left[\frac{\left(3 x^{2}+4\right)\left(4 x^{2}-6\right)\left(5 x^{2}+2\right)}{4 x^{6}+2 x^{4}-1}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow \infty}\left[\frac{(3 x-4)^{3}(4 x+3)^{4}}{(3 x+2)^{7}}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow \infty}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]$
Solution:

Question 4.
$\lim _{x \rightarrow \infty}\left[\frac{(2 x-1)^{20}(3 x-1)^{30}}{(2 x+1)^{50}}\right]$
Solution:

Question 5.
$\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^{2}+5}-\sqrt{x^{2}-3}}{\sqrt{x^{2}+3}-\sqrt{x^{2}+1}}\right]$
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 7.6 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.6 Questions And Answers Maharashtra Board

I. Evaluate the following limits:

Question 1.
$\lim _{x \rightarrow 0}\left[\frac{9^{x}-5^{x}}{4^{x}-1}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow 0}\left[\frac{5^{x}+3^{x}-2^{x}-1}{x}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}-3}{\sin x}\right)$
Solution:

Question 4.
$\lim _{x \rightarrow 0}\left(\frac{6^{x}+5^{x}+4^{x}-3^{x+1}}{\sin x}\right)$
Solution:

Question 5.
$\lim _{x \rightarrow 0}\left(\frac{8^{\sin x}-2^{\tan x}}{e^{2 x}-1}\right)$
Solution:

II. Evaluate the following limits:

Question 1.
$\lim _{x \rightarrow 0}\left[\frac{3^{x}+3^{-x}-2}{x \cdot \tan x}\right]$
Solution:

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

Question 3.
$\lim _{x \rightarrow 0}\left[\frac{5 x+3}{3-2 x}\right]^{\frac{2}{x}}$
Solution:

Question 4.
$\lim _{x \rightarrow 0}\left[\frac{\log (3-x)-\log (3+x)}{x}\right]$
Solution:

Question 5.
$\lim _{x \rightarrow 0}\left[\frac{4 x+1}{1-4 x}\right]^{\frac{1}{x}}$
Solution:

Question 6.
$\lim _{x \rightarrow 0}\left[\frac{5+7 x}{5-3 x}\right]^{\frac{1}{3 x}}$
Solution:

III. Evaluate the following limits:

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

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

Question 3.
$\lim _{x \rightarrow 0}\left[\frac{15^{x}-5^{x}-3^{x}+1}{x \cdot \sin x}\right]$
Solution:

Question 4.
$\lim _{x \rightarrow 0}\left[\frac{(25)^{x}-2(5)^{x}+1}{x \cdot \sin x}\right]$
Solution:

Question 5.
$\lim _{x \rightarrow 0}\left[\frac{(49)^{x}-2(35)^{x}+(25)^{x}}{\sin x \cdot \log (1+2 x)}\right]$
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 7.5 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.5 Questions And Answers Maharashtra Board

I. Evaluate the following:

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

Question 2.
$\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt[5]{x}-\sqrt[5]{a}}$
Solution:

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

Question 4.
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{\cos x-\sqrt{3} \sin x}{\pi-6 x}\right]$
Solution:

Question 5.
$\lim _{x \rightarrow 1}\left[\frac{1-x^{2}}{\sin \pi x}\right]$
Solution:

II. Evaluate the following:

Question 1.
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin x-1}{\pi-6 x}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\sqrt{2}-\cos x-\sin x}{(4 x-\pi)^{2}}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-\sqrt{3} \cos x-\sin x}{(6 x-\pi)^{2}}\right]$
Solution:

Question 4.
$\lim _{x \rightarrow a}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{a})}{x-a}\right]$
Solution:

Question 5.
$\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{\cos 3 x+3 \cos x}{(2 x-\pi)^{3}}\right]$
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 7.4 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.4 Questions And Answers Maharashtra Board

I. Evaluate the following limits:

Question 1.
$\lim _{\theta \rightarrow 0}\left[\frac{\sin (m \theta)}{\tan (n \theta)}\right]$
Solution:

Question 2.
$\lim _{\theta \rightarrow 0}\left[\frac{1-\cos 2 \theta}{\theta^{2}}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow 0}\left[\frac{x \cdot \tan x}{1-\cos x}\right]$
Solution:

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

II. Evaluate the following limits:

Question 1.
$\lim _{x \rightarrow 0}\left[\frac{1-\cos (n x)}{1-\cos (m x)}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-{cosec} x}{\cot ^{2} x-3}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\cos x-\sin x}{\cos 2 x}\right]$
Solution:

III. Evaluate the following limits:

Question 1.
$\lim _{x \rightarrow 0}\left[\frac{\cos (a x)-\cos (b x)}{\cos (c x)-1}\right]$
Solution:

Question 2.
$\lim _{x \rightarrow \pi}\left[\frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin ^{2} x}\right]$
Solution:

Question 3.
$\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\tan ^{2} x-\cot ^{2} x}{\sec x-{cosec} x}\right]$
Solution:

Question 4.
$\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}\right]$
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 7.3 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