Class 11

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

11th Maths Part 2 Limits Miscellaneous Exercise 7 Questions And Answers Maharashtra Board

I. Select the correct answer from the given alternatives.

Question 1.
\(\lim _{x \rightarrow 2}\left(\frac{x^{4}-16}{x^{2}-5 x+6}\right)=\)
(A) 23
(B) 32
(C) -32
(D) -16
Answer:
(C) -32
Hint:

Question 2.
\(\lim _{x \rightarrow-2}\left(\frac{x^{7}+128}{x^{3}+8}\right)=\)
(A) \(\frac{56}{3}\)
(B) \(\frac{112}{3}\)
(C) \(\frac{121}{3}\)
(D) \(\frac{28}{3}\)
Answer:
(B) \(\frac{112}{3}\)
Hint:

Question 3.
\(\lim _{x \rightarrow 3}\left(\frac{1}{x^{2}-11 x+24}+\frac{1}{x^{2}-x-6}\right)=\)
(A) \(-\frac{2}{25}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{7}{25}\)
(D) \(-\frac{7}{25}\)
Answer:
(A) \(-\frac{2}{25}\)
Hint:

Question 4.
\(\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11-2}}\right)=\)
(A) \(\frac{-2}{9}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{2}{9}\)
Answer:
(D) \(\frac{2}{9}\)
Hint:

Question 5.
\(\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^{2} x-3}{\sec ^{3} x-8}\right)=\)
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(\frac{1}{3}\)
Hint:

Question 6.
\(\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^{2}}\right)=\)
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(C) 2
Hint:

Question 7.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{3 \cos x+\cos 3 x}{(2 x-\pi)^{3}}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(\frac{1}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(-\frac{1}{2}\)
Hint:

Question 8.
\(\lim _{x \rightarrow 0}\left(\frac{15^{x}-3^{x}-5^{x}+1}{\sin ^{2} x}\right)=\)
(A) log 15
(B) log 3 + log 5
(C) log 3 . log 5
(D) 3 log 5
Answer:
(C) log 3 . log 5
Hint:

Question 9.
\(\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=\)
(A) e³
(B) e⁶
(C) e⁹
(D) e^-3
Answer:
(A) e³
Hint:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(-\frac{5}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{2}{5}\)
Answer:
(D) \(\frac{2}{5}\)
Hint:

Question 11.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right)=\)
(A) 1
(B) log 3
(C) \(3^{\frac{\pi}{2}}\)
(D) 3 log 3
Answer:
(B) log 3
Hint:

Question 12.
\(\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^{2}}\right]=\)
(A) \(\frac{1}{\mathrm{e}^{9}}\)
(B) \(\frac{1}{\mathrm{e}^{3}}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{3}\)
Answer:
(D) \(\frac{1}{3}\)
Hint:

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^{3}}{\left(3^{x}-1\right) \cdot \tan x \cdot \log (1+x)}\right]=\)
(A) 3 log 3
(B) 2 log 3
(C) (log 3)²
(D) (log 3)³
Answer:
(C) (log 3)²
Hint:

Question 14.
\(\lim _{x \rightarrow 3}\left[\frac{5^{x-3}-4^{x-3}}{\sin (x-3)}\right]=\)
(A) log 5 – 4
(B) log \(\frac{5}{4}\)
(C) \(\frac{\log 5}{\log 4}\)
(D) \(\frac{\log 5}{4}\)
Answer:
(B) log \(\frac{5}{4}\)
Hint:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+3)^{7}(x-5)^{3}}{(2 x-5)^{10}}\right]=\)
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{8}\)
Hint:

(II) Evaluate the following.

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

Question 2.
\(\lim _{x \rightarrow 0}[x]\) ([*] is a greatest integer function.)
Solution:

Question 3.
If f(r) = πr² then find \(\lim _{h \rightarrow 0}\left[\frac{f(r+h)-f(r)}{h}\right]\)
Solution:

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

Question 5.
Find the limit of the function, if it exists, at x = 1
\(f(x)=\left\{\begin{array}{lll}
7-4 x & \text { for } & x<1 \\
x^{2}+2 & \text { for } & x \geq 1
\end{array}\right.\)
Solution:

Question 6.
Given that 7x ≤ f(x) ≤ 3x² – 6 for all x. Determine the value of \(\lim _{x \rightarrow 3} f(x)\)
Solution:

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{\sec x^{2}-1}{x^{4}}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 0}\left[\frac{e^{x}+e^{-x}-2}{x \cdot \tan x}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\cos (6 x)-\cos (4 x)}\right]\)
Solution:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x \cdot \tan x}\right]\)
Solution:

Question 11.
\(\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{x-a}\right]\)
Solution:

Question 12.
\(\lim _{x \rightarrow 2}\left[\frac{\log x-\log 2}{x-2}\right]\)
Solution:

Question 13.
\(\lim _{x \rightarrow 1}\left[\frac{a b^{x}-a^{x} b}{x^{2}-1}\right]\)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{\left(2^{x}-1\right) \log (1+x)}\right]\)
Solution:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^{2}(7 x-3)^{3}}{(5 x+2)^{5}}\right]\)
Solution:

Question 16.
\(\lim _{x \rightarrow a}\left[\frac{x \cos a-a \cos x}{x-a}\right]\)
Solution:

Question 17.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^{2}}{\sqrt{2}-\sin x-\cos x}\right]\)
Solution:

Question 18.
\(\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^{x}+1}{\sin ^{2}(x-1)}\right]\)
Solution:

Question 19.
\(\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^{2}}\right]\)
Solution:

Question 20.
\(\lim _{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]\)
Solution:

Question 21.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)\)
Solution:

Question 22.
\(\lim _{x \rightarrow 1}\left(\frac{x+3 x^{2}+5 x^{3}+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^{n}-n^{2}}{x-1}\right)\)
Solution:

Question 23.
\(\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^{2}}{2}\right)-\cos \left(\frac{x^{4}}{4}\right)+\cos \left(\frac{x^{2}}{2}\right) \cdot \cos \left(\frac{x^{4}}{4}\right)\right]\)
Solution:

Question 24.
\(\lim _{x \rightarrow \infty}\left(\frac{8 x^{2}+5 x+3}{2 x^{2}-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}\)
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 7.7 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