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  • Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 26

Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 26

Answers (1)

Answer:

 The correct option is (b)

Hint:

 If a function f  is continuous at x = a, then 

\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)

\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)=1

\text { (ii) } \lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)=1

Given:

 the function

f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}

is continuous at each point of its domain

Solution:

Step 1: Understand that according to question, f(x) is continuous at x = 0

\begin{aligned} &\therefore \lim f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0) \end{aligned}

\lim _{x \rightarrow \infty} \frac{x\left(2-\frac{\sin ^{-1} x}{x}\right)}{x\left(2+\frac{\tan ^{-1} x}{x}\right)}=f(0)

Simplify further,

\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=f(0)

\frac{2-\lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)}{2+\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)}=f(0)

Apply formula (i) and formula (ii)

\begin{aligned} &\frac{2-1}{2+1}=f(0) \\ &f(0)=\frac{1}{3} \end{aligned}

Hence, the correct answer is option (b)

Posted by

Gurleen Kaur

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