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  • Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 23

Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 23

Answers (1)

Answer:

 The correct option is (d)

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 x}{x}\right)=1

\text { (ii) } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1

\text { (iii) } \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)=\log a

Given:

 The function

f(x)= \begin{cases}\frac{\left(4^{x}-1\right)^{3}}{\sin (x a) \log \left(1+x^{2} 3\right)} &, x \neq 0 \\ 12(\log 4)^{3} & , x=0\end{cases}

Solution:

Step 1: Calculate, f(x) is continuous for x = 0

\begin{aligned} &\therefore \lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0}\left[\frac{\left(4^{x}-1\right)^{3}}{\sin \frac{x}{a} \log \left(1+\frac{x^{2}}{3}\right)}\right]=12(\log 4)^{3} \end{aligned}

\lim _{x \rightarrow 0}\left[\frac{\frac{\left(4^{x}-1\right)^{3}}{x^{3}}}{\frac{\sin \frac{x}{a} \log \left(1+\frac{x^{2}}{3}\right)}{x^{3}}}\right]=12(\log 4)^{3}

\lim _{x \rightarrow 0}\left[\frac{a\left(\frac{4^{x}-1}{x}\right)^{3}}{\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \log \left(1+\frac{x^{2}}{3}\right)} x^{2}\right]=12(\log 4)^{3}

3 a \lim _{x \rightarrow 0}\left[\frac{\left(\frac{4^{x}-1}{x}\right)^{3}}{\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \frac{\log \left(1+\frac{x^{2}}{3}\right)}{\left(\frac{x^{2}}{3}\right)}}\right]=12(\log 4)^{3}

3 a\left[\frac{\lim_{x\rightarrow 0}\left(\frac{4^{x}-1}{x}\right)^{3}}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \lim_{x \rightarrow 0} \frac{\log \left(1+\frac{x^{2}}{3}\right)}{\left(\frac{x^{2}}{3}\right)}}\right]=12(\log 4)^{3}

Apply formula (i), formula (ii) and formula (iii)

\begin{aligned} &\therefore 3 a(\log 4)^{3}=12(\log 4)^{3} \\ &a=4 \end{aligned}

Hence, the correct answer is option (d)

Posted by

Gurleen Kaur

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