#### Explain solution RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 4 maths

The correct option is (C)

Hint:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{x \rightarrow a}\left\{\frac{f(x)}{g(x)}\right\}=\frac{1}{m} \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m\\ &\text { (iii) } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{aligned}

Given:

$f(x)=\left\{\begin{array}{l} \frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}} \\ k \end{array}\right.$      \begin{aligned} , \quad &x \neq 0 \\ , \quad &x=0 \end{aligned}

Solution:

Function f(x) is continuous at x = 0

To calculate the value of k, understand that function is continuous at x = 0

\begin{aligned} &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\sqrt{2}\left[2 \sin ^{2}\left(\frac{x}{4}\right)\right]}\right)=k \end{aligned}

$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{2 \sqrt{2} x^{2} \cdot \frac{\frac{\sin ^{2} x}{4}}{\frac{x^{2}}{16} \times 16}}\right)=k$

$\Rightarrow \lim _{x \rightarrow 0}\left\{\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\frac{\sqrt{2}}{8} x^{2} \times \frac{\sin ^{2} x y}{\frac{x^{2}}{16}}}\right\}=k$

Using formula (ii) and (iii)

\begin{aligned} &\left.\Rightarrow \frac{8 \ln 9 \ln 4}{\sqrt{2}}=k \quad \text { [understand that } \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{a^{x}}\right)=\ln a\right] \\ &\Rightarrow \frac{32 \ln 3 \ln 2}{\sqrt{2}}=k \\ &\Rightarrow k=16 \sqrt{2} \ln 2 \ln 3 \end{aligned}

So correct option is (C)