Solve this
Question:

If $f(x)=\left\{\begin{array}{cl}\frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ k & , x=0\end{array}\right.$ is continuous at $x=0$, then $k$ equals

(a) $16 \sqrt{2} \log 2 \log 3$

(b) $16 \sqrt{2} \ln 6$

(c) $16 \sqrt{2} \ln 2 \ln 3$

(d) none of these

Solution:

(c) $16 \sqrt{2} \ln 2 \ln 3$

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

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k$

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

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

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

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

$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\sqrt{2}\left[1-\cos \left(\frac{x}{2}\right)\right]}\right)=k$

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

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

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

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

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

$\Rightarrow \frac{8}{\sqrt{2}} \times \frac{\ln 9 \times \ln 4}{1}=k \quad\left[\because \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)=a\right]$

$\Rightarrow \frac{8}{\sqrt{2}} \times \frac{2 \ln 3 \times(2 \ln 2)}{1}=k$

$\Rightarrow \frac{32}{\sqrt{2}} \times \frac{\ln 3 \ln 2}{1}=k$

$\Rightarrow \frac{32 \sqrt{2}}{2} \times \frac{\ln 3 \ln 2}{1}=k$

$\Rightarrow k=16 \sqrt{2} \ln 2 \ln 3$