Question

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

Answer 1

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

Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{{36}^x-9^x-4^x+1}{\sqrt{2}-\sqrt{1+\cos x}},x\ne 0\\ k,x=0\end{array}\right.$
If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\left(\frac{{36}^x-9^x-4^x+1}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{9^x{4}^x-9^x-4^x+1}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{9^x\left(4^x-1\right)-1\left(4^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2}-\sqrt{2}\cos \left({\displaystyle \frac{x}{2}}\right)}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2}\left[1-\cos \left({\displaystyle \frac{x}{2}}\right)\right]}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2}\left[2{\sin }^2\left({\displaystyle \frac{x}{4}}\right)\right]}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{8\left(9^x-1\right)\left(4^x-1\right)}{16\sqrt{2}{x}^2\left[{\displaystyle \frac{{\sin }^2\left({\displaystyle \frac{x}{4}}\right)}{x^2}}\right]}\right)=k \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{8\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2}{x}^2\left[{\displaystyle \frac{{\sin }^2\left({\displaystyle \frac{x}{4}}\right)}{\left({\displaystyle \frac{x^2}{16}}\right)}}\right]}\right)=k \\ \Rightarrow \frac{8}{\sqrt{2}}\underset{x\to 0}{\lim }\left(\frac{\left(9^x-1\right)\left(4^x-1\right)}{x^2\left[{\displaystyle \frac{{\sin }^2\left({\displaystyle \frac{x}{4}}\right)}{{\displaystyle {\left(\frac{x}{4}\right)}^2}}}\right]}\right)=k \\ \Rightarrow \frac{8}{\sqrt{2}}\frac{\underset{x\to 0}{\lim }\left({\displaystyle \frac{9^x-1}{x}}\right)\underset{x\to 0}{\lim }\left({\displaystyle \frac{4^x-1}{x}}\right)}{\underset{x\to 0}{\lim }{\left[{\displaystyle \frac{\sin \left({\displaystyle \frac{x}{4}}\right)}{\left({\displaystyle \frac{x}{4}}\right)}}\right]}^2}=k \\ \Rightarrow \frac{8}{\sqrt{2}}\times \frac{\ln 9\times \ln 4}{1}=k\left[\because \underset{x\to 0}{\lim }\left(\frac{a^x-1}{x}\right)=a\right] \\ \Rightarrow \frac{8}{\sqrt{2}}\times \frac{2\ln 3\times \left(2\ln 2\right)}{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 \end{gathered}$$