Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}};& x\ne 0\\ k\log 2\log 3;& x=0\end{array}$. If $f$ is continuous function at $x=0$, then $k=$

• A. $12\sqrt{2}$
• B. $24$
• C. $18\sqrt{2}$
• D. $24\sqrt{2}$

Answer 1

The correct option is D $24\sqrt{2}$

Given function is $f\left(x\right)=\{\begin{array}{cc}\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}};& x\ne 0\\ k\log 2\log 3;& x=0\end{array}$

$f$ is continuous at $x=0$

$\Rightarrow \underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\frac{9^x(8^x-1)-1(8^x-1)}{\sqrt{2}-\sqrt{1+\cos x}}\times \frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\frac{(9^x-1)(8^x-1)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\frac{(9^x-1)}{x}\times \frac{(8^x-1)}{x}\times \frac{x^2}{1-\cos x}\left(\sqrt{2}+\sqrt{1+\cos x}\right)=f\left(0\right)$

$\Rightarrow \log 9\log 8.2(\sqrt{2}+\sqrt{2})=k\log 2\log 3$

$\Rightarrow 2\log 3\times 3\log 2\times 2(\sqrt{2}+\sqrt{2})=k\log 2\log 3$

$\Rightarrow 24\sqrt{2}=k$

Hence, $k=24\sqrt{2}$ is the answer.