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  • If   <img alt="\mathrm{f(x)=\left\{\begin{array}{cc}\frac{36^x-9^x-4^x+1}{\sqrt{2}-\sqrt{1+\cos x},}, x \neq 0 \\ k & x=0\end{array}\right.}" src="https://entrancecorner.oncodecogs.co

If   \mathrm{f(x)=\left\{\begin{array}{cc}\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

Option: 1

16 \sqrt{2} \log 2 \log 3


Option: 2

16 \sqrt{2} \ln 6


Option: 3

16 \sqrt{2} \ln 2 \ln 3


Option: 4

none of these


Answers (1)

best_answer

For f(x) to be continuous at x=0, we must have

\mathrm{\lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2}-\sqrt{2 \cos ^2 x / 2}=k} }
\mathrm{ \Rightarrow \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(4^x-1\right)}{\sqrt{2} \cdot 2 \sin ^2 x / 4}=k \\ }
\mathrm{ \Rightarrow \lim _{x \rightarrow 0} \frac{16 \times\left(\frac{9^x-1}{x}\right)\left(\frac{4^x-1}{x}\right)}{2 \sqrt{2}\left(\frac{\sin x / 2}{x / 4}\right)^2}=k }
\mathrm{ \Rightarrow \frac{16}{2 \sqrt{2}} \log 9 \cdot \log 4=k \\ . }

\mathrm{ \Rightarrow k=4 \sqrt{2} \log 9 \cdot \log 4 \\ . }

\mathrm{ \Rightarrow k=16 \sqrt{2} \log 3 \log 2 . }
 

Posted by

Suraj Bhandari

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