Question

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

• A. $16\sqrt{2}log2log3$
• B. $16\sqrt{2}In6$
• C. $16\sqrt{2}$ In $2$ In $3$
• D. None of these

Answer 1

Answer: (C)

$16\sqrt{2}$ In $2$ In $3$

$f\left(x\right)=\{\begin{array}{c}\frac{{36}^x-9^x-4^x}{\sqrt{2}-\sqrt{1+\cos x}}+1,x\ne 0\\ kx=0\end{array}{lim}_{x\to 0}\frac{9^x\left(4^x-1\right)-\left(4^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}{lim}_{x\to 0}\frac{\left(9^x-1\right)\left(4^x-1\right)\left(\sqrt{2}+\sqrt{1+\cos x}\right)}{\left(1-\cos x\right)}2\sqrt{2}{lim}_{x\to 0}\frac{\left(9^x-1\right)\left(4^x-1\right)}{\frac{\left(1-\cos x\right)}{x^2}{x}^2}4\sqrt{2}{lim}_{x\to 0}\frac{\left(9^x-1\right)}{x}\frac{\left(4^x-1\right)}{x}f\left(0\right)=4\sqrt{2}\ln 9\ln 4f\left(0\right)=16\sqrt{2}In3In2$

Option $C$ is correct.