MCQ
If the function $f(x)=\left\{\begin{array}{cc}\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} & , x \neq 0 \\ a \log _e 2 \log _e 3 & , x=0\end{array}\right.$ is continuous at $x=0$, then the value of $a^2$ is equal to
  • A
    $968$
  • $1152$
  • C
    $746$
  • D
    $1250$

Answer

Correct option: B.
$1152$
b
$ \lim _{x \rightarrow 0} f(x)=a \ln 2 \ln 3 $

$ \lim _{n \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\lim _{x \rightarrow 0} \frac{\left(8^x-1\right)\left(9^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}} $

$ \lim _{n \rightarrow 0}\left(\frac{8^x-1}{x}\right)\left(\frac{9^x-1}{x}\right)\left(\frac{x^2}{1-\cos x}\right)(\sqrt{2}+\sqrt{1+\cos x}) $

$ \therefore \ln 8 \times \ln 9 \times 2 \times 2 \sqrt{2}=24 \sqrt{2} \ln 2 \ln 3 $

$ \therefore a=24 \sqrt{2}, a^2=576 \times 2=1152$

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