JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 4)

If the function

$$f(x)= \begin{cases}\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ a \log _e 2 \log _e 3 & , x=0\end{cases}$$

is continuous at $$x=0$$, then the value of $$a^2$$ is equal to

968
1250
1152
746

Explanation

$$f(x) = \left\{ \matrix{ {{{{72}^x} - {9^x} - {8^x} + 1} \over {\sqrt 2 - \sqrt {1 + \cos x} }},\,x \ne 0 \hfill \cr a{\log _e}2{\log _e}3\,\,\,\,\,\,,\,\,x = 0 \hfill \cr} \right.$$

$$\because f(x)$$ is continuous at $$x=0$$

$$\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} \\ & \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{x^2} \times x^2} \\ & =(\ln 9 \cdot \ln 8)(2 \sqrt{2}) \times 2 \\ & =4 \sqrt{2} \times 2 \times 3 \ln 2 \cdot \ln 3 \\ & 24 \sqrt{2} \cdot \ln 2 \cdot \ln 3 \\ & \Rightarrow \quad a=24 \sqrt{2} \\ & \quad a^2=1152 \end{aligned}$$

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