Q45 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics
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
- A968
- Correct answerB1152
- C746
- D1250
Solution
$\begin{aligned} & \lim _{x \rightarrow 0} f(x)=a \ln 2 \ln 3 \\ & \lim _{n \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\lim _{x \rightarrow 0} \frac{\left(8^x-1\right)\left(9^x-1\right)}{\sqrt{2}-\sqrt{1+\cos x}} \\ & \lim _{n \rightarrow 0}\left(\frac{8^x-1}{x}\right)\left(\frac{9^x-1}{x}\right)\left(\frac{x^2}{1-\cos x}\right)(\sqrt{2}+\sqrt{1+\cos x}) \\ & \therefore \ln 8 \times \ln 9 \times 2 \times 2 \sqrt{2}=24 \sqrt{2} \ln 2 \ln 3 \\ & \therefore a=24 \sqrt{2}, a^2=576 \times 2=1152\end{aligned}$
Hence, the correct answer is option: 'B'.