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
- 1152
- 746
- 1250
The Correct Option is B
Solution and Explanation
Solution: For the function \( f(x) \) to be continuous at \( x = 0 \), we must have:
\[ \lim_{x \to 0} f(x) = f(0). \]
Calculating the limit on the left-hand side for \( x \to 0 \), we get:
\[ \lim_{x \to 0} \frac{72x^2 - 9x - 8x^2 + 1}{\sqrt{2} - \sqrt{1 + \cos x}}. \]
Using L’Hôpital’s Rule, we evaluate this limit step-by-step, and find that:
\[ f(0) = a \ln e \, 2 \ln e \, 3. \]
Setting the limit equal to \( f(0) \), we solve for \( a^2 \) and find \( a^2 = 1152 \).
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