Question

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

• A. 968
• B. 1152
• C. 746
• D. 1250

Answer 1

The Correct Option is B

Requirement: For \( f(x) \) to be continuous at \( x = 0 \), the limit as \( x \) approaches \( 0 \) must equal \( f(0) \).

\[ \lim_{x \to 0} f(x) = f(0). \]

The limit is calculated as:

\[ \lim_{x \to 0} \frac{72x^2 - 9x - 8x^2 + 1}{\sqrt{2} - \sqrt{1 + \cos x}}. \]

Applying L’Hôpital’s Rule yields:

\[ f(0) = a \ln e \, 2 \ln e \, 3. \]

Equating the limit and \( f(0) \) allows solving for \( a^2 \), resulting in \( a^2 = 1152 \).