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

Answer: (B)

1152

Answer 2

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 $a^2 = 1152$.