Question

If $f(x) = \frac{4^{x - \pi} + 4^{\pi - x} - 2}{(x - \pi)^2}$, for $x \ne \pi$, is continuous at $x = \pi$, then $f(\pi)$ is

• A. $2(\log 2)^2$
• B. $4(\log 2)^2$
• C. $(\log 2)^2$
• D. $8(\log 2)^2$

Hint:

Any limit in the form $\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}$ perfectly wraps up into $(\log a)^2$. Recognizing that $a^x + a^{-x} - 2$ is the expansion of $(a^{x/2} - a^{-x/2})^2$ saves a lot of time!

Answer 1

The Correct Option is B

Step 1: Use the continuity rule.
For $f$ to be continuous at $x = \pi$, the value $f(\pi)$ must equal the limit of $f(x)$ as $x \to \pi$.

Step 2: Substitute to simplify.
Let $t = x - \pi$, so $t \to 0$. The function becomes
\[ \frac{4^{t} + 4^{-t} - 2}{t^2} \]
Step 3: Recognise a perfect square.
The top is $4^{t} + 4^{-t} - 2 = (2^{t} - 2^{-t})^2$, since $4^t = (2^t)^2$.
\[ \frac{(2^{t} - 2^{-t})^2}{t^2} = \left(\frac{2^{t} - 2^{-t}}{t}\right)^2 \]
Step 4: Use the standard limit.
As $t \to 0$, $\frac{a^t - 1}{t} \to \log a$. So $\frac{2^t - 2^{-t}}{t} \to \log 2 - (-\log 2) = 2\log 2$.

Step 5: Square the limit.
\[ f(\pi) = (2\log 2)^2 = 4(\log 2)^2 \]
Step 6: Conclusion.
So the value that makes $f$ continuous is $4(\log 2)^2$. \[ \boxed{4(\log 2)^2 \text{ (Option 2)}} \]