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If $f(x)=\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^2}$ for $\neq \pi$, is continuous at $x=\pi$, then $=\mathrm{k} \quad$ for $=\pi$
$\mathrm{k}=$
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$\mathrm{k}=$
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2183 Upvotes
Verified Answer
The correct answer is:
$-4(\log 2)^2$
$$
f(\pi)=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{4^{x-\pi}+4^{\pi-4}-2}{(x-\pi)^2}
$$
Put $\mathrm{x}=\pi+\mathrm{h}$. As $\mathrm{x} \rightarrow \pi, \mathrm{h} \rightarrow 0$
$$
\begin{aligned}
& \therefore \mathrm{f}(\pi)=\lim _{\mathrm{h} \rightarrow 0} \frac{4^{\mathrm{h}}+4^{-\mathrm{h}}-1-1}{\mathrm{~h}^2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\left(4^{\mathrm{h}}-1\right)}{\mathrm{h}} \times \lim _{\mathrm{h} \rightarrow 0} \frac{4^{-\mathrm{h}}-1}{-\mathrm{h}} \times(-1) \\
& =(\log 4)(-\log 4)=(2 \log 2)(-2 \log 2)=-4(\log 2)^2
\end{aligned}
$$
f(\pi)=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{4^{x-\pi}+4^{\pi-4}-2}{(x-\pi)^2}
$$
Put $\mathrm{x}=\pi+\mathrm{h}$. As $\mathrm{x} \rightarrow \pi, \mathrm{h} \rightarrow 0$
$$
\begin{aligned}
& \therefore \mathrm{f}(\pi)=\lim _{\mathrm{h} \rightarrow 0} \frac{4^{\mathrm{h}}+4^{-\mathrm{h}}-1-1}{\mathrm{~h}^2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\left(4^{\mathrm{h}}-1\right)}{\mathrm{h}} \times \lim _{\mathrm{h} \rightarrow 0} \frac{4^{-\mathrm{h}}-1}{-\mathrm{h}} \times(-1) \\
& =(\log 4)(-\log 4)=(2 \log 2)(-2 \log 2)=-4(\log 2)^2
\end{aligned}
$$
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