If f(x) = frac{4^{x-pi} + 4^{pi-x} - 2}{(x-pi | vedclass

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(B) Since $f(x)$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.Let $x - \pi = h$. As $x \rightarrow \pi$,$h \rightarrow

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$(\log 2)^2$

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(B) Since $f(x)$ is continuous at $x = \pi$,we have $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.Let $x - \pi = h$. As $x \rightarrow \pi$,$h \rightarrow 0$.Then $f(\pi) = \lim_{h \rightarrow 0} \frac{4^h + 4^{-h} - 2}{h^2}$.We can rewrite the numerator as $(4^h - 1) - (1 - 4^{-h}) = (4^h - 1) - (1 - \frac{1}{4^h}) = (4^h - 1) - \frac{4^h - 1}{4^h} = (4^h - 1)(1 - 4^{-h})$.Alternatively,using the expansion $a^h = e^{h \ln a} = 1 + h \ln a + \frac{(h \ln a)^2}{2!} + \dots$,we have:$4^h = 1 + h \ln 4 + \frac{(h \ln 4)^2}{2} + O(h^3)$$4^{-h} = 1 - h \ln 4 + \frac{(h \ln 4)^2}{2} + O(h^3)$Adding these: $4^h + 4^{-h} = 2 + (h \ln 4)^2 + O(h^4)$.Thus,$f(\pi) = \lim_{h \rightarrow 0} \frac{2 + (h \ln 4)^2 - 2}{h^2} = \lim_{h \rightarrow 0} \frac{h^2 (\ln 4)^2}{h^2} = (\ln 4)^2$.Since $\ln 4 = \ln(2^2) = 2 \ln 2$,we have $f(\pi) = (2 \ln 2)^2 = 4(\ln 2)^2$.Wait,checking the sign: $4^h + 4^{-h} - 2 = (2^{h/2} - 2^{-h/2})^2$. As $h \rightarrow 0$,$\frac{(2^{h/2} - 2^{-h/2})^2}{h^2} = \left( \frac{2^{h/2} - 2^{-h/2}}{h} \right)^2 = \left( \frac{1}{2} \cdot \frac{2^{h/2} - 1}{h/2} - \frac{1}{2} \cdot \frac{2^{-h/2} - 1}{h/2} \right)^2 = \left( \frac{1}{2} \ln 2 - \frac{1}{2} (-\ln 2) \right)^2 = (\ln 2)^2$.

If $f(x) = \frac{4^{x-\pi} + 4^{\pi-x} - 2}{(x-\pi)^2}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) = k$. Find the value of $k$.

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