If the function f(x) = begin{cases} frac{72^x | vedclass

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(B) For the function to be continuous at $x=0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0)$.First,simplify the numerator: $72^x - 9^x - 8^x + 1 =

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$1152$

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(B) For the function to be continuous at $x=0$,we must have $\lim_{x ightarrow 0} f(x) = f(0)$.First,simplify the numerator: $72^x - 9^x - 8^x + 1 = (9^x - 1)(8^x - 1)$.Next,simplify the denominator: $\sqrt{2} - \sqrt{1 + \cos x} = \sqrt{2} - \sqrt{2 \cos^2(x/2)} = \sqrt{2}(1 - |\cos(x/2)|)$.Since $x ightarrow 0$,$\cos(x/2) > 0$,so $\sqrt{2} - \sqrt{1 + \cos x} = \sqrt{2}(1 - \cos(x/2))$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we get $\sqrt{2}(2 \sin^2(x/4))$.Now,evaluate the limit:$\lim_{x ightarrow 0} \frac{(9^x - 1)(8^x - 1)}{\sqrt{2}(2 \sin^2(x/4))} = \lim_{x ightarrow 0} \frac{(9^x - 1)}{x} \cdot \frac{(8^x - 1)}{x} \cdot \frac{x^2}{2\sqrt{2} \sin^2(x/4)}$.Using $\lim_{x ightarrow 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x ightarrow 0} \frac{\sin(kx)}{x} = k$:$= \ln 9 \cdot \ln 8 \cdot \frac{1}{2\sqrt{2} (1/4)^2} = (2 \ln 3)(3 \ln 2) \cdot \frac{16}{2\sqrt{2}} = 6 \ln 2 \ln 3 \cdot 4\sqrt{2} = 24\sqrt{2} \ln 2 \ln 3$.Equating to $f(0) = a \ln 2 \ln 3$,we get $a = 24\sqrt{2}$.Therefore,$a^2 = (24\sqrt{2})^2 = 576 	imes 2 = 1152$.

If the function $f(x) = \begin{cases} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} & , x \neq 0 \\ a \ln 2 \ln 3 & , x=0 \end{cases}$ is continuous at $x=0$,then the value of $a^2$ is equal to

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