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(A) For the function $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(0) = \ln 3 \ln 4$.Now,evaluate the limit: $\l

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(A) For the function $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Given $f(0) = \ln 3 \ln 4$.Now,evaluate the limit: $\lim_{x 	o 0} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}ight)}$.Numerator: $6^x-3^x-2^x+1 = (3^x-1)(2^x-1)$.Denominator: $1-\cos \left(\frac{x}{a}ight) = 2 \sin^2 \left(\frac{x}{2a}ight)$.So,$\lim_{x 	o 0} \frac{(3^x-1)(2^x-1)}{2 \sin^2 \left(\frac{x}{2a}ight)} = \lim_{x 	o 0} \frac{\left(\frac{3^x-1}{x}ight) \left(\frac{2^x-1}{x}ight) x^2}{2 \left(\frac{\sin(x/2a)}{x/2a}ight)^2 \left(\frac{x}{2a}ight)^2}$.Using $\lim_{x 	o 0} \frac{k^x-1}{x} = \ln k$,we get $\frac{\ln 3 \cdot \ln 2}{2 \cdot \frac{1}{4a^2}} = \frac{\ln 3 \cdot \ln 2 \cdot 4a^2}{2} = 2a^2 \ln 3 \ln 2$.Since $\ln 4 = 2 \ln 2$,the expression is $a^2 \ln 3 \ln 4$.Equating to $f(0)$: $a^2 \ln 3 \ln 4 = \ln 3 \ln 4$.Thus,$a^2 = 1$. Since $a$ is positive,$a=1$.

Let $a$ be a positive real number. If a real valued function $f(x) = \begin{cases} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then $a=$

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