If $\mathrm{f}(x)$ is continuous at point $x=0$ where $\mathrm{f}(x)=\left\{\begin{array}{cl}\frac{3 \sin x+5 \tan x}{a^x-1} & , x < 0 \\ \frac{2}{\log 2} & , x=0 \\ \frac{8 x+2 x \cos x}{\mathbf{b}^x-1} & , x>0\end{array}\right.$ then the values of $a$ and b , respectively, are

The function $f(x)$ is continuous at $x=0$ when $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = \frac{2}{\log 2}$, with $\log$ interpreted as the natural logarithm.

Evaluating the left-hand limit:

$\lim_{x \to 0^-} \frac{3 \sin x + 5 \tan x}{a^x - 1} = \lim_{x \to 0^-} \frac{\frac{3 \sin x}{x} + \frac{5 \tan x}{x}}{\frac{a^x - 1}{x}}$

Using standard limits $\lim_{x \to 0} \frac{\sin x}{x} = 1$, $\lim_{x \to 0} \frac{\tan x}{x} = 1$, and $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$, we obtain

$\frac{8}{\ln a}$.

For the right-hand limit:

$\lim_{x \to 0^+} \frac{8x + 2x \cos x}{b^x - 1} = \lim_{x \to 0^+} \frac{8 + 2 \cos x}{\frac{b^x - 1}{x}}$

Applying $\lim_{x \to 0} \cos x = 1$ and $\lim_{x \to 0} \frac{b^x - 1}{x} = \ln b$ gives

$\frac{10}{\ln b}$.

Equating both limits to $f(0)$:

$\frac{8}{\ln a} = \frac{2}{\ln 2} \Rightarrow 8 \ln 2 = 2 \ln a \Rightarrow \ln a = 4 \ln 2 \Rightarrow a = 16$

$\frac{10}{\ln b} = \frac{2}{\ln 2} \Rightarrow 10 \ln 2 = 2 \ln b \Rightarrow \ln b = 5 \ln 2 \Rightarrow b = 32$

The values are $a = 16$ and $b = 32$, corresponding to option B.