Let $a, b \in \mathbf{R}, b \neq 0$, Define a function
$\mathrm{f}(\mathrm{x})= \begin{cases}\operatorname{asin} \frac{\pi}{2}(\mathrm{x}-1), & \text { for } \mathrm{x} \leq 0 \\ \frac{\tan 2 \mathrm{x}-\sin 2 \mathrm{x}}{\mathrm{bx}^{3}}, & \text { for } \mathrm{x}>0\end{cases}$
If $\mathrm{f}$ is continuous at $\mathrm{x}=0$, then $10-\mathrm{ab}$ is equal to________.
$f(x)= \begin{cases}\operatorname{asin} \frac{\pi}{2}(x-1,) & x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & x>0\end{cases}$
For continuity at ' 0 '
$\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\tan 2 x-\sin 2 x}{b x^{3}}=-a$
$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\frac{8 x^{3}}{3}+\frac{8 x^{3}}{3 !}}{b x^{3}}=-a$
$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\frac{8 x^{3}}{3}+\frac{8 x^{3}}{3 !}}{b x^{3}}=-a$
$\Rightarrow 8\left(\frac{1}{3}+\frac{1}{3 !}\right)=-\mathrm{ab}$
$\Rightarrow 4=-a b$
$\Rightarrow 10-a b=14$
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