Solve this


Let $a, b \in \mathbf{R}, b \neq 0$, Define a function

$f(x)= \begin{cases}\operatorname{asin} \frac{\pi}{2}(x-1), & \text { for } x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & \text { for } x>0\end{cases}$

If $\mathrm{f}$ is continuous at $\mathrm{x}=0$, then $10-\mathrm{ab}$ is equal to_____________


$f(x)= \begin{cases}a \sin \frac{\pi}{2}(x-1), & x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & x>0\end{cases}$

For continuity at ' 0 '

$\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\tan 2 x-\sin 2 x}{b x^{3}}=-a$

$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\frac{8 x^{3}}{3}+\frac{8 x^{3}}{3 !}}{b x^{3}}=-a$

$\Rightarrow 8\left(\frac{1}{3}+\frac{1}{3 !}\right)=-\mathrm{ab}$

$\Rightarrow 4=-\mathrm{ab}$

$\Rightarrow 10-\mathrm{ab}=14$

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