$f(x)= \begin{cases}\operatorname{a} \sin \frac{\pi}{2}(x-1), & \text { for } x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & \text { for } x>0\end{cases}$
If $f$ is continuous at $x=0$, then $10-a b$ is equal to ...... .
For continuity at $' 0 '$
$\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\tan 2 x-\sin 2 x}{b x^{3}}=-a$
$\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\frac{8 x^{3}}{3}+\frac{8 x^{3}}{3 !}}{b x^{3}}=-a$
$\Rightarrow 8\left(\frac{1}{3}+\frac{1}{3 !}\right)=-a b$
$\Rightarrow 4=-a b$
$\Rightarrow 10-a b=14$
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| Height (in $cm$) | $160$ | $150$ | $152$ | $161$ | $156$ | $154$ | $155$ |
| No of students | $12$ | $8$ | $4$ | $4$ | $3$ | $3$ | $7$ |
The median of the distribution is