Question

Let $a , b \in R , b \neq 0$, Define a function
$f(x)= \begin{cases}a \sin \frac{\pi}{2}(x-1), & \text { for } x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & \text { for } x>0\end{cases}$
If $f$ is continuous at $x =0$, then $10- ab$ is equal to ____

Answer 1

$f(x)= \begin{cases}a \sin \frac{\pi}{2}(x-1), & x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & x>0\end{cases}$
For continuity at '$0$'
$\displaystyle\lim _{ x \rightarrow 0^{+}} f ( x )= f (0)$
$\Rightarrow \displaystyle\lim _{ x \rightarrow 0^{+}} \frac{\tan 2 x -\sin 2 x }{ bx ^{3}}=- a $
$\Rightarrow \displaystyle\lim _{ x \rightarrow 0^{+}} \frac{\frac{8 x ^{3}}{3}+\frac{8 x ^{3}}{3 !}}{ bx ^{3}}=- a$
$\Rightarrow 8\left(\frac{1}{3}+\frac{1}{3 !}\right)=- ab$
$\Rightarrow 4=- ab$
$\Rightarrow 10- ab =14$