Question

Let $a,b,\in R,b\ne 0.$ Define a function
$f\left(x\right)=\{\begin{array}{cc}a\sin \frac{\pi }{2}(x-1),& \text{for}x\le 0\\ \frac{\tan 2x-\sin 2x}{b{x}^3},& \text{for}x>0\end{array}.$
If $f$ is continuous at $x=0,$ then $10-ab$ is equal to

Answer 1

LHL at $x=0$ for $f\left(x\right)$
$\underset{x\to 0^{-}}{lim}a\sin \frac{\pi }{2}(x-1)=a\sin (-\frac{\pi }{2})=-a$
RHL at $x=0$ for $f\left(x\right)$
$\underset{x\to 0^{+}}{lim}\frac{\tan 2x-\sin 2x}{b{x}^3}=\underset{x\to 0^{+}}{lim}\frac{\tan 2x}{x}\frac{(1-\cos 2x)}{x^2}\cdot \frac{1}{b}$
$=\frac{1}{b}\underset{x\to 0^{+}}{lim}\frac{\tan 2x}{x}\cdot \frac{1-\cos 2x}{4x^2}\cdot 4$
$=\frac{1}{b}\cdot 2\cdot \frac{1}{2}\cdot 4=\frac{4}{b}$
For continuity, $-a=\frac{4}{b}\Rightarrow ab=-4$
So, $10-ab=14$