Question

The number of integral value(s) of $a$ for which function $f\left(x\right)=\{\begin{array}{cc}\frac{5\sin x}{x},& x<0\\ a,& x=0\\ \frac{e^x-1}{2x},& x>0\end{array}$
is strictly decreasing at $x=0$, is equal to

Answer 1

Given : $f\left(x\right)=\{\begin{array}{cc}\frac{5\sin x}{x},& x<0\\ a,& x=0\\ \frac{e^x-1}{2x},& x>0\end{array}$
is strictly decreasing at $x=0$
$\Rightarrow f(0-h)>f\left(0\right)>f(0+h)$ as $h\to 0^{+}$
Now,
$f(0-h)\Rightarrow \underset{h\to 0^{-}}{lim}\frac{5\sin (0-h)}{0-h}=\underset{h\to 0^{+}}{lim}\frac{5\sin h}{h}=5\times 1=5$
$(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$

$f(0+h)=\underset{h\to 0^{+}}{lim}\frac{e^h-1}{2h}=\frac{1}{2}\times 1=\frac{1}{2}$
$(\underset{x\to 0^{+}}{lim}\frac{e^x-1}{x}=1)$
$\Rightarrow 5>a>\frac{1}{2}$
$\therefore$ there are $4$ possible integral values of $a.$