Question

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-\cos 10x}{x^2},& x<0\\ a,& x=0\\ \frac{\sqrt{x}}{\sqrt{625+\sqrt{x}}-25},& x>0\end{array}\right.$
then the value of a so that f (x) may be continuous at x = 0, is
(a) 25
(b) 50
(c) −25
(d) none of these

Answer 1

(b) 50

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{\mathrm{x}\to 0^{-}}{\lim }f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{h\to 0}{\lim }f\left(-h\right)=f\left(0\right)$

$$\begin{gathered} \Rightarrow \underset{h\to 0}{\lim }\frac{\left(1-\cos \left(-10h\right)\right)}{{\left(-h\right)}^2}=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }\frac{\left(1-\cos \left(10h\right)\right)}{h^2}=f\left(0\right) \\ \Rightarrow \underset{h\to 0}{\lim }\frac{\left(2{\sin }^2\left(5h\right)\right)}{h^2}=a \\ \Rightarrow \underset{h\to 0}{\lim }\frac{2\times 25\left({\sin }^2\left(5h\right)\right)}{25h^2}=a \\ \Rightarrow 50\underset{h\to 0}{\lim }\frac{\left({\sin }^2\left(5h\right)\right)}{{\left(5h\right)}^2}=a \\ \Rightarrow 50\underset{h\to 0}{\lim }{\left(\frac{\sin \left(5h\right)}{5h}\right)}^2=a \\ \Rightarrow a=50 \end{gathered}$$