Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{1-\sin \pi x}{1+\cos 2\pi x},& x<\frac{1}{2}\\ p,& x=\frac{1}{2}\\ \frac{\sqrt{2k+\sqrt{2x-1}}-2}{\sqrt{2x-1}},& x>\frac{1}{2}\end{array}$
If $f$ is continuous at $x=\frac{1}{2},$ then the value of $k+12p$ is

• A. $-1$
• B. $2$
• C. $5$
• D. $0$

Answer 1

The correct option is C $5$

L.H.L. $=\underset{x\to {\frac{1}{2}}^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}\frac{1-\sin \pi (\frac{1}{2}-h)}{1+\cos 2\pi (\frac{1}{2}-h)}$
$=\underset{h\to 0}{lim}\frac{1-\cos \left(\pi h\right)}{1-\cos \left(2\pi h\right)}$ $\left(\frac{0}{0}\right)\text{form}$
By L'Hospital's rule,
L.H.L. $=\underset{h\to 0}{lim}\frac{\sin \left(\pi h\right)\times \pi }{\sin \left(2\pi h\right)\times 2\pi }$

$=\underset{h\to 0}{lim}\frac{1}{4}\times \frac{\frac{\sin \left(\pi h\right)}{\pi h}}{\frac{\sin \left(2\pi h\right)}{2\pi h}}$
$=\frac{1}{4}\times 1=\frac{1}{4}=p$

R.H.L. $=\underset{x\to {\frac{1}{2}}^{+}}{lim}f\left(x\right)$
$=\underset{x\to {\frac{1}{2}}^{+}}{lim}\frac{\sqrt{2k+\sqrt{2x-1}}-2}{\sqrt{2x-1}}$
$=\underset{h\to 0}{lim}\frac{\sqrt{2k+\sqrt{2(\frac{1}{2}+h)-1}}-2}{\sqrt{2(\frac{1}{2}+h)-1}}$
$=\underset{h\to 0}{lim}\frac{\sqrt{2k+\sqrt{2h}}-2}{\sqrt{2h}}$
For existence of above limit,
$\sqrt{2k}-2=0$
$\Rightarrow 2k=4\Rightarrow k=2$
$\therefore k+12p=2+12\times \frac{1}{4}=5$