Question

Examine the continuity and differentiability in $-\infty <x<\infty$ of the following function :

$f\left(x\right)=1$ in $-\infty <x<0$,

$f\left(x\right)=1+\sin x$ in $0\le x\le \pi /2$ ,

$f\left(x\right)=2+{(x-\pi /2)}^2$ in $\pi /2\le x<\infty .$

• A. continuous and differentiable at $x=\frac{\pi }{2}$
• B. continuous but not differentiable at $x=\frac{\pi }{2}$
• C. dis -continuous and not -differentiable at $x=\frac{\pi }{2}$
• D. continuous and differentiable everywhere except at $x=\frac{\pi }{2}$

Answer 1

The correct option is A continuous and differentiable at $x=\frac{\pi }{2}$

We consider $x=0,\pi /2.$
At x=0,we have $f\left(0\right)=1+\sin 0=1$

$f(0-0)=\underset{h\to 0}{lim}1=1$
and $f(0+0)={lim}_{h\to 0}[1+\sin (0+h)]=1.$

Since $f(0+0)=f(0-0)=f\left(0\right),$ the function is continuous at x=0.

Now $L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{1-1}{-h}=0$
and $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{1+\sin (0+h)-1}{h}=\underset{h\to 0}{lim}\frac{\sin h}{h}=1.$

Since $L{f}^{\prime }\left(0\right)\ne R{f}^{\prime }\left(0\right),$ the function is non-differentiable at x=0.
At $x=\pi /2,$ we have $f\left(\pi /2\right)=2+{(\pi /2-\pi /2)}^2=2$, $f(\pi /2-0)=\underset{h\to \to 0}{lim}\{1+\sin (\pi /2-h)\}=\underset{h\to 0}{lim}(1+\cos h)=2,f(\frac{\pi }{2}+0)=\underset{h\to 0}{lim}\{2+{(\frac{\pi }{2}+h-\frac{\pi }{2})}^2\}=2,$

Since $f(\pi /2+0)=f(\pi /2-0)=f\left(\pi /2\right),$ the function is continuous at $x=\pi /2.$

Now $L{f}^{\prime }\frac{\pi }{2}=\underset{h\to 0}{lim}\frac{1+\sin (\pi /2-h)-2}{-h}$ $=\underset{h\to 0}{lim}\frac{1-\cos h}{h}\left[form\frac{0}{0}\right]=\underset{h\to 0}{lim}\frac{\sin h}{1}=0$

And $R{f}^{\prime }\frac{\pi }{2}=\underset{h\to 0}{lim}\frac{2+{(\pi /2+h-\pi /2)}^2-2}{h}=\underset{h\to 0}{lim}h=0.$

Since $R{f}^{\prime }\left(\pi /2\right)=L{f}^{\prime }\left(\pi /2\right),$ the function is differentiable at $x=\pi /2.$