Question

Let $f\left(x\right)=x-x^2$ and $g\left(x\right)=\{\begin{array}{c}maxf\left(t\right),0\le t\le x,0\le ,x\le 1\\ \sin \pi x,x>1\end{array}$, then in the interval $[0,\infty )$

• A. $g\left(x\right)$ is everywhere continuous except at two points
• B. $g\left(x\right)$ is everywhere differentiable except at two points
• C. $g\left(x\right)$ is everywhere differentiable except at $x=1$
• D. None of these

Answer 1

The correct option is C $g\left(x\right)$ is everywhere differentiable except at $x=1$

$f\left(x\right)=x-x^2$

$f\left(t\right)=t-t^2$

max $f\left(t\right)$

$f^{\prime }\left(t\right)=0$

$1-2t=0$

$t=\frac{1}{2}$

$f^{\prime \prime }\left(t\right)<0\Rightarrow$ max at $t=\frac{1}{2}$

$f\left(1/2\right)=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

$g\left(x\right)=\{\begin{array}{c}1/4;0\le x\le 1\\ \sin \pi x;x>1\end{array}$

at $x=1$

${lim}_{x\to 1^{-}}g\left(x\right)=\frac{1}{4}$

${lim}_{x\to 1^{+}}g\left(x\right)={lim}_{x\to 1^{+}}\left(\sin \pi x\right)=0$

LHL $\ne$RHL

$\Rightarrow$Discontinuous

Not differentiable at this point

For all other points $g\left(x\right)$ is continuous and hence differentiable