Question

lf $f\left(x\right)=\frac{x^2}{2}$, if $0\le x\le 1$,$f\left(x\right)=2x^2-3x+\left(3/2\right)$ , lf $1\le x\le 2$ then the function $f^{\prime \prime }\left(x\right)$ is

• A. Continuous
• B. Discontinuous
• C. Differentiable
• D. Not differentiable

Answer 1

The correct option is B Discontinuous

$f\left(x\right)=\frac{x^2}{2},0\le x\le 1$
$f\left(x\right)=2x^2-3x+\frac{3}{2},1\le x\le 2$
$f^{\prime }\left(x\right)=x,0\le x\le 1$
$f^{\prime }\left(x\right)=4x-3,1\le x\le 2$
$f^{\prime \prime }\left(x\right)=1,0\le x\le 1$
$f^{\prime \prime }\left(x\right)=4,1\le x\le 2$
as $f^{\prime \prime }\left(x\right)$is discontinous at $x=1$ so ${f^{\prime }}^{\prime \prime }\left(x\right)$ doesn't exist