Question

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

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

Answer 1

Answer: (B), (D)

Not differentiable
Discontinuous

$f\left(x\right)=\{\begin{array}{c}\frac{x^2}{2},0\le x\le 1\\ 2x^2-3x+\frac{3}{2},1\le x\le 2\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}2x,0\le x\le 1\\ 4x-3,1\le x\le 2\end{array}$

$f^{\prime \prime }\left(x\right)=\{\begin{array}{c}2,0\le x\le 1\\ 4,1\le x\le 2\end{array}$

Clearly $f^{\prime \prime }\left(1^{-}\right)=2\ne f\left(1^{+}\right)=4\Rightarrow$ f'' is discontinuous
Hence it is also non differentiable