Question

$\text{Let}f:[-\frac{1}{2},2]\to R\text{and}g:[-\frac{1}{2},2]\to R$ be functions defined by $f\left(x\right)=[x^2-3]$ and $g\left(x\right)=|x|f\left(x\right)+|4x-7|f\left(x\right),$ where $\left[y\right]$ denotes the greatest integer less than or equal to $y$ for $y\in R.$ Then

• A. $f$ is discontinuous exactly at three points in $[-\frac{1}{2},2]$
• B. $f$ is discontinuous exactly at four points in $[-\frac{1}{2},2]$
• C. $g$ is NOT differentiable exactly at four points in $(-\frac{1}{2},2)$
• D. $g$ is NOT differentiable exactly at five points in $(-\frac{1}{2},2)$

Answer 1

Answer: (B), (C)

$g$ is NOT differentiable exactly at four points in $(-\frac{1}{2},2)$

$f:[-\frac{1}{2},2]\to R$ is defined as
$f\left(x\right)=[x^2-3]=\left[x^2\right]-3$
$f\left(x\right)$ is not continuous where $\left[x^2\right]$ takes integral values.
Let $x^2=t$
Then $f\left(t\right)=\left[t\right]-3\forall t\in [0,4]$
Clearly, $f\left(t\right)$ is continuous at $t=0.$
Hence $f\left(x\right)$ is continuous at $x=0$
$f\left(t\right)$ is discontinuous at $t=1,2,3,4$
Hence $f\left(x\right)$ is discontinuous at $x=1,\sqrt{2},\sqrt{3},2$

So, $f\left(x\right)$ is NOT continuous at $x=1,\sqrt{2},\sqrt{3},2$ in $[-\frac{1}{2},2]$

$g:[-\frac{1}{2},2]\to R$ is defined as
$g\left(x\right)=(|x|+|4x-7|)f\left(x\right)=(|x|+|4x-7|)[x^2-3]$

Since, $f\left(x\right)$ is not continuous at $x=1,\sqrt{2},\sqrt{3}$ in $(-\frac{1}{2},2)$
$\therefore g\left(x\right)$ is not differentiable at $x=1,\sqrt{2},\sqrt{3}$

By the definition of $g\left(x\right),$ we need to check the differentiability of $g\left(x\right)$ at $x=0$ and $x=\frac{7}{4}$ also.

$|x|+|4x-7|$ is not differentiable at $x=0$ and $f\left(x\right)$ is differentiable at $x=0.$
$\therefore g\left(x\right)$ is not differentiable at $x=0$

For $x\in [\sqrt{3},2),f\left(x\right)=0$
$\therefore g\left(x\right)=0\forall x\in [\sqrt{3},2)$
$\therefore g\left(x\right)$ is differentiable at $x=\frac{7}{4}$ as $\frac{7}{4}\in [\sqrt{3},2)$

So, $g\left(x\right)$ is not differentiable at $x=0,1,\sqrt{2},\sqrt{3}$ in $(-\frac{1}{2},2)$