Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f\left(x\right)=\left[x^2\right]-3$
$g\left(x\right)=\left(\left|x\right|+\left|4x-7\right|\right) \left(\left[x^2\right]-3\right)$
$\because$$f$is discontinuous at (when x²is an integer)
$x=1, \sqrt{2}, \sqrt{3}, 2 \mathrm{i}\mathrm{n}\mathrm{ }\left[-\frac{1}{2}, 2\right]$
and$\left|x\right|+\left|4x-7\right|\ne 0$at$x=1, \sqrt{2}, \sqrt{3}, 2$
$\Rightarrow$$g\left(x\right)$is discontinuous at$x=1, \sqrt{2}, \sqrt{3}$in$\left(-\frac{1}{2}, 2\right)$
$\mathrm{In}$$\left(0-\delta , 0+\delta \right)$$$($\delta$is very small)
$\mathrm{\Rightarrow g}\left(x\right)=\left(\left|x\right|+\left|4x-7\right|\right). \left(-3\right)$
$\Rightarrow$'g' is non derivable at$x=0$.as$\left|x\right|$is there which is not diff. at$x=0$
$\mathrm{In}$$\left(\frac{7}{4}-\delta ,\frac{7}{4}+\delta \right)$
$\mathrm{\Rightarrow g}\left(x\right)=0$as$f\left(x\right)=0$
$\Rightarrow$Derivable at$x=\frac{7}{4}$
$\therefore$'g' is non - derivable at$0, 1, \sqrt{2},\sqrt{3}$