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

(a) f is discontinuous exactly at three points in [-1/2 , 2]

(b) f is discontinuous exactly at four points in [-1/2 , 2]

(c) g is not differentiable exactly at four points in (-1/2 , 2)

(d) g is not differentiable exactly at five points in (-1/2 , 2)

(b), (c)

We have $f(x)=\left[x^2-3\right]=\left[x^2\right]-3$

and, $g(x)=(|x|+|4 x-7|) f(x)$

Now, $x \in[-1 / 2,2] \Rightarrow 0 \leq x^2 \leq 4$

So, $f(x)=\left[x^2\right]-3$ is discontinuous and hence nondifferentiable at $x=1, \sqrt{2}, \sqrt{3}$ and 2 in $[-1 / 2,2]$.

Since $g(x)=(|x|+|4 x-7|) f(x)$. So, g(x) is not continuous and hence non-differentiable at $x=1, \sqrt{2}, \sqrt{3}$ in $(-1 / 2,2)$.

In the left neighbourhood of x = 0, we find that

$g(x)=(-5 x+7)(-3)=15 x-21$

In the right neighbourhood of x = 0, we have

$g(x)=(-3 x+7)(-3)=9 x-21$

Clearly, g is not differentiable at x = 0.

In the left neighbourhood of x = 7/4 :

$g(x)=(-3 x+7) \times 0=0$

In the right neighbourhood of x = 7/4 :

$g(x)=(5 x-7) \times 0=0$

So, g(x) is differentiable at x = 0.

Hence, g is not differentiable at $x=0,1, \sqrt{2}, 7 / 4$ in $(-1/2, 2)$.