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(B, C) The function $f(x) = [x^2 - 3]$ is discontinuous where $x^2 - 3$ is an integer.In the interval $[-\frac{1}{2}, 2]$,$x^2$ ranges from $0$ to $4$

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$A, D$

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(B, C) The function $f(x) = [x^2 - 3]$ is discontinuous where $x^2 - 3$ is an integer.In the interval $[-\frac{1}{2}, 2]$,$x^2$ ranges from $0$ to $4$,so $x^2 - 3$ ranges from $-3$ to $1$.The values of $x^2 - 3$ are integers at $x^2 - 3 \in \{-3, -2, -1, 0, 1\}$,which means $x^2 \in \{0, 1, 2, 3, 4\}$.Since $x \in [-\frac{1}{2}, 2]$,the points of discontinuity for $f(x)$ are $x = 0, 1, \sqrt{2}, \sqrt{3}, 2$. However,at $x=2$,the function is defined on a closed interval,so we check the interior points. The points are $x = 0, 1, \sqrt{2}, \sqrt{3}$. Thus,$f$ is discontinuous at $4$ points in $[-\frac{1}{2}, 2]$. Option $B$ is correct.Now consider $g(x) = (|x| + |4x - 7|)f(x)$.$f(x)$ is discontinuous at $x \in \{0, 1, \sqrt{2}, \sqrt{3}, 2\}$.$|x|$ is non-differentiable at $x=0$.$|4x-7|$ is non-differentiable at $x=7/4 = 1.75$.At $x=0$,$f(x) = [-3] = -3 
eq 0$,so $g(x)$ is non-differentiable.At $x=1, \sqrt{2}, \sqrt{3}$,$f(x)$ is discontinuous,so $g(x)$ is non-differentiable.At $x=7/4$,$f(x) = [(7/4)^2 - 3] = [49/16 - 3] = [3.0625 - 3] = [0.0625] = 0$. Since $f(7/4) = 0$,the product $g(x) = (|x| + |4x-7|)f(x)$ becomes differentiable at $x=7/4$ because the jump discontinuity of $f$ is multiplied by $0$.Thus,$g(x)$ is non-differentiable at $x \in \{0, 1, \sqrt{2}, \sqrt{3}\}$,which is $4$ points. Option $C$ is correct.

List-$I$	List-$II$
$a$. If $y=\|x\|+\|x-2\|$, then at $x=2, \frac{dy}{dx}=$	$i$. $2$
$b$. If $f(x)=\|\cos 2x\|$, then $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$, where $[\cdot]$ denotes the greatest integer function, then $f^{\prime}(1-)=$	$iii$. $-2$
$d$. If $f(x)=\log\|x-1\|, x \neq 1$, then $f^{\prime}(\frac{1}{2})=$	$iv$. Does not exist

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

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