Let f:left[-frac{1}{2}, 2right] rightarrow R | vedclass

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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.

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