Consider $f(x) = \{x^2-1\}[|x|]$, then:
[<b>Note:</b> Where $\{y\}$ denotes fractional part function and $[y]$ denotes greatest integer function less than or equal to $y$.]
number of points where $f$ is discontinuous in $[-2, 2]$ is $4$
number of points where $f$ is discontinuous in $[-2, 2]$ is $6$
number of solution(s) of the equation $2f(x) = |x|$ is $5$
number of solution(s) of the equation $2f(x) = |x|$ is more than $5$
Step-by-Step Solution
Step 1: Analyze $f(x) = \{x^2-1\}[|x|]$ on $[-2,2]$. Note $[|x|]$ takes values $0$ for $|x|\in[0,1)$, $1$ for $|x|\in[1,2)$, $2$ for $|x|=2$.
Step 2: For $x\in(-1,1)$: $[|x|]=0$, so $f(x)=0$. No discontinuity from $[|x|]$ here, but $\{x^2-1\}$ could be discontinuous when $x^2-1$ is an integer, i.e., $x^2=0,1,2,...$ In $(-1,1)$: $x^2-1\in(-1,0)$, so $\{x^2-1\}=x^2-1+1=x^2$, continuous. $f(x)=0$ throughout.
Step 3: For $x\in[1,2)$: $[|x|]=1$, so $f(x)=\{x^2-1\}$. Discontinuities of $\{x^2-1\}$ occur when $x^2-1$ is an integer: $x^2=1\Rightarrow x=1$, $x^2=2\Rightarrow x=\sqrt{2}$, $x^2=3\Rightarrow x=\sqrt{3}$. All three are in $[1,2)$.
Step 4: For $x\in(-2,-1]$: $[|x|]=1$, so $f(x)=\{x^2-1\}$. Same discontinuities at $x=-1,-\sqrt{2},-\sqrt{3}$.
Step 5: At $x=\pm 2$: $[|x|]=2$, $f(\pm 2)=\{4-1\}\cdot 2=\{3\}\cdot 2=0$. Check continuity from left at $x=2$: $\lim_{x\to 2^-}\{x^2-1\}\cdot 1 = \{3^-\}=1^-\neq 0$. Discontinuity at $x=2$ and $x=-2$.
Step 6: Count discontinuities in $[-2,2]$: at $x=\pm 1, \pm\sqrt{2}, \pm\sqrt{3}$ (from $\{x^2-1\}$ integer jumps) and at $x=\pm 2$ (from $[|x|]$ jump). But at $x=\pm 1$: $[|x|]$ also jumps from $0$ to $1$. Check: from left of $x=1$, $f\to 0$; at $x=1$, $f(1)=\{0\}\cdot 1=0$; from right, $f\to\{0^+\}\cdot 1=0^+$. So $x=1$ may be continuous. Re-examine: $\{x^2-1\}$ at $x=1^+$: $x^2-1=0^+$, $\{0^+\}=0^+$. At $x=1^-$: $[|x|]=0$, $f=0$. So $f$ is continuous at $x=\pm 1$.
Discontinuities: $x=\pm\sqrt{2}, \pm\sqrt{3}, \pm 2$ — total $6$. Option (b) is correct.
Step 7: Solve $2f(x)=|x|$. For $x\in(-1,1)$: $2\cdot 0=|x|\Rightarrow x=0$. One solution.
For $x\in[1,2)$: $2\{x^2-1\}=x$. On $[1,\sqrt{2})$: $\{x^2-1\}=x^2-1$, so $2(x^2-1)=x\Rightarrow 2x^2-x-2=0\Rightarrow x=\frac{1+\sqrt{17}}{4}\approx 1.28$. One solution.
On $[\sqrt{2},\sqrt{3})$: $\{x^2-1\}=x^2-2$, so $2(x^2-2)=x\Rightarrow 2x^2-x-4=0\Rightarrow x=\frac{1+\sqrt{33}}{4}\approx 1.69$. One solution.
On $[\sqrt{3},2)$: $\{x^2-1\}=x^2-3$, so $2(x^2-3)=x\Rightarrow 2x^2-x-6=0\Rightarrow x=\frac{1+\sqrt{49}}{4}=2$. Not in $[\sqrt{3},2)$.
By symmetry for $x\in(-2,-1]$: $2\{x^2-1\}=-x$ (since $|x|=-x$). Same analysis gives $2$ solutions.
Total: $1+1+1+2=5$ solutions. Option (c) is correct.
Correct Answer: 2, 3