Let $f (x) = | | x | - | x^{2} - 1 | |$

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The absolute difference between the number of points of local minima and local maxima is 1

The number of points at which f is not differentiable is 7

f(x) = k will have 8 distinct real roots if $k \in (0, 1)$

f(x) = k will have 8 distinct real roots if $k \in (1, \infty)$

answer is A, B, C.

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Detailed Solution

We have $| x | = {\begin{cases} - x, & x < 0 \\ x, & x \geq 0 \end{cases}$ and $| x^{2} - 1 | = {\begin{cases} 1 - x^{2}, & - 1 \leq x \leq 1 \\ x^{2} - 1, & x \leq - 1 or x \geq 1 \end{cases}$
Since f is even function, we discuss in $(0, \infty)$
Now $f (x) = {\begin{matrix} | x - 1 + x^{2} | & 0 \leq x \leq 1 \\ | x - x^{2} + 1 | & x \geq 1 \end{matrix} = {\begin{matrix} | x^{2} + x - 1 | & 0 \leq x < 1 \\ x^{2} - x + 1 & x \geq 1 \end{matrix}$
$= {\begin{matrix} - (x^{2} + x - 1) & 0 \leq x < \frac{\sqrt{5} - 1}{2} \\ x^{2} + x - 1 & \frac{\sqrt{5} - 1}{2} \leq x < \frac{\sqrt{5} + 1}{2} \\ - (x^{2} + x - 1) & \frac{\sqrt{5} + 1}{2} \leq x < 1 \\ x^{2} - x - 1 & x \geq 1 \end{matrix}$
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f has 4 points of local minima and 3 points of local maxima.