Question

Let $f ( x )=\begin{cases}2 \ln (- x -1), & - e -1 \leq x \leq-2 \\ \frac{1}{2}\left(4- x ^2\right), & -2< x <0 \\ 0, & x =0 \\ \frac{1}{2}\left( x ^2-4\right), & 0< x <2 \\ \frac{2}{2} \ln ( x -1), & 2 \leq x \leq e +1\end{cases}$

and graph of $f(x)$ is as shown

If the equation $g(x)=k$ has exactly two distinct solutions in $[-e-1, e+1]$ then the sum of all possible integral values of $k$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

Correct answer is (d) 3