Suppose $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{cases}$. Then the set $\{x \in [-2, 2] : x \leq 0 \text{ and } f(|x|) = x\}$ is equal to

Suppose $f:[-2, 2] \to R$ is defined by $f(x) = \begin{cases} -1 & \text{for } -2 \le x \le 0 \\ x - 1 & \text{for } 0 < x \le 2 \end{cases}$,then find the set $\{ x \in (-2, 2) : x \le 0 \text{ and } f(|x|) = x \}$.

Let $f(x) = x^{2}$ and $g(x) = 2x + 1$ be two real functions. Find $(f+g)(x)$,$(f-g)(x)$,$(fg)(x)$,and $(\frac{f}{g})(x)$.

If $f : R \rightarrow R$ is defined by $f(x) = \begin{cases} x + 4, & x < -4 \\ 3x + 2, & -4 \leq x < 4 \\ x - 4, & x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is :
List-$I$
$(A) f(-5) + f(-4)$
$(B) f(|f(-8)|)$
$(C) f(f(-7) + f(3))$
$(D) f(f(f(f(0)))) + 1$
List-$II$
$(i) 14$
$(ii) 4$
$(iii) -11$
$(iv) -1$
$(v) 1$
$(vi) 0$

If $f$ is an even function defined on the interval $(-5, 5),$ then four real values of $x$ satisfying the equation $f(x) = f\left( \frac{x + 1}{x + 2} \right)$ are

Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 2, 3, 4, 5, 6\}$. Then the number of functions $f: A \rightarrow B$ satisfying $f(1) + f(2) = f(4) - 1$ is equal to