$f(x)=ax^2+bx+c$ is an even function and $g(x)=px^3+qx^2+rx$ is an odd function. If $h(x)=f(x)+g(x)$ and $h(-2)=0$,then $8p+4q+2r=$

The function $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$ defined as $f(x) = \frac{x}{1 + x^2}$ is

Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2, & x \leq -1 \\ x^2, & -1 < x < 1 \\ 2 - x, & x \geq 1 \end{cases}$. Then the value of $f(-1.75) + f(0.5) + f(1.5)$ is

If $f: R \to R$ is a continuous function such that $|f(x) - f(y)| \geqslant |e^x - e^y|$ for all $x, y \in R$,then $f(x)$ is:

Let the functions $f$ and $g$ be $f: [0, \frac{\pi}{2}] \rightarrow R$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow R$ given by $g(x) = \cos x$,where $R$ is the set of real numbers. Consider the following statements:
Statement $(I)$: $f$ and $g$ are one-one.
Statement $(II)$: $f+g$ is one-one.
Which of the following is correct?

Let $A = \{x \in R : x \text{ is not a positive integer}\}$. Define a function $f : A \to R$ as $f(x) = \frac{2x}{x - 1}$. Then $f$ is