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

Which one of the following functions is a bijection?

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

Let $R$ be the set of all real numbers. Statement $I$: The function $f: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ defined by $f(x) = \sec x + \tan x$ is a one-one function. Statement $II$: The function $f: [0, \infty) \rightarrow R$ defined by $f(x) = x^2$ is a one-one function. Which of the above statements is(are) true?

If $f: R \rightarrow R$ is defined by $f(x) = x + 2|x + 1| + 2|x - 1|$,then the element in the co-domain,which has a unique pre-image in the domain is

If $f(x)=|x-1|+|x-2|+|x-3|$ for $2 < x < 3$,then $f$ is