If $f(x)$ is the signum function,then in terms of $f(x)$,the constant function $g(x)=1, \forall x \in R$ is

  • A
    $g(x)= \begin{cases}2-f(x), & x < 0 \\ f(x), & x \geq 0\end{cases}$
  • B
    $g(x)= \begin{cases}f(x)+f(-x), & x < 0 \\ f(x) f(-x), & x \geq 0\end{cases}$
  • C
    $g(x)= \begin{cases}1+f(x), & x>0 \\ 1-f(x), & x \leq 0\end{cases}$
  • D
    $g(x)= \begin{cases}f(x)+2, & x < 0 \\ 1+f(x), & x=0 \\ f(x), & x>0\end{cases}$

Explore More

Similar Questions

Define $f: R \rightarrow R$ by $f(x) = \max \{x+1, 1-x, 2\}$. Then,$f$ is

The function $f: (-\infty, \infty) \rightarrow (-\infty, \infty)$ defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is :

The function $f: R \to R$ defined by $f(x) = x|x| + x^3|x|$ is

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

Let $A = \{x \in R \mid x \text{ is not a positive integer}\}$. Let a function $f$ be defined as $f: A \rightarrow R$ such that $f(x) = \frac{2x}{x-1}$. Then $f$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo