Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

  • A
    $R - \{-\frac{5}{2}\}$
  • B
    $R$
  • C
    $R - \{-\frac{7}{4}\}$
  • D
    $R - \{-\frac{5}{2}, -\frac{7}{4}\}$

Explore More

Similar Questions

$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?

If $f(x) = (a - x^n)^{1/n},$ where $a > 0$ and $n$ is a positive integer,then $f[f(x)] = $

If $f(x) = \frac{4x+7}{7x-4}$,then the value of $f\{f[f(2)]\} =$ ?

For $x \in R$,two real-valued functions $f(x)$ and $g(x)$ are such that $g(x) = \sqrt{x} + 1$ and $(f \circ g)(x) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to:

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$. Then,the number of points in $R$ where $(f \circ g)(x)$ is $NOT$ differentiable is equal to

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