Let $f: R \rightarrow R$ be defined by $f(x)=3 x^2-5$ and $g: R \rightarrow R$ by $g(x)=\frac{x}{x^2+1}$,then $g \circ f$ is
- A$\frac{3 x^2-5}{9 x^4-6 x^2+26}$
- B$\frac{3 x^2}{x^4+2 x^2-4}$
- C$\frac{3 x^2}{9 x^4+30 x^2-2}$
- D$\frac{3 x^2-5}{9 x^4-30 x^2+26}$
Explore More
Similar Questions
If $f(x) = \frac{1+x}{1-x}$ where $x \neq 1$,then $f(x) \cdot f(y) = $ . . . . . . .
If $f(x) = e^{2x}$ and $g(x) = \log \sqrt{x}$ $(x > 0)$,then $fog(x)$ is equal to
Medium
View Solution$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=x^2$ and $g(x)=\frac{1}{x^2}$,then $x^4(f \circ g)(x)$ is equal to
$[x]$ represents the greatest integer function. Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -3, & x < 0 \\ 0, & x = 0 \\ 5, & x > 0 \end{cases}$. Then $f(g(x))$ is:
If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in \mathbb{Z} \\ 0, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x)) = $
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo