If $f(x) = e^{2x}$ and $g(x) = \log \sqrt{x}$ $(x > 0)$,then $fog(x)$ is equal to

  • A
    $e^{2x}$
  • B
    $\log \sqrt{x}$
  • C
    $e^{2x} \log \sqrt{x}$
  • D
    $x$

Explore More

Similar Questions

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=5x^2-2$,then the least value of the function $(g \circ f)(x)$ is

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

If $g(x) = x^2 + x - 2$ and $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,then $f(x)$ is

Difficult
View Solution

Two functions $f: R \rightarrow R, g: R \rightarrow R$ are defined as follows: $f(x) = \begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} -1, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases}$. Then,$(f \circ g)(\pi) + (g \circ f)(e)$ is equal to:

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2x+3$ and $g(x)=x^2+7$, then the values of $x$ such that $g(f(x))=8$ are

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