Let $(g \circ f)(x) = \sin x$ and $(f \circ g)(x) = (\sin \sqrt{x})^2$. Then,

  • A
    $f(x) = \sin^2 x, g(x) = x$
  • B
    $f(x) = \sin \sqrt{x}, g(x) = \sqrt{x}$
  • C
    $f(x) = \sin^2 x, g(x) = \sqrt{x}$
  • D
    $f(x) = \sin \sqrt{x}, g(x) = x^2$

Explore More

Similar Questions

If $f: R \rightarrow R$ is defined as $f(x) = (2020 - x^{2019})^{1 / 2019}$,$\forall x \in R$,find $(f \circ f \circ f \circ f) \left( \frac{2019}{2020} \right)$.

Let $f'(x) > 0$ and $g'(x) < 0$ for all $x \in \mathbb{R}$. Then which of the following is true?

Difficult
View Solution

If $f : R \to R$ and $g : R \to R$ are defined as $f(x) = 2x - |x|$ and $g(x) = 2x + |x|$, then

Define the functions $f, g$ and $h$ from $R$ to $R$ such that $f(x) = x^2 - 1, g(x) = \sqrt{x^2 + 1}$ and $h(x) = \begin{cases} 0, & x \leq 0 \\ x, & x \geq 0 \end{cases}$ Consider the following statements:

If $f(x) = \begin{cases} \sin x, & x \ne n\pi, n \in \mathbb{Z} \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \ne 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x))$ 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