Let $f(x) = \sin x$ and $g(x) = \ln |x|$. If the ranges of the composite functions $fog$ and $gof$ are $R_1$ and $R_2$ respectively,then:

  • A
    $R_1 = \{ u: - 1 < u < 1\} , R_2 = \{ v: - \infty < v < 0\}$
  • B
    $R_1 = \{ u: - \infty < u \le 0\} , R_2 = \{ v: - 1 \le v \le 1\}$
  • C
    $R_1 = \{ u: - 1 < u < 1\} , R_2 = \{ v: - \infty < v < 0\}$
  • D
    $R_1 = \{ u: - 1 \le u \le 1\} , R_2 = \{ v: - \infty < v \le 0\}$

Explore More

Similar Questions

Let $f(x) = \sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)$ for all $x \in R$ and $g(x) = \frac{\pi}{2} \sin x$ for all $x \in R$. Let $(f \circ g)(x)$ denote $f(g(x))$ and $(g \circ f)(x)$ denote $g(f(x))$. Then which of the following is (are) true?
$(A)$ Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(B)$ Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$(C)$ $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}$
$(D)$ There is an $x \in R$ such that $(g \circ f)(x) = 1$

If $f: R \rightarrow R$ is defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ is defined by $g(x)=x^{2}$, then the mapping $(g \circ f): R \rightarrow R$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in R$. Which of the following is true?

The relation $R$ is defined as $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then $R^{-1} o R$ is:

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

Difficult
View Solution

If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3x-2$ and $g(x)=x^2+2$, then $[(g \circ f)+(f \circ g)](x) = $

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