If $f(x) = \frac{x - 3}{x + 1}$,then $f[f\{f(x)\}]$ equals

  • A
    $x$
  • B
    $-x$
  • C
    $\frac{x}{2}$
  • D
    $-\frac{1}{x}$

Explore More

Similar Questions

If $f: R \rightarrow R$ is given by $f(x) = (3 - x^{3})^{\frac{1}{3}}$,then $fof(x)$ is ..........

Let $f(x)=3+2x$ and $g_n(x)=(f \circ f \circ f \circ \dots \text{n times})(x)$. For all $n \in N$,if all the lines $y=g_n(x)$ pass through a fixed point $(\alpha, \beta)$,then $\alpha+\beta=$

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 $f(x) = \frac{x}{2-x}$ and $g(x) = \frac{x+1}{x+2}$,then $(g \circ g \circ f)(x) = $

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

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