Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} x^2 - 4x + 3, & \text{if } x < 2 \\ x - 3, & \text{if } x \geq 2 \end{cases}$. Then the number of real numbers $x$ for which $f(x) = 8$ is:

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

Explore More

Similar Questions

Are $f$ and $g$ both necessarily onto,if $g \circ f$ is onto?

Let $A = \{x \in R \mid x \text{ is not a positive integer}\}$. Let a function $f$ be defined as $f: A \rightarrow R$ such that $f(x) = \frac{2x}{x-1}$. Then $f$ is:

The functions which map $[-1, 1]$ to $[0, 2]$ are

Difficult
View Solution

Let $f, g: N - \{1\} \rightarrow N$ be functions defined by $f(a) = \alpha$,where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$,and $g(a) = a + 1$,for all $a \in N - \{1\}$. Then,the function $f + g$ is.

If a function $f: R-\{l\} \to R-\{m\}$ defined by $f(x) = \frac{x+3}{x-2}$ is a bijection,then $3l - 2m =$

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