Q108 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

The function $f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & x \neq 0 \\ 2, & x \neq 0\end{array}\right.$

  • A
    is continuous for $\forall x \in R$
  • Correct answer
    B
    has maximum value 2
  • C
    has neither minimum nor maximum
  • D
    has minimum value 2

Solution

$f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & x \neq 0 \\ 2, & x=0\end{array}=\left\{\begin{array}{cc}2, & x \lt 0 \\ 0, & x\gt0 \\ 2, & x=0\end{array}\right.\right.$ $\therefore$ Maximum value of $f(x)$ is 2 .

Hence, the correct answer is option: 'B'.

Relevant Questions