The real valued function $f: \mathrm{R} \rightarrow\left[\frac{5}{2}, \infty\right)$ defined by $f(x)=|2…

The real valued function $f: \mathrm{R} \rightarrow\left[\frac{5}{2}, \infty\right)$ defined by $f(x)=|2 x+1|+1|x-2|$ is
  1. One-one function but not onto
  2. Onto function but not one-one
  3. Bijection
  4. Neither one-one function nor onto

Solution

$f: \mathbb{R} \rightarrow\left[\frac{5}{2}, \infty\right), f(x)=|2 x+1|+|x-2|$
$\rightarrow f(x)=\left\{\begin{array}{lc}-3 x+1, & x \lt \frac{-1}{2} \\ x+3, & \frac{-1}{2} \lt x \lt 2 \\ 3 x-1, & x\gt2\end{array}\right.$
Clearly, any horizontal line will cut the graph atleast twice. $\therefore f(x)$ is not one-one and clearly, $f(x)$ is onto.

Asked in: AP EAMCET 2024 (21 May Shift 2)

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