Q41 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

Let $f: \mathrm{A} \rightarrow \mathrm{B}$ be defined as $f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be defined $g(x)=\sqrt{3+4 x-4 x^2}$. If A, B, C are subsets of $\mathbb{R}$ and $f$ is an onto function then the range of the function $f(x)$ is

  • A
    $(-\infty, \infty)$
  • B
    $[0, \infty)$
  • Correct answer
    C
    $\left[-\frac{1}{2}, \frac{3}{2}\right]$
  • D
    $[-1,1]$

Solution

Since $f$ is an onto function Thus domain of $\mathrm{g}(\mathrm{x})$ will be range of $\mathrm{f}(\mathrm{x})$ Now, $3+4 \mathrm{x}-4 \mathrm{x}^2 \geq 0$ to define $\mathrm{g}(\mathrm{x})$ $$ \Rightarrow 4 \mathrm{x}^2-4 \mathrm{x}-3 \mathrm{x} 0 $$ On solving we get $\mathrm{x} \in\left[-\frac{1}{2}, \frac{3}{2}\right]$

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

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