Q61 / 180 · KCET 2015 · Mathematics

\[ f(x)=\frac{1}{2}-\tan \left(\frac{\Pi x}{2}\right)-1 < x < 1 \text { and } g(x)=\sqrt{\left(3+4 x-4 x^{2}\right)} \]
Find domain of \( (f+g) \)

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

Solution

Given that, $f(x)=\frac{1}{2}-\tan ^{-1}\left(\frac{\Pi x}{2}\right)$ and $g(x)=\sqrt{3+4 x-4 x^{2}}$ Since, $f x()$ is defined in $(-1,1)$ So, $g(x)$ to defined, we have $3+4 x-4 x^{2} \geq 0$ $4 x^{2}-4 x-3 \leq 0$ $\Rightarrow 4 x^{2}-4 x+1-4 \leq 0$ $\Rightarrow(2 x-1)^{2}-4 \leq 0$ $\Rightarrow(2 x-1)^{2} \leq 4$ $\Rightarrow-2 \leq 2 x-1 \leq 2$ $\Rightarrow-1 \leq 2 x \leq 3$ $\Rightarrow-\frac{1}{2} \leq x \leq \frac{3}{2}$ Therefore, domain of $(f+g)=\left(-\frac{1}{2}, 1\right)$

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

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