Q87 / 180 · KCET 2021 · Mathematics
The function $f(x)=\sqrt{3} \sin 2 x-\cos 2 x+4$ is one-one in the interval
- Correct answerA$\left[\frac{-\pi}{6}, \frac{\pi}{3}\right]$
- B$\left[\frac{\pi}{6}, \frac{-\pi}{3}\right]$
- C$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
- D$\left[\frac{-\pi}{6}, \frac{-\pi}{3}\right]$
Solution
Given function,
$f(x)=\sqrt{3} \sin 2 x-\cos 2 x+4$
$\begin{aligned}
f(x) &=2\left(\sin 2 x \cdot \frac{\sqrt{3}}{2}-\cos 2 x \cdot \frac{1}{2}\right)+4 \\
&=2\left(\sin 2 x \cdot \cos \frac{\pi}{6}-\cos 2 x \cdot \sin \frac{\pi}{6}\right)+4 \\
&=2\left[\sin \left(2 x-\frac{\pi}{6}\right)\right]+4
\end{aligned}$
$f$ is one-one.
$\begin{array}{ll}
& -\frac{\pi}{2} \leq 2 x-\frac{\pi}{6} \leq \frac{\pi}{2} \\
\Rightarrow & -\frac{\pi}{2}+\frac{\pi}{6} \leq 2 x \leq \frac{\pi}{2}+\frac{\pi}{6} \\
\Rightarrow & \frac{-3 \pi+\pi}{6} \leq 2 x \leq \frac{4 \pi}{6} \Rightarrow-\frac{\pi}{6} \leq x \leq \frac{\pi}{3} \\
\therefore & x \in\left[-\frac{\pi}{6}, \frac{\pi}{3}\right]
\end{array}$
Hence, the correct answer is option: 'A'.