Question

Find domain of the given function:

$f\left(x\right)=\frac{\sqrt{\cos x-\frac{1}{2}}}{\sqrt{6+35x-6x^2}}$

• A. $(-\frac{1}{6},\frac{\pi }{3})\cup \left(\frac{5\pi }{3},6\right)$
• B. $[-\frac{\pi }{3},\frac{\pi }{3}]\cup \left[\frac{5\pi }{3},6\right)$
• C. $xϵ\left[-\frac{\pi }{3},\frac{\pi }{3}\right]\cup [\frac{5\pi }{3},\frac{7\pi }{3}]$
• D. None of these

Answer 1

The correct option is B $[-\frac{\pi }{3},\frac{\pi }{3}]\cup \left[\frac{5\pi }{3},6\right)$

$f\left(x\right)=\frac{\sqrt{\cos x-\frac{1}{2}}}{\sqrt{6+35x-6x^2}}$
For $f\left(x\right)$ to be defined $\cos x-\frac{1}{2}\ge 0$
$\Rightarrow x\in \left[-\frac{\pi }{3},\frac{\pi }{3}\right]\cup [\frac{5\pi }{3},\frac{7\pi }{3}]$
Also $6x^2-35x-6<0\Rightarrow -\frac{1}{6}<x<6$
Thus combining above two we get required domain $\Rightarrow x\in [-\frac{\pi }{3},\frac{\pi }{3}]\cup \left[\frac{5\pi }{3},6\right)$