Question

The set of all real values of $x$ for which the function $f\left(x\right)=\sqrt{\sin x+\cos x}+\sqrt{7x-x^2-6}$ takes real values is

• A. $[1,\frac{3\pi }{4}]\cup [\frac{7\pi }{4},6]$
• B. $[1,6]$
• C. $[1,\frac{\pi }{4}]\cup [\frac{7\pi }{4},6]$
• D. $[1,6]-\{\frac{3\pi }{4},\frac{7\pi }{4}\}$

Answer 1

The correct option is A $[1,\frac{3\pi }{4}]\cup [\frac{7\pi }{4},6]$

Given : $f\left(x\right)=\sqrt{\sin x+\cos x}+\sqrt{7x-x^2-6}$
$f\left(x\right)=\sqrt{\sin (\frac{\pi }{4}+x)}\cdot 2^{1/4}+\sqrt{7x-x^2-6}$
Now,
$\sin (\frac{\pi }{4}+x)\ge 0\Rightarrow \frac{\pi }{4}+x\in [2n\pi +0,2n\pi +\pi ],n\in Z\Rightarrow x\in [2n\pi -\frac{\pi }{4},2n\pi +\frac{3\pi }{4}]\cdots \left(1\right)$
And
$7x-x^2-6\ge 0\Rightarrow x^2-7x+6\le 0\Rightarrow (x-1)(x-6)\le 0\Rightarrow x\in [1,6]\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get the domain of $f\left(x\right)$
$x\in [1,\frac{3\pi }{4}]\cup [\frac{7\pi }{4},6]$