Question

The domain of the function $y=\sqrt{\sin x+\cos x}+\sqrt{7x-x^2-6}$ is $[p,\frac{q\pi }{4}]\cup [\frac{r\pi }{4},s]$ then value of $p+q+r+s$ is

Answer 1

It is given that : $y=\sqrt{sinx+cosx}+\sqrt{7x-x^2-6}$

For $y$ to be real,

$(sinx+cosx)\ge 0$

In I Quad, sin x and cos x both are positive.

So, $0\le x\le \pi 2$

In II Quad, sin x is positive while cos x is negative.

$sinx\ge cosx$ So, $\pi /2\le x\le 3\pi /4$

In III Quad, $sinx$ and $cosx$ both are negative.

So, no value of $x$.

In IV Quad, sin x is negative while cos x is positive.

$sinx\le cosx$ So, $7\pi /4\le x\le 2\pi$

$x\in [0,3\pi /4]\cup [7\pi /4,2\pi ]$-----(a)

Now,

$(7x-x^2-6)\ge 0$

$(x-1)(6-x)$ $\ge 0$

This is possible when $1\le x\le 6$-----(b)

From (a) and (b),

$x\in [1,3\pi /4]\cup [7\pi /4,6]$

On comparing with given equation, we got

$p$ $=$ $1$; $q$ $=$ $3$; $r$ $=$ $7$; $s$ $=$ $6$

So, $p+q+r+s=1+3+7+6$ $=17$

$Answer=17$