Question

Solve for $x$, which satisfy $sinx+cosx=\underset{aϵR}{min}\{1,a^2-4a+6\}$

• A. $x=n\pi +(-1{)}^n\frac{\pi }{4}+\frac{\pi }{4}$
• B. $x=n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$
• C. $x=n\pi \pm \frac{\pi }{4}-\frac{\pi }{4}$
• D. $x=n\pi -(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$

Answer 1

The correct option is B $x=n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$

Let $f\left(a\right)=a^2-4a+6$
Now
$f^{\prime }\left(a\right)=2a-4$
Or
$a=2$
$f"\left(2\right)=2$
Hence minima.
Hence
$f\left(2\right)=4-8+6$
$=2$.
Thus $min1,a^2-4a+6$ is 1.
Now $cosx+sinx=1$ Implies
$\sqrt{2}\left[sin\right(x+\frac{\pi }{4}\left)\right]=1$
Or
$sin(x+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$
Or
$x+\frac{\pi }{4}=2k\pi +\frac{\pi }{4}$ and
$x+\frac{\pi }{4}=(2k+1)\pi -\frac{\pi }{4}$
Therefore $x=2k\pi$ and $x=(2k+1)\pi -\frac{\pi }{2}$.
Hence
$x=n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$.