Question

General solution as sin $x + \cos x = min [1, a^2 - 4a + 6]$ is $a \in R$

• A. $n\frac{\pi}{2}+(-1)^n\frac{\pi}{4}$
• B. $2n\pi2+(-1)^n\frac{\pi}{4}$
• C. $n\pi+(-1)^{n+1}\frac{\pi}{4}$
• D. $n\pi+(-1)^{n+1}\frac{\pi}{4}-\frac{\pi}{4}$

Answer 1

Answer: (D)

$n\pi+(-1)^{n+1}\frac{\pi}{4}-\frac{\pi}{4}$

Answer 2

Given that, \sin x + \cos {x =\underset{\text{a $\in$ R }}{ {min}}} $\\{ 1 , a^2 - 4a +6\\}$ Now, $a^2 - 4a + 6 = ( a - 2)^2 + 2$ $\therefore$ {\underset{\text{a $\in$ R }}{ {min}}} $\\{ 1 , a^2 - 4a +6\\}$ = min {1,2} = 1 $\therefore$ $\sin x + \cos x= 1$ $\Rightarrow$ $\sin\left( x+ \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$ $\Rightarrow$ $x+ \frac{\pi}{4} = n \pi + (-1)^n \, \frac{\pi}{4}$ $\Rightarrow$ $x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$