Q. General solution of $\sin{x + \cos{x = \begin{matrix} \min \\ a \in R \end{matrix}}\{ 1,\ a^{2} - 4a + 6\}}$ is
A) $\frac{n\pi}{2} + ( - 1)^{n}\frac{\pi}{4}$
B) $2n\pi + ( - 1)^{n}\frac{\pi}{4}$
C) $n\pi + ( - 1)^{n + 1}\frac{\pi}{4}$
D) $n\pi + ( - 1)^{n}\frac{\pi}{4} - \frac{\pi}{4}$
Explanation
(d)
Given that,
$\sin{x + \cos{x = \begin{matrix}
\min \\
a \in R
\end{matrix}}}\{ 1,\ a^{2} - 4a + 6\}$
Now, $a^{2} - 4a + 6 = (a - 2)^{2} + 2$
$\therefore\ \ \ \begin{matrix}
\min \\
a \in R
\end{matrix}\left\{ 1,\ a^{2} - 4a + 6 \right\} = \min{\left\{ 1,\ 2 \right\} = 1}$
$\therefore\ \ \sin{x + \cos{= 1}}$
$\Rightarrow \ \ \sin{\left( x + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}}$
$\Rightarrow \ \ x + \frac{\pi}{4} = n\pi + ( - 1)^{n}.\frac{\pi}{4}$
$\Rightarrow \ \ x = n\pi + ( - 1)^{n}\frac{\pi}{4} - \frac{\pi}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.