Q45 / 160 · WBJEE 2009 · Mathematics
General solution of $\sin \mathrm{x}+\cos \mathrm{x}=\min _{a \in I R}\left\{1, a^2-4 a+6\right\}$ is
- A$\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}$
- B$2 n \pi+(-1)^n \frac{\pi}{4}$
- C$n \pi+(-1)^{n+1} \frac{\pi}{4}$
- Correct answerD$n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4}$
Solution
Hints: $\sin x+\cos x=\min _{x \in I R}\left\{1, a^2-4 a+6\right\}$
$$
\begin{aligned}
& a^2-4 a+6=(a-2)^2+2 \therefore \min _{a \in I R}\left(a^2-4 a+6\right)=2 \\
& \therefore \min _{a \in I R}\left\{1, a^2-4 a+6\right\}=\min \{1,2\}=1 \\
& \sin x+\cos x=1 \Rightarrow \frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x=\frac{1}{\sqrt{2}} \\
& \Rightarrow \sin \left(x+\frac{\pi}{4}\right)=\sin \frac{\pi}{4}, \Rightarrow x+\frac{\pi}{4}=n \pi+(-1)^n \cdot \frac{\pi}{4}
\end{aligned}
$$
$$
\Rightarrow x=n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4}
$$
Hence, the correct answer is option: 'D'.