Question

General solution of $\sin x+\cos x=mi{n}_{a\in R}\left\{1,a^2-4a+6\right\}$ is

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

Answer 1

The correct option is D

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

Explanation for the correct option:

Step 1. Find the general solution:

$\sin x+\cos x=mi{n}_{a\in R}\left\{1,a^2-4a+6\right\}$

Let $f\left(a\right)=a^2–4a+6$

$f\text{'}\left(a\right)=2a–4$

If $f\text{'}\left(a\right)=0$

⇒ $2a–4=0$

⇒ $a=2>0$

Now, $f\left(2\right)=4–8+6$

$=2$

$\therefore$min $(1,a^2–4a+6)$$=1$

Thus, $\sin x+\cos x=1$

Step 2. Divide it by √2 on both sides,

⇒ $\frac{1}{\surd 2}\sin x+\frac{1}{\surd 2}\cos x=\frac{1}{\surd 2}$

⇒ $\sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}=\frac{1}{\surd 2}$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\right]$

⇒ $\sin \left(x+\frac{\pi }{4}\right)=\frac{1}{\surd 2}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\left(A+B\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{.}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}\right]$

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

⇒ $x=\mathbf{}\mathit{n}\mathit{\pi }\mathbf{}\mathbf{+}\mathbf{}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}\mathbf{}\frac{\pi }{4}\mathbf{}\mathbf{–}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{4}}$

Hence, Option ‘D’ is Correct.