For $a \in \mathrm{R}-\{0\}$, if $a \cos x+a \sin x+a=2 K+1$ has a solution, then $K$ lies in the interval

For $a \in \mathrm{R}-\{0\}$, if $a \cos x+a \sin x+a=2 K+1$ has a solution, then $K$ lies in the interval
  1. $\left[\frac{a-1-a \sqrt{2}}{2}, \frac{a-1+a \sqrt{2}}{2}\right]$
  2. $\left[\frac{a+1-\sqrt{2}}{2}, \frac{a+1+\sqrt{2}}{2}\right]$
  3. $\left[\frac{a-1-\sqrt{2}}{2}, \frac{a-1+\sqrt{2}}{2}\right]$
  4. $\left[-\frac{\sqrt{2 a^2+2 a+1}+1}{2}, \frac{\left(\sqrt{2 a^2+2 a+1}-1\right)}{2}\right]$

Solution

$\begin{aligned} & \text { } a \cos x+a \sin x+a=2 K+1 \\ & a\left[\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+1\right]=2 K+1 \\ & a(1-\sqrt{2}) \leq a\left[\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+1\right] \leq a(1+\sqrt{2}) \\ & a-a \sqrt{2} \leq 2 K+1 \leq a(1+\sqrt{2}) \\ & \frac{a-a \sqrt{2}-1}{2} \leq K \leq \frac{a+a \sqrt{2}-1}{2}\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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