AP EAMCET · Maths · Trigonometric Ratios & Identities
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
- A \(\left[\frac{a-1-a \sqrt{2}}{2}, \frac{a-1+a \sqrt{2}}{2}\right]\)
- B \(\left[\frac{a+1-\sqrt{2}}{2}, \frac{a+1+\sqrt{2}}{2}\right]\)
- C \(\left[\frac{a-1-\sqrt{2}}{2}, \frac{a-1+\sqrt{2}}{2}\right]\)
- D \(\left[-\frac{\sqrt{2 a^2+2 a+1}+1}{2}, \frac{\left(\sqrt{2 a^2+2 a+1}-1\right)}{2}\right]\)
Answer & Solution
Correct Answer
(A) \(\left[\frac{a-1-a \sqrt{2}}{2}, \frac{a-1+a \sqrt{2}}{2}\right]\)
Step-by-step Solution
Detailed explanation
\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}) \\ &…
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