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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Trigonometric Equations

NDA 2016 · Mathematics · Trigonometric Equations

Consider the equation: k x+ 2 x=2 k-7 If the equation possesses...

Consider the equation: ksin⁡x+cos⁡2x=2k−7k \sin x+\cos 2 x=2 k-7 If the equation possesses solution, then what is the minimum value of k ?

Options

  1. A1
  2. B2
  3. C4
  4. D6

Correct answer

B. 2

Step-by-step solution

Ksin⁡x+cos⁡2x=2K−7Ksin⁡x+(1−2sin⁡2x)=(2K−7)2sin⁡2x−Ksin⁡x+(2K−8)=0K \sin x+\cos 2 x=2 K-7K \sin x+\left(1-2 \sin ^2 x\right)=(2 K-7)2 \sin ^2 x-K \sin x+(2 K-8)=0 This is a quadratic equation in sin⁡x\sin x. sin⁡x=−(−K)±K2−4(2)(2K−8)2×2\sin x=\frac{-(-K) \pm \sqrt{K^2-4(2)(2 K-8)}}{2 \times 2} For minimum value of KK, sin⁡x=−1\sin x=-1⇒K±K2−16K+644=−1⇒(±)K2−16K+64=−K−4\begin{aligned} & \Rightarrow \frac{K \pm \sqrt{K^2-16 K+64}}{4}=-1 \\ & \Rightarrow( \pm) \sqrt{K^2-16 K+64}=-K-4 \end{aligned}Squaringbothsides,wegetSquaring both sides, we get\begin{aligned} & K^2-16 K+64=K^2+16+8 K \\ & \Rightarrow 24 K=48 \\ & \Rightarrow K=2 \end{aligned}$$
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