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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2024 · Paper 1 · Mathematics · Trigonometric Equations

Let (π)/(2) < x < π be such that x= -5 11 . Then ( (11 x)/(2))( 6...

Let π2<x<π\frac{\pi}{2} < x < \pi be such that cot⁡x=−511\cot x=\frac{-5}{\sqrt{11}}. Then \left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) is equal to

Options

  1. A11−123\frac{\sqrt{11}-1}{2 \sqrt{3}}
  2. B11+123\frac{\sqrt{11}+1}{2 \sqrt{3}}
  3. C11+132\frac{\sqrt{11}+1}{3 \sqrt{2}}
  4. D11−132\frac{\sqrt{11}-1}{3 \sqrt{2}}

Correct answer

B. 11+123\frac{\sqrt{11}+1}{2 \sqrt{3}}

Step-by-step solution

Given, cot⁡x=−5111−tan⁡2x22tan⁡x2=−511tan⁡x2=11,−111So, tan⁡x2=11, As π4<x2<π2\begin{aligned} & \cot x=-\frac{5}{\sqrt{11}} \\ & \frac{1-\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}}=-\frac{5}{\sqrt{11}} \\ & \tan \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}} \\ & \text{So, } \tan \frac{x}{2}=\sqrt{11}, \text{ As } \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}\end{aligned}As, x∈(π2,π)Now, (sin⁡11x2)(sin⁡6x−cos⁡6x)+(cos⁡11x2)(sin⁡6x+cos⁡6x)={sin⁡6xsin⁡11x2+cos⁡11x2cos⁡6x}=cos⁡(6x−11x2)+sin⁡(6x−11x2)=cos⁡x2+sin⁡x2=123+1123=11+123⇒ Option (2) is correct. \begin{aligned} & \text{As, } x \in\left(\frac{\pi}{2}, \pi\right) \\ & \text{Now, } \left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) \\ & =\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x \right\} \\ & =\cos \left(6 x-\frac{11 x}{2}\right)+\sin \left(6 x-\frac{11 x}{2}\right) \\ & =\cos \frac{x}{2}+\sin \frac{x}{2} \\ & =\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}} \\ & =\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow \text{ Option (2) is correct. }\end{aligned}$
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