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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2010 · Paper 1 · Mathematics · Application of Derivatives

The maximum value of the expression (1)/( ^2 θ+3 θ θ+5 ^2 θ) is

The maximum value of the expression 1sin⁡2θ+3sin⁡θcos⁡θ+5cos⁡2θ\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta} is

Correct answer

2

Step-by-step solution

Let f(θ)=1sin⁡2θ+3sin⁡θcos⁡θ+5cos⁡2θ Again let g(θ)=sin⁡2θ+3sin⁡θcos⁡θ+5cos⁡2θ=1−cos⁡2θ2+5(1+cos⁡2θ2)+32sin⁡2θ=3+2cos⁡2θ+32sin⁡2θ∴g(θ)min⁡=3−4+94=3−52=12∴f(θ)=1g(θ)min⁡=2\begin{aligned}& f(\theta)=\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta}\\ & \text{ Again let }\\ & g(\theta)=\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta\\ & =\frac{1-\cos 2 \theta}{2}+5\left(\frac{1+\cos 2 \theta}{2}\right)+\frac{3}{2} \sin2 \theta\\ & =3+2 \cos 2 \theta+\frac{3}{2} \sin2 \theta\\ & \therefore \quad g(\theta)_{\min }=3-\sqrt{4+\frac{9}{4}}=3-\frac{5}{2}=\frac{1}{2}\\ & \therefore \quad f(\theta)=\frac{1}{g(\theta)_{\min }}=2 \end{aligned}
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