SATHEE: Application of Derivatives 4 Question 65

Application of Derivatives 4 Question 65

68. The maximum value of the expression $\frac{1}{\sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ}$ is ……

(2010)

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Solution:

$f (θ) = \frac{1}{\sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ}$

Again let, $g (θ) = \sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ$

$\begin{aligned} = \frac{1 - \cos 2 θ}{2} + 5 \frac{1 + \cos 2 θ}{2} + \frac{3}{2} \sin 2 θ \\ = 3 + 2 \cos 2 θ + \frac{3}{2} \sin 2 θ \\ ∴ g (θ)_{min} & = 3 - \sqrt{4 + \frac{9}{4}} \\ = 3 - \frac{5}{2} = \frac{1}{2} \end{aligned}$

$∴$ Maximum value of $f (θ) = \frac{1}{1 / 2} = 2$

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Application of Derivatives 4 Question 65

68. The maximum value of the expression 1sin2⁡θ+3sin⁡θcos⁡θ+5cos2⁡θ is ……

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68. The maximum value of the expression $\frac{1}{\sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ}$ is ……