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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

Let

a

be maximum value of

(3\cos\theta - 4\sin\theta)

\theta \neq \frac{n\pi}{2}

\alpha = a \sin^{2}\theta \cdot \cos^{3}\theta

\beta = a \sin^{3}\theta \cdot \cos^{2}\theta

\sqrt{ \frac{(\alpha^{2}+\beta^{2})^{5}}{(\alpha\beta)^{4}} } =

$\frac{ \frac{5\sin\theta}{2\cos^{2}\theta} }{2}$
$-3\sin\theta$
$5$
$16$

Solution:

Maximum value of

3\cos\theta - 4\sin\theta

\therefore

a = \sqrt{3^{2} + (-4)^{2}} = 5

\alpha = 5 \sin^{2}\theta \cos^{3}\theta

\beta = 5 \sin^{3}\theta \cos^{2}\theta

\text{ Now, } \alpha^{2}+\beta^{2} = 5^{2}(\sin^{4}\theta \cos^{6}\theta + \sin^{6}\theta \cos^{4}\theta)

= 25(\cos^{2}\theta + \sin^{2}\theta) \sin^{4}\theta \cos^{4}\theta

= 25 \sin^{4}\theta \cos^{4}\theta

\therefore (\alpha^{2}+\beta^{2})^{5} = (25 \sin^{4}\theta \cos^{4}\theta)^{5}

\Rightarrow (\alpha^{2}+\beta^{2})^{5/2} = (5 \sin^{2}\theta \cos^{2}\theta)^{5}

\alpha\beta = 25 \sin^{5}\theta \cos^{5}\theta

\Rightarrow \sqrt{ \frac{(\alpha^{2}+\beta^{2})^{5}}{(\alpha\beta)^{4}} } = \frac{(5 \sin^{2}\theta \cos^{2}\theta)^{5}}{(5^{2} \sin^{5}\theta \cos^{5}\theta)^{2}}

= \frac{5^{5} \sin^{10}\theta \cos^{10}\theta}{5^{4} \sin^{10}\theta \cos^{10}\theta} = 5

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