Let a be the maximum value of (3 cos theta - | vedclass

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(C) The maximum value of $3 \cos 	heta - 4 \sin 	heta$ is given by $\sqrt{3^2 + (-4)^2} = 5$. So,$a = 5$. Given $\alpha = 5 \sin^2 	heta \cos^3 	h

Vedclass · Accepted Answer

$5$

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(C) The maximum value of $3 \cos 	heta - 4 \sin 	heta$ is given by $\sqrt{3^2 + (-4)^2} = 5$. So,$a = 5$. Given $\alpha = 5 \sin^2 	heta \cos^3 	heta$ and $\beta = 5 \sin^3 	heta \cos^2 	heta$. Then $\alpha^2 + \beta^2 = 25 \sin^4 	heta \cos^6 	heta + 25 \sin^6 	heta \cos^4 	heta = 25 \sin^4 	heta \cos^4 	heta (\cos^2 	heta + \sin^2 	heta) = 25 \sin^4 	heta \cos^4 	heta$. Therefore,$(\alpha^2 + \beta^2)^5 = (25 \sin^4 	heta \cos^4 	heta)^5 = (5^2 \sin^4 	heta \cos^4 	heta)^5 = 5^{10} \sin^{20} 	heta \cos^{20} 	heta$. Also,$\alpha \beta = (5 \sin^2 	heta \cos^3 	heta)(5 \sin^3 	heta \cos^2 	heta) = 25 \sin^5 	heta \cos^5 	heta$. Then $(\alpha \beta)^4 = (25 \sin^5 	heta \cos^5 	heta)^4 = (5^2 \sin^5 	heta \cos^5 	heta)^4 = 5^8 \sin^{20} 	heta \cos^{20} 	heta$. Finally,$\sqrt{\frac{(\alpha^2 + \beta^2)^5}{(\alpha \beta)^4}} = \sqrt{\frac{5^{10} \sin^{20} 	heta \cos^{20} 	heta}{5^8 \sin^{20} 	heta \cos^{20} 	heta}} = \sqrt{5^2} = 5$.

Let $a$ be the maximum value of $(3 \cos \theta - 4 \sin \theta)$ and $\theta \neq \frac{n \pi}{2}$. If $\alpha = a \sin^2 \theta \cos^3 \theta$ and $\beta = a \sin^3 \theta \cos^2 \theta$,then $\sqrt{\frac{(\alpha^2 + \beta^2)^5}{(\alpha \beta)^4}} = $

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