If the range of f(theta) = frac{sin^4 theta + | vedclass

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(B) Let $x = \cos^2 	heta$,where $x \in [0, 1]$.Then $\sin^4 	heta = (1 - x)^2 = 1 - 2x + x^2$.$f(	heta) = \frac{(1 - 2x + x^2) + 3x}{(1 - 2x + x^2

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(B) Let $x = \cos^2 	heta$,where $x \in [0, 1]$.Then $\sin^4 	heta = (1 - x)^2 = 1 - 2x + x^2$.$f(	heta) = \frac{(1 - 2x + x^2) + 3x}{(1 - 2x + x^2) + x} = \frac{x^2 + x + 1}{x^2 - x + 1}$.Let $y = \frac{x^2 + x + 1}{x^2 - x + 1}$.$y(x^2 - x + 1) = x^2 + x + 1 \implies x^2(y - 1) - x(y + 1) + (y - 1) = 0$.For $x$ to be real,the discriminant $D \ge 0$.$D = (y + 1)^2 - 4(y - 1)^2 \ge 0$.$(y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \ge 0$.$(3 - y)(3y - 1) \ge 0 \implies (y - 3)(3y - 1) \le 0$.So,$y \in [1/3, 3]$.Thus,$\alpha = 1/3$ and $\beta = 3$.The common ratio $r = \frac{\alpha}{\beta} = \frac{1/3}{3} = 1/9$.The sum of the infinite $G.P.$ is $S = \frac{a}{1 - r} = \frac{64}{1 - 1/9} = \frac{64}{8/9} = 64 	imes \frac{9}{8} = 72$.

If the range of $f(\theta) = \frac{\sin^4 \theta + 3 \cos^2 \theta}{\sin^4 \theta + \cos^2 \theta}$,$\theta \in R$ is $[\alpha, \beta]$,then the sum of the infinite $G.P.$,whose first term is $64$ and the common ratio is $\frac{\alpha}{\beta}$,is equal to...........

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