यदि f(theta) = frac{sin^4 theta + 3 cos^2 the | vedclass

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

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(B) माना $x = \cos^2 	heta$,जहाँ $x \in [0, 1]$.तब $\sin^4 	heta = (1 - x)^2 = 1 - 2x + x^2$.$f(	heta) = \frac{(1 - 2x + x^2) + 3x}{(1 - 2x + x^2) + x} = \frac{x^2 + x + 1}{x^2 - x + 1}$.माना $y = \frac{x^2 + x + 1}{x^2 - x + 1}$.$y(x^2 - x + 1) = x^2 + x + 1 \implies x^2(y - 1) - x(y + 1) + (y - 1) = 0$.$x$ के वास्तविक होने के लिए,विविक्तकर $D \ge 0$.$D = (y + 1)^2 - 4(y - 1)^2 \ge 0$.$(y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \ge 0$.$(3 - y)(3y - 1) \ge 0 \implies (y - 3)(3y - 1) \le 0$.अतः,$y \in [1/3, 3]$.इस प्रकार,$\alpha = 1/3$ और $\beta = 3$.सार्व अनुपात $r = \frac{\alpha}{\beta} = \frac{1/3}{3} = 1/9$.अनंत $G.P.$ का योग $S = \frac{a}{1 - r} = \frac{64}{1 - 1/9} = \frac{64}{8/9} = 64 	imes \frac{9}{8} = 72$.

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