The minimum value of f(x) = a^{2} cos^{2} x + | vedclass

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(B) Given the function $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$. Using the trigonometric identities $\cos^{2} x = \frac{1 + \cos 2x}{2}$ and $\sin

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$b^{2}$

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(B) Given the function $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$. Using the trigonometric identities $\cos^{2} x = \frac{1 + \cos 2x}{2}$ and $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get: $f(x) = a^{2} \left( \frac{1 + \cos 2x}{2} ight) + b^{2} \left( \frac{1 - \cos 2x}{2} ight)$ $f(x) = \frac{a^{2} + a^{2} \cos 2x + b^{2} - b^{2} \cos 2x}{2}$ $f(x) = \frac{a^{2} + b^{2}}{2} + \left( \frac{a^{2} - b^{2}}{2} ight) \cos 2x$. Since $-1 \leq \cos 2x \leq 1$,the function $f(x)$ attains its minimum value when $\cos 2x = -1$ (given $a^{2} > b^{2}$,so $\frac{a^{2} - b^{2}}{2} > 0$). Substituting $\cos 2x = -1$: $f_{	ext{min}} = \frac{a^{2} + b^{2}}{2} + \left( \frac{a^{2} - b^{2}}{2} ight) (-1)$ $f_{	ext{min}} = \frac{a^{2} + b^{2} - a^{2} + b^{2}}{2} = \frac{2b^{2}}{2} = b^{2}$. Thus,the minimum value is $b^{2}$.

List-$I$	List-$II$
$(I) \ 3 \sin^2 x + 4 \cos^2 x - 2$	$(a) \ [\frac{1}{4}, 1]$
$(II) \ \cos^2 x + \sin^4 x$	$(b) \ [-\frac{1}{4}, \frac{1}{4}]$
$(III) \ \sin^6 x + \cos^6 x$	$(c) \ [1, 2]$
$(IV) \ \cos x \cos(\frac{2 \pi}{3} + x) \cos(\frac{2 \pi}{3} - x)$	$(d) \ [\frac{3}{4}, 1]$
	$(e) \ [0, 1]$

The minimum value of $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$ if $a^{2} > b^{2}$,is

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