Match the ranges of the functions given in Li | vedclass

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(A) $(I) \ 3 \sin^2 x + 4 \cos^2 x - 2 = 3(\sin^2 x + \cos^2 x) + \cos^2 x - 2 = 3 + \cos^2 x - 2 = \cos^2 x + 1$. Since $0 \leq \cos^2 x \leq 1$,the

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$(I) \rightarrow (c), (II) \rightarrow (d), (III) \rightarrow (a), (IV) \rightarrow (b)$

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(A) $(I) \ 3 \sin^2 x + 4 \cos^2 x - 2 = 3(\sin^2 x + \cos^2 x) + \cos^2 x - 2 = 3 + \cos^2 x - 2 = \cos^2 x + 1$. Since $0 \leq \cos^2 x \leq 1$,the range is $[1, 2]$. Thus,$(I) \rightarrow (c)$.$(II) \ \cos^2 x + \sin^4 x = (1 - \sin^2 x) + \sin^4 x = \sin^4 x - \sin^2 x + 1$. Let $t = \sin^2 x$,where $t \in [0, 1]$. The function $f(t) = t^2 - t + 1$ has a vertex at $t = 1/2$. $f(0) = 1$,$f(1) = 1$,$f(1/2) = 1/4 - 1/2 + 1 = 3/4$. Thus,the range is $[3/4, 1]$. Thus,$(II) \rightarrow (d)$.$(III) \ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 - 3 \sin^2 x \cos^2 x = 1 - \frac{3}{4} \sin^2(2x)$. Since $0 \leq \sin^2(2x) \leq 1$,the range is $[1 - 3/4, 1 - 0] = [1/4, 1]$. Thus,$(III) \rightarrow (a)$.$(IV) \ \cos x \cos(\frac{2 \pi}{3} + x) \cos(\frac{2 \pi}{3} - x) = \cos x (\cos^2(\frac{2 \pi}{3}) \cos^2 x - \sin^2(\frac{2 \pi}{3}) \sin^2 x) = \cos x (\frac{1}{4} \cos^2 x - \frac{3}{4} \sin^2 x) = \frac{1}{4} \cos^3 x - \frac{3}{4} \cos x \sin^2 x = \frac{1}{4} \cos 3x$. Since $-1 \leq \cos 3x \leq 1$,the range is $[-1/4, 1/4]$. Thus,$(IV) \rightarrow (b)$.

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

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