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if f_1(x)= \cos (\sin x) and f_2(x)= \sin (\cos x) then which of the following is true for x\in (0,\frac{\pi}{2}) ?

Option: 1

f_2(x)>f_1(x)


Option: 2

f_2(x)<f_1(x)


Option: 3

f_1(x)= f_2(x)


Option: 4

None of these


Answers (1)

best_answer

\\\text{Let f(x)=x-}\sin x \ \ 0<x<\frac{\pi}{2}\\ f'(x)=1-\cos x >0 \ \ \ \ 0<x<\frac{\pi}{2}\\ \text{f(x) is increasing function }\\ x> \sin x \\ \cos x \epsilon (0,1) \ \ \ x\epsilon (0,\frac{\pi}{2})\\ \text{Replace x by cos x}\\ \cos x> \sin (\cos x) .... (i)\\ \text{cos x is decreasing function so }\\ \cos x< \cos(\sin x) ....(ii)\\ \text{From equation (i) and equation (ii)} \\ \sin (\cos x)<\cos x< \cos(\sin x) \\ f_1(x)= \cos (\sin x) \\ f_2(x)= \sin (\cos x) \\ f_2(x)<f_1(x)

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