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Question

Let f(x)=∣ ∣sinx02cosx0sinx02cosx0sinx∣ ∣ where x(0,π). Then total number of local maxima and local minima of f(x) is

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Solution

We have, f(x)=∣ ∣sinx02cosx0sinx02cosx0sinx∣ ∣
f(x)=sin3x2cosx(2sinxcosx)
f(x)=sin3x4sinx(1sin2x)
f(x)=5sin3x4sinx
Differentiating w.r.t. x, we get
f(x)=15sin2xcosx4cosx
f(x)=15cosx(sin2x415)
f(x)=15cosx(sinx215)(sinx+215)
cosx=0 at x=π2
and sinx=215 at two points in (0,π)
Total required points =3

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