Question

Let f'(sin x) < 0 and f"(sin x) > 0 for all x in (0, π/2) and g(x) = f(sin x) + f(cos x), then find the interval where g(x) increases or decreases.

Answer 1

We have, g(x) = f(sinx) + f(cosx)

⇒ g'(x) = f'(sinx) cosx + f'(cosx)(– sinx)

⇒ g'(x) = f'(sin x) cosx – f'(cos x)sinx

⇒ g"(x) = f"(sinx) cos²x – sinxf'(sinx) + f"(cosx)sinx – f'(cosx)cosx

⇒ g" (x) < 0,

since f'(sin x) < 0, f"(sin x) > 0

and f'(cos x) > 0, f"(cosx) < 0

in the interval [ 0, π/2]

⇒ g'(x) is a decreasing function

Now, g'(π/4) = 0

If x < π/4 than g'(x) < (π/4)

⇒g'(x) > 0

⇒ g'(x) is increasing in [ 0, π/4]

If x > π/4 then g'(x) < g'(π/4)

⇒ g'(x) < 0

⇒ g(x) is decreasing in π/4 < x ≤ π/2 .