Question

Prove that f(x)=x²-x sin x is increasing on(0,π/2).

Answer 1

we have to prove that f(x) = x² - x sinx , is increasing on (0, π/2).

concept : any function , y = f(x) is increasing in interval (a, b) only when f'(x) > 0 in interval (a, b).

so first of all, differentiate f(x) with respect to x,

i.e., f'(x) = d(x²)/dx - d(x sinx)/dx

= 2x - [ xcosx + sinx ]

= 2x - xcosx + sinx

hence, f'(x) = 2x - xcosx + sinx

it doesn't clear that , f'(x) is positive or negative in (0, π/2)

so, again differentiate f'(x) with respect to x,

f"(x) = 2 - (-xsinx + cosx) + cosx

= 2 + xsinx - cosx + cosx

= 2 + xsinx

in interval (0, π/2) , f"(x) > 0

hence, f'(x) is increasing function.

now, range of f'(x) :

at x = 0, f'(0) = 0

at x = π/2, f'(π/2) = 2(π/2) - π/2cos(π/2)+sin(π/2) = π + 1

hence, 0 < f'(x) < π + 1

it is also clear that f'(x) > 0

hence, f(x) is also increasing function on (0, π/2)