Solve this

Number 3 can be done by just applying the definition of the derivative from the limiting point of view....

f'(1)=f(1+Δh)-f(1)Δh>f(1), where Δh>0 but →0.....means f(1+Δh)>f(1)....this proves it....

Q1.

Q3) I prove the generalized version: If f(x₀) = 0 and f'(x) > f(x) for all x ≥ x₀ then prove that f(x) > 0 for x >x₀.

g(x) = f(sin x) + f(cos x)

g'(x) = f'(sin x) cos x - f'(cos x) sin x

Consider the function h(x) = f'(sin x). Then h"(x) = f"(sin x) cos x>0 i.e. h(x) is an increasing function

1) x \in \left(0, \frac{\pi}{4} \right)

We have x < \frac{\pi}{2} - x

Then since f'(sin x) is increasing

f'(\sin \left(\frac {\pi}{2} - x \right) \right))> f'( \sin x) or

0> f'(\cos x)> f'( \sin x) \Rightarrow 0<-f'(\cos x) < - f'(\sin x)

Also in this interval 0<\sin x < \cos x

From the above two inequalities, we get

-f'(cos x) sinx <- f'(sin x) cos x or f'(sin x) cosx - f'(cos x) sinx <0

or g'(x) <0. That means g(x) is strictly decreasing in this interval

(ii) x \in \left(\frac{\pi}{4}, \frac{\pi}{2} \right)

Since g(x) = g \left(\frac{\pi}{2} - x \right) we have g'(x) = -g' \left(\frac{\pi}{2} - x \right) which tells us that in this interval g(x) >0 so that g(x) is strictly increasing

At x = π/4 of course g'(x) = 0.