1. Let f be defined on [0,1] be a twice differentiable function such that, \left|f''(x) \right|\leq 1 for all x \in \left[0,1 \right]. If f(0)=f(1) , then show that, \left|f'(x)\right|<1 for all x \in \left[0,1 \right].
2. Show that exactly two real values of x satisfy the equation x2 = xsinx + cosx.
3. Let a, b, c be three real numbers such that a<b<c, f(x) is continuous in [a,c] and differentiable in (a,c). Also f'(x) is strictly increasing in (a,c). Prove that (c-b)f(a)+(b-a)f(c)>(c-a)f(b) .
Please provide hints and observations only. [1]
1for the 2nd,is graph not working????....a rough one[can`t draw an exact one]
for the 1st one,if u integrate an inequality,will the inequality hold for the integrals of the fnctns??sry for asking so many qstns n not actually solving..but if it does hold then i can prove the sum..
30I tried it. It'll be very difficult to get the correct graph. And I get a feeling, a correct graph will be required, because an approximation could give an error! Had it been only xsnx or cosx it would be fine. Its their sum, big problem!
11For the 2nd sum observe that if x is a soln then so is -x.
Now if you put f(x)=x2-xsinx-cosx, f'(x)=2x-xcosx>0 for x>0....implying f(x) is increasing with f(0)<0.
That gives only 1 positive soln....and altogether only two solns.
3) To prove f(a)+b-ac-bf(c)>f(b)+b-ac-bf(b).....which again is equivalent to proving f(a)-f(b)>b-ac-b(f(b)-f(c))
Using LMVT together with the fact that f'(x) is increasing finishes it.
1) I think it is straight forward enough.....
