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!
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]
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5 Answers
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..
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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.....
