# Difficult problem (for me at least) !!!!

If $f: [0,1] \rightarrow (0, \infty)$ such that $\int_0^1 f(x)\,dx = 1$, $\int_0^1 xf(x)\,dx = \alpha$ and $\int_0^1 x^2f(x)\,dx = \alpha^2$, then

A. $f(x)=5$

B. $f(2)=8$

C. $f(x)$ is not possible

D. $f(x)=2x-1$

I didn't post this as a problem because I am not sure whether the answer given in my book is correct or not. Please try to write an explanation rather than only the answer. Note by Maharnab Mitra
5 years, 11 months ago

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Consider the integral $\int_0^1 (x - \alpha)^2 f(x) \: dx.$

- 5 years, 11 months ago

no one of the A, B and D figures [0,1] to (0,+inf).

- 5 years, 11 months ago

@Jon Haussmann what is purpose of considering this integral.bcoz it is 0.as variance is 0.

- 5 years, 6 months ago

answer isn't D. If we put x=0, (2x-1)=-1 which does not lie in the range. option A isn't possible. Putting f(x)=5 does not satisfy the 1st equation.

- 5 years, 6 months ago

A is not true because assuming f(x) is 5 makes the first integral untrue.

B is not certainly true because without being able to directly solve for f(x), we have no idea what the f(2) should be; its out of the integral range

D is not true just by plugging it in again.

C must be the answer. Imagine f(x) and 1, x, and x^2 in perpendicular planes, and as you integrate the functions, you multiply f(x) and 1, x, or x^2, and a small dx to get a 3d volume. Between all of these volumes they all have the same height f(x), therefore the ratios of the volumes are simply the ratios of the widths of the solids. Since the ratio of ∫x^2dx/∫xdx is not equal to ∫xdx/∫dx (with 0,1 as limits), f(x) cannot exist

- 5 years, 6 months ago

Multiply the first term by 4,second by -4,and third by 1 and add up those integrals.You will get RHS = 0 and LHS having integral of (alpha-2)^2*f(x) from 0 to 1. Now f(x) is positive and the other term is also positive so integral of a positive function can never be zero.So no function in possible.

- 5 years, 6 months ago