Finding the function from given data

[utkarshakash]

Homework Statement

If f(x) is differentiable function satisfying $6x \displaystyle \int_0^1 f(tx)dt = 2x^3-3x^2+6x+5$ then select correct options(more than one may be correct)

a)f(x) is symmetric about x=1
b)f(x)=0 has no real roots
c)f(-x)=f(x+1) for all x in R
d) f(x)=1/2 has 2 real and equal roots

The Attempt at a Solution

From the options it is clear that the function f(x) is to be determined first. Now, if I divide both sides by 6,
$x \displaystyle \int_0^1 f(tx)dt = \int (x^2 - x +1) dx + \frac{5}{6}$

I have no idea how to take it ahead.

 

[Curious3141]

Homework Statement

If f(x) is differentiable function satisfying $6x \displaystyle \int_0^1 f(tx)dt = 2x^3-3x^2+6x+5$ then select correct options(more than one may be correct)

a)f(x) is symmetric about x=1
b)f(x)=0 has no real roots
c)f(-x)=f(x+1) for all x in R
d) f(x)=1/2 has 2 real and equal roots

The Attempt at a Solution

From the options it is clear that the function f(x) is to be determined first. Now, if I divide both sides by 6,
$x \displaystyle \int_0^1 f(tx)dt = \int (x^2 - x +1) dx + \frac{5}{6}$

I have no idea how to take it ahead.

Hint: In $6x \displaystyle \int_0^1 f(tx)dt$, substitute $y = tx$ and use Fundamental Theorem of Calculus to differentiate it after transforming the bounds.

EDIT: I'm not at all certain that the constant term (5) belongs in the RHS of that equation. It makes no sense to me. If it is disregarded, the question is easily solvable. Maybe someone else will have an insight into this.

 

[utkarshakash]

Hint: In $6x \displaystyle \int_0^1 f(tx)dt$, substitute $y = tx$ and use Fundamental Theorem of Calculus to differentiate it after transforming the bounds.

A clever approach. Thank You!

 

[Saitama]

A clever approach. Thank You!

I suggest you to make a note of this approach, its going to be helpful a lot of times. Test papers often include problems on FTOC.

 

[Curious3141]

The consensus in the homework help forum is that the question is wrong. The presence of the constant term (5) on the RHS is inconsistent. Putting x = 0 makes the LHS vanish but not the RHS.

*Another* wrong question?

 

[utkarshakash]

I suggest you to make a note of this approach, its going to be helpful a lot of times. Test papers often include problems on FTOC.

Thanks for your suggestion. I'm going to make it right now.