That's a really quick and neat method. The way I did it is essentially the same as yours, just a little slower: Let I be the value of the integral. Then 6 + f(x) = 2f(-x) + 3x^2 * I (let p(x) denote this assertion). Then p(-x) gives 6 + f(-x) = 2f(x) + 3x^2 * I. Solving these simultaneous equations for f(x), we get f(x) = 6 - 3x^2 * I. Integrating gives I = 12 - 2I, so I = 4.
I used a similar way. I made two equations and found that f(x)-f(-x)=0. Now I used that in one equation to find out f(x)=-3Ix^2+6. Replacing x with t and then integrating both sides to find I=-2I+12, and I=4. Put it in f(x) to find f(x)=-12x^2+6. Now Integrate to get 4.
Take derivative twice: f’’(x) = 2f’’(-x) + 6I, 6I = f’’(x) - 2f’’(-x) LHS constant, so RHS constant for all x too. Plug in x = 0 ==> I = -1/6 f’’(0) But note: 6 + f(x) - 2f(-x) = 3x² I = 6 + f(-x) - 2f(x). By adding and subtracting itself, ==> 12 - f(x) - f(-x) = 6x² I (*) & 3f(x) - 3f(-x) = 0 ==> f(x) is even ==> f(x) = ax² + b ==> I = -a/3 find a,b by equating coefficients of (*), note only a quadratic coefficient: Thus b = 6, and -2a = 6*2*(a/3 + 6) = 4a + 72 ==> -6a = 72 ==> a = -12 ==> I = 4 Edit: way longer than it needed to be, it’s 3am and my brain thought it’s better to differentiate both sides rather than integrate lol, I guess the added bonus is that f(x) = 6 - 12x² is fully and uniquely determined
Your approach was really clean! I did it a different (worse) way. I assumed that f(x) had to be a quadratic in the form ax^2 + bx + c, and replaced f(x) in the given expression with that. And letting the definite integral component be k, I ended up with: 6 + a x^2 + b x + c = 2 (a (-x)^2 + b(-x) + c) + 3 x^2 k I rearranged that to match up the coefficients: -a x^2 + 3 b x + 6 = 3 q x^2 + c This means that a = -3 k, b = 0, and c = 6. I plugged those back in to give me a useful expression for f(x): f(x) = -3 k x^2 + 6. I finally integrated that to give me k = 12 - 2 k, so k = 4. It didn't occur to me to just integrate as the first step.
I’ve had experience with similar questions before, just through old exercises that I had done as a student years ago. I don’t remember how I might have approached this question back then, but this time it seemed like a natural thing to do.
For a multiple choice test, this certainly works to get you the right answer. You haven't actually proven that answer is correct, though (which they probably wanted in order to get full marks). It could be that the problem is ill-posed and there are different functions f that satisfy the equation but give different values for the integral. By assuming a form for f, you haven't proven that other forms are either impossible or give the same result.
That's a really quick and neat method. The way I did it is essentially the same as yours, just a little slower:
Let I be the value of the integral. Then 6 + f(x) = 2f(-x) + 3x^2 * I (let p(x) denote this assertion).
Then p(-x) gives 6 + f(-x) = 2f(x) + 3x^2 * I. Solving these simultaneous equations for f(x), we get
f(x) = 6 - 3x^2 * I.
Integrating gives I = 12 - 2I, so I = 4.
Oh interesting! So you just extracted a second equation and solved it, very cool!
I used a similar way. I made two equations and found that f(x)-f(-x)=0. Now I used that in one equation to find out f(x)=-3Ix^2+6. Replacing x with t and then integrating both sides to find I=-2I+12, and I=4. Put it in f(x) to find f(x)=-12x^2+6. Now Integrate to get 4.
Take derivative twice: f’’(x) = 2f’’(-x) + 6I, 6I = f’’(x) - 2f’’(-x)
LHS constant, so RHS constant for all x too. Plug in x = 0
==> I = -1/6 f’’(0)
But note:
6 + f(x) - 2f(-x) = 3x² I = 6 + f(-x) - 2f(x). By adding and subtracting itself,
==> 12 - f(x) - f(-x) = 6x² I (*)
& 3f(x) - 3f(-x) = 0 ==> f(x) is even
==> f(x) = ax² + b ==> I = -a/3
find a,b by equating coefficients of (*), note only a quadratic coefficient:
Thus b = 6, and -2a = 6*2*(a/3 + 6) = 4a + 72
==> -6a = 72 ==> a = -12 ==> I = 4
Edit: way longer than it needed to be, it’s 3am and my brain thought it’s better to differentiate both sides rather than integrate lol, I guess the added bonus is that f(x) = 6 - 12x² is fully and uniquely determined
I like that. I thought about differentiating, but quickly realised it got me nowhere. Differentiating twice didn't occur to me...
@thomasdalton1508 it’s cool that it works out here!
Your approach was really clean! I did it a different (worse) way. I assumed that f(x) had to be a quadratic in the form ax^2 + bx + c, and replaced f(x) in the given expression with that. And letting the definite integral component be k, I ended up with:
6 + a x^2 + b x + c = 2 (a (-x)^2 + b(-x) + c) + 3 x^2 k
I rearranged that to match up the coefficients:
-a x^2 + 3 b x + 6 = 3 q x^2 + c
This means that a = -3 k, b = 0, and c = 6. I plugged those back in to give me a useful expression for f(x):
f(x) = -3 k x^2 + 6. I finally integrated that to give me k = 12 - 2 k, so k = 4.
It didn't occur to me to just integrate as the first step.
I’ve had experience with similar questions before, just through old exercises that I had done as a student years ago. I don’t remember how I might have approached this question back then, but this time it seemed like a natural thing to do.
For a multiple choice test, this certainly works to get you the right answer. You haven't actually proven that answer is correct, though (which they probably wanted in order to get full marks). It could be that the problem is ill-posed and there are different functions f that satisfy the equation but give different values for the integral. By assuming a form for f, you haven't proven that other forms are either impossible or give the same result.