TargetIIT

Seems to me that we cannot find such polynomials.

We will use the fact that for sufficiently large magnitudes of x, the sign of P(x) coincides with the sign of the leading coefficient.

For convenience I will write A instead of A(x).

Let G = |A|-|B|+|C|

There are two cases to be considered:

I. A and C have the same signs for the leading coefficient.

Then we see that [im]http://codecogs.izyba.com/gif.latex?G%28x%29%20%3D%20-G%28-x%29[/im] for sufficiently large magnitudes of x and this implies that [im]http://codecogs.izyba.com/gif.latex?%7CG%28x%29%7C%20%3D%20%7CG%28-x%29%7C[/im]

But we get a contradiction here as [im]http://codecogs.izyba.com/gif.latex?x%20%5Crightarrow%20%5Cinfty%20%5CRightarrow%20%7CG%28x%29%7C%20%5Crightarrow%20%5Cinfty[/im] and [im]http://codecogs.izyba.com/gif.latex?x%20%5Crightarrow%20-%5Cinfty%20%5CRightarrow%20%7CG%28x%29%7C%3D1[/im]

2. WLOG A and B have the same sign and leading coeff of C is -ve

Then we get the equations

A-B-C = -2x+2

-A+B+C = -1

which are true for infinitely many values of x

so that C ≡ x + 3/2 contradicting that leading coefficient of C is -ve

So no such polynomials A,B, C exist

If you take A,B, C as given by you, then for x> 1/2, you get |A(x)| = A(x), |B(x)| = B(x) and |C(x)| = x - 1/2

So, |A(x)|-|B(x)| + |C(x)| = 1. So the answer provided by you doesnt fulfil the conditions.