Let f: ℝ → ℝ be a function defined by f(x)=(2+3 a) x²+((a+2)/(a-1)) x+b, a ≠ 1. If f(x+y)=f(x)+f(y)+1-2/7 x y, then the value of 28 Σᵢ₌₁⁵|f(i)| is

Solution

Maths sir · 5 steps

Expand f(x+y) and match coefficients: the xy term fixes a, the constant term fixes b.

1. Expand both sides. Write f(x) = px² + qx + b with p = 2 + 3a and q = (a+2)/(a-1). Then f(x+y) = px² + py² + 2pxy + qx + qy + b, while the right side is px² + py² + qx + qy + 2b + 1 - 2/7xy.
2. Match coefficients. The xy terms give 2p = -2/7, so p = -1/7. The constants give b = 2b + 1, so b = -1.
3. Find a and q. 2 + 3a = -1/7 gives a = -5/7. Then q = (-5/7 + 2)/(-5/7 - 1) = (9/7)/(-12/7) = -3/4.

   28f(x) = -4x² - 21x - 28

4. Remove the modulus. For positive x every term of 28f(x) is negative, so 28|f(i)| = 4i² + 21i + 28. Using Σ i² = 55 and Σ i = 15 for i = 1 to 5, the sum is 4 × 55 + 21 × 15 + 28 × 5.
5. So. 28Σᵢ₌₁⁵|f(i)| = 220 + 315 + 140 = 675, which is option (b).

The idea to keep. When a functional equation is given for a polynomial, expand both sides and compare coefficients of x², xy, x and the constant. Each comparison gives one equation.

Answer from the NTA final answer key. Solution by PrepWiser.