Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \left( 2 + 3a \right)x^2 + \left( \frac{a+2}{a-1} \right)x + b, a \neq 1 \). If \[ f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy, \] then the value of \( 28 \sum_{i=1}^5 f(i) \) is:

• A. 715
• B. 675
• C. 545
• D. 735

Hint:

Be mindful of the properties of functions when solving for unknowns in functional equations.

Answer 1

The Correct Option is B

Given the functional equation \( f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy \) and the form \( f(x) = (2+3a)x^2 + \frac{a+2}{a-1}x + b \), we aim to determine \( 28 \sum_{i=1}^5 f(i) \).

1. Substitution into the functional equation:
Substituting the expression for \(f(x)\) into the given functional equation yields:
\( (2+3a)(x+y)^2 + \frac{a+2}{a-1}(x+y) + b = (2+3a)x^2 + \frac{a+2}{a-1}x + b + (2+3a)y^2 + \frac{a+2}{a-1}y + b + 1 - \frac{2}{7}xy \)

2. Expansion and simplification:
Expanding and simplifying the equation results in:
\( (2+3a)x^2 + 2(2+3a)xy + (2+3a)y^2 + \frac{a+2}{a-1}x + \frac{a+2}{a-1}y + b = (2+3a)x^2 + (2+3a)y^2 + \frac{a+2}{a-1}x + \frac{a+2}{a-1}y + 2b + 1 - \frac{2}{7}xy \)

3. Coefficient comparison:
Comparing the coefficients of the \(xy\) term and the constant terms provides two equations:
\( 2(2+3a) = -\frac{2}{7} \) and \( b = 2b + 1 \)

4. Determination of \(a\) and \(b\):
Solving \( 2(2+3a) = -\frac{2}{7} \) gives \( 2+3a = -\frac{1}{7} \), which simplifies to \( 3a = -\frac{15}{7} \), and thus \( a = -\frac{5}{7} \).
Solving \( b = 2b + 1 \) yields \( b = -1 \).

5. Calculation of \(\frac{a+2}{a-1}\):
Substituting \( a = -\frac{5}{7} \) into the expression \(\frac{a+2}{a-1}\) gives:
\( \frac{a+2}{a-1} = \frac{-\frac{5}{7} + 2}{-\frac{5}{7} - 1} = \frac{\frac{9}{7}}{-\frac{12}{7}} = -\frac{9}{12} = -\frac{3}{4} \)

6. Explicit form of \(f(x)\):
With \( a = -\frac{5}{7} \), \( b = -1 \), and \(\frac{a+2}{a-1} = -\frac{3}{4}\), the explicit form of \(f(x)\) is:
\( f(x) = (2 + 3(-\frac{5}{7}))x^2 - \frac{3}{4}x - 1 = (2 - \frac{15}{7})x^2 - \frac{3}{4}x - 1 = -\frac{1}{7}x^2 - \frac{3}{4}x - 1 \)

7. Summation of \(f(i)\) from \(i=1\) to \(5\):
The summation is computed as follows:
\( \sum_{i=1}^5 f(i) = \sum_{i=1}^5 \left( -\frac{1}{7}i^2 - \frac{3}{4}i - 1 \right) = -\frac{1}{7} \sum_{i=1}^5 i^2 - \frac{3}{4} \sum_{i=1}^5 i - \sum_{i=1}^5 1 \) \( = -\frac{1}{7} \left( \frac{5(5+1)(2(5)+1)}{6} \right) - \frac{3}{4} \left( \frac{5(5+1)}{2} \right) - 5 \) \( = -\frac{1}{7} (55) - \frac{3}{4} (15) - 5 = -\frac{55}{7} - \frac{45}{4} - 5 = -\frac{220 + 315 + 140}{28} = -\frac{675}{28} \)

8. Final calculation of \(28 \sum_{i=1}^5 f(i)\):
Multiplying the sum by 28 yields:
\( 28 \sum_{i=1}^5 f(i) = 28 \left( -\frac{675}{28} \right) = 675 \)

Final Answer:
The value of \( 28 \sum_{i=1}^5 f(i) \) is \( {675} \).