Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a function defined by f(x)=(2+3a)x2+(a+2a−1)x+b,a≠1f(x) = (2 + 3a) x^2 + \left( \frac{a + 2}{a - 1} \right) x + b, a \neq 1. If f(x+y)=f(x)+f(y)+1−27xyf(x + y) = f(x) + f(y) + 1 - \frac{2}{7} xy, then the value of 28∑i=15∣f(i)∣28 \sum_{i=1}^{5} |f(i)| is

Answer: 667

Step-by-step solution

Step 1: Analyze the given functional equation

The problem provides a functional equation relating f(x+y)f(x+y) to f(x)f(x) and f(y)f(y). This equation is key to determining the coefficients of the quadratic function f(x)f(x). We will substitute x=0x=0 and y=0y=0 into this equation to find the value of bb. Then, we will substitute the general form of f(x)f(x) into the functional equation to find the values of aa and bb.

Step 2: Determine the constant term 'b'

Substitute x=0x=0 and y=0y=0 into the functional equation f(x+y)=f(x)+f(y)+1−27xyf(x+y) = f(x) + f(y) + 1 - \frac{2}{7} xy. This simplifies to f(0)=f(0)+f(0)+1f(0) = f(0) + f(0) + 1, which implies f(0)=−1f(0) = -1. From the definition of f(x)f(x), we know that f(0)=bf(0) = b. Therefore, we find that b=−1b = -1.

Step 3: Substitute f(x)f(x) into the functional equation

Now, substitute the expression for f(x)f(x) into the functional equation. Expand both sides and simplify. This will allow us to compare the coefficients of xyxy and other terms to find the value of aa.

Step 4: Solve for 'a'

Expand the left side of the equation. After simplification, the terms involving x2x^2, y2y^2, xx, and yy cancel out on both sides. We are left with a comparison of the xyxy terms. Equating the coefficients of xyxy gives 2(2+3a)=−272(2+3a) = -\frac{2}{7}. Solving this equation for aa yields a=−57a = -\frac{5}{7}.

Step 5: Determine the final form of f(x)f(x)

Now that we have the values of a=−57a = -\frac{5}{7} and b=−1b = -1, we can substitute them back into the original definition of f(x)f(x). Calculate the coefficients (2+3a)(2+3a) and (a+2a−1)\left(\frac{a+2}{a-1}\right) to get the complete expression for f(x)f(x).

Step 6: Calculate ∑i=15∣f(i)∣\sum_{i=1}^{5} |f(i)|

Now we need to calculate f(i)f(i) for i=1,2,3,4,5i=1, 2, 3, 4, 5. Since all the coefficients of f(x)f(x) are negative and the constant term is negative, f(x)f(x) will be negative for positive values of xx. Therefore, ∣f(i)∣=−f(i)|f(i)| = -f(i). Calculate each f(i)f(i) and sum their absolute values.

Step 7: Calculate the final value

Finally, multiply the sum of the absolute values by 28 to get the required answer.

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