Let be a function defined by . , then the value of is [2025]

Question

Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ . If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5}$ $| f (i) |$ is [2025]

Smart Achievers · Accepted Answer

(1)

Given, $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ ... (i)

Also, $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ ... (ii)

Put x = y = 0 in (ii), we get

f(0) = f(0) + f(0) + 1 – 0 $\Rightarrow$ f(0) = –1 ... (iii)

Using (i), f(0) = b = –1

Put y = – x in (ii), we get

$f (x - x) = f (x) + f (- x) + 1 + \frac{2}{7} x^{2}$

$\Rightarrow - 1 = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x - 1 + (2 + 3 a) x^{2} - (\frac{a + 2}{a - 1}) x - 1 + 1 + \frac{2}{7} x^{2}$

$\Rightarrow (2 (2 + 3 a) + \frac{2}{7}) x^{2} = 0 \Rightarrow a = \frac{- 5}{7}$

$∴ f (x) = \frac{- 1}{7} x^{2} - \frac{3}{4} x - 1 = \frac{- 1}{28} (4 x^{2} + 21 x + 28)$

Now, $28 \sum_{i = 1}^{5} | f (i) | = 28 [| f (1) | + | f (2) | + . . . + | f (5) |]$

$= 28 \times \frac{1}{28} [4 (1^{2} + 2^{2} + . . . + 5^{2}) + 21 (1 + 2 + . . . + 5) + 28 (5)]$

= 675

Solution

Q.
Let $f : R \to R$ be a function defined by $f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b, a \neq 1$ . If $f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y$ , then the value of $28 \sum_{i = 1}^{5}$ $| f (i) |$ is [2025]

1 675

2 545

3 735

4 715

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