Let f: R rightarrow R be a function defined b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(NONE) Given $f(x) = (3a+2)x^2 + \left(\frac{a+2}{a-1}\right)x + b$.
Putting $x=y=0$ in $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$,we get $f(0) = 2f(

Vedclass · Accepted Answer

$266$

Vedclass · Answer

(NONE) Given $f(x) = (3a+2)x^2 + \left(\frac{a+2}{a-1}\right)x + b$.
Putting $x=y=0$ in $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$,we get $f(0) = 2f(0) + 1$,which implies $f(0) = -1$.
Since $f(0) = b$,we have $b = -1$.
Now,$f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$.
Comparing the coefficients of $xy$ in the expansion of $f(x+y) = A(x+y)^2 + B(x+y) + C$,we have $f(x+y) = A(x^2+2xy+y^2) + B(x+y) + C = Ax^2 + Ay^2 + 2Axy + Bx + By + C$.
Given $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$,we equate $2A = -\frac{2}{7}$,so $A = -\frac{1}{7}$.
Since $A = 3a+2$,we have $3a+2 = -\frac{1}{7} \Rightarrow 3a = -\frac{1}{7} - 2 = -\frac{15}{7} \Rightarrow a = -\frac{5}{7}$.
Now,$B = \frac{a+2}{a-1} = \frac{-5/7 + 2}{-5/7 - 1} = \frac{9/7}{-12/7} = -\frac{3}{4}$.
Thus,$f(x) = -\frac{1}{7}x^2 - \frac{3}{4}x - 1$.
We need to calculate $28 \sum_{i=1}^3 |f(i)|$.
$f(1) = -\frac{1}{7} - \frac{3}{4} - 1 = \frac{-4-21-28}{28} = -\frac{53}{28}$.
$f(2) = -\frac{4}{7} - \frac{6}{4} - 1 = -\frac{4}{7} - \frac{3}{2} - 1 = \frac{-8-21-14}{14} = -\frac{43}{14} = -\frac{86}{28}$.
$f(3) = -\frac{9}{7} - \frac{9}{4} - 1 = \frac{-36-63-28}{28} = -\frac{127}{28}$.
$28 \sum_{i=1}^3 |f(i)| = 28 \left( \frac{53}{28} + \frac{86}{28} + \frac{127}{28} \right) = 53 + 86 + 127 = 266$.

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

Similar Questions

If $f:R \to R$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 7$,then $\sum_{r = 1}^n f(r)$ is

Let $a, b, c \in \mathbb{R}$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy, \forall x, y \in \mathbb{R}$,then $\sum_{n=1}^{10} f(n)$ is equal to:

If $f(x + y) = f(x)f(y)$ for all $x$ and $y$ and $f(5) = 2$,$f'(0) = 3$,then $f'(5)$ will be

$f: R \rightarrow R$ is defined by $f(x+y)=f(x)+12y, \forall x, y \in R$. If $f(1)=6$,then $\sum_{r=1}^n f(r)=$

If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x = 1}^n {f(x) = 120}$,find the value of $n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module