Functions Ques 49
- Let $a, b, c \in R$. If $f(x)=a x^{2}+b x+c$ be such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+x y, \forall x, y \in R$, then $\sum _{n=1}^{10} f(n)$ is equal to
(2017 Main)
(a) 330
(b) 165
(c) 190
(d) 255
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Answer:
Correct Answer: 49.(a)
Solution:
Formula:
- We have, $f(x)=a x^{2}+b x+c$
Now, $f(x+y)=f(x)+f(y)+x y$
Put $\quad y=0 \Rightarrow f(x)=f(x)+f(0)+0$
$$ \begin{array}{cc} \Rightarrow & f(0)=0 \\ \Rightarrow & c=0 \end{array} $$
Again, put $\quad y=-x$
$$ \begin{aligned} & \therefore & f(0) & =f(x)+f(-x)-x^{2} \\ \Rightarrow & & 0 & =a x^{2}+b x+a x^{2}-b x-x^{2} \end{aligned} $$
$$ \begin{aligned} \Rightarrow & 2 a x^{2}-x^{2} & =0 \\ \Rightarrow & a & =\frac{1}{2} \end{aligned} $$
Also, $a+b+c=3$
$\Rightarrow \quad \frac{1}{2}+b+0=3 \Rightarrow b=\frac{5}{2}$
$\therefore \quad f(x)=\frac{x^{2}+5 x}{2}$
Now, $\quad f(n)=\frac{n^{2}+5 n}{2}=\frac{1}{2} n^{2}+\frac{5}{2} n$
$\therefore \quad \sum _{n=1}^{10} f(n)=\frac{1}{2} \sum _{n=1}^{10} n^{2}+\frac{5}{2} \sum _{n=1}^{10} n$
$$ \begin{aligned} & =\frac{1}{2} \cdot \frac{10 \times 11 \times 21}{6}+\frac{5}{2} \times \frac{10 \times 11}{2} \\ & =\frac{385}{2}+\frac{275}{2}=\frac{660}{2}=330 \end{aligned} $$
BETA
