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AP EAPCET 04-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 11 of 160

Marks: +1, -0

f(x)=a x^{2}+b x+c

a, b, c \in C

a+b+c=3

f(x+y)=f(x)+f(y)+xy \forall x, y \in C

\sum\limits_{n=1}^{10} f(n)=

Solution: 👈: Video Solution

f(x)=a x^2+b x+c

a+b+c\;=3 \;\text{ and }\; f(x+y)=f(x)+f(y)+xy

f(x)\;=a x^2+b x+c

f(1)\;=a+b+c=3

f(2)\;=f(1+1)=f(1)+f(1)+1=2 f(1)+1

f(3)\;=f(2+1)=f(2)+f(1)+2

\;=2 f(1)+1+f(1)+2=3 f(1)+3

f(4)\;=f(3+1)=f(3)+f(1)+3

=3 f(1)+3+f(1)+3=4 f(1)+6

\sum\limits_{n=1}^{10} f(n)=f(1)+f(2)+f(3)+\dots+f(10)

=f(1)+2 f(1)+1+\;3 f(1)+3+4 f(1)

\;+6+\dots+10 f(1)+45

=\;f(1)[1+2+3+\dots+10]+

\;{[1+3+6+10+15+21+28+36+45]}

=\;f(1) \left[\;\frac{10 \cdot (10+1)}{2}\right] +[165]

=\;55 f(1)+165=55 \times 3+165

=\;165+165=330

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