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Functions Part 2

Section: Mathematics

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Question : 29 of 73

Marks: +1, -0

Let

f: \mathbb{R} \longrightarrow \mathbb{R}

be a function defined by

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

f(x+y)=

f(x)+f(y)+1-\frac{2}{7}xy

, then the value of

28 \sum\limits_{i=1}^{5} |f(i)|

[28 Jan 2025 Shift 1]

715
675
735
545

Solution:

Put

y=0

f(x)=f(0)+f(x)+1-0

f(0)=-1

f(0)=0+0+b

\Rightarrow b=-1

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

f(0)=f(-1)+f(1)+\frac{9}{7}

-1=(2+3a)+\frac{a+2}{a-1}(-1)+b+(2+3a)

+1=4+6a-2+\frac{9}{7}

-1=2+\frac{9}{7}+6a

6a=-1-2-\frac{9}{7}

a=\frac{-5}{7}

f(x)=\frac{-x^2}{7}+\frac{\frac{9}{7}}{\frac{-12}{7}}x-1

f(x)=\frac{-x^2}{7}-\frac{3}{4}x-1

\sum\limits_{i=1}^{5} f(i)=-\frac{1}{7}\left(\frac{5 \times 6 \times 11}{6}\right)-\frac{3}{4}\left(\frac{5 \times 6}{2}\right)-5

=\frac{-55}{7}-\frac{45}{4}-5

=\frac{675}{28}

\Rightarrow 28 \left| \sum\limits_{i=1}^{5} f(i) \right| =675

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