Test Index

Functions Part 1

Section: Mathematics

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Question : 28 of 100

Marks: +1, -0

Let

f: N \rightarrow R

be a function such that

f(x+y)=2 f(x) f(y)

for natural numbers

x

y

f(1)

=2

, then the value of

\alpha

\sum\limits_{k=1}^{10} f(\alpha+k)=\;\frac{512}{3}(2^{20}-1)

[25-Jun-2022-Shift-1]

2
3
4
6

Solution: 👈: Video Solution

Given,

f(x+y)=2 f(x) f(y)

and

f(1)=2

For

x=1

and

y=1

,

f(1+1)=2 f(1) f(1)

\Rightarrow f(2)=2(f(1))^2=2(2)^2=2^3

For

x=1, y=2

f(1+2)=2 f(1) y(2)

\Rightarrow f(3)=2 \cdot 2 \cdot 2^3=2^5

For

x=1, y=3

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

\Rightarrow f(4)=2 \cdot 2 \cdot 2^5=2^7

For

x=1, y=4

f(1+4)=2 f(1) f(4)

\Rightarrow f(5)=2.2 \cdot 2^7=2^9 \dots

Also given

\sum\limits_{k=1}^{10} f(\alpha+k)=\;\frac{512}{3}(2^{20}-1)

\Rightarrow f(\alpha+1)+f(\alpha+2)+f(\alpha+3)+\dots+f(\alpha+10)=\;\frac{512}{3}(2^{20}-1)

\Rightarrow f(\alpha+1)+f(\alpha+2)+f(\alpha+3)+\dots+f(\alpha+10)=\;\frac{2^9((2^2)^{10}-1)}{2^2-1}

This represent a G.P with first term

=2^9

and common ratio

=2^2

\therefore

First term

=f(\alpha+1)=2^9 \dots (2)

From equation

(1), f(5)=2^9

\therefore

From (1) and (2), we get

f(\alpha+1)=2^9=f(5)

\Rightarrow f(\alpha+1)=f(5)

\Rightarrow f(\alpha+1)=f(4+1)

Comparing both sides we get,

\alpha=4

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