JEE Main 25 June 2022 Shift 1 Solved Paper

Given,
f(x+y)=2f(x)f(y)
and f(1)=2
For x=1 and y=1,
f(1+1)=2f(1)f(1)
⇒f(2)=2(f(1))2=2(2)2=23
For x=1,y=2
f(1+2)=2f(1)y(2)
⇒f(3)=2⋅2⋅23=25
For x=1,y=3
f(1+3)=2f(1)f(3)
⇒f(4)=2⋅2⋅25=27
For x=1,y=4
f(1+4)=2f(1)f(4)
⇒f(5)=2.2⋅27=29....
Also given

10

∑

k=1

f(α+k)=‌

512

3

(220−1)
⇒f(α+1)+f(α+2)+f(α+3)+...+f(α+10)=‌

512

3

(220−1)
⇒f(α+1)+f(α+2)+f(α+3)+...+f(α+10)=‌

29((22)10−1)

22−1

This represent a G.P with first term =29 and common ratio =22
∴ First term =f(α+1)=29....(2)
From equation (1),f(5)=29
∴ From (1) and (2), we get

f(α+1)=29=f(5)
⇒f(α+1)=f(5)
⇒f(α+1)=f(4+1)
Comparing both sides we get, α=4