JEE Main 2022 — Functions Question with Solution

Given,

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

and $$f(1)=2$$

For x = 1 and y = 1,

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

$$\Rightarrow f(2)=2{\left(f(1)\right)}^2=2(2{)}^2=2^3$$

For x = 1, y = 2,

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

$$\Rightarrow f(3)=2.2.2^3=2^5$$

For x = 1, y = 3,

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

$$\Rightarrow f(4)=2.2.2^5=2^7$$

For x = 1, y = 4,

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

$$\Rightarrow f(5)=2.2.2^7=2^9$$ ..... (1)

Also given

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

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

$$\Rightarrow f(\alpha +1)+f(\alpha +2)+f(\alpha +3)+....+f(\alpha +10)=\frac{2^9\left({(2^2)}^{10}-1\right)}{2^2-1}$$

This represent a G.P with first term = 2⁹ and common ratio = 2²

$$\therefore$$ First term $$=f(\alpha +1)=2^9$$ ..... (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,

$$\alpha =4$$