Let f : N \to R be a function such that f(x + y) = 2f(x)f(y) for natural numbers x and y. If f(1) = 2, then the value of \alpha for which \sum\limits_k = 1^10 f(\alpha + k) = 512 \over 3(2^20 - 1) holds, is :

Functions PYQ — JEE Main 2022 · 25 Jun · Shift 1 · Q25 with Solution

View written solutionFree

Correct answer: C

We are given $f(x+y)=2f(x)f(y), \quad x,y\in \mathbb N$ and $f(1)=2.$

We compute successively:

For $y=1$ , $f(x+1)=2f(x)f(1)=2f(x)\cdot 2=4f(x).$

So the function satisfies the recurrence $f(n+1)=4f(n).$

Since $f(1)=2$ , we get $f(2)=4\cdot 2=8,$ $f(3)=4\cdot 8=32,$ and in general, $f(n)=2\cdot 4^{n-1}=2^{2n-1}.$

Check: $2f(x)f(y)=2\cdot 2^{2x-1}\cdot 2^{2y-1}=2^{2x+2y-1}=2^{2(x+y)-1}=f(x+y),$ so this is correct.

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

Now, $f(\alpha+k)=2^{2(\alpha+k)-1}=2^{2\alpha+2k-1}.$

Hence

Factor out $2^{2\alpha+1}$ since $2^{2\alpha+2k-1}=2^{2\alpha+1}\cdot 2^{2k-2}=2^{2\alpha+1}\cdot 4^{k-1}.$ Thus,

This is a GP: $\sum_{k=1}^{10}4^{k-1}=1+4+4^2+\cdots+4^9=\frac{4^{10}-1}{4-1}=\frac{2^{20}-1}{3}.$

Therefore, $\sum_{k=1}^{10} f(\alpha+k)=2^{2\alpha+1}\cdot \frac{2^{20}-1}{3}.$

Given, $2^{2\alpha+1}\cdot \frac{2^{20}-1}{3}=\frac{512}{3}(2^{20}-1).$

Cancel the common factor $\frac{2^{20}-1}{3}$ : $2^{2\alpha+1}=512=2^9.$

So, $2\alpha+1=9$ $2\alpha=8$ $\alpha=4.$

So the correct option is C.