Let $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{R}$ be a function such that $\mathrm{f}(\mathrm{x}+\mathrm{y})=2 \mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})$ for natural numbers x and y . If $f(1)=2$, then the value of $\alpha$ for which \[ \sum_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right) \] holds, is

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Sets Relations and Functions

Let

f : N \to R

be a function such that

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

for natural numbers x and y . If

f (1) = 2

, then the value of

α

for which

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

holds, is

Answer: (C)

From the functional equation

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

with

f (1) = 2

, we find

f (2) = 2 f^{2} (1) = 2^{3}

f (3) = 2 f (2) f (1) = 2^{5}

, and in general

f (n) = 2^{2 n - 1}

. Thus

\sum_{k = 1}^{10} f (α + k) = 2 f (α) \sum_{k = 1}^{10} f (k) = 2 \cdot 2^{2 α - 1} (2 + 2^{3} + \dots + 2^{19}) = 2^{2 α} \cdot 2 \cdot \frac{2 ^{20} - 1}{3}

. Equating to

\frac{512}{3} (2^{20} - 1) = \frac{2 ^{9}}{3} (2^{20} - 1)

, we get

2^{2 α + 1} = 2^{9}

, so

2 α + 1 = 9

and

α = 4

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option