Let f : N rightarrow R be a function such tha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x+y) = 2f(x)f(y)$ and $f(1) = 2$.Let $g(x) = 2f(x)$. Then $g(x+y) = 2f(x+y) = 4f(x)f(y) = g(x)g(y)$.Since $g(1) = 2f(1) = 4 = 2^2$,we hav

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given $f(x+y) = 2f(x)f(y)$ and $f(1) = 2$.Let $g(x) = 2f(x)$. Then $g(x+y) = 2f(x+y) = 4f(x)f(y) = g(x)g(y)$.Since $g(1) = 2f(1) = 4 = 2^2$,we have $g(x) = 2^{2x}$.Thus,$f(x) = \frac{1}{2} g(x) = \frac{1}{2} \cdot 2^{2x} = 2^{2x-1}$.Now,$\sum_{k=1}^{10} f(\alpha+k) = \sum_{k=1}^{10} 2^{2(\alpha+k)-1} = 2^{2\alpha-1} \sum_{k=1}^{10} 2^{2k} = 2^{2\alpha-1} \cdot 4 \cdot \frac{4^{10}-1}{4-1} = 2^{2\alpha+1} \cdot \frac{2^{20}-1}{3}$.We are given $\sum_{k=1}^{10} f(\alpha+k) = \frac{512}{3}(2^{20}-1) = \frac{2^9}{3}(2^{20}-1)$.Equating the two expressions: $2^{2\alpha+1} \cdot \frac{2^{20}-1}{3} = \frac{2^9}{3}(2^{20}-1)$.$2^{2\alpha+1} = 2^9 \implies 2\alpha+1 = 9 \implies 2\alpha = 8 \implies \alpha = 4$.

Let $f : N \rightarrow R$ 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 $\alpha$ for which $\sum_{k=1}^{10} f(\alpha+k)=\frac{512}{3}(2^{20}-1)$ holds,is

Similar Questions

If $f(x + y) = f(x)f(y)$ for all $x$ and $y$ and $f(5) = 2$,$f'(0) = 3$,then $f'(5)$ will be

Find $\sum_{t=1}^{39} f(t)$ if $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$.

Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in R$. If $f(x)$ is differentiable at $x=0$,then which of the following statements are true?

If $f(x+2y, x-2y) = xy$, then $f(x, y)$ is equal to

Let a function $f : R \rightarrow R$ be defined such that $3f(2x^2 - 3x + 5) + 2f(3x^2 - 2x + 4) = x^2 - 7x + 9$ for all $x \in R$. Then the value of $f(5)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module