Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

  • A
    $4$
  • B
    $16$
  • C
    $2$
  • D
    $3$

Explore More

Similar Questions

Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{2x+1}{3}$. If $\alpha$ is an element in the domain of $f$ whose image is $\frac{1}{\alpha}$,then the sum of all possible values of such $\alpha$ is

Let $f : R \rightarrow R$ be a continuous function such that $f(3x) - f(x) = x$. If $f(8) = 7$,then $f(14)$ is equal to.

Let $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f\left( \frac{x}{y} \right) = f(x) - f(y)$ for all $x, y > 0$ and $f(e) = 1$. Then:

Given the function $f(x) = \frac{a^x + a^{-x}}{2}, (a > 2)$,then $f(x + y) + f(x - y)$ is equal to

Let $f : R - \{0, 1\} \rightarrow R$ be a function such that $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. Then $f(2)$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo