QJEEM20Mae701: JEE Main 2020 Mathematics - Class 11 JEEM20Mae701 Solution: Suppose that a function f : R R satisfies f(x + y) = f(x)f(y)…

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

put x = y = 1
$\therefore$ f(2) = (ƒ(1))² = 3²

put x = 2, y = 1
$\therefore$ f(3) = (ƒ(1))³ = 3³

Similarly f(x) = 3^x
$\Rightarrow$ f(i) = 3ⁱ

Given, $\sum\limits_{i = 1}^n {f(i)} = 363$

$\Rightarrow$ 3 + 3² + 3³ +.... + 3ⁿ = 363

$\Rightarrow$ ${{3\left( {{3^n} - 1} \right)} \over {3 - 1}}$ = 363

$\Rightarrow$ 3ⁿ - 1 = ${{363 \times 2} \over 3}$ = 242

$\Rightarrow$ 3ⁿ = 243 = 3⁵

$\Rightarrow$ n = 5