JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Relations And Functions XII

Suppose

f : R \to (0, \infty)

be a differentiable function such that

5 f (x + y) = f (x) \cdot f (y), \forall x, y \in R

. If

f (3) = 320

, then

\sum_{n = 0}^{5} f (n)

is equal to:

Answer: (C)

Solution

Putting

x = y = 0

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

gives

5 f (0) = f (0)^{2}

, so

f (0) = 5

. Setting

y = 1

gives

f (x + 1) = \frac{f ( 1 )}{5} f (x)

, a geometric progression with ratio

r = \frac{f ( 1 )}{5}

, hence

f (n) = 5 r^{n}

. From

f (3) = 5 r^{3} = 320

r^{3} = 64

and

r = 4

, so

f (n) = 5 \cdot 4^{n}

. Therefore

n = 0 \sum 5 f (n) = 5 n = 0 \sum 5 4^{n} = 5 \cdot \frac{4 ^{6} - 1}{3} = 5 \cdot 1365 = 6825.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option

Let $f : R - {0} \to (- \infty, 1)$ be a polynomial of degree 2 , satisfying $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ . If $f (K) = - 2 K$ , then the sum of squares of all possible values of …
JEE Main 2025View solution →
Let $f (x) = \frac{2 ^{x + 2} + 16}{2 ^{2 x + 1} + 2 ^{x + 4} + 32}$ . Then the value of $8 (f (\frac{1}{15}) + f (\frac{2}{15}) + \dots + f (\frac{59}{15}))$ is equal to
JEE Main 2025View solution →
If $f (x) = \frac{( t a n 1 ^{o} ) x + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{o} )}, x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is
JEE Main 2023View solution →
Let $f : R \to R$ be a function defined by $f (x) = (2 (1 - \frac{x ^{25}}{2}) (2 + x^{25}))^{\frac{1}{50}}$ . If the function $g (x) = f (f (f (x))) + f (f (x))$ , then the greatest integer less than or equal to …
JEE Main 2022View solution →
Let $a, b, c \in R$ . If $f (x) = a x^{2} + b x + c$ is such that $a + b + c = 3$ and
$f (x + y) = f (x) + f (y) + x y, \forall x, y \in R$ , then $\sum_{n = 1}^{10} f (n)$ is equal …
JEE Main 2017View solution →