We are given the functional equation: $f (x)+3f (\frac{24}{x}) = 4x$, where $x \neq 0$. We need to find the value of $f (3) + f (8)$.
Substitute $x = 3$ into the equation:
$f (3)+3f (\frac{24}{3}) = 4(3)$
$f (3)+3f (8) = 12 \quad \cdots (1)$
Substitute $x = 8$ into the equation:
$f (8)+3f (\frac{24}{8}) = 4(8)$
$f (8)+3f (3) = 32 \quad \cdots (2)$
We now have a system of two linear equations with two variables, $f(3)$ and $f(8)$:
Multiply Equation (1) by 3:
$3(f(3) + 3f(8)) = 3(12)$
$3f(3) + 9f(8) = 36 \quad \cdots (3)$
Subtract Equation (2) from Equation (3):
$(3f(3) + 9f(8)) - (3f(3) + f(8)) = 36 - 32$
$8f(8) = 4
$f(8) = \frac{4}{8} = \frac{1}{2}
Substitute the value of $f(8)$ back into Equation (1):
$f(3) + 3(\frac{1}{2}) = 12$
$f(3) + \frac{3}{2} = 12$
$f(3) = 12 - \frac{3}{2} = \frac{24}{2} - \frac{3}{2} = \frac{21}{2}
We need to find $f(3) + f(8)$.
$f(3) + f(8) = \frac{21}{2} + \frac{1}{2} = \frac{22}{2} = 11
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.