To solve this problem, we need to find the value of \(f(g(2))\).
\(g(2) = 3(2)^2 + 2(2) - 3 = 3(4) + 4 - 3 = 12 + 4 - 3 = 13\)
\(4(3(f(x))^2 + 2f(x) - 3) = 3x^2 - 32x + 72\)
\(12(f(x))^2 + 8f(x) - 12 = 3x^2 - 32x + 72\)
Divide by 4:
\(3(f(x))^2 + 2f(x) - 3 = \frac{1}{4}(3x^2 - 32x + 72)\)
Simplify the right-hand side:
\(3(f(x))^2 + 2f(x) - 3 = \frac{3}{4}x^2 - 8x + 18\)
Set \(f(x) = \frac{3}{4}x^2 - 8x + 18\) and evaluate at \(x = 13\).
\(f(13) = \frac{3}{4}(13)^2 - 8(13) + 18\)
\(f(13) = \frac{3}{4}(169) - 104 + 18\)
\(f(13) = \frac{507}{4} - 104 + 18\)
\(f(13) = \frac{507}{4} - \frac{416}{4} + \frac{72}{4}\)
\(f(13) = \frac{507 - 416 + 72}{4}\)
\(f(13) = \frac{579}{4}\)
\(f(13) = \frac{25}{6}\)
Therefore, \(f(g(2)) = \frac{25}{6}\).
The correct answer is \(\frac{25}{6}\).
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 ________.