The function \( f_1(x) \) is given as:
\[ f_1(x) = \frac{3x + 2}{2x + 3} \]
First Iteration:
\[ f_1(x) = \frac{3x + 2}{2x + 3} \]
Second Iteration:
\[ f_2(x) = f_1(f_1(x)) = \frac{13x + 12}{12x + 13} \]
Notice how the numerator and denominator coefficients evolve as the function is iterated.
Third Iteration:
\[ f_3(x) = f_1(f_2(x)) = \frac{63x + 62}{62x + 63} \]
The pattern becomes clearer as we proceed further. Observe the symmetry in the coefficients.
Fifth Iteration:
\[ f_5(x) = \frac{1563x + 1562}{1562x + 1563} \]
This results from applying the function iteratively, maintaining the structure of coefficients in the numerator and denominator.
Given Condition:
The condition provided is:
\[ a + b = 3125 \]
Here, \( a \) and \( b \) are the coefficients of \( x \) and the constant term in the numerator of \( f_5(x) \), respectively.
Conclusion:
We have \( a = 1563 \) and \( b = 1562 \), so:
\[ a + b = 1563 + 1562 = 3125 \]
Thus, the given condition is satisfied.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
