\(f(3x)–f(x)=x\) …(1)
\(x→\frac x3\)
\(f(x)−f(\frac x3)=\frac x3\) ⋯(2)
Again \(x→\frac x3\)
\(f(\frac x3)−f(\frac x9)=\frac {x}{3^2}\) ⋯(3)
Similarly \(f(\frac {x}{3^{n−2}})−f(\frac {x}{3^{n−1}})=\frac {x}{3^{n−1}}\) ⋅⋅⋅⋅⋅⋅⋅(n)
Adding all these and applying \(n→∞\)
\(\lim\limits _{n→∞}(f(3x)−f(\frac {x}{3^{n−1}}))=x(1+\frac 13+\frac {1}{3^2}+⋯)\)
\(f(3x)−f(0)=\frac {3x}{2}\)
Putting \(x=\frac 83\)
\(f(8) – f(0) = 4\)
\(⇒ f(0) = 3\)
Putting \(x=\frac {14}{3}\)
\(f(14)–3=7\)
\(f(14)=10\)
So, the correct option is (B): \(10\)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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.