\(\frac {1}{49}\)
\(\frac {3}{49}\)
\(\frac {43}{49}\)
\(\frac {91}{49}\)
\(ƒ(x) = 3x^2 + 1\)
\(ƒ^′(x)\) is bijective function
and \(ƒ(g(x)) = x⇒g(x\)) is inverse of \(ƒ(x)\)
\(g(ƒ(x)) = x\)
\(g^′(f(x)).f^′(x) = 1\)
\(g^′(f(x)) = \frac {1}{3x^2+1}\)
Put \(x = 4\) we get
\(g^′(63)=\frac {1}{49}\)
So, the correct option is (A): \(\frac {1}{49}\)
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:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ 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.