\(\frac {1}{49}\)
\(\frac {3}{49}\)
\(\frac {43}{49}\)
\(\frac {91}{49}\)
\(ƒ(x) = 3x^2 + 1\)
\(ƒ^′(x)\) 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 $ A = \{0, 1, 2, 3, 4, 5, 6\} $ and $ R_1 = \{(x, y): \max(x, y) \in \{3, 4 \}$. Consider the two statements:
Statement 1: Total number of elements in $ R_1 $ is 18.
Statement 2: $ R $ is symmetric but not reflexive and transitive.
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $
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.