The correct answer is (C) : \(\max f(x)>\max g(x)\)
\(f'(x)=2+\frac{1}{1+x^2},\ g'(x)=\frac{1}{\sqrt{1+x^2}}\)
Both does not have critical values
\(f(0)=0,f(3)=6+\tan^{-1}(3)\)
\(g(0)=0,g(3)=\log(3+\sqrt{10})\)
Let h(x) = f(x) - g(x)
\(h'(x) > 0∀x∈(0,3)\)
∴ h(x) is increasing function
Computing derivatives: \[ g'(x) = \frac{1}{\sqrt{1 + x^2} + x} \] Since \( g(x) \) is always increasing in \( [0,3] \), \[ \min g'(x) = \frac{1}{\sqrt{10} + 3}, \quad \max g'(x) = \frac{1}{\sqrt{1} + 0} = 1 \] For \( f(x) \), \[ f(x) = 2x + \tan^{-1} x \] Since \( f(x) \) is increasing, \[ \max f(x) = 6 + \tan^{-1} 3 \] For \( g(x) \), \[ \max g(x) = \ln (3 + \sqrt{10}) \] \[ 6 + \tan^{-1} 3 > \ln (3 + \sqrt{10}) \]
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.
