Let
\(f(x)=\frac{x−1}{x+1},x∈R− \left\{0,−1,1\right\}\)
If ƒn+1(x) = ƒ(ƒn(x)) for all n∈N, then ƒ6(6) + ƒ7(7) is equal to :
\(\frac{7}{6}\)
\(-\frac{3}{2}\)
\(\frac{7}{12}\)
\(-\frac{11}{12}\)
The correct answer is (B) : \(-\frac{3}{2}\)
\(f(x) = \frac{x - 1}{x + 1} \Rightarrow f(f(x)) = \frac{\left(\frac{x - 1}{x + 1}\right) - 1}{\left(\frac{x - 1}{x + 1}\right) + 1} = -\frac{1}{x}\)
\( \Rightarrow f^3(x) = -\frac{x + 1}{x - 1} \Rightarrow f^4(x) = -\frac{\left(\frac{x - 1}{x + 1} + 1\right)}{\left(\frac{x - 1}{x + 1} - 1\right)} = x\)
So, ƒ6(6) + ƒ7(7) = ƒ2(6) + ƒ3(7)
\(= -\frac{1}{6}-\frac{7+1}{7-1}= - \frac{9}{6}=-\frac{3}{2}\)
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 : \(A_2 \rightleftharpoons 2A\)
The composite function refers to the resultant value of two specified functions. When the output derived from the application of a function with a second independent variable function becomes the input of the third function, then it is called a composite function. Also, whose scope includes the values of the independent variable for which the result of the first function is placed in the domain of the second.
In Mathematics, the composition of a function is a process, where two functions say f and g create a new function say h in such a way that h (x) = g (f (x)). Here, we can see function g applies to the function of x i.e., f (x)
Let f: A → B and g: B → C are two functions.
So, the composition of f and g, denoted by gof, is known as the function:
g of: A → C given by gof (x) = g (f (x)), A x ∈ A.