Question

Let $f(x) = \dfrac{{Kx}}{{x + 1}}\left( {x \ne - 1} \right)$ then the value of K for which (fof) (x) =x is
A. 1
B. – 1
C. 2
D. $\sqrt 2$

Answer 1

Hint : Here, given function is a composite function, so first find the value of composite function and equate it with value given i.e., x. then find the value f x and hence K. Here, function itself become a domain i.e., the function with replace x and by simplifying we can get the value of composite function.

Complete step-by-step answer :
Given $f(x) = \dfrac{{Kx}}{{x + 1}}\left( {x \ne - 1} \right)$
Here, range or output one f(x) becomes the domain of f in finding fof (x).
fof(x) = $f\left( {\dfrac{{Kx}}{{x + 1}}} \right) = \dfrac{{K\left( {\dfrac{{Kx}}{{x + 1}}} \right)}}{{\left( {\dfrac{{Kx}}{{x + 1}}} \right) + 1}} = \dfrac{{{K^2}x}}{{Kx + x + 1}}$
Also given that (fof) (x) = x
$\dfrac{{{K^2}x}}{{Kx + x + 1}} = x$
On cross multiplying
$K{x^2} = K{x^2} + {x^2} + x$
This is a quadratic equation, cancel the common terms of both sides
$\Rightarrow {x^2} + x = 0$
Factorisation method
⇒ x(x + 1) = 0
⇒ x = 0 or x = − 1
But given that x ≠ − 1, so x = 0
For x = 0, K can be any real number.

Note : In these types of questions, apply the concept of composite function. In a composite function, the range of one function becomes the domain of another function.
Composite function: A function whose values are obtained from two given functions containing same independent variable, by applying one function as an independent variable and then applying the second function to the result and whose domain consists of those values of the independent variable for which the result yielded by the first function lies in the domain of the second. It should be noted that two functions must be in some relation. If both functions have different variables and their variables have no relation then their composite function can be determined.