Problem 84 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\frac{1}{1-x}\), then the points of discontinuity of the function \(f^{3 n}(x)\), where \(f^{n}=\) fof \(\ldots\) of \((n\) times \()\), are (A) \(x=2\) (B) \(x=0\) (C) \(x=1\) (D) continuous everywhere

Step-by-Step Solution

Answer

The function is discontinuous at \(x = 1\).

Step 1: Understanding Composition

The function \(f(x) = \frac{1}{1-x}\) needs to be composed \(n\) times to form \(f^n(x)\). The composition \(f^n(x)\) implies applying the function \(f(x)\) repeatedly on itself \(n\) times.

Step 2: First Composition Example

To illustrate, let's compute the first iteration: \(f^2(x) = f(f(x)) = f\left(\frac{1}{1-x}\right) = \frac{1}{1-\left(\frac{1}{1-x}\right)}\). We simplify this to \(f^2(x) = \frac{1-x}{-x}\).

Step 3: Identifying the Denominators

The points of discontinuity arise when the denominator of the expression equals zero, making the function undefined. Therefore, for each composition, observe where the denominator (a function of \(x\)) can be zero.

Step 4: General Form of Discontinuity

When we calculate the denominator over repeated compositions, each iteration includes a factor of \(1-x\). Since \(x\) is in the denominator for each step, the further compositions like \(f^3, f^4, \ldots, f^{3n}\) will always include another factor of \(1-x\).

Step 5: Discontinuity Check Finalization

Realize that for each \(f^n(x)\), if \(x = 1\), then the entire denominator eventually simplifies to zero. Hence, for all \(f^n(x)\), \(x = 1\) makes the function undefined.

Key Concepts

Function Composition

Function composition is the process of applying one function to the results of another function. It is denoted as \((f \, \circ \, g)(x)\), meaning you first apply function \(g\) to \(x\), then function \(f\) to the result of \(g(x)\). It requires performing the operations from right to left.

• In the given problem, we're dealing with the composition of a single function multiple times, termed as \(f^n(x)\). The notation signifies applying the function \(f\) to its own output repeatedly \(n\) times.
• For example, if \(f(x) = \frac{1}{1-x}\), then composing it twice, \(f^2(x)\), involves first calculating \(f(x) = \frac{1}{1-x}\) and then plugging this back into \(f\) to get \(f(f(x))\).

A key aspect of function composition is understanding how the operations of one function affect the next, particularly when dealing with functions like rational functions which have undefined points that lead to discontinuities.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. They are generally of the form \(R(x) = \frac{P(x)}{Q(x)}\). The domain of a rational function excludes values that make the denominator zero, as these would make the function undefined.

• In the function \(f(x) = \frac{1}{1-x}\), we see that the denominator polynomial is \(Q(x) = 1-x\). This setup automatically highlights values of \(x\) that would create division by zero errors, specifically \(x = 1\).
• While studying rational functions, be mindful of the expressions underneath to trace where they might become problematic. Even complex compositions still abide by the fundamental rules of rational expressions.

By understanding how these polynomials interact, especially through operations like multiplication or subtraction involved in the composition, one can predict and locate these points of discontinuity which is crucial when dealing with compositions.

Points of Discontinuity

A point of discontinuity in a function is a specific value of \(x\) where the function is not defined. For rational functions, these occur where the denominator equals zero.

• In our exercise, we determined the points of discontinuity by setting the denominator of \(f^n(x)\) to zero. As \(f(x) = \frac{1}{1-x}\), the discontinuity at \(x = 1\) is clear from the original function.
• With repeated compositions like \(f^2(x)\) or \(f^{3n}(x)\), the denominator continues to incorporate powers or factors of \(1-x\) . Thus, the point \(x=1\) creates a zero denominator in all compositions, leading to discontinuity.
• Assessing these through simplifying and calculating denotations step by step reveals that despite repeated compositions or alterations, the core issue of division by zero caused by \(x=1\) persists.

Ensuring a strong grasp of where functions become undefined can help preemptively address potential issues seen in problem scenarios or solutions.