Problem 33 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=1+3 x^{2}+3^{2} x^{4}+\ldots+3^{30} \cdot x^{60} .\) Then \(f(x)\) has (A) atleast one maximum (B) exactly one maximum (C) atleast one minimum (D) exactly one minimum

Step-by-Step Solution

Answer

Option (D) exactly one minimum.

Step 1: Identify the Type of Series

The function given is a power series, where each term is of the form \(3^n \cdot x^{2n}\) for \(n = 0, 1, 2, \, ... , \, 15\). Notice that this is a geometric series with the first term \(a = 1\) and the common ratio \(r = 3x^2\).

Step 2: Formula for the Geometric Series

The sum of a geometric series \(a + ar + ar^2 + \, ... \, + ar^n\) can be calculated using the formula: \[ S_n = \frac{a(1 - r^{n+1})}{1 - r} \]with the condition \(|r| < 1\). For this series, \(a = 1\) and \(r = 3x^2\), and there are 16 terms (from \(n = 0\) to \(n = 15\)).

Step 3: Calculate the sum of the series

Substitute \(a = 1\), \(r = 3x^2\), and \(n = 15\) into the geometric series sum formula to find:\[f(x) = \frac{1 - (3x^2)^{16}}{1 - 3x^2}\]This function defines \(f(x)\) as long as \(|3x^2| < 1\).

Step 4: Analyze Behavior of the Series

Because the denominator is \(1 - 3x^2\), this is a rational function with vertical asymptotes at \(3x^2 = 1\) or \(x = \pm\frac{1}{\sqrt{3}}\). Between these asymptotes, the series behaves similar to quadratic functions with a parabolic shape.

Step 5: Identify Maximum or Minimum Points

To find critical points, differentiate the function \(f(x)\) and set \(f'(x) = 0\). Solving for \(x\) will give the critical points. Due to symmetry and the nature of the problem, there is typically exactly one minimum at \(x = 0\) as the function is derived from squares of terms, thus always non-negative.

Step 6: Conclusion: Presence of Maximum or Minimum

The function \(f(x)\), based on its geometric series formation and parabolic nature, reaches a minimum value at \(x = 0\), which is the only place where the function achieves this least value, proving option (D) correct.

Key Concepts

Power Series

A power series is essentially a series of the form \( a_0 + a_1x + a_2x^2 + \ldots \) where each term involves a power of \( x \). The series provided in the exercise is special because the terms are scaled by a power of 3 and it involves even powers of \( x \), specifically \( 3^n \cdot x^{2n} \).
This specific setup is important because it reveals how the series behaves over different values of \( x \).
Power series can remind us of polynomials but with potentially infinite terms. However, our series here ends at the term \( 3^{15} \cdot x^{30} \), making it a finite power series.

• The use of powers of both \( 3 \) and \( x^2 \) makes it a geometric series.
• Such a series converges within certain limits of \( x \), specifically when \( |3x^2| < 1 \).

Understanding these aspects helps us identify where the function increases, decreases, or remains stable.

Rational Function

A rational function is any function that can be expressed as the quotient of two polynomials.
In this problem, after summing the power series, we end up with the function \( f(x) = \frac{1 - (3x^2)^{16}}{1 - 3x^2} \). This structure is why it's considered a rational function.
Like other rational functions, \( f(x) \) has properties that cause its behavior to change at specific points, known as asymptotes.

• The denominator, \( 1 - 3x^2 \), indicates that there are vertical asymptotes where this expression equals zero.
• This leads to the critical values \( x = \pm \frac{1}{\sqrt{3}} \), where the function is undefined.

These asymptotes divide the function into intervals where it must be analyzed separately.

Critical Points

Critical points are values of \( x \) where the derivative of the function \( f'(x) \) equals zero or where the derivative does not exist.
These points help identify where the function might have local minima or maxima, or possibly inflection points.
To find them here, we differentiate the function derived from the series. Solving \( f'(x) = 0 \) reveals potential minima or maxima.

• The setup and squaring in our function suggest that the critical point at \( x = 0 \) will be a minimum.
• Given the non-negative nature of powers of \( x^2 \), the potential for negative values is nullified.

Understanding critical points aids in determining the overall shape and extremities of the function.

Parabolic Shape

A parabolic shape is commonly related to quadratic functions, which exhibit a specific "U" shaped curve.
In our function, the terms like \( x^{2n} \) resemble quadratic components because they involve even powers. This mirrors the behavior of a parabola.
Especially between the vertical asymptotes, the function has characteristics similar to a parabola.

• The function is symmetric around the y-axis due to the even powers in each \( x^{2n} \) term.
• The overall shape indicates minima at \( x = 0 \), which aligns with the finding of an exact minimum in the series problem.

Understanding this shape provides insights into how the function will behave across its domain, guiding predictions of its maximum and minimum positions.