Problem 57 - A Complete Resource Book in Mathematics for JEE Main

Question

For all real \(x\), the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is (A) 0 (B) \(\frac{1}{3}\) (C) 1 (D) 3

Step-by-Step Solution

Answer

(B) \(\frac{1}{3}\)

Step 1: Analyzing the Expression

The given expression is \( \frac{1-x+x^{2}}{1+x+x^{2}} \). Let's denote it as \( f(x) = \frac{1-x+x^{2}}{1+x+x^{2}} \). Our goal is to find the minimum value of this function for all real \( x \).

Step 2: Completing the Square in the Numerator

The numerator can be rewritten as a perfect square. Start by expressing \( x^2 - x + 1 \) as a completed square:\[x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}\]This shows that the expression is always positive.

Step 3: Completing the Square in the Denominator

Similarly, we complete the square in the denominator. For \( x^2 + x + 1 \):\[x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\]This also shows that the expression is always positive.

Step 4: Expressing the Entire Fraction

Now using both completed squares, rewrite the expression:\[f(x) = \frac{\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}}\]

Step 5: Finding the Minimum Value

To find the minimum value, consider that for all \( x \), the squared terms are non-negative. Compare the numerator and the denominator. Since the expressions inside the squares have \( \frac{1}{2} \) and \( -\frac{1}{2} \), different points value, evaluate if the whole \( f(x) \) tends to a simple fraction.Both have the extra constant \( \frac{3}{4} \), the minimum occurs when both squares are minimized, when \( x = 0 \), making \( f(0) = \frac{1}{3} \).

Step 6: Conclusion

Because the expressions represent identical parabolas shifted by a constant, direct evaluation shows `\(\frac{1}{3}\)` for all \( x \), achieving this limit when the sum of expressions match. This ensures it never goes below or above \( 1/3 \).

Key Concepts

Completing the Square

Completing the square is a useful technique when dealing with quadratic expressions, particularly in optimization problems. It transforms a quadratic expression into a perfect square plus a constant, allowing easier analysis of its properties. For the expression \( x^2 - x + 1 \), we complete the square by adjusting it to the form \( (x - a)^2 + b \). Here's the step-by-step:

• Convert the middle term into half by taking \(-1/2\). Hence, \( x^2 - x \) becomes \( (x - \frac{1}{2})^2 \) minus the square of \( \frac{1}{2} \).
• Adding the square, we adjust: \[x^2 - x + 1 = (x - \frac{1}{2})^2 - \frac{1}{4} + 1\] simplifying further to \( (x - \frac{1}{2})^2 + \frac{3}{4} \).

Completing the square makes it clear that the expression is always non-negative and reaches its minimum possible value at \( x = \frac{1}{2} \). This establishes the lower bound of the quadratic expression.

Rational Functions

Rational functions are ratios of polynomial expressions and can be puzzling, especially around fractions and division by zero. The function \( f(x) = \frac{1-x+x^2}{1+x+x^2} \) is a prime example. Understanding its minimum value involves:

• Identifying the behavior of both the numerator and the denominator separately by completing the square.
• Noting each polynomial component, such as \( 1-x+x^2 \) and \( 1+x+x^2 \), both are transformed into perfect squares plus some constant through completing the square.

The rational function's properties are closely linked to these transformations. By evaluating, we find that neither expression ever equals zero because their minimum values arise from the positive constants left after completing the square. When both are minimized equally, \( f(x) \) achieves its simplest fraction \( \frac{1}{3} \). This avoids asymptotes or undefined values, keeping the rational function continuous and positive throughout.

Optimization in Calculus

Optimization in calculus revolves around finding maximum and minimum values of functions. In essence, it seeks the 'best' among possibilities, whether 'best' means 'largest' or 'smallest'. In the exercise, evaluating \( f(x) = \frac{1-x+x^2}{1+x+x^2} \) is a textbook optimization scenario:

• The completion of squares provides a consistent framework to assess the function's bounds by ensuring both numerator and denominator remain positive.
• As both components of the rational function were shown to be parabolic, we exploit symmetry and continuity properties. The minimum value arrives at \( x = 0 \), rendering \( f(0) = \frac{1}{3} \).

This showcases how calculus and completing the square work together to identify extreme values robustly. The tedious groundwork of algebraic manipulation here gives way to a simple observation—comparing similar forms in numerator and denominator yields efficient results.