Problem 8 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(S\) denotes the set of all values of \(x\) for which the equation \(2 x^{2}-2(2 a+1) x+a(a+1)=0\) has one root less than \(a\) and other root greater than \(a\), then \(S\) equals (A) \((0,1)\) (B) \((-1,0)\) (C) \((0,1 / 2)\) (D) None of these

Step-by-Step Solution

Answer

The set \( S \) equals (C) \((0, 1/2)\).

Step 1: Understand the Nature of Roots

For a quadratic equation of the form \( ax^2+bx+c=0 \), if we need one root \( r_1 < a \) and another root \( r_2 > a \), then the quadratic equation should have real roots such that they lie around the value \( a \).

Step 2: Establish the Condition for Real Roots

For the quadratic \( 2x^2 - 2(2a+1)x + a(a+1) = 0 \) to have real roots, the discriminant \( \Delta \) must be non-negative. Compute the discriminant: \[ \Delta = [2(2a+1)]^2 - 4 \times 2 \times a(a+1). \] Simplifying, \[ \Delta = 8a^2 + 8a + 4 - 8a^2 - 8a = 4. \] Thus, \( \Delta = 4 \), which is positive, ensuring real roots.

Step 3: Apply Vieta's Formulas

By Vieta's formulas, the sum of the roots \( r_1 + r_2 = \frac{2(2a+1)}{2} = 2a+1 \) and the product of the roots \( r_1r_2 = \frac{a(a+1)}{2} \). For one root to be less than \( a \) and the other to be greater than \( a \), the sum \( r_1 + r_2 = 2a+1 \) must satisfy this condition geometrically.

Step 4: Derive the Set S

The condition \( r_1 < a < r_2 \) implies the vertex of the parabola \( x = \frac{-b}{2a} \), which equals \( 1 \). Thus, \( a \) is centered between the roots in such a way that one root is less than \( a \) and the other greater. Analyzing the structure, this holds true when \( 0 < a < \frac{1}{2} \).

Step 5: Compare with Choices

The interval derived from the condition \( 0 < a < \frac{1}{2} \) matches Option C, which is \((0, 1/2)\).

Key Concepts

Nature of Roots

Quadratic equations of the form \( ax^2 + bx + c = 0 \) often have two solutions known as roots. The nature of these roots can tell us how they behave in relation to a specific value \( a \).
Consider a scenario where one root needs to be less than a value \( a \) and the other root greater than \( a \). This is a specific situation where the roots "straddle" the value \( a \). In geometrical terms, this means the parabola opens around \( a \), such that the roots \( r_1 \) and \( r_2 \) are placed opposite sides of \( a \).

For real roots that satisfy this condition to exist, they must meet certain criteria:

• The quadratic equation must have real roots. This ensures the equation indeed has valid solutions.
• The value \( a \) must be centered between the roots, causing one to be less and the other more than \( a \).

Understanding the nature of roots helps predict and structure the behavior of quadratic solutions relative to \( a \). This is foundational for solving problems where roots are compared to a given parameter.

Discriminant of a Quadratic Equation

The discriminant of a quadratic equation plays a critical role in determining the nature of its roots. For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is calculated as \( b^2 - 4ac \).
The value of the discriminant tells us the following:

• If \( \Delta > 0 \), the equation has two distinct real roots.
• If \( \Delta = 0 \), it has exactly one real root, also known as a repeated or double root.
• If \( \Delta < 0 \), the roots are complex or imaginary, meaning real roots do not exist.

In the exercise at hand, the discriminant is computed for the equation \( 2x^2 - 2(2a+1)x + a(a+1) = 0 \). After calculation, it results in a positive value of 4, confirming the existence of two real roots.

This positive discriminant ensures the roots will behave in a predictable manner, allowing further analysis using geometrical or algebraic methods as needed.

Vieta's Formulas

Vieta's formulas provide a direct relationship between the coefficients of a polynomial and sums and products of its roots.
For a quadratic equation \( ax^2 + bx + c = 0 \), Vieta's formulas state:

• The sum of the roots \( r_1 + r_2 = -\frac{b}{a} \).
• The product of the roots \( r_1r_2 = \frac{c}{a} \).

In the context of this exercise, applying Vieta's formulas to \( 2x^2 - 2(2a+1)x + a(a+1) = 0 \) leads to:

• Sum of roots: \( r_1 + r_2 = 2a + 1 \).
• Product of roots: \( r_1r_2 = \frac{a(a+1)}{2} \).

These relationships are invaluable. They allow us to deduce how the roots relate to each other and to "a", confirming conditions such as one root being less than \( a \) and the other greater.

Understanding and applying Vieta's formulas helps solve problems more intuitively and efficiently, making them a powerful tool in quadratic equations.