Problem 76 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(S\) denote the set of all values of \(S\) 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) \(\left(0, \frac{1}{2}\right)\) (D) None of these

Step-by-Step Solution

Answer

The set \( S \) is \( (0, 1) \), so the answer is (A).

Step 1: Identify the condition for one root less than \( a \) and the other root greater

The equation given is a quadratic equation of the form \( f(x) = 2x^2 - 2(2a+1)x + a(a+1) = 0 \). We need one root to be less than \( a \) and the other to be greater than \( a \). This condition implies that \( x = a \) lies between the two roots of the equation.

Step 2: Use Vieta’s formulas for roots

Let the roots of the quadratic equation be \( p \) and \( q \). By Vieta’s formulas, we have \( p + q = \frac{2(2a+1)}{2} = 2a+1 \) and \( pq = \frac{a(a+1)}{2} \). The condition \( p < a < q \) implies that \( a \) must be between the roots.

Step 3: Arrange the inequality for \( x = a \) between the roots

For \( a \) to be between \( p \) and \( q \), the inequality \( (p-a)(q-a) < 0 \) must be satisfied. Rearranging it yields \( pq - a(p+q) + a^2 < 0 \). Substituting the values from Vieta's, we get \[ \frac{a(a+1)}{2} - a(2a+1) + a^2 < 0 \].

Step 4: Simplify the inequality

Simplify the inequality to find \( S \):\[ \frac{a(a+1)}{2} - 2a^2 - a + a^2 < 0 \]\[ \frac{a^2 + a - 4a^2 - 2a + 2a^2}{2} < 0 \]\[ \frac{-a^2 - a}{2} < 0 \]This implies \( a^2 + a > 0 \), or \( a(a+1) > 0 \).

Step 5: Determine the solution set for \( a(a+1) > 0 \)

Consider \( a(a+1) > 0 \):- \( a(a+1) > 0 \) is satisfied for \( a > 0 \) and \( a < -1 \) since the factorization \( a(a+1) = 0 \) yields roots \( a = 0 \) and \( a = -1 \), dividing the real number line into three intervals.- Only the interval \( 0 < a < 1 \) satisfies the condition of having one root less than \( a \) and the other greater than \( a \), leading to the correct interval for \( S \) as \( (0, 1) \).

Key Concepts

Vieta's Formulas

Vieta's formulas provide a nifty way to connect the coefficients of a polynomial to the sums and products of its roots. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), if \( p \) and \( q \) are roots, then:

• Sum of roots: \( p + q = -\frac{b}{a} \)
• Product of roots: \( pq = \frac{c}{a} \)

In the equation \( 2x^2 - 2(2a+1)x + a(a+1) = 0 \), applying Vieta's formulas gives us:

• Sum of roots: \( p + q = 2a + 1 \).
• Product of roots: \( pq = \frac{a(a+1)}{2} \).

These formulas are crucial for deriving other properties of the roots, such as inequalities, which help us find sets of \( a \) that satisfy certain conditions.

Root Inequalities

Root inequalities involve conditions that the roots of a quadratic must satisfy. In this problem, we want one root to be less than \( a \) and the other to be greater than \( a \). This creates a situation where \( a \) is placed between the two roots \( p \) and \( q \). To express this, we use the inequality \( (p - a)(q - a) < 0 \). This means that if \( a \) is between the roots, one term is positive and the other is negative, making their product negative.

To work with this, we substitute the expressions from Vieta's formulas into the inequality \( pq - a(p+q) + a^2 < 0 \). This step allows us to simplify and identify conditions that must hold for \( a \). Root inequalities are very useful in specifying the nature and position of the roots relative to a specific point, such as \( a \) in this case.

Quadratic Inequality Solutions

Solving quadratic inequalities can be a bit tricky but is essential to find intervals where specific conditions hold. In our quadratic equation example, the solution revolves around solving the inequality:

• \( \frac{a^2 + a - 4a^2 - 2a + 2a^2}{2} < 0 \)
• Simplifying to: \( \frac{-a^2 - a}{2} < 0 \)
• Resulting inequality: \( a^2 + a > 0 \)

Now, to solve \( a(a+1) > 0 \), observe the critical points: \( a = 0 \) and \( a = -1 \). These points divide the number line into intervals. The inequality holds true for \( a > 0 \) or \( a < -1 \). However, only the interval \( (0, 1) \) satisfies the original condition where \( one \) root is less than \( a \) and the other is greater.

This method of analyzing intervals is a powerful way to determine feasible solutions to quadratic inequalities and ensures that all conditions in the problem statement are met.