Problem 130 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(k\) be any point such that \(k \in R\) and \(\alpha, \beta\) are the roots of the quadratic equation \(f(x)=a x^{2}+b x+c=0\). If \(k\) lies outside and is less than both the roots then the equation must have real and distinct roots and the sign of \(f(k)\) is same as the sign of ' \(a\) '. Also, \(k\) is less than the \(x\)-coordinate of the vertex of the parabola \(y=a x^{2}+b x+c\). If \(k\) lies between both the roots, then the sign of \(f(k)\) is opposite to the sign of ' \(a\) '. If \(k\) lies outside and is greater than both the roots, then the sign of \(f(k)\) is same as the sign of ' \(a\) '. Also, \(k\) is greater than the \(x\)-coordinate of the vertex of the parabola \(y=a x^{2}+b x+c\). If both the roots of the equation lie between two real numbers \(k_{1}\) and \(k_{2}\), then equation must have real and distinct roots and the sign of \(f\left(k_{1}\right)\) and \(f\left(k_{2}\right)\) is same as the sign of \(a\). Also, the \(x\)-coordinate of the vertex of the parabola \(y=a x^{2}+b x+c\) lies between \(k_{1}\) and \(k_{2}\). The values of \(a\) for which each one of the roots of \(x^{2}-4 a x+2 a^{2}-3 a+5=0\) is greater than 2, are (A) \(a \in(1, \infty)\) (B) \(a=1\) (C) \(a \in(-\infty, 1)\) (D) \(a \in(9 / 2, \infty)\)

Step-by-Step Solution

Answer

The values of \(a\) are \((1, \infty)\).

Step 1: Identify the Quadratic Equation

The quadratic equation given is \(x^2 - 4ax + (2a^2 - 3a + 5) = 0\). We need to find conditions on \(a\) such that both roots of this equation are greater than 2.

Step 2: Transform Roots Conditions

We set the conditions \(\alpha > 2\) and \(\beta > 2\), where \(\alpha\) and \(\beta\) are the roots. This leads to two inequalities \(\alpha - 2 > 0\) and \(\beta - 2 > 0\). Since both roots must satisfy the inequalities, we can consider shifting the equation by substituting \(x = x + 2\).

Step 3: Substitute in Transformed Equation

Substitute \(x = y + 2\) into the quadratic equation to get \((y+2)^2 - 4a(y+2) + (2a^2 - 3a + 5) = 0\). Simplify to get a new quadratic in \(y\). This manipulation ensures that we examine conditions for \(y > 0\) which corresponds to both roots of the original equation being greater than 2.

Step 4: Simplify and Collect Like Terms

After substitution and simplification, we obtain a new quadratic equation in \(y\). Group like terms to form \(y^2 + (-4a + 4)y + (4 - 8a + 2a^2 - 3a + 5) = 0\) and further simplify the quadratic.

Step 5: Analyze Discriminant Condition

For the original quadratic to have real roots, we need the discriminant to be non-negative: \(b^2 - 4ac \geq 0\). For the translated equation, maintain this condition while ensuring that both roots \(y\) are positive.

Step 6: Determine the Correct Range for \(a\)

Apply the conditions \(\alpha > 2\) and \(\beta > 2\) on \(b^2 - 4ac \geq 0\) and hence solve for possible values of \(a\). Further, check the original condition that would then imply the transformed roots (\(y\)) satisfy \(y > 0\). By verifying these conditions, we find \(a \in (1, \infty)\).

Step 7: Verify with Options

Among the given options, verify that only option (A) satisfies our derived condition for \(a\). Thus, given the roots are greater than 2, \(a\) must lie within the interval \((1, \infty)\).

Key Concepts

Roots of Quadratic Equations

A quadratic equation is typically expressed in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are known as the coefficients. The solutions, or roots, of this equation are the values of \( x \) that make the equation true. These roots can be calculated using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The expression \( b^2 - 4ac \) is called the discriminant and it's essential in determining the nature of the roots.

• If \( b^2 - 4ac > 0 \), the roots are real and distinct.
• If \( b^2 - 4ac = 0 \), the roots are real and equal.
• If \( b^2 - 4ac < 0 \), the roots are complex and conjugate.

Knowing the nature of the roots is important, especially when considering where a certain value \( k \) lies in relation to these roots. For example, if a quadratic has real and distinct roots, and if \( k \) is less than both roots, the sign of the quadratic at \( k \) matches the sign of \( a \), the coefficient of the \( x^2 \) term.

Inequalities Involving Roots

The placement of the roots within the number line can determine the solution to several inequalities. For instance, consider the inequality \( f(x) < 0 \) or \( f(x) > 0 \). If represented graphically as a parabola, this equation's sign at different intervals depends on its roots.

• If \( x \) lies between the roots, \( f(x) \) will take on a sign opposite to the coefficient \( a \).
• If \( x \) lies outside the roots, either to the left of both or to the right, then \( f(x) \) matches the sign of \( a \).

Understanding these inequalities assists in graph comprehension and informing solutions to related mathematical problems. For example, evaluating the sign of \( f(k) \) at different placements on the number line while considering the parity of the coefficient \( a \), grants insight into quadratic inequalities, a critical step in solving many real-world problems.

Parabola Vertex

In the context of quadratic functions, the graph is a parabola. The parabola is symmetric around its vertex. The vertex can be found using the formula \( x = -\frac{b}{2a} \).

• For \( y = ax^2 + bx + c \), if \( a > 0 \), the parabola opens upward, making the vertex a minimum point.
• If \( a < 0 \), the parabola opens downward, and the vertex represents a maximum point.

The location of the vertex is pivotal because it determines the axis of symmetry and the direction in which the parabola opens. Therefore, for a given quadratic equation, observing whether a point \( k \) is before or after the vertex allows for predictions about the parabola's behavior. For instance, in constraint scenarios where a point must be strictly less or greater than both roots such that \( k \) is not among them, understanding the vertex's location offers valuable clarity.