Problem 127 - 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 the roots of the equation \((a+1) x^{2}-3 a x+4 a=0(a \neq-1)\) to be greater than unity are (A) \(\frac{-16}{7} \leq a<-1\) (B) \(-2

Step-by-Step Solution

Answer

(D) None of these.

Step 1: Find Conditions for Roots Greater Than One

For the roots of the quadratic equation \( (a+1)x^2 - 3ax + 4a = 0 \) to be greater than one, we need to consider bounds on the roots. If both roots, say \ \alpha \ and \ \beta, \ are greater than 1, then the vertex's x-coordinate of the parabola must be less than 1. This enforces conditions both on the sum and the product of these roots.

Step 2: Use Sum and Product of Roots

For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of roots \( \alpha + \beta = \frac{-b}{a} \) and the product of roots \( \alpha \beta = \frac{c}{a} \). Thus, for \( (a+1)x^2 - 3ax + 4a \), we have: - \( \alpha + \beta = \frac{3a}{a+1} \) - \( \alpha \beta = \frac{4a}{a+1} \)

Step 3: Establish Inequalities with the Quadratic Formula

For both roots to be greater than 1, solve the inequalities:\[ \frac{3a}{a+1} > 2 \] and \[ \frac{4a}{a+1} > 1 \]Solve these inequalities to find the range of values for \( a \).

Step 4: Solve First Inequality

Solve \( \frac{3a}{a+1} > 2 \): \[ 3a > 2(a+1) \]\[ 3a > 2a + 2 \]\[ a > 2 \]

Step 5: Solve Second Inequality

Solve \( \frac{4a}{a+1} > 1 \): \[ 4a > a + 1 \]\[ 3a > 1 \]\[ a > \frac{1}{3} \]

Step 6: Combine Inequality Results

Combine results from solving inequalities. The simultaneous solution to \( a > 2 \) and \( a > \frac{1}{3} \) provides \( a > 2 \).

Step 7: Check Validity with Given Options

The derived solution \( a > 2 \) doesn't fit any given range of (A) \(\frac{-16}{7} \leq a < -1\), (B) \( -2 < a < -1\), (C) \(0 < a < 1\). Hence, the correct answer is (D) None of these.

Key Concepts

Roots of Quadratic Equations

Quadratic equations play a crucial role in algebra, especially when determining the points where a parabola intersects the x-axis. The solutions to a quadratic equation, also known as roots, are the x-values where the expression evaluates to zero. For a general quadratic equation of the form \( ax^2 + bx + c = 0 \), the roots can be found using the quadratic formula: \

• \( \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\Depending on the value of the discriminant \( b^2 - 4ac \), the nature of the roots varies:

• 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.

\Understanding these conditions is vital as they inform us about the behavior of the parabola related to the x-axis, which can be insightful when analyzing graphical representations or solving real-world problems.

Inequalities in Quadratics

Inequalities involving quadratic expressions often arise in mathematical analysis and various applications. Solving these involves finding intervals where the quadratic expression is positive or negative. Consider a quadratic expression \( ax^2 + bx + c \). The sign of the expression between and beyond its roots \( \alpha \) and \( \beta \) is closely linked to the leading coefficient \( a \). Let's examine this more closely:

• If \( a > 0 \), the parabola opens upwards, and the expression is positive outside the interval \( (\alpha, \beta) \) and negative within.
• If \( a < 0 \), the parabola opens downwards, and the expression is negative outside \( (\alpha, \beta) \) and positive within.

For determining where a quadratic is greater or less than zero, it's important to:

• Identify the roots of the equation using the quadratic formula.
• Analyze the sign changes in the intervals split by these roots.

This is particularly useful when analyzing conditions such as profit calculations, motion, optimization problems, and more, where such inequalities frequently appear.

Parabola Vertex

The vertex of a parabola is a point where the curve turns, indicating its maximum or minimum value depending on the orientation. For the quadratic expression \( ax^2 + bx + c \), you find the vertex using:

• The x-coordinate: \( x = \frac{-b}{2a} \)
• The y-value at the vertex: plug \( x \) back into the expression to find \( ax^2 + bx + c \).

Understanding the vertex is crucial when interpreting and graphing quadratics. It offers insights into the graph's symmetry and can help in solving optimization problems where the maximum or minimum value of the quadratic is required.
For example, if \( a > 0 \), the vertex represents the minimum point of the parabola, while if \( a < 0 \), it represents the maximum point. Knowing this helps determine crucial aspects of the parabola, especially when looking into problems that involve maximizing or minimizing certain quantities.