Problem 96 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find all the values of \(m\) for which both roots of the function \(2 x^{2}+m x+m^{2}-5\) i. are less than 1. ii. exceed \(-1\).

Step-by-Step Solution

Answer

The values of m for which both roots of the function \(2x^2 + mx + m^2 - 5\) are less than 1 and exceed -1 are in the interval \(-4 < m < 4\).

Step 1: Standard Form of Quadratic Equation

We are given the quadratic function: \(2x^2 + mx + m^2 - 5\). To find the roots, we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) where a, b, and c are the coefficients of the quadratic equation in the form \(ax^2 + bx + c = 0\). In this case: a = 2, b = m, and c = \(m^2 - 5\)

Step 2: Substitute the Coefficients

Insert the quadratic coefficients into the formula: \(x = \frac{-m \pm \sqrt{m^2 - 4(2)(m^2 - 5)}}{2(2)}\)

Step 3: Simplify the Equation

Simplify the roots expression: \(x = \frac{-m \pm \sqrt{m^2 - 8m^2 + 40}}{4}\) Since we are looking for both roots to be less than 1 and exceeding -1, we can set up two inequalities: \(x < 1\) and \(x > -1\)

Step 4: Identify the Vertex of the Parabola

Since the parabola will either open upwards or downwards, and we are given that a = 2 (positive), the parabola will open upwards. Thus, the vertex of the parabola will be the highest (least) value of x that we want to find. The vertex can be found using the formula: \(x_{vertex} = \frac{-b}{2a}\) In this case: \(x_{vertex} = \frac{-m}{2(2)} = \frac{-m}{4}\)

Step 5: Set up Inequalities

Now we have the vertex, so we plug it into the inequalities and solve for m: \(\frac{-m}{4} < 1\) and \(\frac{-m}{4} > -1\) Multiply both sides of each inequality by 4: \(-m < 4\) and \(-m > -4\) Divide both sides of each inequality by -1 (and remember to flip the inequality signs): \(m > -4\) and \(m < 4\)

Step 6: Combine the Inequalities

The final step is to combine both inequalities to find the range of values of m: \(-4 < m < 4\) So, m takes all values in the interval \((-4, 4)\) for both roots of the function to be less than 1 and exceed -1.

Key Concepts

Quadratic Inequalities

When dealing with quadratic inequalities, we aim to find the range of values that satisfy a certain inequality constraint. For instance, with the function \(2x^2 + mx + m^2 - 5\), we need to evaluate conditions like both roots being less than 1, or both exceeding -1. These inequalities guide us to an interval for \(m\), which will ensure our conditions have been met. Quadratic inequalities are often solved by first determining the roots, which serve as boundary points to test within intervals determined by these roots.

• Set the inequality according to the conditions given.
• Calculate the roots using the quadratic formula.
• Determine which intervals are valid for the inequality and evaluate the inequality within these intervals.

This process helps in comprehending the behavior of quadratic equations under specified conditions.

Roots of Equations

Understanding the roots of an equation is crucial, particularly in quadratic equations. Roots represent the values of \(x\) that make the equation equal to zero. For the equation \(2x^2 + mx + m^2 - 5 = 0\), the roots can be calculated using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula helps us find points where the parabola touches or crosses the x-axis. To solve, we identify the coefficients \(a = 2\), \(b = m\), and \(c = m^2 - 5\). Plug these coefficients into the formula to find the roots.

• The roots can be real or complex, depending on the discriminant \(b^2 - 4ac\).
• If the discriminant is positive, we get two distinct real roots.
• If it's zero, there's exactly one real root (vertex touches x-axis).
• If negative, the roots are complex.

In our case, calculating valid \(m\) values helps determine when both roots are within certain thresholds.

Vertex of a Parabola

The vertex of a parabola provides key information about the shape and direction of the parabola. For any quadratic equation in the form \(ax^2 + bx + c\), the vertex can be found using the formula:\[ x_{vertex} = \frac{-b}{2a} \]This helps us find the axis of symmetry of the parabola. Knowing that \(a = 2\) and \(b = m\), the vertex \(x\)-coordinate in our specific equation is \(\frac{-m}{4}\). Because \(a\) is positive, the parabola opens upwards, meaning the vertex is its lowest point.

• By determining the vertex, we understand how the parabola behaves between its roots.
• The vertex's \(y\)-value can show the lowest or highest value of the quadratic, depending on the parabola's opening direction.

For constraints involving the vertex, we use it to check boundaries given in inequalities, ensuring the roots fall within a specific range by analyzing the \(x\)-coordinate of the vertex.