Problem 6 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The number of integral values of \(m\) for which the equation \(\left(1+m^{2}\right) x^{2}-2(1+3 m) x+(1+8 m)=0\) has no real root is : (a) 1 (b) 2 (c) infinitely many (d) 3

Step-by-Step Solution

Answer

(a) 1

Step 1: Understanding the Discriminant Condition

For a quadratic equation \(ax^2 + bx + c = 0\), the roots are real if the discriminant \(\Delta = b^2 - 4ac\) is greater than or equal to zero. For the equation to have no real roots, the discriminant should be less than zero.

Step 2: Identify Coefficients and Setup the Discriminant

Identify the coefficients from the given equation \((1+m^2) x^2 - 2(1+3m) x + (1+8m) = 0\): \(a = 1 + m^2\), \(b = -2(1+3m)\), and \(c = 1+8m\). The discriminant \(\Delta\) is given by: \[\Delta = b^2 - 4ac\].

Step 3: Compute the Discriminant Expression

Substitute the coefficients into the discriminant formula: \[\Delta = \left(-2(1+3m)\right)^2 - 4(1+m^2)(1+8m)\]. Simplify this expression: \[\Delta = 4(1 + 6m + 9m^2) - 4((1+m^2)(1+8m))\].

Step 4: Simplify the Discriminant Expression

Simplify further: \[\Delta = 4(1 + 6m + 9m^2) - 4(1 + 8m + m^2 + 8m^3)\]. Expanding gives: \[\Delta = 4 + 24m + 36m^2 - 4 - 32m - 4m^2 - 32m^3\]. Hence, \[\Delta = 32m - 32m^3\].

Step 5: Solve the Inequality

Since we need \(\Delta < 0\), solve the inequality: \[-32m^3 + 32m < 0\], which simplifies to \[32m(m^2 - 1) < 0\], or \[32m(m-1)(m+1) < 0\].

Step 6: Determine the Critical Points and Solution Intervals

Find the critical points by solving \(m(m-1)(m+1) = 0\), which gives \(m = 0\), \(m = 1\), and \(m = -1\). Use a sign chart to determine where \((m-1)(m+1)m < 0\). The intervals are: \((-\infty, -1)\), \((-1, 0)\), \((0, 1)\), and \((1, \infty)\). The solution where the product is negative is \((0, 1)\).

Step 7: Count the Integral Solutions

The only interval where \(m(m-1)(m+1) < 0\) is between 0 and 1. The integral value of \(m\) in this interval is \(m = 0\). Thus, there is 1 integral value that causes the equation to have no real roots.

Key Concepts

Discriminant and Its Role in Quadratic Equations

The discriminant in a quadratic equation is a crucial factor in determining the nature of the roots. For a given quadratic equation in the form \(ax^2 + bx + c = 0\), the discriminant \(\Delta\) is calculated as \(b^2 - 4ac\). It helps us understand the types of roots we can expect without actually solving the equation.

• If \(\Delta > 0\), the equation has two distinct real roots.
• If \(\Delta = 0\), the equation has exactly one real root, which is a repeated root.
• If \(\Delta < 0\), the equation has no real roots, only complex ones.

This makes the discriminant very powerful in categorizing equations quickly. In this exercise, we are dealing with the condition \(\Delta < 0\), implying that we are looking to ensure the equation has no real roots. The discriminant approach allows us to establish when the roots switch from being real to non-real.

Integral Solutions: Finding Whole Number Roots

When solving inequalities that arise from discriminant conditions, finding integral solutions involves identifying the integer values that satisfy these inequalities. For the quadratic equation \[(1+m^2) x^2 - 2(1+3m) x + (1+8m) = 0\]with its discriminant given by:\[\Delta = 32m(m^2 - 1)\]we want this to be less than zero to ensure no real roots exist.

To solve for integral values, we:

• Determine the values of \(m\) such that \(\Delta < 0\). It simplifies to \(m(m-1)(m+1) < 0\).
• Identify the critical points from the expression: \(m = -1\), \(m = 0\), and \(m = 1\).
• Create intervals between these critical points: - \((-\infty, -1)\) - \((-1, 0)\) - \((0, 1)\) - \((1, \infty)\)

The task is to find which intervals satisfy the inequality \(\Delta < 0\). The interval \((0, 1)\) is valid, where only the integer \(m = 0\) satisfies the condition for \(\Delta < 0\). Hence, there is one integral value for \(m\) that keeps the roots of the quadratic equation non-real.

Understanding Inequalities in Quadratic Equations

Inequalities are used extensively to evaluate where certain conditions are met within a range of values. When applied to quadratic equations, the goal is to ascertain intervals where inequalities hold true.

To illustrate this, consider a simplified inequality derived from a quadratic equation's discriminant:\[32m(m-1)(m+1) < 0\]
This means we are interested in the intervals of \(m\) where the product changes from positive to negative. Such intervals are examined using a sign chart approach:

• Identify the critical points where the product changes sign: \(m = -1, 0, 1\).
• Break the number line at these points, generating intervals to test: - \((-\infty, -1)\) - \((-1, 0)\) - \((0, 1)\) - \((1, \infty)\)
• Determine the sign within each interval. Only the interval \((0, 1)\) produces a negative sign in the inequality \(m(m-1)(m+1) < 0\).

Utilizing this method ensures that we correctly identify intervals fulfilling the inequality requirement, optimal for problems where specific conditions like "no real roots" must be upheld. With this understanding, evaluating inequalities becomes a systemic process, serving well in mathematical analysis.