Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Given \(l x^{2}-m x+5=0\) does not have two distinct real roots, the minimum value of \(5 l+m\) is (A) 5 (B) \(-5\) (C) 1 (D) \(-1\)

Step-by-Step Solution

Answer

The minimum value of \(5l + m\) is -1.

Step 1: Understanding the Quadratic Condition

To determine if the quadratic equation \( l x^{2} - m x + 5 = 0 \) has distinct real roots, we need to analyze the discriminant, \( b^2 - 4ac \). For this equation, \( a=l \), \( b=-m \), and \( c=5 \). Two distinct real roots exist only if the discriminant is positive.

Step 2: Setting the Discriminant Condition

We use the discriminant \((-m)^2 - 4 \cdot l \cdot 5 \), simplifying to \( m^2 - 20l \). To ensure there are no distinct real roots, the discriminant must be less than or equal to zero: \( m^2 - 20l \leq 0 \).

Step 3: Solving the Inequality for l and m

Rewriting the inequality \( m^2 \leq 20l \) gives the condition for \( m \) and \( l \) such that no two distinct real roots exist. From this, we deduce \( l \geq \frac{m^2}{20} \).

Step 4: Minimizing the Expression 5l + m

We need to minimize the expression \(5l + m\). Substitute the boundary condition \( l = \frac{m^2}{20} \) into the expression: \( 5\left(\frac{m^2}{20}\right) + m \). This simplifies to \( \frac{m^2}{4} + m \).

Step 5: Finding the Critical Point

Consider \( f(m) = \frac{m^2}{4} + m \) to find its minimum. Take the derivative \( f'(m) = \frac{m}{2} + 1 \) and set it to zero: \( \frac{m}{2} + 1 = 0 \). Solve this to find \( m = -2 \).

Step 6: Calculating the Minimum Value

Substitute \( m = -2 \) back into \( l = \frac{m^2}{20} \), so \( l = \frac{4}{20} = \frac{1}{5} \). Calculate \( 5l + m = 5\left(\frac{1}{5}\right) - 2 = 1 - 2 = -1 \).

Step 7: Verifying the Inequality Condition

Verify that \( m = -2 \) and \( l = \frac{1}{5} \) satisfy \( m^2 \leq 20l \). Checking: 4 = 4, so this condition holds.

Key Concepts

Discriminant

In quadratic equations like \( l x^2 - m x + 5 = 0 \), the discriminant is a key concept to determine the nature of the roots. The discriminant is given by the formula \( \Delta = b^2 - 4ac \), where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation. If the discriminant is positive \((\Delta > 0)\), the equation has two distinct real roots. If it equals zero \((\Delta = 0)\), the equation has exactly one real root or a repeated root. When negative \((\Delta < 0)\), no real roots exist. In our context, we need the discriminant \( m^2 - 20l \) to be less than or equal to zero to ensure that the roots are not distinct real roots.

Real Roots

Real roots of a quadratic equation refer to the values of \( x \) that satisfy the equation and are real numbers. When we say an equation "does not have two distinct real roots," it implies that either the equation has a repeated root or complex roots. To prevent two distinct real roots in the equation \( l x^2 - m x + 5 = 0 \), the discriminant \( m^2 - 20l \) must not be positive. This situation will result in either a single real root or two complex conjugate roots.

Inequality

In optimizing equations, inequalities play a crucial role in defining the relationships between variables. Here, the inequality \( m^2 \leq 20l \) was derived from requiring the discriminant to be non-positive, ensuring no distinct real roots. We further transformed this inequality to find a condition on \( l \) by expressing \( l \) as \( l \geq \frac{m^2}{20} \). This condition helps in searching for the minimum of the expression \(5l + m\) under the given constraints. Inequalities guide us in finding permissible values that satisfy problem requirements.

Optimization

Optimization in this problem involves finding the minimum value of the expression \( 5l + m \) with respect to the constraint \( m^2 \leq 20l \). By substituting \( l = \frac{m^2}{20} \) into the expression, we reformulate it as \( \frac{m^2}{4} + m \). To find the minimum, take the derivative \((f'(m) = \frac{m}{2} + 1)\) and set it to zero: this gives \( m = -2 \). Substitute back into \( l = \frac{m^2}{20} \) to get \( l = \frac{1}{5} \). Calculating \( 5l + m \) yields \(-1\) as the minimum value. This process illustrates how derivatives help in locating extrema. Always reconcile such results with the original inequality to ensure validity.