Problem 71 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a x^{2}+b x+6=0\) does not have two distinct real roots \(a \in R, b \in R\), then the least value of \(3 a+b\) is (A) 4 (B) \(-1\) (C) 1 (D) \(-2\)

Step-by-Step Solution

Answer

The least value of \( 3a + b \) is \(-2\).

Step 1: Understand the condition for the roots

For the quadratic equation \( ax^2 + bx + c = 0 \) to not have two distinct real roots, the discriminant \( D = b^2 - 4ac \) must be less than or equal to zero.

Step 2: Apply the discriminant condition

In this exercise, \( a x^2 + b x + 6 = 0 \) with \( c = 6 \). The discriminant is \( D = b^2 - 4a imes 6 = b^2 - 24a \). To ensure no two distinct real roots, \( b^2 - 24a \leq 0 \).

Step 3: Solve for the condition on \( a \) and \( b \)

From the condition \( b^2 \leq 24a \), we express \( a \) in terms of \( b \): \( a \geq \frac{b^2}{24} \).

Step 4: Determine the expression for \( 3a + b \)

Using \( a = \frac{b^2}{24} \), we find \( 3a + b = 3\left(\frac{b^2}{24}\right) + b = \frac{b^2}{8} + b \).

Step 5: Minimize \( \frac{b^2}{8} + b \)

Rewrite the expression as \( f(b) = \frac{b^2}{8} + b \). Take the derivative with respect to \( b \), \( f'(b) = \frac{b}{4} + 1 \). Set \( f'(b) = 0 \) to find the critical points: \( \frac{b}{4} + 1 = 0 \) which gives \( b = -4 \).

Step 6: Evaluate \( 3a + b \) at the critical point

Substitute \( b = -4 \) into \( a = \frac{b^2}{24} = \frac{(-4)^2}{24} = \frac{16}{24} = \frac{2}{3} \). Then calculate \( 3a + b = 3 \times \frac{2}{3} + (-4) = 2 - 4 = -2 \).

Key Concepts

Quadratic Equation

A quadratic equation is generally written in the form \( ax^2 + bx + c = 0 \). This equation represents a parabola when graphed on the coordinate system. The coefficients \( a \), \( b \), and \( c \) are real numbers, with \( a \) not equal to zero. The quadratic equation is an essential tool in algebra, as it helps in finding the values of \( x \) that satisfy the equation. These values are known as roots or solutions.

Here are key characteristics of quadratic equations:

• The graph of a quadratic equation is a parabola, which opens upwards if \( a > 0 \) and downwards if \( a < 0 \).
• The vertex of the parabola is the highest or lowest point of the graph depending on its orientation.
• The axis of symmetry of the parabola runs vertically through the vertex, dividing the graph into two mirror images.
• The roots of the quadratic equation can be found using various methods such as factoring, completing the square, or the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Understanding quadratic equations is fundamental as they appear in numerous applications in real-life scenarios, such as physics problems involving acceleration, area calculations, and financial mathematics.

Real Roots

Real roots of a quadratic equation refer to the solutions that are real numbers. The nature of the roots of a quadratic equation is determined by its discriminant, represented by \( D = b^2 - 4ac \). The discriminant gives us insight into the type and number of roots the equation has:

- If \( D > 0 \), there are two distinct real roots. This means the parabola intersects the x-axis at two different points.- If \( D = 0 \), there is exactly one real root, also known as a repeated or double root. In this case, the vertex of the parabola touches the x-axis.- If \( D < 0 \), there are no real roots, indicating the solutions are complex numbers, and the parabola does not touch the x-axis.In this particular exercise, the condition \( b^2 - 24a \leq 0 \) ensures the quadratic equation does not have two distinct real roots. This implies that the discriminant is less than or equal to zero, confirming the roots are either complex or a single real root.

Critical Points

Critical points are the values of the variable that result in a zero derivative of a function, signifying a potential minimum or maximum value for that function. In the context of this exercise, finding critical points helps in determining the minimum value of the expression \( 3a + b \).

To find the critical point, the derivative of the expression \( f(b) = \frac{b^2}{8} + b \) is taken. The derivative, given by \( f'(b) = \frac{b}{4} + 1 \), is then set to zero to solve for \( b \):

• \( \frac{b}{4} + 1 = 0 \)
• Solving this gives \( b = -4 \)

This critical point \( b = -4 \) is then substituted back into the expression to find the minimum value of \( 3a + b \). Evaluating \( a = \frac{b^2}{24} \) at \( b = -4 \), gives \( a = \frac{2}{3} \). With \( 3a + b = 2 - 4 \), the minimum value is confirmed as \(-2\). Critical points are important in optimization problems, allowing one to determine the most extreme values a function can take on a certain interval.