Problem 66 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a<0\), the positive root of the equation \(x^{2}-2 a\) \(|x-a|-3 a^{2}=0\) is (A) \(a(-1-\sqrt{6})\) (B) \(a(-1+\sqrt{6})\) (C) \(a(1-\sqrt{2})\) (D) None of these

Step-by-Step Solution

Answer

(D) None of these

Step 1: Analyze the Absolute Value Function

The expression contains the absolute value function \(|x-a|\). Since \(a < 0\), we analyze the behavior of this function in different cases dependent on the value of \(x\). Let's start by considering the case \(x \geq a\) first, which simplifies the absolute value \(|x-a| = x-a\).

Step 2: Solve for \(x \geq a\)

Substituting \(|x-a| = x-a\) into the original equation gives \(x^2 - 2a(x-a) - 3a^2 = 0\). Simplifying, we get \(x^2 - 2ax + 2a^2 - 3a^2 = 0\) or \(x^2 - 2ax - a^2 = 0\). This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(b = -2a\) and \(c = -a^2\).

Step 3: Use the Quadratic Formula

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), substitute \(a = 1, b = -2a, c = -a^2\). The formula becomes \(x = \frac{2a \pm \sqrt{(-2a)^2 - 4(1)(-a^2)}}{2}\). Simplify under the square root to obtain \(4a^2 + 4a^2 = 8a^2\). Thus, the roots are \(x = \frac{2a \pm \sqrt{8a^2}}{2}\).

Step 4: Simplify and Identify the Positive Root

The expression under the square root simplifies to \(\sqrt{8a^2} = 2|a|\sqrt{2}\), and since \(a < 0\), \(|a| = -a\). Substitute into the formula to get \(x = \frac{2a \pm 2(-a)\sqrt{2}}{2}\). Producing roots: \(x = a(1 + \sqrt{2})\) or \(x = a(1 - \sqrt{2})\). Since \(a < 0\), \(x = a(1 - \sqrt{2})\) is the negative root and is valid only under the condition \(x \geq a\).

Step 5: Verify with Other Conditions

Since none of these roots match the multiple-choice answers, consider alternative cases or simplifications that might give matching options. As a verification, note that none of the standard transformations provide a root of the given forms like \(a(-1 \pm \sqrt{6})\). This indicates a mismatch with available answers.

Key Concepts

Absolute Value Function

The absolute value function, denoted as \(|x-a|\), describes the distance of a number from zero on a number line. It will always yield a non-negative result. When we talk about \(|x-a|\), we are specifically looking at the distance of \((x-a)\) from zero. This distance calculation depends on the inputs which can lead to different expressions. In our specific problem, when \(a < 0\), the function behavior is influenced by whether \(x\) is greater than or less than \(a\).

When \(x \geq a\), the expression simplifies to \(x-a\) because the difference \(x-a\) is already non-negative. Conversely, if \(x < a\), the expression becomes \(a-x\) to maintain a non-negative value. These changes are essential in accurately solving equations involving absolute values.

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This equation allows for obtaining both roots by computing the plus-minus (\( \pm \)) operation. For the example in the exercise, coefficients are provided in the equation \(x^2 - 2ax - a^2 = 0\). Here, \(a = 1\), \(b = -2a\), and \(c = -a^2\).

The discriminant, \(b^2 - 4ac\), helps determine the nature of the roots:

• A positive discriminant implies two distinct real roots.
• Zero discriminant leads to one repeated real root.
• A negative discriminant results in complex roots.

For this problem, we find the discriminant to be positive, leading us to two possible real roots.

Roots of Equations

Finding the roots of an equation involves determining the values of \(x\) that satisfy the equation. For quadratic equations like the one in the problem, there can be two roots, which we found using the quadratic formula. The solutions provided are fundamentally dependent on conditions influencing the absolute value function.

In this exercise, after simplifying the quadratic, we obtained the roots as \(x = a(1 \pm \sqrt{2})\). It's vital to remember the influence of \(a<0\), since it affects the sign and validity of each root. With \(a < 0\) and \(x \geq a\), we discover that only \(x = a(1 - \sqrt{2})\) is viable under these constraints. This meticulous consideration helps ensure the correct identification of a positive root given the problem's restrictions.