Problem 14 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(^{4} x\) ' satisfies \(\left|x^{2}-3 x+2\right|+|x-1|=x-3\), then (A) \(x \in \phi\) (B) \(x \in[1,2]\) (C) \(x \in[3, \infty)\) (D) \(x \in(-\infty, \infty)\)

Step-by-Step Solution

Answer

(A) \(x \in \phi\) - No solution.

Step 1: Understand the Problem

We need to find the values of \(x\) that satisfy the equation \(|x^2-3x+2| + |x-1| = x-3\). To solve this, we'll evaluate the absolute values based on the nature of the quadratic equation and the linear function within the absolute value.

Step 2: Analyze the Expression Inside Absolute Values

The expression \(x^2-3x+2\) can be factored as \((x-1)(x-2)\). The expression changes sign at \(x=1\) and \(x=2\). The absolute value function \(|x-1|\) changes sign at \(x=1\). We need to consider different cases based on these points: \(x<1\), \(1 \leq x < 2\), and \(x \geq 2\).

Step 3: Solve for Case 1: \(x < 1\)

When \(x < 1\), \(x^2-3x+2 = (x-1)(x-2)\) is positive because both factors are negative and the product is positive, so \(|x^2-3x+2| = x^2-3x+2\). Also, \(|x-1| = 1-x\). Substituting into the equation gives \(x^2 - 3x + 2 + 1 - x = x - 3\). Simplifying, we get \(x^2 - 4x + 3 = x - 3\). Therefore \(x^2 - 5x + 6 = 0\). This factors to \((x-2)(x-3) = 0\), giving solutions \(x = 2\) or \(x = 3\). However, \(x < 1\), so there is no valid solution in this range.

Step 4: Solve for Case 2: \(1 \leq x < 2\)

In this case, \(x^2-3x+2 = -((x-1)(x-2))\) as the expression evaluates to negative. Therefore, \(|x^2-3x+2| = -(x^2-3x+2) = 3x - x^2 - 2\) and \(|x-1| = x-1\). Substituting gives \((3x - x^2 - 2) + (x - 1) = x - 3\). Simplifying, we get \(-x^2 + 4x - 3 = x - 3\), leading to \(-x^2 + 3x = 0\), or \(x(x-3) = 0\). Solving gives \(x=0\) or \(x=3\). However, within \(1 \leq x < 2\), there are no solutions.

Step 5: Solve for Case 3: \(x \geq 2\)

In this range, \(x^2-3x+2 = (x-1)(x-2)\) which is non-negative, and \(|x-1| = x-1\). Substituting into the equation gives \(x^2 - 3x + 2 + x - 1 = x - 3\). Simplify to get \(x^2 - 2x + 1 = x - 3\), or \(x^2 - 3x + 4 = 0\). This quadratic equation has the discriminant \(b^2 - 4ac = 9 - 16 = -7\) which is negative, indicating no real roots in this range of \(x\). Thus, there are no valid solutions for \(x \geq 2\).

Step 6: Final Conclusion

Based on the evaluations made in all different scenarios for \(x\), we find no values of \(x\) that satisfy the original equation. Therefore, the set of solutions is empty.

Key Concepts

Understanding Absolute Value Equations

Absolute value equations involve expressions where you take the absolute value of an expression, which essentially measures its distance from zero on the number line. This means that inside the absolute value, the expression could either be positive or negative, but the absolute value itself will always be positive or zero.

In the equation \(|x^2 - 3x + 2| + |x-1| = x-3\), you have two absolute value operations.

• The expression \(x^2 - 3x + 2\) can switch its sign at the points where it equals 0, which are its roots.
• For \(|x-1|\), the switch occurs at \(x = 1\).
• Therefore, when solving such equations, it's crucial to explore different scenarios or "cases" where the expressions inside the absolute values change sign.

In practice, solving an absolute value equation means breaking it down into several cases. You evaluate each piece by considering its behavior (positive or negative) in different intervals, thus allowing for accurate solutions.

Quadratic Equation Solving

Solving quadratic equations, such as \(x^2 - 5x + 6 = 0\), is a fundamental part of algebra. A quadratic equation is generally in the form \(ax^2 + bx + c = 0\). This specific equation is actually a simplified form encountered during solving the original exercise.

Here’s how you can solve a quadratic equation:

• **Factoring:** Start by seeing if the quadratic can be factored into simpler terms like \((x-2)(x-3) = 0\). If so, you have found the roots directly as \(x = 2\) and \(x = 3\).
• **Completing the square:** This is useful when the quadratic doesn't factor neatly. It involves rewriting the equation in the form of \((x - p)^2 = q\).
• **Quadratic formula:** For any quadratic equation, the roots can be found using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

During the solution process, also note the discriminant \(b^2 - 4ac\), which indicates the nature of the roots:

• If positive, there are two real roots.
• If zero, there is one real root.
• If negative, there are no real roots.

Understanding these methods can greatly help in tackling quadratic equations effectively.

Tackling JEE Main Mathematics

JEE Main Mathematics is an entrance examination essential for students aiming to pursue engineering in India. It's well-known for its challenging nature.

To excel in JEE Main Mathematics, you should focus on building a strong foundation in various mathematical concepts. Here’s a brief strategy for dealing with problems related to absolute values and quadratics:

• **Understand Absolute Values:** Before diving into problems, take time to get a strong grip on what absolute values mean, and practice switching between positive and negative scenarios.
• **Master Quadratics:** Quadratic equations often form the backbone of more complex problems. Invest time in learning how to solve quadratics using different approaches like factoring, completing squares, and the quadratic formula.
• **Practice Regularly:** Regular practice helps in recognizing patterns and solution paths more quickly, and is crucial for success in competitive exams like JEE Main.
• **Time Management:** In an exam scenario, effective time management can make a significant difference. Practice under timed conditions.

Remember, understanding the concepts is pivotal to solving JEE Main problems efficiently. Focus on clarity, practice, and precision to improve your mathematical abilities.