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Quadratic Equations appeared 26 times across 3 years — 3% of Mathematics. This question is from Equations with Modulus.

Year 2026 2025 2024 Total
Questions 11 10 5 26

The number of real roots of the equation x | x - 2 | + 3 | x - 3 | + 1 = 0 is :

Solution & Explanation

Related Formula

The definition of modulus function handles sub-intervals via critical points:

|x - a| = cases x - a & if x ≥ a -(x - a) & if x < a cases
Core Logic

The critical points are x = 2 and x = 3. We check the three distinct structural intervals:

Case I: x < 2

x(-(x - 2)) + 3(-(x - 3)) + 1 = 0 -x² + 2x - 3x + 9 + 1 = 0 x² + x - 10 = 0 x = -1 ± √(1 + 40)2 = -1 ± √(41)2

Checking domain constraint x < 2: -1 - √(41)2 ≈ (-1 - 6.4)/(2) = -3.7 < 2 (Valid root) -1 + √(41)2 ≈ (-1 + 6.4)/(2) = 2.7 < 2 (Rejected)

Step 1: Intermediate Interval Check

Case II: 2 ≤ x < 3

x(x - 2) + 3(-(x - 3)) + 1 = 0 x² - 2x - 3x + 9 + 1 = 0 x² - 5x + 10 = 0

Discriminant check: D = (-5)² - 4(1)(10) = 25 - 40 = -15 < 0. No real roots exist in this interval.

Step 2: Upper Interval Check

Case III: x ≥ 3

x(x - 2) + 3(x - 3) + 1 = 0 x² - 2x + 3x - 9 + 1 = 0 x² + x - 8 = 0 x = -1 ± √(1 + 32)2 = -1 ± √(33)2

Checking domain constraint x ≥ 3: -1 + √(33)2 ≈ (-1 + 5.74)/(2) = 2.37 < 3 (Rejected) -1 - √(33)2 < 0 (Rejected)

Thus, only 1 valid real root satisfies the conditional layout across all ranges.

Pattern Recognition

Always perform case-by-case boundaries checks on algebraic roots found inside absolute modulus problems to discard ghost solutions quickly.

Chapter Mix

Class 11 Mathematics: Quadratic Equations

Reference Study Guides

More Quadratic Equations Previous-Year Questions — Page 6

Q20 jee_main_2024_31_jan_morning Sign of Quadratic Expressions
Let S be the set of positive integral values of a for which (ax² + 2(a + 1)x + 9a + 4)/(x² - 8x + 32) < 0, x in R. Then, the number of elements in S is:
  • A. 1
  • B. 0
  • C. ∞
  • D. 3

Solution

Core Logic

For the denominator x² - 8x + 32, D = 64 - 128 < 0 and a = 1 > 0. Thus, x² - 8x + 32 > 0 x in R.

Step 1: Constraint on Numerator

Since the denominator is always positive, the numerator must be strictly negative for all x in R.

ax² + 2(a + 1)x + 9a + 4 < 0 x in R

This requires a < 0 and D < 0.

Step 2: Conclusion

Since a must be strictly less than 0, there are no positive integral values of a that satisfy the condition. Hence, S is an empty set. Number of elements is 0.

Chapter Mix

Class 11 Maths: Quadratic Equations

More Quadratic Equations Questions — jee_main_2025_07_april_evening

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)