Maths Q3 – JEE-MAIN 2025

Home Maths Complex & QuadraticQuestion #94

The number of real roots of the equation

x | x - 2 | + 3 | x - 3 | + 1 = 0

is:

Clue 1 of 3: Interval Analysis

Break the problem into intervals based on the critical points where the expressions inside the absolute values change sign.

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Ninja Strategy•Analyze Function Positivity

By analyzing the function's behavior for $x \geq 2$ , we can quickly determine that no roots exist in that region, limiting the possibilities to at most one root from the $x < 2$ region.

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Step 1: Define Intervals based on Critical Points✦ Active

The critical points for the absolute value expressions $| x - 2 |$ and $| x - 3 |$ are $x = 2$ and $x = 3$ . These points divide the number line into three intervals: $x < 2$ , $2 \leq x < 3$ , and $x \geq 3$ . We analyze the equation $x | x - 2 | + 3 | x - 3 | + 1 = 0$ in each interval.

Step 2: Solve in Each Interval○ Expand

Case 1: For $x < 2$ , $| x - 2 | = - (x - 2)$ and $| x - 3 | = - (x - 3)$ . The equation becomes:

x (- (x - 2)) + 3 (- (x - 3)) + 1 = 0 - x^{2} + 2 x - 3 x + 9 + 1 = 0 - x^{2} - x + 10 = 0 x^{2} + x - 10 = 0

Using the quadratic formula, $x = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 10)}}{2} = \frac{- 1 \pm \sqrt{41}}{2}$ . The roots are $x_{1} = \frac{- 1 + \sqrt{41}}{2} \approx 2.70$ and $x_{2} = \frac{- 1 - \sqrt{41}}{2} \approx - 3.70$ . Only $x_{2} \approx - 3.70$ satisfies $x < 2$ . So, one root: $x = \frac{- 1 - \sqrt{41}}{2}$ .

Case 2: For $2 \leq x < 3$ , $| x - 2 | = x - 2$ and $| x - 3 | = - (x - 3)$ . The equation becomes:

x (x - 2) + 3 (- (x - 3)) + 1 = 0 x^{2} - 2 x - 3 x + 9 + 1 = 0 x^{2} - 5 x + 10 = 0

The discriminant is $D = (- 5)^{2} - 4 (1) (10) = 25 - 40 = - 15 < 0$ . There are no real roots in this interval.

Case 3: For $x \geq 3$ , $| x - 2 | = x - 2$ and $| x - 3 | = x - 3$ . The equation becomes:

x (x - 2) + 3 (x - 3) + 1 = 0 x^{2} - 2 x + 3 x - 9 + 1 = 0 x^{2} + x - 8 = 0

Using the quadratic formula, $x = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 8)}}{2} = \frac{- 1 \pm \sqrt{33}}{2}$ . The roots are $x_{3} = \frac{- 1 + \sqrt{33}}{2} \approx 2.37$ and $x_{4} = \frac{- 1 - \sqrt{33}}{2} \approx - 3.37$ . Neither $x_{3}$ nor $x_{4}$ satisfies $x \geq 3$ . So, no roots in this interval.

💡 Teacher's Secret Hint

Remember to always check if the calculated roots fall within the specific interval for which the absolute value expressions were simplified.

Step 3: Count Total Real Roots○ Expand

Combining the results from all intervals, we found only one real root: $x = \frac{- 1 - \sqrt{41}}{2}$ .

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Maths Question 3 – JEE-MAIN 2025

Clue 1 of 3: Interval Analysis

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