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

Solution & Explanation

### Related Formula The definition of modulus function handles sub-intervals via critical points: |x - a| = begincases x - a & textif x ge a \\ -(x - a) & textif x < a endcases ### 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 -x^2 + 2x - 3x + 9 + 1 = 0 implies x^2 + x - 10 = 0 x = frac-1 pm sqrt1 + 402 = frac-1 pm sqrt412 Checking domain constraint x < 2: frac-1 - sqrt412 approx frac-1 - 6.42 = -3.7 < 2 (Valid root) frac-1 + sqrt412 approx frac-1 + 6.42 = 2.7 not< 2 (Rejected) ### Step 1: Intermediate Interval Check **Case II: 2 le x < 3** x(x - 2) + 3(-(x - 3)) + 1 = 0 x^2 - 2x - 3x + 9 + 1 = 0 implies x^2 - 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 ge 3** x(x - 2) + 3(x - 3) + 1 = 0 x^2 - 2x + 3x - 9 + 1 = 0 implies x^2 + x - 8 = 0 x = frac-1 pm sqrt1 + 322 = frac-1 pm sqrt332 Checking domain constraint x ge 3: frac-1 + sqrt332 approx frac-1 + 5.742 = 2.37 < 3 (Rejected) frac-1 - sqrt332 < 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations

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

Solution

### Core Logic For the denominator x^2 - 8x + 32, D = 64 - 128 < 0 and a = 1 > 0. Thus, x^2 - 8x + 32 > 0 forall x in mathbbR. ### Step 1: Constraint on Numerator Since the denominator is always positive, the numerator must be strictly negative for all x in mathbbR. ax^2 + 2(a + 1)x + 9a + 4 < 0 quad forall x in mathbbR 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations

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