JEE Main202622 January 2026Morning ShiftMathematicsFunctionsActual

The number of distinct real solutions of the equation x|x+4|+3|x+2|+10=0 is

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B. 1

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Equation: x|x+4| + 3|x+2| + 10 = 0 . Critical points: x = -4, -2 . Case 1: x -2 : x(x+4) + 3(x+2) + 10 = 0 x^2+7x+16=0 . Discriminant = 49-64 = -15 Case 2: -4 x x^2+x+4=0 . Discriminant = 1-16 = -15 Case 3: x x^2+7x-4=0 . x = -7 65 2 . x = -7+ 65 2 0.53 (rejected, not in x Number of distinct real solutions = 1 .

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