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Continuity and Differentiability · Question 18

JEE Main 2025 (Online) 28th January Morning ShiftNumericalMedium📈 Topic in 16% papers+4
1

Let f(x) = cases 3x, & x < 0 \ \1 + x + [x], x + 2[x]\, & 0 x 2 \ 5, & x > 2 cases where [.] denotes greatest integer function. If and are the number of points, where f is not continuous and is not differentiable, respectively, then + equals .

Answer

5

Solution

0 x < 1: [x] = 0 f(x) = \1 + x, x\ = x

1 x < 2: [x] = 1 f(x) = \2 + x, x + 2\ = x + 2

x = 2: [x] = 2 f(2) = \5, 6\ = 5

At x = 0: f(0-) = 0 = f(0) = f(0+) continuous;

LHD = 3, RHD = 1 not differentiable.

At x = 1: f(1-) = 1, f(1) = 3 discontinuous.

At x = 2: f(2-) = 4, f(2) = 5 = f(2+) discontinuous.

Elsewhere f is linear or constant, hence differentiable.

= 2 (x = 1, 2) and = 3 (x = 0, 1, 2; a discontinuous point is also non-differentiable)

+ = 5

Answer: 5

JEE Main 2025 (Online) 28th January Morning Shift

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3

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8

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9

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