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Integrals · Question 42

JEE Main 2025 (Online) 3rd April Morning ShiftMCQMedium📈 Topic in 71% papers+4−1
1

Let the domain of the function f(x) = 2 4 6 (3 + 4x - x2) be (a, b). If 0b-a [x2] dx = p - q - r, p, q, r N, (p, q, r) = 1, where [ ] is the greatest integer function, then p + q + r is equal to

  1. 10
  2. 11
  3. 9
  4. 8

Answer

(1) 10

Solution

Domain: 4 6 (3 + 4x - x2) > 0

⇒ 6 (3 + 4x - x2) > 1

⇒ 3 + 4x - x2 > 6 ⇒ x2 - 4x + 3 < 0 ⇒ 1 < x < 3

⇒ a = 1, b = 3, b - a = 2

02 [x2] dx = 01 0 dx + 1 2 1 dx + 2 3 2 dx + 32 3 dx

= ( 2 - 1) + 2( 3 - 2) + 3(2 - 3)

= 5 - 2 - 3

⇒ p = 5, q = 2, r = 3

∴ p + q + r = 10

Answer: (1)

JEE Main 2025 (Online) 3rd April Morning Shift

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  3. 2
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3

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5

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  1. 2 - 2 - e(1 + 2)
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6

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7

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8

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  1. 13
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  3. 9
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9

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  1. 4 ( e 2 - 2 )
  2. e2 2 + 2
  3. 2 - e2
  4. 4 ( e 2 + 2 )
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10

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  3. 310
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JEE Main 2025 (Online) 28th January Evening Shift

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