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Integrals

JEE Main 2024 (Online) 8th April Evening ShiftMCQMedium📈 Topic in 24% papers+4−1
Ques. 1JEE Main 2024, 8 April (Evening Shift)

Let e 4 dx ex - 1 = 6. Then e and e- are the roots of the equation :

  1. 2x2 - 5x + 2 = 0
  2. x2 - 2x - 8 = 0
  3. 2x2 - 5x - 2 = 0
  4. x2 + 2x - 8 = 0

Answer

(1) 2x2 - 5x + 2 = 0

Solution

Put ex - 1 = t2 exdx = 2t dt dx = 2t dt1 + t2

dx ex - 1 = 2 dt1 + t2 = 2 -1t

Limits: t = 3 at x = e 4 and t = e - 1 at x =

2 ( 3 - -1 e - 1 ) = 6 -1 e - 1 = 4

e = 2, e- = 12: sum = 52, product = 1

x2 - 52x + 1 = 0 2x2 - 5x + 2 = 0

Answer: (1)

Keep practising — Integrals

Ques. 2JEE Main 2024, 8 April (Evening Shift)

If 1 [5](x - 1)4(x + 3)6 dx = A ( x - 1 x + 3 )B + C, where C is the constant of integration, then the value of + + 20AB is .

Ques. 3JEE Main 2024, 6 April (Morning Shift)

0 /4 2 x 2 x ( 3 x + 3 x )2dx is equal to

  1. 1/9
  2. 1/6
  3. 1/3
  4. 1/12
Ques. 4JEE Main 2024, 6 April (Morning Shift)

Let rk = 01 (1 - x7 )k dx 01 (1 - x7 )k+1 dx, k N. Then the value of k=110 17 (rk - 1 ) is equal to .

Ques. 5JEE Main 2024, 6 April (Evening Shift)

If 1a2 2 x + b2 2 x dx = 112 -1(3 x) + constant, then the maximum value of a x + b x, is :

  1. 41
  2. 39
  3. 40
  4. 42
Ques. 6JEE Main 2024, 6 April (Evening Shift)

Let [t] denote the largest integer less than or equal to t. If 03 ( [x2 ] + [ x22 ] )dx = a + b 2 - 3 - 5 + c 6 - 7, where a, b, c Z, then a + b + c is equal to .

Ques. 7JEE Main 2024, 5 April (Morning Shift)

The integral 0 /4 136 x3 x + 5 x dx is equal to :

  1. 3 - 50 e 2 + 20 e 5
  2. 3 - 25 e 2 + 10 e 5
  3. 3 - 10 e(2 2) + 10 e 5
  4. 3 - 30 e 2 + 20 e 5
Ques. 8JEE Main 2024, 5 April (Morning Shift)

The value of - 2y(1 + y)1 + 2 y dy is :

  1. 2
  2. 2
  3. 22
  4. 2 2
Ques. 9JEE Main 2024, 5 April (Evening Shift)

Let (m, n) = 01 xm - 1(1 - x)n - 1 dx, m, n > 0. If 01 (1 - x10 )20 dx = a (b, c), then 100(a + b + c) equals .

  1. 2012
  2. 1021
  3. 1120
  4. 2120
Ques. 10JEE Main 2024, 5 April (Evening Shift)

If f(t) = 0 2x dx1 - 2 t 2 x, 0 < t < , then the value of 0 2 2 dtf(t) equals .

Ques. 11JEE Main 2024, 4 April (Morning Shift)

Let f(x)= cases-2, & -2 x 0 \ x-2, & 0<x 2 cases and h(x)=f(|x|)+|f(x)|. Then -22 h(x) dx is equal to:

  1. 2
  2. 6
  3. 4
  4. 1

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