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Differential Equations · Question 28

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

Let f : [0, ) R be a differentiable function such that f(x) = 1 - 2x + 0x ex-t f(t) dt for all x [0, ). Then the area of the region bounded by y = f(x) and the coordinate axes is

  1. 5
  2. 2
  3. 2
  4. 12

Answer

(4) 12

Solution

f(x) = 1 - 2x + ex 0x e-t f(t) dt ⇒ f(0) = 1

Differentiating (Leibnitz rule):

f'(x) = -2 + ex 0x e-t f(t) dt + f(x)

= -2 + (f(x) - 1 + 2x ) + f(x)

⇒ f'(x) - 2f(x) = 2x - 3 (linear, I.F. = e-2x)

⇒ ddx (f e-2x ) = (2x - 3)e-2x

⇒ f e-2x = (2x - 3)e-2x dx = (1 - x)e-2x + C

⇒ f(x) = 1 - x + Ce2x; f(0) = 1 ⇒ C = 0

∴ f(x) = 1 - x, a line cutting the axes at (1, 0) and (0, 1).

Area = 12 1 1 = 12

Answer: (4)

JEE Main 2025 (Online) 4th April Morning Shift

Keep practising — Differential Equations

2

If a curve y = y(x) passes through the point (1, 2 ) and satisfies the differential equation (7x4 y - ex cosec y ) dxdy = x5, x 1, then at x = 2, the value of y is :

  1. 2e2 + e64
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3

Let g be a differentiable function such that 0x g(t) dt = x - 0x t g(t) dt, x 0 and let y = y(x) satisfy the differential equation dydx - y x = 2(x+1) x g(x), x [0, 2 ). If y(0) = 0, then y ( 3 ) is equal to

  1. 4 3
  2. 2 3
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4

Let y = y(x) be the solution of the differential equation dydx + 3( 2 x)y + 3y = 2 x, y(0) = 13 + e3. Then y ( 4 ) is equal to :

  1. 43
  2. 23 + e3
  3. 43 + e3
  4. 23
JEE Main 2025 (Online) 3rd April Evening Shift
5

Let f : R R be a thrice differentiable odd function satisfying f'(x) 0, f''(x) = f(x), f(0) = 0, f'(0) = 3. Then 9f( e 3) is equal to .

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6

Let f : [1, ) [2, ) be a differentiable function. If 10 1x f(t) dt = 5xf(x) - x5 - 9 for all x 1, then the value of f(3) is :

  1. 22
  2. 26
  3. 32
  4. 18
JEE Main 2025 (Online) 2nd April Evening Shift
7

Let y = y(x) be the solution of the differential equation dydx + 2y 2 x = 2 2 x + 3 x 2 x such that y(0) = 54. Then 12 (y ( 4 ) - e-2 ) is equal to .

JEE Main 2025 (Online) 2nd April Evening Shift
8

Let y = y(x) be the solution of the differential equation : x( e( x))2 dy + ( x - 3y x e( x))dx = 0, x (0, 2 ). If y ( 4 ) = - 1 e 2, then y ( 6 ) is equal to :

  1. 2 e(3) - e(4)
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  3. 1 e(4) - e(3)
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9

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10

Let for some function y = f(x), 0x t f(t) dt = x2 f(x), x > 0 and f(2) = 3. Then f(6) is equal to

  1. 1
  2. 6
  3. 2
  4. 3
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11

If y = y(x) is the solution of the differential equation, 4 - x2 dydx = ( ( -1 ( x2 ) )2 - y ) -1 ( x2 ), -2 x 2, y(2) = 2 - 84, then y2(0) is equal to .

JEE Main 2025 (Online) 28th January Evening Shift

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