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Application of Derivatives

JEE Main 2025 (Online) 24th January Evening ShiftMCQMedium📈 Topic in 16% papers+4−1
Ques. 1JEE Main 2025, 24 January (Evening Shift)

Let (2,3) be the largest open interval in which the function f(x)=2 e(x-2)-x2+ax+1 is strictly increasing and (b,c) be the largest open interval, in which the function g(x)=(x-1)3(x+2-a)2 is strictly decreasing. Then 100(a+b-c) is equal to :

  1. 360
  2. 420
  3. 160
  4. 280

Answer

(1) 360

Solution

For x>2: f'(x)= 2x-2-2x+a and f''<0 f' falls, so the largest interval ends where f'=0:

f'(3)=0 2-6+a=0 a=4

g'(x)=(x-1)2(x-2)(5x-8) — sign of g':

[figure]

g'<0 on ( 85,2 ) b= 85, c=2

100(a+b-c)=100 (4+ 85-2 )=360

Answer: (1)

Keep practising — Application of Derivatives

Ques. 2JEE Main 2025, 23 January (Morning Shift)

If the set of all values of a, for which the equation 5x3-15x-a=0 has three distinct real roots, is the interval ( , ), then -2 is equal to .

Ques. 3JEE Main 2025, 23 January (Evening Shift)

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of 81 cm3/min and the thickness of the ice-cream layer decreases at the rate of 14 cm/min. The surface area (in cm2) of the chocolate ball (without the ice-cream layer) is :

  1. 128
  2. 196
  3. 225
  4. 256
Ques. 4JEE Main 2025, 22 January (Evening Shift)

Let f(x)= 0x2 t2-8t+15et dt, x . Then the numbers of local maximum and local minimum points of f, respectively, are :

  1. 3 and 2
  2. 2 and 2
  3. 2 and 3
  4. 1 and 3

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