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Application of Derivatives

JEE Main 2025 (Online) 29th January Evening ShiftMCQMedium📈 Topic in 13% papers+4−1
Ques. 1JEE Main 2025, 29 January (Evening Shift)

Let f(x)= 0x t (t2-9t+20 )dt, 1 x 5. If the range of f is [ , ], then 4( + ) equals :

  1. 253
  2. 157
  3. 154
  4. 125

Answer

(2) 157

Solution

f'(x)=x(x-4)(x-5) ⇒ f'>0 on (1,4), f'<0 on (4,5) ⇒ maximum at x=4

[figure]

f(x)= x44-3x3+10x2: f(1)= 294, f(4)=32, f(5)= 1254

Min = 294 ⇒ = 294, =32

∴ 4( + )=29+128=157

Answer: (2)

Keep practising — Application of Derivatives

Ques. 2JEE Main 2025, 28 January (Morning Shift)

The sum of all local minimum values of the function f(x) = cases 1 - 2x, & x < -1 \ 13(7 + 2|x|), & -1 x 2 \ 1118(x - 4)(x - 5), & x > 2 cases is

  1. 16772
  2. 15772
  3. 17172
  4. 13172
Ques. 3JEE Main 2025, 24 January (Morning Shift)

Consider the region R= \(x,y):x y 9- 113x2, x 0 \. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in R, is:

  1. 821123
  2. 567121
  3. 730119
  4. 625111
Ques. 4JEE 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
Ques. 5JEE 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. 6JEE 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. 7JEE 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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