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Application of Derivatives

JEE Main 2024 (Online) 9th April Evening ShiftNumericalMedium📈 Topic in 33% papers+4
Ques. 1JEE Main 2024, 9 April (Evening Shift)

Let the set of all values of p, for which f(x) = (p2 - 6p + 8 ) ( 2 2x - 2 2x ) + 2(2 - p)x + 7 does not have any critical point, be the interval (a, b). Then 16ab is equal to .

Answer

252

Solution

2 2x - 2 2x = - 4x, so f(x) = -(p2 - 6p + 8) 4x + 2(2 - p)x + 7

f'(x) = 4(p - 2)(p - 4) 4x - 2(p - 2) = 2(p - 2) [2(p - 4) 4x - 1 ]

f is differentiable everywhere, so it has no critical point only if f'(x) 0 for every x.

p = 2 makes f'(x) = 0 for all x, so p 2 and we need 2(p - 4) 4x 1 for all x.

Since 2(p - 4) 4x takes every value in [-2|p - 4|, 2|p - 4| ], we need 2|p - 4| < 1

72 < p < 92, so 16ab = 16 72 92 = 252

Answer: 252

Keep practising — Application of Derivatives

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

Let f(x) = 4 3 x + 3 3 2 x - 10. The number of points of local maxima of f in interval (0, 2 ) is

  1. 1
  2. 3
  3. 4
  4. 2
Ques. 3JEE Main 2024, 8 April (Morning Shift)

The number of critical points of the function f(x) = (x - 2)2/3(2x + 1) is

  1. 2
  2. 1
  3. 0
  4. 3
Ques. 4JEE Main 2024, 8 April (Morning Shift)

For the function f(x) = ( x) - x + 1, x R, between the following two statements (S1) f(x) = 0 for only one value of x in [0, ]. (S2) f(x) is decreasing in [0, 2 ] and increasing in [ 2, ].

  1. Both (S1) and (S2) are incorrect.
  2. Only (S1) is correct.
  3. Only (S2) is correct.
  4. Both (S1) and (S2) are correct.
Ques. 5JEE Main 2024, 8 April (Evening Shift)

If the function f(x) = 2x3 - 9ax2 + 12a2x + 1, a > 0 has a local maximum at x = and a local minimum at x = 2, then and 2 are the roots of the equation :

  1. x2 - 6x + 8 = 0
  2. 8x2 - 6x + 1 = 0
  3. 8x2 + 6x - 1 = 0
  4. x2 + 6x + 8 = 0
Ques. 6JEE Main 2024, 8 April (Evening Shift)

Let A be the region enclosed by the parabola y2 = 2x and the line x = 24. Then the maximum area of the rectangle inscribed in the region A is .

Ques. 7JEE Main 2024, 6 April (Morning Shift)

The interval in which the function f(x) = xx, x > 0, is strictly increasing is

  1. (0, )
  2. (0, 1e ]
  3. [ 1e2, 1 )
  4. [ 1e, )
Ques. 8JEE Main 2024, 6 April (Evening Shift)

If the function f(x) = ( 1x )2x; x > 0 attains the maximum value at x = 1e then :

  1. e < e
  2. e2 < (2 )e
  3. (2e) > (2e)
  4. e > e
Ques. 9JEE Main 2024, 5 April (Morning Shift)

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a + b)2 is equal to :

  1. 64
  2. 80
  3. 60
  4. 72
Ques. 10JEE Main 2024, 5 April (Morning Shift)

For the function f(x) = x + 3x - 2 (x2 + x ), where x [0, 2 ], consider the following two statements : (I) f is increasing in (0, 2 ). (II) f' is decreasing in (0, 2 ). Between the above two statements,

  1. only (I) is true.
  2. both (I) and (II) are true.
  3. only (II) is true.
  4. neither (I) nor (II) is true.
Ques. 11JEE Main 2024, 4 April (Evening Shift)

Let f(x) = 3 x - 2 + 4 - x be a real valued function. If and are respectively the minimum and the maximum values of f, then 2 + 2 2 is equal to

  1. 42
  2. 38
  3. 24
  4. 44

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