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JEE Advanced Level Test: Application of Derivative- 2 - JEE MCQ


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30 Questions MCQ Test - JEE Advanced Level Test: Application of Derivative- 2

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JEE Advanced Level Test: Application of Derivative- 2 - Question 1

The value of ‘a’ for which x3 - 3x + a = 0 has two distinct roots in [0, 1] is given by

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 1

Let α, β ∈[0,1].f (x) is continuous on [a,b] & differentiable on (a,b) and f (α) = f (β) = 0
∴ c ∈ (α, β) such that f' (c) = 0 ⇒ c = ±1∉ (0,1)

JEE Advanced Level Test: Application of Derivative- 2 - Question 2

The value of ‘c’ in Lagrange’s mean value theorem for f (x) = x (x- 2)2 in [0, 1]

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 2

f '(c) = 0 2c(c - 2) + (c - 2)2 = 0
c = 2,2/3
∴ c = 2/3 (c ≠2)

JEE Advanced Level Test: Application of Derivative- 2 - Question 3

For the function f (x) = x3 - 6x2 + ax + b, if Roll’s theorem holds in [1, 3] with 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 3

f (1) = f (3) ⇒ a = 11

JEE Advanced Level Test: Application of Derivative- 2 - Question 4

Find Value of ‘c’ by using  Rolle’s theorem for f (x) = log (x2 + 2) - log 3 on [-1,1]

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 4

JEE Advanced Level Test: Application of Derivative- 2 - Question 5

The chord joining the points where x = p and x = q on the curve y = ax2 + bx + c is parallel to the tangent at the point on the curve whose abscissa is 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 5

Apply Lagrange’s theorm

JEE Advanced Level Test: Application of Derivative- 2 - Question 6

The least value of k for which the function f(x) = x2 + kx + 1 is a increasing  function in the interval 1 < x < 2

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 6


JEE Advanced Level Test: Application of Derivative- 2 - Question 7

The interval in  which f (x) = x3 - 3x2 - 9x + 20 is strictly decreasing 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 7

Given f (x) = x3 - 3x2 - 9x + 20
⇒ f '(x) = 3x2 -6x -9
⇒ f '(x) = 3(x - 3)(x +1)

Thus, f (x) is strictly increasing for
x ∈ (-∞,-1) U (3, ∞) and strictly decreasing for x ∈ (-1, 3)

JEE Advanced Level Test: Application of Derivative- 2 - Question 8

The critical points of

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 8


f ' (x) = 0 ⇒ x = 1; f1 (x) does not exist at x = 2
∴ x = 1 and x = 2 are two critical points

JEE Advanced Level Test: Application of Derivative- 2 - Question 9

The number of stationary points of f (x) = sin x in [0,2π] are

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 9

f (x) = sinx ⇒ f '(x) = cosx ⇒ f '(x) = 0
Therefore number of stationary points of f (x) in [0, 2π] is 2.

JEE Advanced Level Test: Application of Derivative- 2 - Question 10

Local minimum values of the function 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 10

AM > GM

JEE Advanced Level Test: Application of Derivative- 2 - Question 11

If the function has maximum at x =-3, then the value of ‘a’ is 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 11

since f (x) has local maximum at x = -3 ⇒ f ' (-3) = 0 and f 11 (-3)< 0

JEE Advanced Level Test: Application of Derivative- 2 - Question 12

The point at which f (x) = (x- 1)4 assumes local maximum or local minimum value are

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 12


Therefore n = iv is even and fiv (1) = 24> 0
Therefore f (x) has local minimum at x = 1.

JEE Advanced Level Test: Application of Derivative- 2 - Question 13

The global maximum and global minimum of f (x) = 2x3 - 9x2 + 12x + 6 in [0, 2]

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 13


Therefore global maximum M1 = max{f (0), f (1), f (2)}= 11
Global minimum
M2 = max{f (0), f (1), f (2)}= 6

JEE Advanced Level Test: Application of Derivative- 2 - Question 14

The approximate value of 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 14

JEE Advanced Level Test: Application of Derivative- 2 - Question 15

The approximate value of 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 15


JEE Advanced Level Test: Application of Derivative- 2 - Question 16

If the percentage error in the surface area of sphere is k, then the percentage error in its volume is 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 16

JEE Advanced Level Test: Application of Derivative- 2 - Question 17

If an error of  is made in measuring the radius of a sphere then percentage error in its volume is

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 17

V% = 3(S%)

JEE Advanced Level Test: Application of Derivative- 2 - Question 18

The height of a cylinder is equal to its radius. If an error of 1 % is made in its height. Then the percentage error in its volume is

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 18

h = r and v = ph3; V% = 3( h%)

JEE Advanced Level Test: Application of Derivative- 2 - Question 19

The slope of the normal to the curve given by 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 19

JEE Advanced Level Test: Application of Derivative- 2 - Question 20

The line   is a tangent to the curve  then n ∈

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 20

Calculate slope

JEE Advanced Level Test: Application of Derivative- 2 - Question 21

The points on the curve  at which the tangent is perpendicular to x-axis are

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 21

dy/dx is not defined.

JEE Advanced Level Test: Application of Derivative- 2 - Question 22

The point on the curve   at which the tangent drawn is

JEE Advanced Level Test: Application of Derivative- 2 - Question 23

The sum of the squares of the intercepts on the axes of the tangent at any point on the curve x 2/3 + y2/3= a2/3 is

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 23

Equation of the tangent at p (θ) to 

JEE Advanced Level Test: Application of Derivative- 2 - Question 24

If the straight line x cos α + y sinα = p touches the curve  at the point (a, b) on it, then

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 24

Find dy/dx and the equation of the tangent

JEE Advanced Level Test: Application of Derivative- 2 - Question 25

If the curves x = y² and xy = k cut each other orthogonally then k² = 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 25

m1.m2 =-1

JEE Advanced Level Test: Application of Derivative- 2 - Question 26

The angle between the curves y = x³ and 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 26

Find dy/dx to the two curves at (1, 1) they are m1 and m2. Then 

JEE Advanced Level Test: Application of Derivative- 2 - Question 27

If the curves ay + x² = 7 and x³ = y cut orthogonally at (1, 1) then a = 

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 27

Slope of the first curve at (1, 1) is  Slope of the second curve at (1, 1) is m2 = 3

JEE Advanced Level Test: Application of Derivative- 2 - Question 28

A particle moves along a line is given by then the distance travelled by the particle before it first comes to rest is

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 28

JEE Advanced Level Test: Application of Derivative- 2 - Question 29

A particle is moving along a line such that s = 3t3 - 8t + 1. Find the time ‘t’ when the distance ‘S’ travelled by the particle increases.

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 29


JEE Advanced Level Test: Application of Derivative- 2 - Question 30

A particle moves along a line by S = t3 - 9t2 + 24t the time when its velocity decreases.

Detailed Solution for JEE Advanced Level Test: Application of Derivative- 2 - Question 30

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