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JEE Advanced Level Test: Indefinite Integration- 1 - JEE MCQ


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30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced Level Test: Indefinite Integration- 1

JEE Advanced Level Test: Indefinite Integration- 1 for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced Level Test: Indefinite Integration- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Indefinite Integration- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Indefinite Integration- 1 below.
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JEE Advanced Level Test: Indefinite Integration- 1 - Question 1

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 15 cm, then the rate at which the thickness of ice decreases, is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 1

JEE Advanced Level Test: Indefinite Integration- 1 - Question 2

Area of the greatest rectangle that can be inscribed in the ellipse   is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 2

Let a point on the ellipse be P (a cosθ, b sinθ)
So, In the rectangle formed as this point as one of the vertices,
(Length)/2 = a cosθ
(Width)/2 = b sinθ
Area of the rectangle = (2a cosθ)(2b sinθ)
= 4ab(sinθ)(cosθ)
= 2ab sin(2θ)
Max (Area of the rectangle) = 2ab

JEE Advanced Level Test: Indefinite Integration- 1 - Question 3

Let f(x) be differentiable for all x. If f (1) =-2 and f '(x)³ ≥2 x ∈ (1/6)  then

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 3

By LMVT

JEE Advanced Level Test: Indefinite Integration- 1 - Question 4

The normal to the curve x = a ( cos θ + θ sin θ) , y = a (sin θ - θ cos θ) at any point ' θ' is such that

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 4

Equation of normal is 
(y -θ sinq- cosθ )

JEE Advanced Level Test: Indefinite Integration- 1 - Question 5

The function f (x) = x/2+2/x has a local minimum at 

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 5

Signs of f ' (x )

f ' (x ) changes sign as x crosses 2.

f (x ) has minima at x = 2.

 

JEE Advanced Level Test: Indefinite Integration- 1 - Question 6

A value of c for which the conclusion of Mean value Theorem holds for the function f ( x ) = loge x on the interval [1, 3] is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 6

JEE Advanced Level Test: Indefinite Integration- 1 - Question 7

The function f ( x ) = tan-1 (sin x cos x)  is an increasing function in

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 7

JEE Advanced Level Test: Indefinite Integration- 1 - Question 8

Suppose the cubic x3 - px + q = 0 has three distinct real roots where p > 0 and q > 0. Then, which one of the following holds?

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 8

Graph of  y = x2 - px + q cuts x-axis at three distinct pints

Maximum at  x = -√p/3 , minima at  x = √p/3

JEE Advanced Level Test: Indefinite Integration- 1 - Question 9

The shortest distance between the line y - x= 1 and the curve x =y2 is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 9

Condition for shortest distance is slope of tangent to x = y² must be same as slope of line y = x + 1

shortest distance = 

JEE Advanced Level Test: Indefinite Integration- 1 - Question 10

Given p ( x ) = x 4 + ax3 + bx 2 + cx + d such that x =0 is the only real root of p ' ( x ) = 0 .

 If p ( -1) < p (1) , then in the interval [-1,1]

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 10

p ' (x ) has only one change of sign. x = 0 is a point of minima.

P(-1) =1- a + b + d P(0) = d

P(1) =1+ a + b + d

P ( -1) < P (1) , P ( 0) < P (1) , P ( -1) > P ( 0)
P (-1) is not minimum but P(1) is maximum

JEE Advanced Level Test: Indefinite Integration- 1 - Question 11

The equation of the tangent to the curve y= x+4/x2 that is parallel to the x-axis is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 11

(2, 3) is point of contact thus y = 3 is tangent 

JEE Advanced Level Test: Indefinite Integration- 1 - Question 12

Let  be defined by   If f has a local minimum at x = - 1, then a possible value of k is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 12

∴f has a local minimum at x = - 1 

Possible value of k is – 1

JEE Advanced Level Test: Indefinite Integration- 1 - Question 13

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 13

As f(x) is continuous and 1/3 belongs to range

JEE Advanced Level Test: Indefinite Integration- 1 - Question 14

The normal to the curve  x2 + 2xy - 3y2 = 0 at 1,1

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 14

slope of normal at (1, 1) = - 1

∴ Equation of normal y -1 = -1(x -1)

y +x= 2
Solving we get point as (1, 1) and (1/-1/3)

So in forth quadrant 

JEE Advanced Level Test: Indefinite Integration- 1 - Question 15

Consider the function, f ( x ) =| x - 2 | + | x - 5 | .x∈ R 

Statement -1 : f ' ( 4) = 0

Statement -2 : f is continous in [2, 5], differentiable in (2, 5) and f ( 2) = f ( 5)

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 15

f (x ) is constant is [2, 5]

JEE Advanced Level Test: Indefinite Integration- 1 - Question 16

A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 16

JEE Advanced Level Test: Indefinite Integration- 1 - Question 17

Let a, b ∈ R be such that the function f given by f ( x ) = In | x | +bx 2 + ax, x ≠ 0 has extreme values at x =-1 and x = 2.

Statement – 1: f has local maximum at x = - 1 and at x = 2.

Statement-2 : a=1/2 and =-1/4

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 17

JEE Advanced Level Test: Indefinite Integration- 1 - Question 18

The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1]

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 18

f ' ( x ) = 6 x2+ 3
⇒ so f(x) is increasing equation cannot have 2 distinct roots.

JEE Advanced Level Test: Indefinite Integration- 1 - Question 19

The function f ( x ) = sin4 x+ cos4 x is increasing if 

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 19


JEE Advanced Level Test: Indefinite Integration- 1 - Question 20

​Let  dt then f decreasing in the interval

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 20

JEE Advanced Level Test: Indefinite Integration- 1 - Question 21

The two curves x3 - 3xy2 + 2 = 0 and 3x 2 y - y3 - 2 = 0

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 21

mm = -1 ⇒ Curves are orthogonal

JEE Advanced Level Test: Indefinite Integration- 1 - Question 22

The greatest value of f (x) = (x + 1)1/3 - (x - 1)1/3 on [0, 1] is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 22

is decreasing on [0, 1].
Greatest value of f (x ) is f ( 0) = 1 - ( -1) = 2

JEE Advanced Level Test: Indefinite Integration- 1 - Question 23

The function f ( x ) = cot -1 x+ x increases in the interval

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 23

JEE Advanced Level Test: Indefinite Integration- 1 - Question 24

The real number x (x > 0) when added to its reciprocal gives the minimum sum at x equals

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 24

Signs of f ' (x )

Minimum occurs at x = 1

JEE Advanced Level Test: Indefinite Integration- 1 - Question 25

If the function f ( x ) = 2x3 - 9ax2 + 12a 2x + 1, where a >0 , attains its maximum and minimum at p and q respectively such that p² = q, then 'a' equals

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 25

JEE Advanced Level Test: Indefinite Integration- 1 - Question 26

If 2a + 3b + 6c = 0, then at least one root of the equation ax 2 + bx + c = 0 lies in the interval

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 26

f ( 0) = df (1) (∵ 2a + 3b + 6x = 0)

By Rolls theorem, at least one root of f ' (x ) =0 lies in (0, 1)

JEE Advanced Level Test: Indefinite Integration- 1 - Question 27

A point on the parabola y2 = 18x at which the ordinate increases at twice the rate of the abscissa, is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 27

JEE Advanced Level Test: Indefinite Integration- 1 - Question 28

The normal to the curve  x = a (1 + cos θ) , y = a sin θ at 'θ ' always passes through the fixed point

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 28

x = a (1 + cosθ ) , ya = sinθ 

Equation of normal at point ( a (1 + cosθ ) ,a sinθ ) is
y - a sinθ = sinθ/cosθ (x-a(1+cosθ))
y cosθ = x- a ) sinθ

It is clear that normal passes through fixed point (a, 0)

JEE Advanced Level Test: Indefinite Integration- 1 - Question 29

Angle between the tangents to the curve y = x 2 - 5x + 6 at the points (2, 0) and (3, 0) is

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 29

Product of slopes = -1 ⇒ angle between tangents is π/2

JEE Advanced Level Test: Indefinite Integration- 1 - Question 30

A function is matched below against an interval where it is supposed to be increasing, Which of the following pairs is incorrectly matched?

     

Detailed Solution for JEE Advanced Level Test: Indefinite Integration- 1 - Question 30

 Let  f (x) = x2 + 6x +6 ⇒ f (x) = 3x (x +4)

Increasing in (-∞,-4) Let ( ) 2 2
f2 (x) = 3x2 - 2x + 1

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