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JEE Exam  >  JEE Tests  >  Mathematics (Maths) for JEE Main & Advanced  >  Test: Rate of Change And Approximations - JEE MCQ

Test: Rate of Change And Approximations - JEE MCQ


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10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Rate of Change And Approximations

Test: Rate of Change And Approximations for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Rate of Change And Approximations questions and answers have been prepared according to the JEE exam syllabus.The Test: Rate of Change And Approximations MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Rate of Change And Approximations below.
Solutions of Test: Rate of Change And Approximations questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Rate of Change And Approximations solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Rate of Change And Approximations | 10 questions in 10 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Test: Rate of Change And Approximations - Question 1
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The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.​

Test: Rate of Change And Approximations - Question 2
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The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 5x2 + 22x + 35. Find the marginal revenue, when x = 7, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant

Detailed Solution for Test: Rate of Change And Approximations - Question 2

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Test: Rate of Change And Approximations - Question 3
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Find the approximate change in total surface area of a cube of side x metre caused by increase in side by 1%.​

Test: Rate of Change And Approximations - Question 4
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The volume of cube is increasing at the constant rate of 3 cm3/s. Find the rate of change of edge of the cube when its edge is 5 cm.​
Detailed Solution for Test: Rate of Change And Approximations - Question 4

Let V be the instantaneous volume of the cube.
dV/dt = 3 cm3/s
Let x be the side of the cube.
V = x3
dV/dt = 3x2 * (dx/dt)
3 = 3*(52)*(dx/dt)
So, dx/dt = 1/25 cm/s

Test: Rate of Change And Approximations - Question 5
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The total cost associated with the production of x units of a product is given by C(x) = 5x2 + 14x + 6. Find marginal cost when 5 units are produced
Detailed Solution for Test: Rate of Change And Approximations - Question 5

C(x) = 5x2 + 14x + 6
Marginal Cost M(x) = C’(x)
M(x) = 10x +14
So, M(5) = 50 + 14
= 64 Rs

Test: Rate of Change And Approximations - Question 6
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At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.
Detailed Solution for Test: Rate of Change And Approximations - Question 6

Tan(z) = h⁄x

Test: Rate of Change And Approximations - Question 7
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A stone is dropped into a quiet lake and waves move in circles at a speed of 2cm per second. At the instant, when the radius of the circular wave is 12 cm, how fast is the enclosed area changing ?​
Detailed Solution for Test: Rate of Change And Approximations - Question 7

Rate of increase of radius dr/dt = 2 cm/s
Area of circle A = πr2
dA/dt = π*(2r)*(dr/dt)
= π*(24)*2
= 48π cm2/s
Rate of increase of area is 48π cm2/s (increasing as it is positive).

Test: Rate of Change And Approximations - Question 8
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Using approximation find the value of  ​
Detailed Solution for Test: Rate of Change And Approximations - Question 8

Let x=4, Δx=0.01

y=x^½ = 2

y+Δy = (x+ Δx)^½ = (4.01)^½

Δy = (dy/dx) * Δx

Δy = (x^(-1/2))/2 * Δx

Δy = (½)*(½) * 0.01

Δy = 0.25 * 0.01

Δy = 0.0025

So, (4.01)^½ = 2 + 0.0025 = 2.0025

Test: Rate of Change And Approximations - Question 9
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Find the approximate value of f(10.01) where f(x) = 5x2 +6x + 3​
Detailed Solution for Test: Rate of Change And Approximations - Question 9

f(x) = 5x2 +6x + 3
f(10.01) = 5*(10.01)2 + 6*(10.01) + 3
To find (10.01)2
Let p=10, Δp=0.01
y=p2 = 100
y+Δy = (p+ Δp)2 = (10.01)2
Δy = (dy/dp) * Δp
Δy = 2*p* Δx
Δy = 2*10* 0.01
Δy = 20 * 0.01
Δy = 0.2
So, (10.01)2 = y + Δy
= 100.2
So,
f(10.01) = 5*(100.2) + 6*(10.01) + 3
= 501 + 60.06 + 3
= 564.06

Test: Rate of Change And Approximations - Question 10
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Given a function y = f(x) . Let Δx be the very small change in the value of x , then the corresponding change in the value of y that is Δy is approximately given by
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