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JEE Exam  >  JEE Tests  >  Mathematics (Maths) Class 12  >  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) Class 12 - Test: Rate of Change And Approximations

Test: Rate of Change And Approximations for JEE 2025 is part of Mathematics (Maths) Class 12 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 2025 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) Class 12 for JEE & Test: Rate of Change And Approximations solutions in Hindi for Mathematics (Maths) Class 12 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) Class 12 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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