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JEE Exam  >  JEE Tests  >  Mathematics (Maths) Class 12  >  Test: Application of Derivatives- Case Based Type Questions- 1 - JEE MCQ

Test: Application of Derivatives- Case Based Type Questions- 1 - JEE MCQ


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15 Questions MCQ Test Mathematics (Maths) Class 12 - Test: Application of Derivatives- Case Based Type Questions- 1

Test: Application of Derivatives- Case Based Type Questions- 1 for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The Test: Application of Derivatives- Case Based Type Questions- 1 questions and answers have been prepared according to the JEE exam syllabus.The Test: Application of Derivatives- Case Based Type Questions- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Application of Derivatives- Case Based Type Questions- 1 below.
Solutions of Test: Application of Derivatives- Case Based Type Questions- 1 questions in English are available as part of our Mathematics (Maths) Class 12 for JEE & Test: Application of Derivatives- Case Based Type Questions- 1 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: Application of Derivatives- Case Based Type Questions- 1 | 15 questions in 30 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: Application of Derivatives- Case Based Type Questions- 1 - Question 1
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Read the following text and answer the following questions. On the basis of the same:

An open box is to be made out of a piece of cardboard measuring (24 cm × 24 cm) by cutting of equal squares from the corners and turning up the sides.

Find the value of dV/dx ?

Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 1

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 2
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Read the following text and answer the following questions, on the basis of the same:

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed by the following equation y = 4x – 1/2x2. where x is the number of days exposed to sunlight.

What is the maximum height of the plant?

Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 2

We have, number of days for maximum height of plant = 4 days

∴ Maximum height of plant

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Test: Application of Derivatives- Case Based Type Questions- 1 - Question 3
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Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.

Find the second order derivative of the function at x = 5.

Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 3
f(x) = 6(2x4 - x2)

f’(x) = 6[8x3 - 2x]

f”(x) = 6[24x2 - 2]

f”(5) = 6[24 x 25 - 2]

= 6[600 - 2]

= 3588

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 4
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Read the following text and answer the following questions on the basis of the same:

P(x) = –5x2 + 125x + 37500 is the total profit function of a company, where x is the production of the company.

What will be the production when the profit is maximum?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 4

We, have

p(x) =-5x2 + 125x+ 37500

p’(x) = -10x + 125

For maximum profit

p’(x) = 0

-10x + 125 = 0

-10x = -125

x = 125/10

= 12.5

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 5
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Read the following text and answer the following questions, on the basis of the same:

An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:

The maximum value of area A is :
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 5

For maxima,

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 6
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Read the following text and answer the following questions on the basis of the same:

A right circular cylinder is inscribed in a cone.

S = Curved Surface Area of Cylinder.

r/r1 = ?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 6

In ?DEC and ? OBC

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 7
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Read the following text and answer the following questions, on the basis of the same:

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed by the following equation y = 4x – 1/2x2. where x is the number of days exposed to sunlight.

The rate of growth of the plant with respect to sunlight is ______ .
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 7

∴ rate of growth of the plant with respect to sunlight

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 8
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Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.

At which of the following intervals will f(x) be increasing?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 8

For increasing

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 9
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Read the following text and answer the following questions. On the basis of the same:

An open box is to be made out of a piece of cardboard measuring (24 cm × 24 cm) by cutting of equal squares from the corners and turning up the sides.

Find the value of x other than 12?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 9
The length x is cut from every corner of the paper so that we can fold it and it looks like an open box.

It means the height of the box is x cm.

Then the volume of the box will be (24 − 2x)2x.

On solving it further we get the volume V as:

V = (24 − 2x)2x

On differentiating the above volume with respect to x we get,

We will use here uv rule to differentiate the term.

When we differentiate uv that is (uv)’ we get (uv)’ = u’v + uv’.

On equating it to zero to get the value of x,

We got the value of x = 12, 4.

The value of x as 12 will be neglected because it will not be a box if we cut the length of 12cm from both the sides. So, the value of x is 4cm.

And the maximum volume of box will be ⇒ (24 − 2(4))2(4) = 1024cm3

We need to find the height when the volume is maximum. If we observe the figure we can clearly see that x is the height itself and it is 4cm.

If we want to check that the maximum volume will be at 4 or not we will double differentiate the volume at 4 and if it is negative it means this point is point of maxima so, we do,

On putting the value of x = 4 in the above equation we get,

Hence the value is negative so, point 4 is the point of maxima.

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 10
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Read the following text and answer the following questions on the basis of the same:

P(x) = –5x2 + 125x + 37500 is the total profit function of a company, where x is the production of the company.

Check in which interval the profit is strictly increasing .
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 10
Profit is strictly increasing where

P’(x) > 0

-10x + 125 >0

125 > 10x

10x < 125

X < 12.5

So, profit is strictly increasing for x ∈ (0, 12.5)

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 11
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Read the following text and answer the following questions, on the basis of the same:

An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:

If x and y represents the length and breadth of the rectangular region, then the relation between the variables is :
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 11

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 12
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Read the following text and answer the following questions on the basis of the same:

A right circular cylinder is inscribed in a cone.

S = Curved Surface Area of Cylinder.

r/r1 = ?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 12

In ?DEC and ? OBC

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 13
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Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.

Which value from the following may be abscissa of critical point?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 13
Critical point is point where f’(x) = 0

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 14
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Read the following text and answer the following questions on the basis of the same:

A right circular cylinder is inscribed in a cone.

What is the value of dS/dh?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 14

Test: Application of Derivatives- Case Based Type Questions- 1 - Question 15
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Read the following text and answer the following questions. On the basis of the same:

An open box is to be made out of a piece of cardboard measuring (24 cm × 24 cm) by cutting of equal squares from the corners and turning up the sides.

Volume is maximum at what height of that open box?
Detailed Solution for Test: Application of Derivatives- Case Based Type Questions- 1 - Question 15

For maximum value,

dV/dx = 0

i.e., 12(x2 - 16x + 48) = 0

X2 - 16 x + 48 = 0

x2 - 4x - 12x + 48 = 0

x(x - 4) - 12(x - 4) = 0

(x - 4)(x - 12) = 0

X = 4, 12

V(x = 4) = (24 - 2 x 4)(24 - 2 x 4) x 4

= 16 x 16 x 4

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