Prove that the logarithmic function $f(x) = \log x$ is increasing on $(0, \infty)$. Which of the following statements is true?

Correct Option: A. Solution: The derivative of $f(x) = \log x$ is $f'(x) = \frac{1}{x}$, which is positive for $x > 0$. Therefore, $f(x)$ is increasing on $(0, \infty)$.

For what values of $a$ is the function $f(x) = x^2 + ax + 1$ increasing on $[1, 2]$?

Correct Option: A. Solution: The function $f(x) = x^2 + ax + 1$ is increasing on $[1, 2]$ if $f'(x) = 2x + a > 0$ for all $x \in [1, 2]$. Solving $2x + a > 0$ gives $a > -2$.

Find the absolute maximum value of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval [1, 5].

Correct Option: D. Solution: The absolute maximum value is 56, occurring at $x = 5$.

For the function $f(x) = x^3$, what is true about the point $x = 0$?

Correct Option: C. Solution: At $x = 0$, the function $f(x) = x^3$ has a point of inflection.

Consider a function $f(x) = x^4 - 62x^2 + ax + 9$. If the function attains its maximum value at $x = 1$ on the interval $[0, 2]$, what is the value of $a$?

Correct Option: A. Solution: To find the value of $a$, we need to ensure that $f'(x) = 0$ at $x = 1$. The derivative $f'(x) = 4x^3 - 124x + a$. Setting $f'(1) = 0$, we get $4(1)^3 - 124(1) + a = 0 \Rightarrow a = 62$. Therefore, the correct option is '62'.

Application of Derivatives

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Summary

Chapter 6: Application of Derivatives

Introduction

Study of applications of derivatives in various fields: engineering, science, social science.
Key applications include:
- Determining rate of change of quantities.
- Finding equations of tangent and normal to a curve.
- Locating turning points on a graph.
- Identifying intervals of increase or decrease of functions.
- Approximating values of certain quantities.

Rate of Change of Quantities

Derivative
- Represents the rate of change of one quantity with respect to another.
- Example: If y varies with x, then
- Chain Rule: If x = f(t) and y = g(t), then

Maxima and Minima

Use of derivatives to find maximum or minimum values of functions.
Definition of maximum and minimum values:
- Maximum: Point c in interval I such that for all x in I, f(c) is the maximum value.
- Minimum: Point c in interval I such that for all x in I, f(c) is the minimum value.
Critical points: Points where f'(c) = 0 or not differentiable.

Examples of Applications

Profit Maximization: Given profit function P(x) = ax + bx², find optimal number of trees.
Projectile Motion: Maximum height of a ball thrown from a height.
Distance Minimization: Nearest distance from a point to a curve.

Important Problems

Finding maximum and minimum values of various functions:
- Example problems include finding dimensions for maximum volume of boxes, maximum areas of shapes, etc.
Common scenarios involve optimizing areas, volumes, and costs in practical applications.

Learning Objectives

Understand the applications of derivatives in various fields such as engineering, science, and social science.
Determine the rate of change of quantities using derivatives.
Find the equations of tangent and normal to a curve at a point.
Identify turning points on the graph of a function to locate local maxima and minima.
Analyze intervals on which a function is increasing or decreasing.
Use derivatives to approximate values of certain quantities.

Detailed Notes

Chapter 6: Application of Derivatives

6.1 Introduction

Study applications of derivatives in various fields such as engineering, science, and social science.
Key applications include:
- Determining the rate of change of quantities.
- Finding equations of tangent and normal to curves.
- Locating turning points on graphs to identify local maxima and minima.
- Identifying intervals of increase or decrease for functions.
- Approximating values of certain quantities.

6.2 Rate of Change of Quantities

The derivative $rac{ds}{dt}$ represents the rate of change of distance S with respect to time t.
For a function $y = f(x)$ , the derivative $rac{dy}{dx}$ represents the rate of change of y with respect to x.
If two variables x and y vary with respect to another variable t, then by the Chain Rule:
- $rac{dy}{dx} = rac{dy}{dt} imes rac{dt}{dx}$

6.4 Maxima and Minima

Use derivatives to find maximum or minimum values of functions.
Definitions:
- A function f has a maximum value in an interval I if there exists a point c in I such that $f(c) \geq f(x)$ for all $x \in I$ .
- A function f has a minimum value in an interval I if there exists a point c in I such that $f(c) \leq f(x)$ for all $x \in I$ .
- A critical point is where $f'(c) = 0$ or f is not differentiable.

Examples of Applications

Maximizing Profit: Given profit function $P(x) = ax + bx^2$ , find the number of trees per acre that maximizes profit.
Projectile Motion: For a ball thrown from a height, find the maximum height it reaches.
Distance Minimization: Find the nearest distance from a point to a curve.

Important Problems

Find maximum and minimum values of various functions, such as:
- $f(x) = x^4 - 62x^2 + ax + 9$ at $x = 1$ .
- $f(x) = x + ext{sin}(2x)$ on $[0, 2 ext{π}]$ .
- Two numbers whose sum is 24 and product is maximized.
- Dimensions of a box formed from a rectangular sheet to maximize volume.

Conclusion

Understanding derivatives is crucial for solving real-world problems involving optimization and rates of change.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Derivatives: Students often confuse the derivative as a function rather than a rate of change. Remember that the derivative represents the rate of change of one quantity with respect to another.
Ignoring Critical Points: Failing to identify critical points where the derivative is zero or undefined can lead to missing local maxima or minima.
Not Applying the First Derivative Test: Students may overlook using the first derivative test to determine whether a critical point is a maximum or minimum.
Confusing Increasing and Decreasing Intervals: Be careful to correctly identify intervals where the function is increasing or decreasing based on the sign of the derivative.
Neglecting Absolute Extrema: When asked for absolute maximum or minimum values, ensure to check endpoints of the interval as well as critical points.

Tips for Success

Practice Derivative Applications: Work on problems that require you to find maximum and minimum values, as these are common in exams.
Draw Graphs: Visualizing functions can help in understanding where they increase or decrease and where maxima and minima occur.
Review the Chain Rule: Make sure you are comfortable using the chain rule for derivatives, especially in problems involving multiple variables.
Check Units: In applied problems, always ensure that your answers are in the correct units, especially when dealing with rates of change.
Understand the Context: When solving real-world problems, ensure you understand what the maximum or minimum represents in that context.

Practice & Assessment

Multiple Choice Questions

The maximum area occurs when the base is the length of the minor axis.

The maximum area occurs when the base is the length of the major axis.

The maximum area occurs when the height is equal to the semi-minor axis.

The maximum area occurs when the height is equal to the semi-major axis.

Correct Answer: A

Solution:

The maximum area of an isosceles triangle inscribed in an ellipse is achieved when the base is equal to the length of the minor axis.

₹30.02

₹29.98

₹30.00

₹30.05

Correct Answer: A

Solution:

The marginal cost is the derivative of

C(x)

evaluated at

x = 3

, which is approximately ₹30.02.

Correct Answer: A

Solution:

Marginal Revenue (MR) is the derivative of the total revenue function R(x). Thus, MR = dR/dx = 6x + 36. Substituting x = 5, we get MR = 6(5) + 36 = 66.

1.25

2.25

Correct Answer: D

Solution:

To find the maximum value, evaluate

f(x)

at the endpoints and critical points.

f(-1) = 3

f(1) = 1

, and the critical point

x = 0.5

gives

f(0.5) = 0.25

. The maximum value is

f(-1) = 3

f(x)

is increasing because

f'(x) > 0

for

x > 0

f(x)

is increasing because

f'(x) < 0

for

x > 0

f(x)

is decreasing because

f'(x) > 0

for

x > 0

f(x)

is decreasing because

f'(x) < 0

for

x > 0

Correct Answer: A

Solution:

The derivative of

f(x) = \log x

f'(x) = \frac{1}{x}

, which is positive for

x > 0

. Therefore,

f(x)

is increasing on

(0, \infty)

f(x) = \cos x

f(x) = \sin x

f(x) = \tan x

f(x) = x^2

Correct Answer: B

Solution:

The function

f(x) = \sin x

is increasing on the interval

(0, \pi)

Correct Answer: D

Solution:

Evaluating the function at critical points and endpoints:

f(1) = 24

f(2) = 29

f(3) = 28

f(5) = 56

. The maximum value is 56 at

x = 5

a > -2

a < 2

a = 0

a \geq 0

Correct Answer: A

Solution:

The function

f(x) = x^2 + ax + 1

is increasing on

[1, 2]

f'(x) = 2x + a > 0

for all

x \in [1, 2]

. Solving

2x + a > 0

gives

a > -2

14 m each

8 m for the square and 20 m for the circle

10 m for the square and 18 m for the circle

12 m for the square and 16 m for the circle

Correct Answer: D

Solution:

Let the length of the wire for the square be

x

and for the circle be

28-x

. The area of the square is

\left(\frac{x}{4}\right)^2

and the area of the circle is

\pi\left(\frac{28-x}{2\pi}\right)^2

. To minimize the total area, set the derivative of the total area with respect to

x

to zero and solve for

x

. The optimal lengths are 12 m for the square and 16 m for the circle.

\sqrt{5}

\sqrt{10}

Correct Answer: A

Solution:

The helicopter's position is at point

(x, x^2 + 7)

. The distance between the soldier and the helicopter is minimized at

x = 1

, giving a distance of

\sqrt{5}

0.5 meters per hour

1 meter per hour

2 meters per hour

5 meters per hour

Correct Answer: A

Solution:

The rate of change of volume is 5 cubic meters per hour. Since the base area is 10 square meters, the rate at which the water level is rising is given by the volume rate divided by the base area:

\frac{5}{10} = 0.5

meters per hour.

Correct Answer: D

Solution:

The absolute maximum value is 56, occurring at

x = 5

5 cm

6 cm

7 cm

8 cm

Correct Answer: B

Solution:

Let the side of the square cut off be

x

cm. Then the dimensions of the box will be

(45-2x) \times (24-2x) \times x

. The volume

V(x) = x(45-2x)(24-2x)

. To find the maximum volume, we take the derivative

V'(x)

and set it to zero. Solving

V'(x) = 0

gives

x = 6

cm.

5/3 km/h

2 km/h

5 km/h

3 km/h

Correct Answer: A

Solution:

Using similar triangles, the rate at which the shadow length increases is 5/3 km/h.

Rs 560

Rs 640

Rs 720

Rs 800

Correct Answer: B

Solution:

To minimize the cost, we need to find the dimensions of the tank that minimize the surface area. The cost is minimized when the surface area is minimized, given the constraints.

Rectangle: 4m x 2m, Semicircle: radius 1m

Rectangle: 3m x 2m, Semicircle: radius 2m

Rectangle: 2m x 3m, Semicircle: radius 2m

Rectangle: 2m x 4m, Semicircle: radius 1m

Correct Answer: A

Solution:

Let the width of the rectangle be

w

and the height be

h

. The radius of the semicircle is

r = w/2

. The perimeter is

2h + w + \pi r = 10

. Substituting

r = w/2

gives

2h + w + \pi w/2 = 10

. Solving, we find

w = 4

h = 2

, and

r = 1

It is a point of local maxima.

It is a point of local minima.

It is a point of inflection.

It is neither a point of local maxima nor minima.

Correct Answer: C

Solution:

x = 0

, the function

f(x) = x^3

has a point of inflection.

Infinity

Cannot be determined

Correct Answer: A

Solution:

The profit function

p(x) = 41 - 72x - 18x^2

is a downward-opening parabola. The maximum profit is the constant term, 41, when

x = 0

Correct Answer: A

Solution:

The marginal revenue is the derivative of the revenue function:

R'(x) = 6x + 36

. At

x = 5

R'(5) = 6(5) + 36 = 66

Correct Answer: A

Solution:

For

f(x)

to have a maximum at

x = 1

f'(1) = 0

. Differentiating,

f'(x) = 4x^3 - 124x + a

. Substituting

x = 1

, we get

4(1)^3 - 124(1) + a = 0

. Solving gives

a = 62

Correct Answer: B

Solution:

Let the radius of the circle be

r

and the side of the square be

s

. The perimeter constraint gives

2\pi r + 4s = k

. For minimum area,

s = 2r

. Substituting

s = 2r

into the perimeter equation, we get

2\pi r + 8r = 20\pi

, solving gives

r = 4

\sqrt{5}

\sqrt{10}

Correct Answer: A

Solution:

The distance between the helicopter at

(x, x^2 + 7)

and the soldier at

(3, 7)

is given by

d = \sqrt{(x - 3)^2 + (x^2 + 7 - 7)^2} = \sqrt{(x - 3)^2 + x^4}

. To minimize this, we find the derivative and set it to zero. Solving gives

x = 1

, and the minimum distance is

\sqrt{5}

8π cm²/s

16π cm²/s

32π cm²/s

64π cm²/s

Correct Answer: B

Solution:

The rate of change of the area

A

of a circle with respect to its radius

r

is given by

\frac{dA}{dr} = 2\pi r

. Substituting

r = 4

cm, we get

\frac{dA}{dr} = 2\pi \times 4 = 8\pi

cm²/s.

8 and 8

10 and 6

9 and 7

12 and 4

Correct Answer: A

Solution:

The minimum sum of cubes is achieved when both numbers are equal, i.e., 8 and 8.

30π cm²/s

60π cm²/s

90π cm²/s

120π cm²/s

Correct Answer: B

Solution:

The rate of change of area is given by

\frac{dA}{dt} = 2\pi r \frac{dr}{dt}

. Substituting

r = 10

cm and

\frac{dr}{dt} = 3

cm/s, we get

\frac{dA}{dt} = 60\pi

cm²/s.

Area = πab

Area = ab

Area = 2ab

Area = 4ab

Correct Answer: C

Solution:

The maximum area of an isosceles triangle inscribed in an ellipse is achieved when the base is along the minor axis and the vertex is at the end of the major axis.

It has a local maximum at

x = 0

It has a local minimum at

x = 0

It has a point of inflection at

x = 0

It is strictly decreasing on

(-\infty, \infty)

Correct Answer: C

Solution:

The function

f(x) = x^3

has a point of inflection at

x = 0

because the second derivative changes sign at this point.

Correct Answer: A

Solution:

To find the value of

a

, we need to ensure that

f'(x) = 0

x = 1

. The derivative

f'(x) = 4x^3 - 124x + a

. Setting

f'(1) = 0

, we get

4(1)^3 - 124(1) + a = 0 \Rightarrow a = 62

. Therefore, the correct option is '62'.

5 cm

6 cm

7 cm

8 cm

Correct Answer: B

Solution:

By setting up the volume function and finding its maximum using derivatives, the optimal side length is found to be 6 cm.

40π cm²/s

50π cm²/s

60π cm²/s

70π cm²/s

Correct Answer: B

Solution:

The area

A

of the circle is given by

A = πr^2

. The rate of change of the area with respect to time is

\frac{dA}{dt} = 2πr \frac{dr}{dt}

. Given

\frac{dr}{dt} = 5

cm/s and

r = 8

cm,

\frac{dA}{dt} = 2π(8)(5) = 80π

cm²/s.

Width = 2 m, Height = 3 m

Width = 3 m, Height = 2 m

Width = 2.5 m, Height = 2.5 m

Width = 3.5 m, Height = 1.5 m

Correct Answer: B

Solution:

To maximize the light, the area of the window should be maximized. Using the perimeter constraint, the dimensions can be calculated.

Rs 560

Rs 640

Rs 720

Rs 800

Correct Answer: B

Solution:

Let the length of the base be

l

and the width be

w

. The volume constraint gives

l \times w \times 2 = 8

, so

l \times w = 4

. The cost function is

C = 70lw + 45(2l + 2w) \times 2 = 70 \times 4 + 180(l + w)

. Using

lw = 4

, we minimize

C = 280 + 180(l + \frac{4}{l})

. By setting the derivative to zero and solving, we find the minimum cost is Rs 640.

₹30.02

₹30.04

₹30.06

₹30.08

Correct Answer: B

Solution:

The marginal cost is the derivative

C'(x) = 0.015x^2 - 0.04x + 30

. Substituting

x = 4

, we get

0.015(4)^2 - 0.04(4) + 30 = 30.04

3 cm

4 cm

5 cm

6 cm

Correct Answer: B

Solution:

Let the side of the square cut off be

x

. The volume of the box is

V = (45 - 2x)(24 - 2x)x

. Taking the derivative of

V

with respect to

x

and setting it to zero gives the critical points. Solving gives

x = 4

cm for maximum volume.

0.25 m/h

0.5 m/h

0.75 m/h

1 m/h

Correct Answer: B

Solution:

The semi-vertical angle

\alpha = \tan^{-1}(0.5)

implies

\tan(\alpha) = 0.5 = \frac{r}{h}

, giving

r = 0.5h

. The volume

V

of the cone is

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (0.5h)^2 h = \frac{1}{12}\pi h^3

. Differentiating with respect to time

t

\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}

. Given

\frac{dV}{dt} = 5

, we substitute

h = 4

to find

\frac{dh}{dt} = \frac{5}{\pi \times 4^2} = \frac{5}{16\pi} \approx 0.5

m/h.

True

False

Depends on the value of k

Cannot be determined

Correct Answer: A

Solution:

By setting up the equations for the perimeter and areas, and using calculus to find the minimum, it can be shown that the condition holds true.

Infinity

100

Does not exist

Correct Answer: C

Solution:

The function f(x) = x³ - 3x² + 3x - 100 is a cubic polynomial. Cubic polynomials do not have absolute maximum or minimum values over the entire real line. Thus, the maximum value does not exist.

18 cm²/s

6 cm²/s

9 cm²/s

12 cm²/s

Correct Answer: A

Solution:

The area

A

of a circle is

\pi r^2

. The rate of change of area with respect to radius is

\frac{dA}{dr} = 2\pi r

. At

r = 3

, this rate is

2\pi(3) = 6\pi \approx 18

cm²/s.

It has a local maximum at

x = 1

It has a local minimum at

x = 1

It is always increasing.

It is always decreasing.

Correct Answer: C

Solution:

The derivative

f'(x) = 3x^2 - 6x + 3

simplifies to

3(x-1)^2

, which is always non-negative. Therefore, the function is always increasing.

14 m each

18 m for the square, 10 m for the circle

20 m for the square, 8 m for the circle

22 m for the square, 6 m for the circle

Correct Answer: A

Solution:

The combined area is minimized when the wire is divided equally, giving 14 m each for the square and the circle.

Correct Answer: A

Solution:

To find the value of

C

, we first find the derivative

f'(x) = 3x^2 - 12x + 9

. Setting

f'(x) = 0

gives

3x^2 - 12x + 9 = 0

. Solving this quadratic equation, we get

x = 1

and

x = 3

. However, the problem states that there is a local maximum at

x = 2

. Thus,

f(x)

must be flat at

x = 2

, meaning

f'(2) = 0

. Substituting

x = 2

into

f'(x)

3(2)^2 - 12(2) + 9 = 0

, which simplifies to

12 - 24 + 9 = -3

. Therefore,

C

must adjust this equation to be zero, meaning

C = 8

Has a local maximum

Has a local minimum

Has both a local maximum and minimum

Has neither a local maximum nor minimum

Correct Answer: D

Solution:

The function

f(x) = x^3 - 3x^2 + 3x - 100

is a cubic polynomial. Its derivative

f'(x) = 3x^2 - 6x + 3

does not change sign, indicating no local maxima or minima.

Strictly increasing

Strictly decreasing

Neither strictly increasing nor decreasing

Constant

Correct Answer: C

Solution:

The derivative

f'(x) = 2x - 1

changes sign over the interval

(-1, 1)

, indicating that the function is neither strictly increasing nor decreasing on this interval.

(2√2, 4)

(2√2, 0)

(0, 0)

(2, 2)

Correct Answer: A

Solution:

The distance between a point (x, y) on the curve and (0, 5) is minimized by finding the point where the derivative of the distance function is zero. Solving gives the point (2√2, 4).

2.5 km/h

3.0 km/h

3.5 km/h

4.0 km/h

Correct Answer: A

Solution:

Using similar triangles, the length of the shadow

s

and the distance of the man from the lamp post

x

are related by

\frac{s}{x} = \frac{2}{6}

. Thus,

s = \frac{x}{3}

. Differentiating with respect to time,

\frac{ds}{dt} = \frac{1}{3} \frac{dx}{dt} = \frac{1}{3} \times 5 = 2.5

km/h.

Maximum: 2π, Minimum: 0

Maximum: 2π + 1, Minimum: 0

Maximum: 2π + 1, Minimum: 1

Maximum: 2π, Minimum: 1

Correct Answer: B

Solution:

The function

f(x) = x + \sin 2x

reaches its maximum at

x = 2\pi

and its minimum at

x = 0

\frac{ab}{2}

ab

\frac{a^2b}{2}

\frac{a^2b^2}{2}

Correct Answer: A

Solution:

The maximum area of an isosceles triangle inscribed in an ellipse with its vertex at one end of the major axis is given by

\frac{ab}{2}

. This is derived by considering the geometry of the ellipse and the properties of the isosceles triangle.

Correct Answer: C

Solution:

The maximum profit occurs at the vertex of the parabola given by

p(x)

. The vertex

x = -\frac{b}{2a}

for

p(x) = ax^2 + bx + c

gives

x = -\frac{-72}{2(-18)} = 2

. Substituting

x = 2

into

p(x)

gives

p(2) = 25

(a^2 + b^2)^{1/2}

(a^3 + b^3)^{1/3}

(a^3 + b^3)^{1/2}

(a^2 + b^2)^{1/3}

Correct Answer: B

Solution:

Using the properties of right triangles, the minimum length of the hypotenuse when a point is at distances

a

and

b

from the two legs is given by

(a^3 + b^3)^{1/3}

. This is derived from the geometric mean of the distances, considering the configuration of the triangle and the Pythagorean theorem.

₹30.00

₹30.02

₹30.015

₹30.05

Correct Answer: C

Solution:

The marginal cost is the derivative of the cost function,

C'(x) = 0.015x^2 - 0.04x + 30

. Substituting

x = 3

, we get

C'(3) = 0.015(3)^2 - 0.04(3) + 30 = 30.015

√5

2√2

Correct Answer: A

Solution:

The distance between the soldier at (3, 7) and the helicopter at

(x, x^2 + 7)

is minimized when

x = 1

, giving the minimum distance as

\sqrt{(1-3)^2 + ((1)^2 + 7 - 7)^2} = \sqrt{4} = 2

. Thus, the nearest distance is

\sqrt{5}

Correct Answer: D

Solution:

To find the absolute maximum, evaluate

f(x)

at critical points and endpoints:

x = 1, 2, 3, 5

. Calculate

f(1) = 24

f(2) = 29

f(3) = 28

f(5) = 56

. The maximum value is 56 at

x = 5

₹30.02

₹30.015

₹30.00

₹29.98

Correct Answer: A

Solution:

The marginal cost is the derivative of the cost function evaluated at

x = 3

, which is ₹30.02.

10 m for the square, 10 m for the circle

12 m for the square, 8 m for the circle

8 m for the square, 12 m for the circle

15 m for the square, 5 m for the circle

Correct Answer: C

Solution:

Let

x

be the length of the wire used for the square, then

20-x

is used for the circle. The area of the square is

\left(\frac{x}{4}\right)^2

and the area of the circle is

\pi \left(\frac{20-x}{2\pi}\right)^2

. To minimize the total area, differentiate and set to zero, solving gives

x = 8

m for the square and

20-x = 12

m for the circle.

12 and 12

10 and 14

8 and 16

6 and 18

Correct Answer: A

Solution:

The product of two numbers is maximized when the numbers are equal, so the optimal numbers are 12 and 12.

True

False

Depends on the triangle

Cannot be determined

Correct Answer: A

Solution:

Using calculus and the properties of triangles, it can be shown that the minimum length of the hypotenuse is given by the expression.

123

164

Correct Answer: A

Solution:

To find the maximum profit, we need to find the critical points by setting the derivative

P'(x) = 41 - 144x - 54x^2 = 0

. Solving this quadratic equation gives

x = -\frac{41}{54}

and

x = \frac{41}{54}

. Evaluating the profit function at these critical points and endpoints, we find that the maximum profit occurs at

x = 0

with

P(0) = 41

. Thus, the maximum profit is 41.

It has a local maximum at

x = 1

It is increasing for all real

x

It has a local minimum at

x = 0

It is decreasing for all real

x

Correct Answer: B

Solution:

The derivative

f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2

is always non-negative and zero only at

x = 1

. Hence,

f(x)

is increasing for all real

x

Correct Answer: C

Solution:

To find the absolute maximum, we evaluate

f(x)

at the critical points and the endpoints of the interval. The derivative

f'(x) = 3x^2 - 6x + 2

. Setting

f'(x) = 0

, we solve

3x^2 - 6x + 2 = 0

to find

x = 1

and

x = \frac{2}{3}

. Evaluating

f(x)

x = 0

f(0) = 0

; at

x = 1

f(1) = 0

; at

x = \frac{2}{3}

f(\frac{2}{3}) = \frac{4}{27}

; and at

x = 3

f(3) = 0

. The maximum value is

2

x = 2

. Thus, the absolute maximum value is

f(2) = 2

\sqrt{5}

\sqrt{10}

\sqrt{13}

\sqrt{20}

Correct Answer: A

Solution:

The position of the helicopter is given by (x, x^2 + 7). The distance between the helicopter and the soldier is

d = \sqrt{(x - 3)^2 + (x^2 + 7 - 7)^2} = \sqrt{(x - 3)^2 + x^4}

. To find the minimum distance, differentiate with respect to

x

, set the derivative to zero, and solve for

x

. The minimum distance is found to be

\sqrt{5}

0.5 m/h

1.0 m/h

1.5 m/h

2.0 m/h

Correct Answer: B

Solution:

Using the relationship between the volume and height, the rate at which the water level is rising is 1.0 m/h.

(1, 8)

(3, 7)

(2, 11)

(0, 7)

Correct Answer: A

Solution:

The distance between the soldier and the helicopter is minimized when

x = 1

, giving the point (1, 8) on the curve.

s = 2r

s = r

s = 4r

s = r/2

Correct Answer: A

Solution:

Given the perimeter constraint

4s + 2\pi r = k

, we minimize the area

s^2 + \pi r^2

. Using the method of Lagrange multipliers or substitution, we find that the minimum occurs when

s = 2r

a > -2

a < -2

a = -2

None of these

Correct Answer: A

Solution:

The function is increasing if its derivative

f'(x) = 2x + a > 0

for all

x

[1, 2]

. Solving

2x + a > 0

for

x = 1

gives

a > -2

True or False

Correct Answer: False

Solution:

A point where the derivative of a function is zero need not be a point of local maxima or minima. For example, for the function

f(x) = x^3

, the derivative

f'(x) = 3x^2

is zero at

x = 0

, but

x = 0

is neither a point of local maxima nor minima.

Correct Answer: True

Solution:

The point

x = 0

is a point of inflection for the function

f(x) = x^3

, as the curvature changes concavity at this point.

Correct Answer: False

Solution:

The derivative

f'(x) = 2x - 1

changes sign in the interval

(-1, 1)

, indicating that the function is neither strictly increasing nor decreasing on this interval.

Correct Answer: False

Solution:

The function

f(x) = x^2 - x + 1

is neither strictly increasing nor decreasing on the interval (-1, 1) because its derivative changes sign within this interval.

Correct Answer: True

Solution:

The derivative

f'(x) = 3(x - 1)^2 + 1

is always positive, indicating that the function is increasing on the entire real line.

Correct Answer: False

Solution:

A point where the derivative is zero is called a critical point, but it is not necessarily a point of local maxima or minima. It could also be a point of inflection.

Correct Answer: True

Solution:

The derivative represents the rate of change of a quantity with respect to another, such as distance with respect to time.

Correct Answer: True

Solution:

If a function is continuous at a point where its derivative is zero, it is differentiable in an interval around that point.

Correct Answer: False

Solution:

For the function

f(x) = x^3

, the derivative

f'(x) = 3x^2

is zero at

x = 0

, but

x = 0

is not a point of local maxima or minima. It is a point of inflection.

Correct Answer: False

Solution:

The condition

f'(x) = 0

x = c

indicates a critical point, but it does not guarantee a local maxima or minima. For example,

f(x) = x^3

has

f'(0) = 0

, but

x = 0

is not a local maxima or minima.

Correct Answer: True

Solution:

The derivative of a function provides information about the slope of the tangent to the curve. If

f'(x) > 0

, the function is increasing; if

f'(x) < 0

x = 5

Correct Answer: True

Solution:

The derivative of the logarithmic function is positive on

(0, \infty)

, indicating that it is increasing.

Correct Answer: True

Solution:

The derivative of the logarithmic function is positive on the interval (0, ∞), indicating that it is increasing on this interval.

Correct Answer: True

Solution:

The derivative of

f(x) = rac{1}{x}

f'(x) = - rac{1}{x^2}

, which is negative for

x > 0

. Therefore,

f(x)

is decreasing on

(0, rac{1}{2})

Correct Answer: False

Solution:

A point where the derivative is zero is a critical point, but it is not necessarily a point of local maxima or minima. For example, the function

f(x) = x^3

has a derivative of zero at

x = 0

, but it is not a point of local maxima or minima.

Correct Answer: True

Solution:

Derivatives are used to determine the rate of change of quantities, which is applicable in fields like engineering and science.

Correct Answer: True

Solution:

The excerpt states that the graph begins at the origin (0, 0) and ends near (1, 1), excluded due to open circles.

Correct Answer: True

Solution:

The derivative is used to determine the rate of change of quantities, such as distance with respect to time.

Correct Answer: True

Solution:

The excerpt provides a proof showing that the function

f(x) = x^3 - 3x^2 + 4x

is increasing on

extbf{R}

by demonstrating that

f'(x) = 3(x - 1)^2 + 1 > 0

for all

x

extbf{R}

Correct Answer: True

Solution:

Theorem 2 in the excerpt states that if a function has a local maxima or minima at a point, then either the derivative is zero or the function is not differentiable at that point.

Correct Answer: True

Solution:

f

is continuous at

c

and

f'(c) = 0

, then there exists an

h > 0

such that

f

is differentiable in the interval

(c - h, c + h)

Correct Answer: True

Solution:

The excerpt states that if

f'(x)

does not change sign as

x

increases through a critical point

c

, then

c

is neither a point of local maxima nor a point of local minima, but rather a point of inflection.

Correct Answer: False

Solution:

The graph has open circles at the endpoints, indicating that the function is not defined at those points, hence it is not continuous over the interval [0, 1].

Correct Answer: True

Solution:

Derivatives can be used to find the maximum area of geometric figures by determining the critical points where the area function reaches its maximum.

Correct Answer: True

Solution:

It is shown that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius

r

\frac{4r}{3}

Correct Answer: True

Solution:

When the side of the square is double the radius of the circle, the sum of their areas is minimized for a given constant perimeter.

Correct Answer: False

Solution:

The maximum volume of a right circular cone inscribed in a sphere is achieved when the altitude is not equal to the diameter of the base. Specific conditions must be met for maximum volume, which do not involve the altitude being equal to the diameter.

Correct Answer: False

Solution:

For

f(x) = x^3

f'(x) = 3x^2

and

f'(0) = 0

, but 0 is neither a point of local maxima nor minima; it is a point of inflection.

Correct Answer: True

Solution:

If the derivative of a function is positive over an interval, it indicates that the function is increasing over that interval.

Correct Answer: True

Solution:

If a function is differentiable and its derivative is positive over an interval, then the function is increasing on that interval.

Correct Answer: True

Solution:

A point

c

in the domain of a function where

f'(c) = 0

f

is not differentiable is called a critical point of

f

. If

f

is continuous at

c

and

f'(c) = 0

, then

c

is a critical point.

Correct Answer: True

Solution:

The derivative of a function at a point gives the slope of the tangent to the curve at that point. Thus, it can be used to find the equation of the tangent line.

Correct Answer: False

Solution:

A point where the derivative is zero is not necessarily a point of local maxima or minima. For example, for the function

f(x) = x^3

f'(0) = 0

, but 0 is neither a local maxima nor a local minima.

Correct Answer: True

Solution:

The function

f(x) = x^2 - x + 1

does not have a consistent increase or decrease over the interval

(-1, 1)

, hence it is neither strictly increasing nor decreasing.

Correct Answer: False

Solution:

The graph ends near (1, 1) but excludes it due to open circles, indicating that (1, 1) is not included.

Correct Answer: True

Solution:

Derivatives can be used to find the maximum area of geometric figures, such as an isosceles triangle inscribed in an ellipse, by determining the critical points.

Correct Answer: True

Solution:

The derivative of a function provides information about the slope of the tangent line to the function at any given point. If the derivative is positive over an interval, the function is increasing on that interval. If the derivative is negative, the function is decreasing.

Correct Answer: False

Solution:

The rate of change of the area of a circle with respect to its radius is not constant; it is given by the derivative of the area with respect to the radius, which is

2\pi r

. This rate depends on the radius

r

Correct Answer: False

Solution:

For

f(x) = x^3

f'(x) = 3x^2

and

f'(0) = 0

, but

x = 0

is not a point of local maxima or minima; it's a point of inflection.

Correct Answer: True

Solution:

The derivative provides the slope of the tangent line at a point on the curve, which can be used to find both the tangent and normal lines.

Correct Answer: True

Solution:

Among all rectangles inscribed in a circle, the square has the maximum area.

Correct Answer: True

Solution:

If a function

f

is continuous at

c

and

f'(c) = 0

, then there exists an

h > 0

such that

f

is differentiable in the interval

(c - h, c + h)

Correct Answer: True

Solution:

According to the excerpt, if

f

is continuous at

c

and

f'(c) = 0

, then there exists an

h > 0

such that

f

is differentiable in the interval

(c - h, c + h)

. This implies that differentiability is assured in a neighborhood around the critical point.

Application of Derivatives

AI Learning Assistant

Summary

Chapter 6: Application of Derivatives

Introduction

Rate of Change of Quantities

Example: If y varies with x, then

Maxima and Minima

Examples of Applications

Important Problems

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 6: Application of Derivatives

6.1 Introduction

6.2 Rate of Change of Quantities

6.4 Maxima and Minima

Examples of Applications

Important Problems

Conclusion

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.Find the maximum area of an isosceles triangle inscribed in an ellipse with its vertex at one end of the major axis.

Correct Answer: A

Q2.If the total cost C(x)C(x)C(x) in Rupees for producing xxx units is given by C(x)=0.005x3−0.02x2+30x+5000C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000C(x)=0.005x3−0.02x2+30x+5000, what is the marginal cost when 3 units are produced?

Correct Answer: A

Q3.The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. Find the marginal revenue when x = 5.

Correct Answer: A

Q4.Find the maximum value of the function f(x)=x2−x+1f(x) = x^2 - x + 1f(x)=x2−x+1 on the interval [−1,1][-1, 1][−1,1].

Correct Answer: D

Q5.Prove that the logarithmic function f(x)=log⁡xf(x) = \log xf(x)=logx is increasing on (0,∞)(0, \infty)(0,∞). Which of the following statements is true?

Correct Answer: A

Q6.Which of the following functions is increasing on the interval (0,π)(0, \pi)(0,π)?

Correct Answer: B

Q7.What is the maximum value of the function f(x)=2x3−15x2+36x+1f(x) = 2x^3 - 15x^2 + 36x + 1f(x)=2x3−15x2+36x+1 on the interval [1,5][1, 5][1,5]?

Correct Answer: D

Q8.For what values of aaa is the function f(x)=x2+ax+1f(x) = x^2 + ax + 1f(x)=x2+ax+1 increasing on [1,2][1, 2][1,2]?

Correct Answer: A

Q9.A wire of length 28 m is to be cut into two pieces. One piece is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Correct Answer: D

Q10.A soldier is positioned at point (3, 7). An Apache helicopter is flying along the curve given by y=x2+7y = x^2 + 7y=x2+7. What is the nearest distance between the soldier and the helicopter?

Correct Answer: A

Q11.A rectangular tank with a base area of 10 square meters is being filled with water at a rate of 5 cubic meters per hour. What is the rate at which the water level is rising?

Correct Answer: A

Q12.Find the absolute maximum value of the function f(x)=2x3−15x2+36x+1f(x) = 2x^3 - 15x^2 + 36x + 1f(x)=2x3−15x2+36x+1 on the interval [1, 5].

Correct Answer: D

Q13.A rectangular sheet of metal 45 cm by 24 cm is to be made into an open-top box by cutting squares from each corner and folding up the sides. What should be the side length of the square to be cut off to maximize the volume of the box?

Correct Answer: B

Q14.A man of height 2 meters walks away from a lamp post 6 meters high at a speed of 5 km/h. At what rate is the length of his shadow increasing?

Correct Answer: A

Q15.A tank with a rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m³. If building the tank costs Rs 70 per square meter for the base and Rs 45 per square meter for the sides, what is the cost of the least expensive tank?

Correct Answer: B

Q16.A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. What are the dimensions of the window to admit maximum light through the whole opening?

Correct Answer: A

Q17.For the function f(x)=x3f(x) = x^3f(x)=x3, what is true about the point x=0x = 0x=0?

Correct Answer: C

Q18.What is the maximum profit a company can make if the profit function is p(x)=41−72x−18x2p(x) = 41 - 72x - 18x^2p(x)=41−72x−18x2?

Correct Answer: A

Q19.Find the marginal revenue when x=5x = 5x=5 for the total revenue function R(x)=3x2+36x+5R(x) = 3x^2 + 36x + 5R(x)=3x2+36x+5.

Correct Answer: A

Q20.Find the value of aaa for which the function f(x)=x4−62x2+ax+9f(x) = x^4 - 62x^2 + ax + 9f(x)=x4−62x2+ax+9 attains its maximum value at x=1x = 1x=1 on the interval [0,2][0, 2][0,2].

Correct Answer: A

Q21.The sum of the perimeter of a circle and a square is kkk, where kkk is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle. What is the value of the radius of the circle if k=20πk = 20\pik=20π?

Correct Answer: B

Q22.A helicopter is flying along the curve given by y=x2+7y = x^2 + 7y=x2+7. A soldier is positioned at (3,7)(3, 7)(3,7). What is the nearest distance between the soldier and the helicopter?

Correct Answer: A

Q23.What is the rate of change of the area of a circle with respect to its radius rrr when r=4r = 4r=4 cm?

Correct Answer: B

Q24.Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Correct Answer: A

Q25.The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Correct Answer: B

Q26.Find the maximum area of an isosceles triangle inscribed in an ellipse with its vertex at one end of the major axis.

Correct Answer: C

Q27.Which of the following is true about the function f(x)=x3f(x) = x^3f(x)=x3?

Correct Answer: C