Class 12 Maths Chapter 6 Important Questions – Applications of Derivative

Applications of Derivatives explains how derivatives are used to study rates of change, the behaviour of functions and maximum or minimum values. The chapter covers related rates, increasing and decreasing functions, local extrema and real-life optimisation problems.

Important Questions for Class 12 Maths Chapter 6 help students practise the main concepts covered in Applications of Derivatives. These include rate of change of quantities, increasing and decreasing functions, maxima and minima, and simple optimisation problems.

Students should revise basic differentiation rules before solving these questions. In rate-of-change problems, identify the changing variables and differentiate with respect to time. In maxima and minima questions, first form the required function and then apply the first or second derivative test.

Key Takeaways

• A derivative represents the instantaneous rate of change of one quantity with respect to another.
• If f′(x) > 0 in an interval, the function is strictly increasing there.
• If f′(x) < 0 in an interval, the function is strictly decreasing there.
• At a local maximum or minimum, f′(x) is generally equal to zero.
• Optimisation problems require expressing the quantity to be maximised or minimised as a function of one variable.

Important Applications of Derivatives Formulas

Rate of change of y with respect to x:

dy/dx

If x and y are functions of time t:

dy/dt = (dy/dx)(dx/dt)

Area of a circle:

A = πr²

Rate of change of area:

dA/dt = 2πr(dr/dt)

Volume of a sphere:

V = (4/3)πr³

Rate of change of volume:

dV/dt = 4πr²(dr/dt)

Surface area of a sphere:

S = 4πr²

Rate of change of surface area:

dS/dt = 8πr(dr/dt)

Volume of a cube:

V = x³

Rate of change of volume:

dV/dt = 3x²(dx/dt)

Surface area of a cube:

S = 6x²

Rate of change of surface area:

dS/dt = 12x(dx/dt)

Volume of a cone:

V = (1/3)πr²h

Increasing function:

f′(x) > 0

Decreasing function:

f′(x) < 0

Second derivative test:

If f′(a) = 0 and f″(a) < 0, then f has a local maximum at x = a.

If f′(a) = 0 and f″(a) > 0, then f has a local minimum at x = a.

Access Class 12 Maths Chapter 6 Important Questions in 30 Minutes

• First 10 minutes: Revise rate of change, chain rule and related-rate formulas.
• Next 10 minutes: Practise increasing and decreasing functions using the sign of f′(x).
• Final 10 minutes: Solve local maxima, local minima and optimisation questions.

Need more help with rate-of-change, increasing–decreasing functions and optimisation questions?

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Multiple Choice and Very Short Answer Questions – 1 Mark

Q1. What does dy/dx represent?

Answer: dy/dx represents the instantaneous rate of change of y with respect to x.

Q2. When is a differentiable function strictly increasing?

Answer: A differentiable function is strictly increasing in an interval when:

f′(x) > 0

throughout that interval.

Q3. When is a differentiable function strictly decreasing?

Answer: A differentiable function is strictly decreasing in an interval when:

f′(x) < 0

throughout that interval.

Q4. State the necessary condition for a differentiable function to have a local extremum at x = a.

Answer:

f′(a) = 0

However, this condition alone is not sufficient.

Q5. What does a negative value of dx/dt indicate?

Answer: A negative value of dx/dt indicates that x is decreasing with time.

Q6. If f′(x) changes from positive to negative at x = a, what happens at x = a?

Answer: The function has a local maximum at x = a.

Q7. If f′(x) changes from negative to positive at x = a, what happens at x = a?

Answer: The function has a local minimum at x = a.

Q8. If f′(a) = 0 and f″(a) = 0, can the second derivative test be used?

Answer: No. The second derivative test is inconclusive in this case.

Q9. What is marginal cost?

Answer: Marginal cost is the rate of change of total cost with respect to the number of units produced.

Marginal cost = dC/dx

Q10. The radius of a circle increases at 2 cm/s. Write the rate-of-change formula for its area.

Answer:

dA/dt = 2πr(dr/dt)

Therefore:

dA/dt = 4πr cm²/s

Short Answer Questions – 2 or 3 Marks

Q11. The total revenue from selling x units is R(x) = 3x² + 36x + 5. Find the marginal revenue when x = 5.

Solution:

R(x) = 3x² + 36x + 5

Marginal revenue:

MR = dR/dx

MR = 6x + 36

At x = 5:

MR = 6(5) + 36

MR = 30 + 36

MR = 66

Therefore, the marginal revenue is ₹66.

Q12. The total cost of producing x units is C(x) = 0.005x³ − 0.02x² + 30x + 5000. Find the marginal cost when x = 3.

Solution:

C(x) = 0.005x³ − 0.02x² + 30x + 5000

MC = dC/dx

MC = 0.015x² − 0.04x + 30

At x = 3:

MC = 0.015(3²) − 0.04(3) + 30

MC = 0.135 − 0.12 + 30

MC = 30.015

Therefore, the marginal cost is approximately ₹30.02.

Q13. A stone is dropped into a quiet lake and circular waves move at 4 cm/s. Find the rate at which the enclosed area is increasing when the radius is 10 cm.

Solution:

Area of the circle:

A = πr²

Differentiating with respect to time:

dA/dt = 2πr(dr/dt)

Given:

r = 10 cm

dr/dt = 4 cm/s

Therefore:

dA/dt = 2π(10)(4)

dA/dt = 80π cm²/s

Hence, the enclosed area is increasing at 80π cm²/s.

Q14. The radius of a circular disc increases at 0.05 cm/s. Find the rate at which its area increases when the radius is 3.2 cm.

Solution:

A = πr²

dA/dt = 2πr(dr/dt)

Substituting:

dA/dt = 2π(3.2)(0.05)

dA/dt = 0.32π cm²/s

Therefore, the area is increasing at 0.32π cm²/s.

Q15. The edge of a cube increases at 3 cm/s. Find the rate at which its volume increases when the edge is 10 cm.

Solution:

Let the edge be x cm.

Volume:

V = x³

Differentiating with respect to time:

dV/dt = 3x²(dx/dt)

Given:

x = 10 cm

dx/dt = 3 cm/s

Therefore:

dV/dt = 3(10²)(3)

dV/dt = 900 cm³/s

Hence, the volume is increasing at 900 cm³/s.

Q16. The length of a rectangle decreases at 5 cm/min while its width increases at 4 cm/min. Find the rate of change of its perimeter.

Solution:

Let the length be x and width be y.

Given:

dx/dt = −5 cm/min

dy/dt = 4 cm/min

Perimeter:

P = 2(x + y)

Differentiating:

dP/dt = 2(dx/dt + dy/dt)

dP/dt = 2(−5 + 4)

dP/dt = −2 cm/min

Therefore, the perimeter is decreasing at 2 cm/min.

Q17. Show that f(x) = x³ − 3x² + 6x − 100 is strictly increasing for all real x.

Solution:

f(x) = x³ − 3x² + 6x − 100

Differentiating:

f′(x) = 3x² − 6x + 6

f′(x) = 3(x² − 2x + 2)

f′(x) = 3[(x − 1)² + 1]

Since:

(x − 1)² ≥ 0

Therefore:

(x − 1)² + 1 > 0

Hence:

f′(x) > 0

for every real x.

Therefore, f is strictly increasing on R.

Q18. Show that f(x) = 4x³ − 18x² + 27x − 7 is increasing on R.

Solution:

f(x) = 4x³ − 18x² + 27x − 7

Differentiating:

f′(x) = 12x² − 36x + 27

f′(x) = 3(4x² − 12x + 9)

f′(x) = 3(2x − 3)²

Since:

(2x − 3)² ≥ 0

Therefore:

f′(x) ≥ 0

for every real x.

Hence, the function is increasing on R.

Increasing and Decreasing Functions – 3 or 4 Marks

Q19. Find the intervals in which f(x) = −2x³ − 9x² − 12x + 1 is strictly increasing or decreasing.

Solution:

f(x) = −2x³ − 9x² − 12x + 1

Differentiating:

f′(x) = −6x² − 18x − 12

f′(x) = −6(x² + 3x + 2)

f′(x) = −6(x + 1)(x + 2)

Set:

f′(x) = 0

Therefore:

x = −2 or x = −1

These values divide the real line into:

(−∞, −2), (−2, −1), (−1, ∞)

For x < −2:

f′(x) < 0

For −2 < x < −1:

f′(x) > 0

For x > −1:

f′(x) < 0

Therefore, f is:

• Strictly increasing on (−2, −1)
• Strictly decreasing on (−∞, −2) and (−1, ∞)

Q20. Find the intervals in which f(x) = 3x⁴ − 4x³ − 12x² + 5 is increasing or decreasing.

Solution:

f(x) = 3x⁴ − 4x³ − 12x² + 5

Differentiating:

f′(x) = 12x³ − 12x² − 24x

f′(x) = 12x(x² − x − 2)

f′(x) = 12x(x + 1)(x − 2)

Set:

f′(x) = 0

Critical points are:

x = −1, 0, 2

Check the sign of f′(x):

• For x < −1, f′(x) < 0
• For −1 < x < 0, f′(x) > 0
• For 0 < x < 2, f′(x) < 0
• For x > 2, f′(x) > 0

Therefore, f is:

• Strictly increasing on (−1, 0) and (2, ∞)
• Strictly decreasing on (−∞, −1) and (0, 2)

Q21. Find the intervals in which f(x) = 20 − 9x + 6x² − x³ is strictly increasing or decreasing.

Solution:

f(x) = 20 − 9x + 6x² − x³

Differentiating:

f′(x) = −9 + 12x − 3x²

f′(x) = −3(x² − 4x + 3)

f′(x) = −3(x − 1)(x − 3)

Set:

f′(x) = 0

Therefore:

x = 1 or x = 3

Sign of f′(x):

• f′(x) < 0 when x < 1
• f′(x) > 0 when 1 < x < 3
• f′(x) < 0 when x > 3

Therefore, f is:

• Strictly increasing on (1, 3)
• Strictly decreasing on (−∞, 1) and (3, ∞)

Q22. Find the intervals in which f(x) = sin x + cos x is increasing or decreasing for 0 ≤ x ≤ 2π.

Solution:

f(x) = sin x + cos x

Differentiating:

f′(x) = cos x − sin x

Set:

f′(x) = 0

cos x − sin x = 0

tan x = 1

Within 0 ≤ x ≤ 2π:

x = π/4 and 5π/4

Check the sign of f′(x):

• f′(x) > 0 on (0, π/4)
• f′(x) < 0 on (π/4, 5π/4)
• f′(x) > 0 on (5π/4, 2π)

Therefore, f is:

• Increasing on [0, π/4] and [5π/4, 2π]
• Decreasing on [π/4, 5π/4]

Maxima and Minima Questions – 4 or 5 Marks

Q23. Find all the points of local maxima and local minima of:

f(x) = (−3/4)x⁴ − 8x³ − (45/2)x² + 105

Solution:

f(x) = (−3/4)x⁴ − 8x³ − (45/2)x² + 105

Differentiating:

f′(x) = −3x³ − 24x² − 45x

f′(x) = −3x(x² + 8x + 15)

f′(x) = −3x(x + 3)(x + 5)

Set:

f′(x) = 0

Therefore:

x = 0, −3, −5

Second derivative:

f″(x) = −9x² − 48x − 45

At x = 0:

f″(0) = −45 < 0

Therefore, x = 0 is a point of local maximum.

At x = −3:

f″(−3) = 18 > 0

Therefore, x = −3 is a point of local minimum.

At x = −5:

f″(−5) = −30 < 0

Therefore, x = −5 is a point of local maximum.

Function values:

f(0) = 105

f(−3) = 3/4

f(−5) = 155/4

Therefore:

• Local maxima occur at (−5, 155/4) and (0, 105)
• Local minimum occurs at (−3, 3/4)

Q24. Find the local maximum and minimum values of f(x) = sin x − cos x for 0 ≤ x ≤ 2π.

Solution:

f(x) = sin x − cos x

Differentiating:

f′(x) = cos x + sin x

Set:

f′(x) = 0

cos x + sin x = 0

tan x = −1

Within the interval:

x = 3π/4 and 7π/4

Second derivative:

f″(x) = −sin x + cos x

At x = 3π/4:

f″(3π/4) = −√2 < 0

Therefore, x = 3π/4 gives a local maximum.

Maximum value:

f(3π/4) = √2

At x = 7π/4:

f″(7π/4) = √2 > 0

Therefore, x = 7π/4 gives a local minimum.

Minimum value:

f(7π/4) = −√2

Q25. Show that f(x) = log x / x has a maximum at x = e.

Solution:

Let:

f(x) = log x / x

Here, log x represents the natural logarithm.

Using the quotient rule:

f′(x) = [x(1/x) − log x] / x²

f′(x) = (1 − log x) / x²

Set:

f′(x) = 0

1 − log x = 0

log x = 1

x = e

For 0 < x < e:

log x < 1

Therefore:

f′(x) > 0

For x > e:

log x > 1

Therefore:

f′(x) < 0

Thus, f′(x) changes from positive to negative at x = e.

Hence, f has a maximum at x = e.

Maximum value:

f(e) = 1/e

Optimisation Questions – 5 or 6 Marks

Q26. Find two positive numbers whose sum is 15 and whose sum of squares is minimum.

Solution:

Let the numbers be x and y.

Given:

x + y = 15

Therefore:

y = 15 − x

Let the sum of their squares be S.

S = x² + y²

S = x² + (15 − x)²

S = x² + 225 − 30x + x²

S = 2x² − 30x + 225

Differentiating:

dS/dx = 4x − 30

Set:

dS/dx = 0

4x − 30 = 0

x = 15/2

Second derivative:

d²S/dx² = 4 > 0

Therefore, S is minimum when:

x = 15/2

Also:

y = 15 − 15/2

y = 15/2

Hence, the two numbers are:

15/2 and 15/2

Q27. An open tank with a square base has a fixed volume V. Show that the surface area is minimum when its depth is half the width of its base.

Solution:

Let x be the side of the square base and h be the depth.

Volume:

V = x²h

Therefore:

h = V/x²

Since the tank is open, its surface area is:

S = x² + 4xh

Substituting h:

S = x² + 4x(V/x²)

S = x² + 4V/x

Differentiating:

dS/dx = 2x − 4V/x²

Set:

dS/dx = 0

2x − 4V/x² = 0

2x³ = 4V

x³ = 2V

Using:

h = V/x²

and:

V = x³/2

we get:

h = (x³/2)/x²

h = x/2

Second derivative:

d²S/dx² = 2 + 8V/x³

This is positive for x > 0.

Therefore, the surface area is minimum when:

Depth = Width/2

Q28. A rectangular tank with an open top has depth 2 m and volume 8 m³. The base costs ₹70 per m² and the sides cost ₹45 per m². Find the least construction cost.

Solution:

Let the length be x m and breadth be y m.

Depth = 2 m

Volume:

2xy = 8

xy = 4

Therefore:

y = 4/x

Cost of base:

70xy = 70(4)

= ₹280

Area of four sides:

2(x + y)(2)

= 4(x + y)

Cost of sides:

45 × 4(x + y)

= 180(x + y)

Total cost:

C = 280 + 180(x + y)

Substituting y = 4/x:

C = 280 + 180(x + 4/x)

Differentiating:

dC/dx = 180(1 − 4/x²)

Set:

dC/dx = 0

1 − 4/x² = 0

x² = 4

x = 2

Since x is positive:

x = 2 m

Then:

y = 4/2

y = 2 m

Second derivative:

d²C/dx² = 1440/x³

At x = 2:

d²C/dx² > 0

Therefore, the cost is minimum.

Minimum cost:

C = 280 + 180(2 + 2)

C = 280 + 720

C = ₹1000

Q29. A window consists of a rectangle surmounted by a semicircle. Its perimeter is 10 m. Find its dimensions when its area is maximum.

Solution:

Let the radius of the semicircle be x m and the height of the rectangular part be y m.

Width of the rectangle = 2x

Perimeter:

2y + 2x + πx = 10

Therefore:

2y = 10 − (π + 2)x

y = [10 − (π + 2)x]/2

Area:

A = Area of rectangle + Area of semicircle

A = 2xy + (1/2)πx²

Substituting y:

A = 2x[10 − (π + 2)x]/2 + (1/2)πx²

A = 10x − (π + 2)x² + (1/2)πx²

A = 10x − [(π + 4)/2]x²

Differentiating:

dA/dx = 10 − (π + 4)x

Set:

dA/dx = 0

x = 10/(π + 4)

Second derivative:

d²A/dx² = −(π + 4) < 0

Therefore, the area is maximum.

Radius of semicircle:

x = 10/(π + 4) m

Width of rectangle:

2x = 20/(π + 4) m

Height:

y = 10/(π + 4) m

Hence, the width is twice the height of the rectangular part.

Practice Questions for Class 12 Maths Chapter 6

1. The radius of a circle increases at 3 cm/s. Find the rate at which its area increases when the radius is 10 cm.
2. The volume of a cube increases at 8 cm³/s. Find the rate at which its surface area increases when its edge is 12 cm.
3. A spherical balloon is inflated at 900 cm³/s. Find the rate at which its radius increases when the radius is 15 cm.
4. Find the intervals in which f(x) = 2x³ + 9x² + 12x + 20 is increasing or decreasing.
5. Find the intervals in which f(x) = x⁴ − 8x³ + 22x² − 24x + 21 is increasing or decreasing.
6. Find the local maximum and minimum values of f(x) = x³ − 6x² + 9x + 15.
7. Find two positive numbers whose product is 64 and whose sum is minimum.
8. Find the point on y² = 4x that is nearest to the point (2, −8).
9. Show that a closed cuboid with a square base and fixed volume has minimum surface area when it is a cube.
10. Find the maximum area of a rectangle that can be inscribed in a circle of radius r.

Important Questions Class 12 Maths Chapter-Wise