Application of Derivatives — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of Derivatives" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of Derivatives" — 7 important questions with detailed answers for CBSE…
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NCERT Solutions for Class 12 Mathematics — Application of Derivatives. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Application of Derivatives — All 7 Questions Solved
Q1. Find the rate of change of the area of a circle with respect to its radius.
Let A be the area of a circle and r be its radius.
We know: A = πr²
The rate of change of area with respect to radius is dA/dr.
dA/dr = d/dr(πr²)
= π × d/dr(r²)
= π × 2r
= 2πr
CONCLUSION: dA/dr = 2πr
Interpretation: When the radius is r, a small increase dr in the radius causes an increase of approximately 2πr × dr in the area. This makes geometric sense—2πr is the circumference of the circle.
Q2. A ladder of length 10 m leans against a wall. If the base moves away from the wall at 2 m/s, find the rate at which the top of the ladder slides down the wall when the base is 6 m from the wall.
Let x = distance of base from wall, y = height of top of ladder on the wall.
Given: ladder length = 10 m (constant), dx/dt = 2 m/s
By Pythagoras: x² + y² = 10²
x² + y² = 100
Differentiating both sides with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
x(dx/dt) + y(dy/dt) = 0
dy/dt = -x(dx/dt)/y
When x = 6 m:
From x² + y² = 100:
36 + y² = 100
y² = 64
y = 8 m
Substituting values:
dy/dt = -(6)(2)/8
= -12/8
= -3/2
= -1.5 m/s
CONCLUSION: The top of the ladder slides down at 1.5 m/s. The negative sign indicates the height is decreasing.
Q3. Find the maximum and minimum values of f(x) = x³ - 3x on the interval [-2, 2].
Given: f(x) = x³ - 3x on [-2, 2]
Step 1: Find critical points
f'(x) = 3x² - 3 = 3(x² - 1) = 3(x-1)(x+1)
Setting f'(x) = 0:
3(x-1)(x+1) = 0
Critical points: x = 1 and x = -1
Both critical points lie in [-2, 2].
Step 2: Evaluate f at critical points and endpoints
f(-2) = (-2)³ - 3(-2) = -8 + 6 = -2
f(-1) = (-1)³ - 3(-1) = -1 + 3 = 2
f(1) = (1)³ - 3(1) = 1 - 3 = -2
f(2) = (2)³ - 3(2) = 8 - 6 = 2
Step 3: Identify maximum and minimum
Maximum value: 2 (occurs at x = -1 and x = 2)
Minimum value: -2 (occurs at x = -2 and x = 1)
CONCLUSION: Maximum value = 2, Minimum value = -2
Q4. Find the equations of the tangent and normal lines to the curve y = x² - 2x at the point (1, -1).
Given: y = x² - 2x, point (1, -1)
Step 1: Verify the point lies on the curve
y(1) = 1² - 2(1) = 1 - 2 = -1 ✓
Step 2: Find the slope of the tangent
dy/dx = 2x - 2
At x = 1: dy/dx = 2(1) - 2 = 0
Slope of tangent = 0
Step 3: Find equation of tangent line
Using point-slope form: y - y₁ = m(x - x₁)
y - (-1) = 0(x - 1)
y + 1 = 0
y = -1
Tangent line: y = -1 (horizontal line)
Step 4: Find equation of normal line
Slope of normal = -1/(slope of tangent) = -1/0 = undefined
A normal with undefined slope is a vertical line.
Normal line: x = 1
CONCLUSION: Tangent line: y = -1; Normal line: x = 1
Q5. The cost function for a company is C(x) = 100 + 5x + 0.1x², where x is the number of units. Find the marginal cost when x = 10.
Given: C(x) = 100 + 5x + 0.1x²
The marginal cost is the rate of change of cost with respect to units produced, which is dC/dx.
Step 1: Find the marginal cost function
dC/dx = d/dx(100 + 5x + 0.1x²)
= 0 + 5 + 0.1(2x)
= 5 + 0.2x
Step 2: Evaluate marginal cost at x = 10
Marginal cost at x = 10 = 5 + 0.2(10)
= 5 + 2
= 7
CONCLUSION: The marginal cost when x = 10 is 7 (currency units per unit produced).
Interpretation: When the company is producing 10 units, the cost of producing one additional unit is approximately 7 currency units.
Q6. A particle moves along a line according to s(t) = t³ - 6t² + 9t, where s is in metres and t is in seconds. Find the velocity and acceleration when t = 2.
Given: s(t) = t³ - 6t² + 9t
Step 1: Find velocity
Velocity v(t) = ds/dt
v(t) = d/dt(t³ - 6t² + 9t)
= 3t² - 12t + 9
At t = 2:
v(2) = 3(2)² - 12(2) + 9
= 3(4) - 24 + 9
= 12 - 24 + 9
= -3 m/s
Step 2: Find acceleration
Acceleration a(t) = dv/dt = d²s/dt²
a(t) = d/dt(3t² - 12t + 9)
= 6t - 12
At t = 2:
a(2) = 6(2) - 12
= 12 - 12
= 0 m/s²
CONCLUSION: At t = 2:
- Velocity = -3 m/s (moving in negative direction)
- Acceleration = 0 m/s² (no change in velocity at this instant)
Q7. A rectangular box has a square base with side length x and height h. The total surface area is fixed at 240 m². Express the volume as a function of x and find the dimensions that maximize the volume.
Let the base have side x and height h.
Step 1: Set up the constraint (surface area = 240)
Surface area = 2x² + 4xh = 240
(2 square bases + 4 rectangular sides)
From this: 2x² + 4xh = 240
4xh = 240 - 2x²
h = (240 - 2x²)/(4x) = (120 - x²)/(2x)
Step 2: Express volume as a function of x
V(x) = x²h = x² × (120 - x²)/(2x)
= x(120 - x²)/2
= (120x - x³)/2
= 60x - x³/2
Step 3: Find critical points
dV/dx = 60 - 3x²/2
Setting dV/dx = 0:
60 - 3x²/2 = 0
3x²/2 = 60
x² = 40
x = 2√10 m (taking positive value)
Step 4: Check second derivative
d²V/dx² = -3x = -3(2√10) = -6√10 < 0
Since d²V/dx² < 0, x = 2√10 gives a maximum.
Step 5: Find h
h = (120 - x²)/(2x) = (120 - 40)/(2 × 2√10) = 80/(4√10) = 20/√10 = 2√10 m
Note: h = x, so the optimal box is actually a cube!
Step 6: Find maximum volume
V = 60(2√10) - (2√10)³/2
= 120√10 - (8 × 10√10)/2
= 120√10 - 40√10
= 80√10 m³
CONCLUSION: Maximum volume occurs when x = h = 2√10 m ≈ 6.32 m, with V_max = 80√10 m³ ≈ 252.98 m³
Application of Derivatives — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A spherical balloon's radius is increasing at 2 cm/s. Find the rate of change of its volume when the radius is 10 cm.
- A. 800π cm³/s
- B. 400π cm³/s
- C. 200π cm³/s
- D. 100π cm³/s
Answer: A. 800π cm³/s — The volume of a sphere is V = (4/3)πr³. Differentiating with respect to time t, we get dV/dt = 4πr² (dr/dt). Substituting r=10 cm and dr/dt=2 cm/s gives dV/dt = 4π(10)²(2) = 800π cm³/s.
MCQ 2. Find the interval in which the function f(x) = x² - 4x + 6 is strictly increasing.
- A. (2, ∞)
- B. (-∞, 2)
- C. (0, 2)
- D. (-∞, ∞)
Answer: A. (2, ∞) — To find where f(x) is strictly increasing, we find the first derivative: f'(x) = 2x - 4. For f(x) to be strictly increasing, f'(x) > 0, which means 2x - 4 > 0, so 2x > 4, or x > 2. Thus, the interval is (2, ∞).
MCQ 3. The slope of the tangent to the curve y = x³ - x at x = 2 is:
- A. 11
- B. 10
- C. 12
- D. 9
Answer: A. 11 — The slope of the tangent is given by the derivative dy/dx. For y = x³ - x, dy/dx = 3x² - 1. At x = 2, the slope is 3(2)² - 1 = 3(4) - 1 = 12 - 1 = 11.
MCQ 4. Using differentials, approximate the value of √(25.3).
- A. 5.03
- B. 5.02
- C. 5.01
- D. 5.00
Answer: A. 5.03 — Let y = √x. We take x = 25 and Δx = 0.3. The derivative dy/dx = 1/(2√x). The approximate change Δy ≈ (dy/dx)Δx = (1/(2√25)) * 0.3 = (1/10) * 0.3 = 0.03. So, √(25.3) ≈ √25 + Δy = 5 + 0.03 = 5.03.
MCQ 5. Find the minimum value of f(x) = x² - 6x + 5.
- A. -4
- B. -2
- C. 1
- D. 5
Answer: A. -4 — First, find the derivative f'(x) = 2x - 6. Set f'(x) = 0 to find critical points: 2x - 6 = 0, so x = 3. The second derivative f''(x) = 2, which is positive, indicating a minimum. The minimum value is f(3) = 3² - 6(3) + 5 = 9 - 18 + 5 = -4.
MCQ 6. The radius of a circle is increasing uniformly at the rate of 3 cm/s. At what rate is the area of the circle increasing when the radius is 10 cm?
- A. 60π cm²/s
- B. 30π cm²/s
- C. 90π cm²/s
- D. 100π cm²/s
Answer: A. 60π cm²/s — The area of a circle is A = πr². Differentiating with respect to time t, dA/dt = 2πr (dr/dt). Given dr/dt = 3 cm/s and r = 10 cm, dA/dt = 2π(10)(3) = 60π cm²/s.
MCQ 7. The function f(x) = sin x - cos x is strictly increasing in the interval:
- A. (-π/4, 3π/4)
- B. (0, π/2)
- C. (π/2, π)
- D. (π/4, 5π/4)
Answer: A. (-π/4, 3π/4) — First, find the derivative: f'(x) = cos x + sin x. For f(x) to be strictly increasing, f'(x) > 0. We can write cos x + sin x as √2 sin(x + π/4). So, √2 sin(x + π/4) > 0, which implies sin(x + π/4) > 0. This inequality holds when 2nπ < x + π/4 < (2n+1)π for an integer n. For n=0, we have 0 < x + π/4 < π, which simplifies to -π/4 < x < 3π/4.
MCQ 8. The equation of the tangent to the curve y² = 4ax at the point (at², 2at) is:
- A. ty = x + at²
- B. y = tx + at²
- C. tx = y + at²
- D. x + ty = at²
Answer: A. ty = x + at² — Differentiating y² = 4ax implicitly with respect to x gives 2y (dy/dx) = 4a, so dy/dx = 2a/y. At the point (at², 2at), the slope of the tangent m = 2a/(2at) = 1/t. The equation of the tangent is y - 2at = (1/t)(x - at²). Multiplying by t gives ty - 2at² = x - at², which rearranges to ty = x + at².
MCQ 9. The maximum value of (1/x)^x is:
- A. e^(1/e)
- B. e
- C. 1
- D. (1/e)^e
Answer: A. e^(1/e) — Let y = (1/x)^x = x^(-x). Taking the natural logarithm, ln y = -x ln x. Differentiating with respect to x, (1/y) (dy/dx) = -(ln x + 1). Setting dy/dx = 0 gives ln x + 1 = 0, so ln x = -1, which means x = e⁻¹ = 1/e. Substituting x = 1/e into the original function gives y = (1/e)^(-(1/e)) = e^(1/e). The second derivative test confirms this is a maximum.
MCQ 10. For the curve y = x³, find the x-coordinate(s) at which the y-coordinate is changing 12 times as fast as the x-coordinate.
- A. ±2
- B. 2
- C. -2
- D. ±4
Answer: A. ±2 — We are given y = x³. Differentiating with respect to time t, dy/dt = 3x² (dx/dt). The problem states that dy/dt = 12 (dx/dt). Substituting this into the equation, we get 12 (dx/dt) = 3x² (dx/dt). Assuming dx/dt ≠ 0, we can divide by dx/dt to get 12 = 3x², which simplifies to x² = 4. Therefore, x = ±2.
MCQ 11. Find the intervals where f(x) = x³ - 3x² - 9x + 7 is strictly decreasing.
- A. (-1, 3)
- B. (-∞, -1) U (3, ∞)
- C. (-3, 1)
- D. (0, ∞)
Answer: A. (-1, 3) — First, find the derivative: f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x-3)(x+1). For f(x) to be strictly decreasing, f'(x) < 0. This occurs when (x-3)(x+1) < 0, which is true for x values between the roots -1 and 3. So, the interval is (-1, 3).
MCQ 12. The equation of the normal to the curve y = x² at the point (1, 1) is:
- A. x + 2y - 3 = 0
- B. 2x - y - 1 = 0
- C. x - 2y + 1 = 0
- D. x + y - 2 = 0
Answer: A. x + 2y - 3 = 0 — The derivative dy/dx = 2x. At the point (1, 1), the slope of the tangent is 2(1) = 2. The slope of the normal is the negative reciprocal, -1/2. Using the point-slope form y - y₁ = m(x - x₁), the equation of the normal is y - 1 = (-1/2)(x - 1). Multiplying by 2 gives 2y - 2 = -x + 1, which rearranges to x + 2y - 3 = 0.
MCQ 13. If y = x⁴ - 10 and x changes from 2 to 1.99, what is the approximate change in y?
- A. -0.32
- B. 0.32
- C. -0.16
- D. 0.16
Answer: A. -0.32 — The approximate change in y, Δy, is given by (dy/dx)Δx. Here, dy/dx = 4x³. At x = 2, dy/dx = 4(2)³ = 32. The change in x is Δx = 1.99 - 2 = -0.01. So, Δy ≈ 32 * (-0.01) = -0.32.
MCQ 14. The two positive numbers whose sum is 16 and the sum of whose squares is minimum are:
- A. 8, 8
- B. 6, 10
- C. 7, 9
- D. 4, 12
Answer: A. 8, 8 — Let the two numbers be x and y. Given x + y = 16, so y = 16 - x. We want to minimize the sum of squares, S = x² + y² = x² + (16 - x)². Differentiating with respect to x, S'(x) = 2x + 2(16 - x)(-1) = 2x - 32 + 2x = 4x - 32. Setting S'(x) = 0 gives 4x = 32, so x = 8. Then y = 16 - 8 = 8. The second derivative S''(x) = 4 > 0, confirming a minimum.
MCQ 15. The volume of a cube is increasing at a rate of 9 cm³/s. How fast is the surface area increasing when the length of an edge is 10 cm?
- A. 3.6 cm²/s
- B. 1.8 cm²/s
- C. 0.9 cm²/s
- D. 7.2 cm²/s
Answer: A. 3.6 cm²/s — Let 'a' be the length of an edge. Volume V = a³, so dV/dt = 3a² (da/dt). Given dV/dt = 9 cm³/s, so 9 = 3a² (da/dt), which means da/dt = 3/a². The surface area S = 6a², so dS/dt = 12a (da/dt). Substituting da/dt, dS/dt = 12a (3/a²) = 36/a. When a = 10 cm, dS/dt = 36/10 = 3.6 cm²/s.
MCQ 16. The function f(x) = |x| is:
- A. neither increasing nor decreasing on R
- B. increasing on R
- C. decreasing on R
- D. increasing on (0, ∞) and decreasing on (-∞, 0)
Answer: A. neither increasing nor decreasing on R — The derivative f'(x) = 1 for x > 0 and f'(x) = -1 for x < 0. Since the derivative changes sign, the function is not strictly increasing or strictly decreasing over its entire domain R. It is increasing on (0, ∞) and decreasing on (-∞, 0), but overall it is neither on R.
MCQ 17. The point on the curve y = x² - 3x + 2 where the tangent is parallel to the x-axis is:
- A. (3/2, -1/4)
- B. (1, 0)
- C. (2, 0)
- D. (0, 2)
Answer: A. (3/2, -1/4) — For the tangent to be parallel to the x-axis, its slope must be 0. The slope is given by dy/dx = 2x - 3. Setting 2x - 3 = 0 gives x = 3/2. Substitute x = 3/2 into the original equation to find y: y = (3/2)² - 3(3/2) + 2 = 9/4 - 9/2 + 2 = 9/4 - 18/4 + 8/4 = -1/4. So the point is (3/2, -1/4).
MCQ 18. The approximate value of (81)^(1/4) is:
- A. 3
- B. 3.001
- C. 2.999
- D. 3.0001
Answer: A. 3 — The value 81^(1/4) means the fourth root of 81. Since 3⁴ = 81, the exact value of 81^(1/4) is 3. No approximation is needed.
MCQ 19. The absolute maximum value of f(x) = (x-1)² + 5 on the interval [-3, 2] is:
- A. 21
- B. 5
- C. 6
- D. 1
Answer: A. 21 — First, find critical points by setting f'(x) = 0. f'(x) = 2(x-1). So, 2(x-1) = 0 implies x = 1. This critical point is within the interval [-3, 2]. Evaluate f(x) at the critical point and endpoints: f(1) = (1-1)² + 5 = 5. f(-3) = (-3-1)² + 5 = (-4)² + 5 = 16 + 5 = 21. f(2) = (2-1)² + 5 = 1² + 5 = 6. Comparing these values (5, 21, 6), the absolute maximum is 21.
MCQ 20. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
- A. 8/3 cm/s
- B. 4/3 cm/s
- C. 2/3 cm/s
- D. 1/3 cm/s
Answer: A. 8/3 cm/s — Let x be the distance of the foot of the ladder from the wall and y be the height on the wall. By Pythagoras theorem, x² + y² = 5² = 25. Differentiating with respect to time t, 2x (dx/dt) + 2y (dy/dt) = 0. Given dx/dt = 2 cm/s. When x = 4 m = 400 cm, then y² = 500² - 400² = 250000 - 160000 = 90000, so y = 300 cm. Substituting these values: 2(400)(2) + 2(300)(dy/dt) = 0. This gives 1600 + 600 (dy/dt) = 0, so dy/dt = -1600/600 = -8/3 cm/s. The height is decreasing at a rate of 8/3 cm/s.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.