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Applications of Derivatives
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Class 12MathsChapter 6NCERTCBSE
Applications of Derivatives – Class 12 Maths Chapter 6
📌 Quick Overview: Chapter 6 applies all the differentiation skills from Chapter 5 to real-world problems. You will learn how derivatives are used to find rates of change, determine where functions are increasing or decreasing, find equations of tangents and normals, and find maximum and minimum values of functions. This chapter carries very high weightage in CBSE board exams — especially maxima-minima and tangent-normal questions.
Topics Covered:
- > Rate of Change
- > Increasing Functions
- > Decreasing Functions
- > Tangents to a Curve
- > Normals to a Curve
- > Approximations
- > Maxima and Minima
- > First Derivative Test
- > Second Derivative Test
- > Absolute Maxima/Minima
- > Critical Points
- > Word Problems (Optimization)
1. Rate of Change of Quantities
Concept: If y = f(x), then dy/dx represents the rate of change of y with respect to x.
If x = f(t) and y = g(t), then dy/dt = rate of change of y with respect to time t.
Key Formula: dy/dx = (dy/dt) / (dx/dt)
If x = f(t) and y = g(t), then dy/dt = rate of change of y with respect to time t.
Key Formula: dy/dx = (dy/dt) / (dx/dt)
⭐ Important Sign Convention:
> If dy/dt > 0 → y is increasing with time
> If dy/dt < 0 → y is decreasing with time
> Rate of change is always with respect to the variable mentioned.
> If dy/dt > 0 → y is increasing with time
> If dy/dt < 0 → y is decreasing with time
> Rate of change is always with respect to the variable mentioned.
Example 1: The radius of a circle is increasing at the rate of 0.7 cm/s. Find the rate of increase of its area when r = 5 cm.
A = pi r²
dA/dt = 2pi r (dr/dt)
= 2pi (5)(0.7)
= 7pi cm²/s
A = pi r²
dA/dt = 2pi r (dr/dt)
= 2pi (5)(0.7)
= 7pi cm²/s
Example 2: A ladder 5m long is leaning against a wall. The bottom slides away at 1 m/s. How fast is the top sliding down when the bottom is 4m from the wall?
x² + y² = 25
Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0
When x = 4: y = sqrt(25-16) = 3
2(4)(1) + 2(3)(dy/dt) = 0
dy/dt = -8/6 = -4/3 m/s (negative means sliding down)
x² + y² = 25
Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0
When x = 4: y = sqrt(25-16) = 3
2(4)(1) + 2(3)(dy/dt) = 0
dy/dt = -8/6 = -4/3 m/s (negative means sliding down)
Example 3: A balloon is pumped at 900 cm³/s. Find the rate of increase of radius when radius = 15 cm.
V = (4/3)pi r³
dV/dt = 4pi r²(dr/dt)
900 = 4pi(15)²(dr/dt)
dr/dt = 900/(4pi x 225) = 900/900pi = 1/pi cm/s
V = (4/3)pi r³
dV/dt = 4pi r²(dr/dt)
900 = 4pi(15)²(dr/dt)
dr/dt = 900/(4pi x 225) = 900/900pi = 1/pi cm/s
2. Increasing and Decreasing Functions ⭐
Definition:
A function f is increasing on an interval (a,b) if:
x₁ < x₂ in (a,b) implies f(x₁) < f(x₂)
A function f is decreasing on an interval (a,b) if:
x₁ < x₂ in (a,b) implies f(x₁) > f(x₂)
A function f is increasing on an interval (a,b) if:
x₁ < x₂ in (a,b) implies f(x₁) < f(x₂)
A function f is decreasing on an interval (a,b) if:
x₁ < x₂ in (a,b) implies f(x₁) > f(x₂)
Test using Derivatives:
> f'(x) > 0 for all x in (a,b) → f is strictly increasing on (a,b)
> f'(x) < 0 for all x in (a,b) → f is strictly decreasing on (a,b)
> f'(x) = 0 for all x in (a,b) → f is constant on (a,b)
> f'(x) > 0 for all x in (a,b) → f is strictly increasing on (a,b)
> f'(x) < 0 for all x in (a,b) → f is strictly decreasing on (a,b)
> f'(x) = 0 for all x in (a,b) → f is constant on (a,b)
Steps to Find Increasing/Decreasing Intervals:
Step 1: Find f'(x).
Step 2: Solve f'(x) = 0 to get critical points.
Step 3: Mark critical points on the number line — they divide it into intervals.
Step 4: Check sign of f'(x) in each interval.
Step 5: f'(x) > 0 → increasing; f'(x) < 0 → decreasing.
Step 1: Find f'(x).
Step 2: Solve f'(x) = 0 to get critical points.
Step 3: Mark critical points on the number line — they divide it into intervals.
Step 4: Check sign of f'(x) in each interval.
Step 5: f'(x) > 0 → increasing; f'(x) < 0 → decreasing.
Example: Find intervals where f(x) = x³ - 6x² + 9x + 15 is increasing or decreasing.
f'(x) = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x-1)(x-3)
Critical points: x = 1 and x = 3
Intervals: (-inf, 1), (1, 3), (3, inf)
> x < 1: f'(0) = 3(1)(3) > 0 → Increasing on (-inf, 1)
> 1 < x < 3: f'(2) = 3(1)(-1) < 0 → Decreasing on (1, 3)
> x > 3: f'(4) = 3(3)(1) > 0 → Increasing on (3, inf)
f'(x) = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x-1)(x-3)
Critical points: x = 1 and x = 3
Intervals: (-inf, 1), (1, 3), (3, inf)
> x < 1: f'(0) = 3(1)(3) > 0 → Increasing on (-inf, 1)
> 1 < x < 3: f'(2) = 3(1)(-1) < 0 → Decreasing on (1, 3)
> x > 3: f'(4) = 3(3)(1) > 0 → Increasing on (3, inf)
3. Tangents and Normals ⭐
Slope of Tangent: For curve y = f(x) at point (x₁, y₁):
m_tangent = f'(x₁) = (dy/dx) at (x₁, y₁)
Slope of Normal:
m_normal = -1/m_tangent = -1/f'(x₁) (if f'(x₁) ≠ 0)
m_tangent = f'(x₁) = (dy/dx) at (x₁, y₁)
Slope of Normal:
m_normal = -1/m_tangent = -1/f'(x₁) (if f'(x₁) ≠ 0)
EQUATIONS:
Tangent at (x1, y1): y - y1 = m_t (x - x1)
Normal at (x1, y1): y - y1 = m_n (x - x1)
SPECIAL CASES:
If f'(x1) = 0 → tangent is horizontal (y = y1), normal is vertical (x = x1)
If f'(x1) = inf → tangent is vertical (x = x1), normal is horizontal (y = y1)
Example 1: Find the equation of tangent to y = x³ at (1, 1).
dy/dx = 3x²; at (1,1): slope = 3(1)² = 3
Tangent: y - 1 = 3(x - 1)
y - 1 = 3x - 3
y = 3x - 2
dy/dx = 3x²; at (1,1): slope = 3(1)² = 3
Tangent: y - 1 = 3(x - 1)
y - 1 = 3x - 3
y = 3x - 2
Example 2: Find the equation of normal to y = x² + 4x at (-1, -3).
dy/dx = 2x + 4; at (-1,-3): slope of tangent = 2(-1)+4 = 2
Slope of normal = -1/2
Normal: y - (-3) = -1/2(x - (-1))
y + 3 = -1/2(x + 1)
2y + 6 = -x - 1
x + 2y + 7 = 0
dy/dx = 2x + 4; at (-1,-3): slope of tangent = 2(-1)+4 = 2
Slope of normal = -1/2
Normal: y - (-3) = -1/2(x - (-1))
y + 3 = -1/2(x + 1)
2y + 6 = -x - 1
x + 2y + 7 = 0
Example 3: Find the point on y = x³ - 11x + 5 where the tangent is y = x - 11.
Slope of given line = 1
dy/dx = 3x² - 11 = 1 → 3x² = 12 → x² = 4 → x = 2 or x = -2
At x = 2: y = 8 - 22 + 5 = -9
Check: point (2, -9) on y = x - 11 → -9 = 2 - 11 = -9 ✓
At x = -2: y = -8 + 22 + 5 = 19
Check: 19 = -2 - 11 = -13 ✗
∴ Point is (2, -9)
Slope of given line = 1
dy/dx = 3x² - 11 = 1 → 3x² = 12 → x² = 4 → x = 2 or x = -2
At x = 2: y = 8 - 22 + 5 = -9
Check: point (2, -9) on y = x - 11 → -9 = 2 - 11 = -9 ✓
At x = -2: y = -8 + 22 + 5 = 19
Check: 19 = -2 - 11 = -13 ✗
∴ Point is (2, -9)
4. Approximations using Derivatives
Formula: If y = f(x), then for small change dx in x:
dy ≈ f'(x) . dx
Approximate value: f(x + dx) ≈ f(x) + dy = f(x) + f'(x).dx
dy ≈ f'(x) . dx
Approximate value: f(x + dx) ≈ f(x) + dy = f(x) + f'(x).dx
Example: Find the approximate value of sqrt(36.4).
Let f(x) = sqrt(x), x = 36, dx = 0.4
f'(x) = 1/(2 sqrt(x)); f'(36) = 1/12
dy = f'(x).dx = (1/12)(0.4) = 0.4/12 = 0.033
sqrt(36.4) ≈ sqrt(36) + 0.033 = 6 + 0.033 = 6.033
Let f(x) = sqrt(x), x = 36, dx = 0.4
f'(x) = 1/(2 sqrt(x)); f'(36) = 1/12
dy = f'(x).dx = (1/12)(0.4) = 0.4/12 = 0.033
sqrt(36.4) ≈ sqrt(36) + 0.033 = 6 + 0.033 = 6.033
5. Maxima and Minima ⭐ (Most Important)
Local Maximum: f(c) is a local maximum if f(c) >= f(x) for all x near c.
Local Minimum: f(c) is a local minimum if f(c) <= f(x) for all x near c.
Critical Point: A point c where f'(c) = 0 or f'(c) does not exist is called a critical point.
Local Minimum: f(c) is a local minimum if f(c) <= f(x) for all x near c.
Critical Point: A point c where f'(c) = 0 or f'(c) does not exist is called a critical point.
5.1 First Derivative Test ⭐
Steps:
Step 1: Find f'(x) and solve f'(x) = 0 to get critical points.
Step 2: For each critical point c, check sign of f'(x) on left and right of c:
> f' changes from + to - at c → Local Maximum at c
> f' changes from - to + at c → Local Minimum at c
> f' does not change sign → Neither (Point of Inflection)
Step 1: Find f'(x) and solve f'(x) = 0 to get critical points.
Step 2: For each critical point c, check sign of f'(x) on left and right of c:
> f' changes from + to - at c → Local Maximum at c
> f' changes from - to + at c → Local Minimum at c
> f' does not change sign → Neither (Point of Inflection)
5.2 Second Derivative Test ⭐
Steps:
Step 1: Find f'(x) and solve f'(x) = 0 to get critical points.
Step 2: Find f''(x) and evaluate at each critical point c:
> f''(c) < 0 → Local Maximum at c
> f''(c) > 0 → Local Minimum at c
> f''(c) = 0 → Test fails → use First Derivative Test
Step 1: Find f'(x) and solve f'(x) = 0 to get critical points.
Step 2: Find f''(x) and evaluate at each critical point c:
> f''(c) < 0 → Local Maximum at c
> f''(c) > 0 → Local Minimum at c
> f''(c) = 0 → Test fails → use First Derivative Test
Example 1: Find local max/min of f(x) = x³ - 3x² + 3x + 1.
f'(x) = 3x² - 6x + 3 = 3(x-1)²
f'(x) = 0 → x = 1 (only critical point)
f''(x) = 6x - 6; f''(1) = 6(1) - 6 = 0 → Second derivative test fails.
Use first derivative test:
f'(0) = 3(1)² > 0 (positive before x=1)
f'(2) = 3(1)² > 0 (positive after x=1)
Sign does NOT change → Neither maximum nor minimum at x = 1 (point of inflection).
f'(x) = 3x² - 6x + 3 = 3(x-1)²
f'(x) = 0 → x = 1 (only critical point)
f''(x) = 6x - 6; f''(1) = 6(1) - 6 = 0 → Second derivative test fails.
Use first derivative test:
f'(0) = 3(1)² > 0 (positive before x=1)
f'(2) = 3(1)² > 0 (positive after x=1)
Sign does NOT change → Neither maximum nor minimum at x = 1 (point of inflection).
Example 2: Find local max/min of f(x) = sin x + cos x, x in [0, 2pi].
f'(x) = cos x - sin x = 0 → tan x = 1 → x = pi/4 or x = 5pi/4
f''(x) = -sin x - cos x
f''(pi/4) = -sin(pi/4) - cos(pi/4) = -1/sqrt2 - 1/sqrt2 = -sqrt2 < 0 → Local Maximum at x = pi/4
f(pi/4) = sin(pi/4) + cos(pi/4) = sqrt2
f''(5pi/4) = -sin(5pi/4) - cos(5pi/4) = 1/sqrt2 + 1/sqrt2 = sqrt2 > 0 → Local Minimum at x = 5pi/4
f(5pi/4) = -sqrt2
f'(x) = cos x - sin x = 0 → tan x = 1 → x = pi/4 or x = 5pi/4
f''(x) = -sin x - cos x
f''(pi/4) = -sin(pi/4) - cos(pi/4) = -1/sqrt2 - 1/sqrt2 = -sqrt2 < 0 → Local Maximum at x = pi/4
f(pi/4) = sin(pi/4) + cos(pi/4) = sqrt2
f''(5pi/4) = -sin(5pi/4) - cos(5pi/4) = 1/sqrt2 + 1/sqrt2 = sqrt2 > 0 → Local Minimum at x = 5pi/4
f(5pi/4) = -sqrt2
5.3 Absolute (Global) Maxima and Minima on Closed Interval [a, b]
Steps (Closed Interval Method):
Step 1: Find all critical points of f in (a, b) — solve f'(x) = 0.
Step 2: Evaluate f at all critical points AND at endpoints a and b.
Step 3: The largest value is the Absolute Maximum; the smallest is the Absolute Minimum.
Step 1: Find all critical points of f in (a, b) — solve f'(x) = 0.
Step 2: Evaluate f at all critical points AND at endpoints a and b.
Step 3: The largest value is the Absolute Maximum; the smallest is the Absolute Minimum.
Example: Find absolute max and min of f(x) = x³ - 3x on [-2, 2].
f'(x) = 3x² - 3 = 3(x²-1) = 0 → x = 1 or x = -1 (both in [-2,2])
f(-2) = -8 + 6 = -2
f(-1) = -1 + 3 = 2
f(1) = 1 - 3 = -2
f(2) = 8 - 6 = 2
Absolute Maximum = 2 (at x = -1 and x = 2)
Absolute Minimum = -2 (at x = -2 and x = 1)
f'(x) = 3x² - 3 = 3(x²-1) = 0 → x = 1 or x = -1 (both in [-2,2])
f(-2) = -8 + 6 = -2
f(-1) = -1 + 3 = 2
f(1) = 1 - 3 = -2
f(2) = 8 - 6 = 2
Absolute Maximum = 2 (at x = -1 and x = 2)
Absolute Minimum = -2 (at x = -2 and x = 1)
6. Optimization Word Problems ⭐ (5 Mark Questions)
General Steps for Optimization:
Step 1: Read carefully — identify what to maximize/minimize.
Step 2: Express the quantity as a function of ONE variable (use constraints to eliminate other variables).
Step 3: Differentiate and set f'(x) = 0.
Step 4: Use second derivative test to confirm max/min.
Step 5: Find the value and state the answer clearly.
Step 1: Read carefully — identify what to maximize/minimize.
Step 2: Express the quantity as a function of ONE variable (use constraints to eliminate other variables).
Step 3: Differentiate and set f'(x) = 0.
Step 4: Use second derivative test to confirm max/min.
Step 5: Find the value and state the answer clearly.
Example 1: Find two positive numbers whose sum is 15 and whose product is maximum.
Let numbers be x and (15 - x).
P = x(15 - x) = 15x - x²
dP/dx = 15 - 2x = 0 → x = 7.5
d²P/dx² = -2 < 0 → Maximum ✓
Numbers are 7.5 and 7.5, Product = 56.25
Let numbers be x and (15 - x).
P = x(15 - x) = 15x - x²
dP/dx = 15 - 2x = 0 → x = 7.5
d²P/dx² = -2 < 0 → Maximum ✓
Numbers are 7.5 and 7.5, Product = 56.25
Example 2: A farmer has 120m of fencing to fence a rectangular field. Find dimensions for maximum area.
Let length = l, breadth = b. Perimeter: 2(l+b) = 120 → l+b = 60 → b = 60 - l
A = l.b = l(60-l) = 60l - l²
dA/dl = 60 - 2l = 0 → l = 30
d²A/dl² = -2 < 0 → Maximum ✓
b = 60 - 30 = 30
Square with side 30m gives maximum area = 900 m²
Let length = l, breadth = b. Perimeter: 2(l+b) = 120 → l+b = 60 → b = 60 - l
A = l.b = l(60-l) = 60l - l²
dA/dl = 60 - 2l = 0 → l = 30
d²A/dl² = -2 < 0 → Maximum ✓
b = 60 - 30 = 30
Square with side 30m gives maximum area = 900 m²
Example 3: Find the dimensions of a cylinder of maximum volume inscribed in a sphere of radius R.
Let radius of cylinder = r, height = 2h (symmetric).
Constraint: r² + h² = R²
V = pi r² (2h) = 2pi r² h
Substitute r² = R² - h²:
V = 2pi(R² - h²)h = 2pi(R²h - h³)
dV/dh = 2pi(R² - 3h²) = 0 → h = R/sqrt3
r² = R² - R²/3 = 2R²/3 → r = R sqrt(2/3)
Height = 2R/sqrt3, Radius = R sqrt(2/3)
Let radius of cylinder = r, height = 2h (symmetric).
Constraint: r² + h² = R²
V = pi r² (2h) = 2pi r² h
Substitute r² = R² - h²:
V = 2pi(R² - h²)h = 2pi(R²h - h³)
dV/dh = 2pi(R² - 3h²) = 0 → h = R/sqrt3
r² = R² - R²/3 = 2R²/3 → r = R sqrt(2/3)
Height = 2R/sqrt3, Radius = R sqrt(2/3)
7. Important Board Exam Questions
Q1. Show that the function f(x) = x³ - 3x is increasing on (-inf,-1) U (1, inf) and decreasing on (-1,1).
f'(x) = 3x² - 3 = 3(x² - 1) = 3(x-1)(x+1)
f'(x) > 0 when x < -1 or x > 1 → Increasing on (-inf,-1) U (1,inf) ✓
f'(x) < 0 when -1 < x < 1 → Decreasing on (-1,1) ✓
f'(x) > 0 when x < -1 or x > 1 → Increasing on (-inf,-1) U (1,inf) ✓
f'(x) < 0 when -1 < x < 1 → Decreasing on (-1,1) ✓
Q2. Find the equation of tangent and normal to the curve y = x⁴ - 6x³ + 13x² - 10x + 5 at (1, 3).
dy/dx = 4x³ - 18x² + 26x - 10
At x = 1: dy/dx = 4 - 18 + 26 - 10 = 2
Tangent: y - 3 = 2(x - 1) → y = 2x + 1
Normal: y - 3 = -1/2(x - 1) → 2y - 6 = -x + 1 → x + 2y = 7
At x = 1: dy/dx = 4 - 18 + 26 - 10 = 2
Tangent: y - 3 = 2(x - 1) → y = 2x + 1
Normal: y - 3 = -1/2(x - 1) → 2y - 6 = -x + 1 → x + 2y = 7
Q3. Find the maximum and minimum values of f(x) = 3x⁴ - 8x³ + 12x² - 48x + 25 on [0, 3].
f'(x) = 12x³ - 24x² + 24x - 48 = 12(x³ - 2x² + 2x - 4)
= 12[x²(x-2) + 2(x-2)] = 12(x-2)(x²+2)
f'(x) = 0 → x = 2 (since x² + 2 > 0 always)
f(0) = 25
f(2) = 48 - 64 + 48 - 96 + 25 = -39
f(3) = 243 - 216 + 108 - 144 + 25 = 16
Absolute Maximum = 25 at x = 0
Absolute Minimum = -39 at x = 2
= 12[x²(x-2) + 2(x-2)] = 12(x-2)(x²+2)
f'(x) = 0 → x = 2 (since x² + 2 > 0 always)
f(0) = 25
f(2) = 48 - 64 + 48 - 96 + 25 = -39
f(3) = 243 - 216 + 108 - 144 + 25 = 16
Absolute Maximum = 25 at x = 0
Absolute Minimum = -39 at x = 2
Q4. A window is in the form of a rectangle surmounted by a semicircular opening. Total perimeter = 10m. Find dimensions for maximum light admission.
Let width = 2r, height of rectangle = h. Semicircle radius = r.
Perimeter: 2r + 2h + pi r = 10 → h = (10 - 2r - pi r)/2
Area = 2rh + (1/2)pi r²
= 2r[(10 - 2r - pi r)/2] + (pi r²)/2
= r(10 - 2r - pi r) + (pi r²)/2
= 10r - 2r² - pi r² + (pi r²)/2
= 10r - 2r² - (pi r²)/2
dA/dr = 10 - 4r - pi r = 0
r = 10/(4 + pi)
d²A/dr² = -4 - pi < 0 → Maximum ✓
Width = 20/(4+pi), Height = 10/(4+pi)
Perimeter: 2r + 2h + pi r = 10 → h = (10 - 2r - pi r)/2
Area = 2rh + (1/2)pi r²
= 2r[(10 - 2r - pi r)/2] + (pi r²)/2
= r(10 - 2r - pi r) + (pi r²)/2
= 10r - 2r² - pi r² + (pi r²)/2
= 10r - 2r² - (pi r²)/2
dA/dr = 10 - 4r - pi r = 0
r = 10/(4 + pi)
d²A/dr² = -4 - pi < 0 → Maximum ✓
Width = 20/(4+pi), Height = 10/(4+pi)
Q5. Prove that the volume of the largest cone inscribed in a sphere of radius R is 8/27 of the volume of the sphere.
Let cone have radius r and height h. Centre of sphere to base = h - R.
r² + (h-R)² = R² → r² = R² - (h-R)² = 2Rh - h²
V = (1/3)pi r² h = (1/3)pi(2Rh - h²)h = (pi/3)(2Rh² - h³)
dV/dh = (pi/3)(4Rh - 3h²) = 0 → h(4R - 3h) = 0 → h = 4R/3
r² = 2R(4R/3) - (4R/3)² = 8R²/3 - 16R²/9 = 8R²/9
V_max = (pi/3)(8R²/9)(4R/3) = 32piR³/81
V_sphere = (4/3)piR³
Ratio = (32piR³/81) / (4piR³/3) = (32/81) x (3/4) = 8/27 ✓
r² + (h-R)² = R² → r² = R² - (h-R)² = 2Rh - h²
V = (1/3)pi r² h = (1/3)pi(2Rh - h²)h = (pi/3)(2Rh² - h³)
dV/dh = (pi/3)(4Rh - 3h²) = 0 → h(4R - 3h) = 0 → h = 4R/3
r² = 2R(4R/3) - (4R/3)² = 8R²/3 - 16R²/9 = 8R²/9
V_max = (pi/3)(8R²/9)(4R/3) = 32piR³/81
V_sphere = (4/3)piR³
Ratio = (32piR³/81) / (4piR³/3) = (32/81) x (3/4) = 8/27 ✓
8. Key Formulas at a Glance
RATE OF CHANGE:
dy/dt = (dy/dx).(dx/dt)
INCREASING/DECREASING:
f'(x) > 0 → Increasing
f'(x) < 0 → Decreasing
TANGENT AND NORMAL:
Slope of tangent m_t = dy/dx at (x1,y1)
Slope of normal m_n = -1/m_t
Tangent: y - y1 = m_t(x - x1)
Normal: y - y1 = m_n(x - x1)
APPROXIMATION:
f(x + dx) ≈ f(x) + f'(x).dx
FIRST DERIVATIVE TEST:
f' changes + to - at c → Local Max
f' changes - to + at c → Local Min
SECOND DERIVATIVE TEST:
f'(c) = 0 and f''(c) < 0 → Local Max
f'(c) = 0 and f''(c) > 0 → Local Min
f''(c) = 0 → Test inconclusive
CLOSED INTERVAL [a,b]:
Evaluate f at critical points + endpoints
Largest = Absolute Max, Smallest = Absolute Min
9. MCQ Practice (1 Mark)
1. The function f(x) = 3x + cos 3x is:
(a) Increasing (b) Decreasing (c) Neither (d) Constant
Answer: (a) Increasing — f'(x) = 3 - 3 sin 3x; since |sin 3x| ≤ 1, f'(x) ≥ 0 always.
2. The slope of the tangent to y = x³ - x at x = 2 is:
(a) 11 (b) 10 (c) 12 (d) 9
Answer: (a) 11 — f'(x) = 3x² - 1; f'(2) = 12 - 1 = 11
3. The point on the curve y = x² where the rate of change of x equals the rate of change of y is:
(a) (1/2, 1/4) (b) (1, 1) (c) (0, 0) (d) (2, 4)
Answer: (a) — dy/dt = dx/dt means dy/dx = 1; 2x = 1 → x = 1/2, y = 1/4
4. If f(x) = x³ - 9x² + 24x + c has a local maximum at x = a and local minimum at x = b, then:
(a) a = 2, b = 4 (b) a = 4, b = 2 (c) a = -2, b = -4 (d) a = -4, b = -2
Answer: (a) a = 2, b = 4 — f'(x) = 3x² - 18x + 24 = 3(x-2)(x-4) = 0
5. The maximum value of sin x + cos x is:
(a) 1 (b) 2 (c) sqrt(2) (d) sqrt(3)
Answer: (c) sqrt(2)
10. Exam Tips
- For rate of change: write the formula relating the two quantities, then differentiate with respect to time t.
- In increasing/decreasing questions: always state the open intervals — closed brackets are wrong for strictly increasing/decreasing.
- For tangent: find slope first (dy/dx at the given point), then substitute into point-slope form.
- Always verify which test to use: Second Derivative Test is faster, but if f''(c) = 0, switch to First Derivative Test.
- In optimization (word problems): always confirm maximum/minimum using the second derivative — don't just stop at f'(x) = 0.
- For closed interval [a,b] questions, NEVER forget to check the endpoints — the absolute max/min is often at an endpoint, not a critical point.
- Common mistake: confusing local max/min with absolute max/min — they are different concepts.
✅ Summary: Applications of Derivatives is one of the most application-heavy chapters in Class 12. The 5-mark optimization problems (cylinders, cones, rectangles) and tangent-normal questions are guaranteed every year in CBSE boards. Master the step-by-step method for optimization and always verify with the second derivative. This chapter directly earns you marks if you practice the word problem types consistently.
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