Class 12 Mathematics - NCERT Mathematics Part I
NCERT Solutions for Class 12 Maths Chapter 6: Application of Derivatives
Chapter 6 covers the applications of derivatives in real-world problems. Students learn rate of change of quantities, increasing and decreasing functions, tangents and normals to curves, approximations using differentials, and finding maxima and minima of functions. This chapter is part of the Calculus unit which carries 35 marks in the CBSE Class 12 board exam 2027. Application of Derivatives typically contributes 8-10 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 6 has 5 exercises plus a Miscellaneous Exercise. Exercise 6.1 (18 Qs) covers rate of change. Exercise 6.2 (19 Qs) covers increasing/decreasing functions. Exercise 6.3 (27 Qs) covers tangents and normals. Exercise 6.4 (9 Qs) covers approximations. Exercise 6.5 (29 Qs) covers maxima and minima. The Miscellaneous Exercise has 24 questions. Total ~126 questions.
Table of Contents
6.1Exercise 6.1 - Rate of Change of Quantities
8 key questions shown below. The remaining 10 questions follow similar patterns.
Q1Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm.
Area of circle A = πr².
dA/dr = 2πr.
(a) When r = 3 cm:
dA/dr = 2π(3) = 6π cm²/cm
The area is increasing at the rate of 6π cm² per cm increase in radius.
(b) When r = 4 cm:
dA/dr = 2π(4) = 8π cm²/cm
The area is increasing at the rate of 8π cm² per cm increase in radius.
Q2The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Let x be the length of an edge of the cube.
Volume V = x³, Surface area S = 6x².
Given: dV/dt = 8 cm³/s. Find: dS/dt when x = 12 cm.
Step 1: Differentiate V = x³ with respect to t:
dV/dt = 3x²(dx/dt)
8 = 3(12)²(dx/dt)
8 = 3(144)(dx/dt)
8 = 432(dx/dt)
dx/dt = 8/432 = 1/54 cm/s
Step 2: Differentiate S = 6x² with respect to t:
dS/dt = 12x(dx/dt)
= 12(12)(1/54)
= 144/54
= 8/3 cm²/s
The surface area is increasing at the rate of 8/3 cm²/s.
Q5A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Let r be the radius of the circular wave.
Given: dr/dt = 5 cm/s. Find: dA/dt when r = 8 cm.
Area A = πr².
dA/dt = 2πr(dr/dt)
= 2π(8)(5)
= 80π cm²/s
The enclosed area is increasing at the rate of 80π cm²/s.
Q7The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter (b) the area of the rectangle.
Given: dx/dt = −5 cm/min (decreasing), dy/dt = 4 cm/min.
(a) Perimeter P = 2(x + y):
dP/dt = 2(dx/dt + dy/dt)
= 2(−5 + 4)
= 2(−1)
= −2 cm/min
The perimeter is decreasing at the rate of 2 cm/min.
(b) Area A = xy:
dA/dt = x(dy/dt) + y(dx/dt)
= 8(4) + 6(−5)
= 32 − 30
= 2 cm²/min
The area is increasing at the rate of 2 cm²/min.
Q9A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with the radius when the radius is 10 cm.
Volume V = (4/3)πr³.
dV/dr = 4πr².
When r = 10 cm:
dV/dr = 4π(10)² = 4π(100) = 400π cm³/cm
The volume is increasing at the rate of 400π cm³ per cm increase in radius.
Q11A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Given: dy/dt = 8(dx/dt).
Differentiate 6y = x³ + 2 with respect to t:
6(dy/dt) = 3x²(dx/dt)
Substitute dy/dt = 8(dx/dt):
6 × 8(dx/dt) = 3x²(dx/dt)
48(dx/dt) = 3x²(dx/dt)
Since dx/dt ≠ 0, divide both sides:
48 = 3x²
x² = 16
x = ±4
When x = 4: 6y = 64 + 2 = 66, so y = 11. Point: (4, 11)
When x = −4: 6y = −64 + 2 = −62, so y = −31/3. Point: (−4, −31/3)
Q15The 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 = 15.
Marginal Revenue = dR/dx = d/dx(3x² + 36x + 5) = 6x + 36.
When x = 15:
MR = 6(15) + 36 = 90 + 36 = 126
The marginal revenue when 15 units are sold is Rs 126.
Q18The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is: (A) 116 (B) 96 (C) 126 (D) 136
MR = dR/dx = 6x + 36.
At x = 15: MR = 6(15) + 36 = 90 + 36 = 126.
Answer: (C) 126
6.5Exercise 6.5 - Maxima and Minima
8 key questions shown below covering local extrema, absolute extrema, and optimization.
Q1Find the maximum and minimum values, if any, of the following functions: (i) f(x) = (2x − 1)² + 3 (ii) f(x) = 9x² + 12x + 2 (iii) f(x) = −(x − 1)² + 10 (iv) f(x) = x³ + 1
(i) f(x) = (2x − 1)² + 3:
Since (2x − 1)² ≥ 0 for all x, we have f(x) ≥ 3.
Minimum value = 3 at x = 1/2. No maximum value (f → ∞ as x → ±∞).
(ii) f(x) = 9x² + 12x + 2 = 9(x + 2/3)² − 2:
Since 9(x + 2/3)² ≥ 0, we have f(x) ≥ −2.
Minimum value = −2 at x = −2/3. No maximum value.
(iii) f(x) = −(x − 1)² + 10:
Since −(x − 1)² ≤ 0, we have f(x) ≤ 10.
Maximum value = 10 at x = 1. No minimum value (f → −∞).
(iv) f(x) = x³ + 1:
f'(x) = 3x² = 0 at x = 0. But f''(0) = 0 (inconclusive - this is a point of inflection, not an extremum).
No maximum or minimum value (f ranges from −∞ to +∞).
Q3Find the local maxima and local minima of the following functions: (i) f(x) = x² (ii) g(x) = x³ − 3x (iii) h(x) = sin x + cos x, 0 < x < π/2 (iv) f(x) = x³ − 6x² + 9x + 15
(i) f(x) = x²:
f'(x) = 2x = 0 at x = 0.
f''(x) = 2 > 0 → Local minimum = 0 at x = 0. No local maximum.
(ii) g(x) = x³ − 3x:
g'(x) = 3x² − 3 = 3(x − 1)(x + 1) = 0 at x = 1, −1.
g''(x) = 6x.
g''(1) = 6 > 0 → Local min = 1 − 3 = −2 at x = 1.
g''(−1) = −6 < 0 → Local max = −1 + 3 = 2 at x = −1.
(iii) h(x) = sin x + cos x:
h'(x) = cos x − sin x = 0 → tan x = 1 → x = π/4.
h''(x) = −sin x − cos x.
h''(π/4) = −1/√2 − 1/√2 = −√2 < 0.
Local max = sin(π/4) + cos(π/4) = 1/√2 + 1/√2 = √2 at x = π/4.
(iv) f(x) = x³ − 6x² + 9x + 15:
f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0 at x = 1, 3.
f''(x) = 6x − 12.
f''(1) = −6 < 0 → Local max = 1 − 6 + 9 + 15 = 19 at x = 1.
f''(3) = 6 > 0 → Local min = 27 − 54 + 27 + 15 = 15 at x = 3.
Q5Find the absolute maximum and minimum values of the following functions in the given intervals: (i) f(x) = x³, x ∈ [−2, 2] (ii) f(x) = sin x + cos x, x ∈ [0, π] (iii) f(x) = 4x − (1/2)x², x ∈ [−2, 9/2] (iv) f(x) = (x − 1)² + 3, x ∈ [−3, 1]
(i) f(x) = x³ on [−2, 2]:
f'(x) = 3x² = 0 at x = 0.
f(−2) = −8, f(0) = 0, f(2) = 8.
Absolute max = 8 at x = 2. Absolute min = −8 at x = −2.
(ii) f(x) = sin x + cos x on [0, π]:
f'(x) = cos x − sin x = 0 → x = π/4.
f(0) = 0 + 1 = 1, f(π/4) = 1/√2 + 1/√2 = √2 ≈ 1.414, f(π) = 0 + (−1) = −1.
Absolute max = √2 at x = π/4. Absolute min = −1 at x = π.
(iii) f(x) = 4x − x²/2 on [−2, 9/2]:
f'(x) = 4 − x = 0 → x = 4.
f(−2) = −8 − 2 = −10, f(4) = 16 − 8 = 8, f(9/2) = 18 − 81/8 = 144/8 − 81/8 = 63/8 = 7.875.
Absolute max = 8 at x = 4. Absolute min = −10 at x = −2.
(iv) f(x) = (x − 1)² + 3 on [−3, 1]:
f'(x) = 2(x − 1) = 0 → x = 1.
f(−3) = (−4)² + 3 = 16 + 3 = 19, f(1) = 0 + 3 = 3.
Absolute max = 19 at x = −3. Absolute min = 3 at x = 1.
Q14Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Let the two numbers be x and (16 − x), where 0 < x < 16.
We need to minimize S = x³ + (16 − x)³.
Step 1: Differentiate:
dS/dx = 3x² − 3(16 − x)²
= 3[x² − (16 − x)²]
= 3(x + 16 − x)(x − 16 + x)
= 3(16)(2x − 16)
= 96(x − 8)
Step 2: Set dS/dx = 0:
96(x − 8) = 0 → x = 8
Step 3: Verify using second derivative:
d²S/dx² = 96 > 0 → x = 8 gives a minimum.
The two numbers are 8 and 8.
Minimum sum of cubes = 8³ + 8³ = 512 + 512 = 1024.
Q18A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
Let side of square cut off = x cm.
After cutting and folding:
Length = 45 − 2x, Width = 24 − 2x, Height = x.
Volume V = x(45 − 2x)(24 − 2x)
Step 1: Expand:
V = x[(45)(24) − 45(2x) − 24(2x) + 4x²]
= x[1080 − 90x − 48x + 4x²]
= x[4x² − 138x + 1080]
= 4x³ − 138x² + 1080x
Step 2: Differentiate:
dV/dx = 12x² − 276x + 1080
= 12(x² − 23x + 90)
= 12(x − 5)(x − 18)
Step 3: Set dV/dx = 0:
x = 5 or x = 18.
Since width = 24 − 2x > 0 requires x < 12, we reject x = 18.
Step 4: Verify:
d²V/dx² = 24x − 276.
At x = 5: d²V/dx² = 120 − 276 = −156 < 0 → maximum ✓
Maximum volume = 5(45 − 10)(24 − 10) = 5 × 35 × 14 = 2450 cm³.
Side of square = 5 cm.
Q22A 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?
Let the piece for the circle have length x metres. Then piece for square = (28 − x) metres.
Circle: circumference = x → 2πr = x → r = x/(2π).
Area of circle = πr² = π × x²/(4π²) = x²/(4π).
Square: perimeter = 28 − x → side = (28 − x)/4.
Area of square = [(28 − x)/4]² = (28 − x)²/16.
Total area A = x²/(4π) + (28 − x)²/16.
Step 1: Differentiate:
dA/dx = 2x/(4π) + 2(28 − x)(−1)/16
= x/(2π) − (28 − x)/8
Step 2: Set dA/dx = 0:
x/(2π) = (28 − x)/8
8x = 2π(28 − x)
8x = 56π − 2πx
8x + 2πx = 56π
x(8 + 2π) = 56π
x = 56π/(8 + 2π) = 28π/(4 + π)
Piece for square = 28 − 28π/(4 + π) = 112/(4 + π).
Step 3: Verify:
d²A/dx² = 1/(2π) + 1/8 > 0 → minimum confirmed ✓
Length for circle = 28π/(4 + π) m ≈ 12.32 m
Length for square = 112/(4 + π) m ≈ 15.68 m
Q25Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan⁻¹(√2).
Let l = slant height (fixed), h = height, r = radius, α = semi-vertical angle.
From the right triangle: l² = r² + h², so r² = l² − h².
Also: r = l sin α, h = l cos α.
Volume V = (1/3)πr²h = (π/3)(l² − h²)h = (π/3)(l²h − h³).
Method 1: Differentiate with respect to h:
dV/dh = (π/3)(l² − 3h²) = 0
l² − 3h² = 0
h² = l²/3 → h = l/√3
r² = l² − l²/3 = 2l²/3 → r = l√(2/3)
tan α = r/h = [l√(2/3)]/(l/√3) = √(2/3) × √3 = √2
∴ α = tan⁻¹(√2) ∎
d²V/dh² = (π/3)(−6h) = −2πh < 0 → maximum confirmed. ✓
Q29The maximum value of [x(x − 1) + 1]^(1/3), 0 ≤ x ≤ 1 is: (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0
f(x) = [x(x − 1) + 1]^(1/3) = [x² − x + 1]^(1/3).
Let g(x) = x² − x + 1.
g'(x) = 2x − 1 = 0 → x = 1/2.
g(1/2) = 1/4 − 1/2 + 1 = 3/4.
g(0) = 0 − 0 + 1 = 1.
g(1) = 1 − 1 + 1 = 1.
Since f is an increasing function of g:
Max of g on [0, 1] = g(0) = g(1) = 1 > g(1/2) = 3/4.
Max f = (1)^(1/3) = 1.
Answer: (C) 1
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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 the height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Let x = distance of foot from wall, y = height on wall.
x² + y² = 25 (Pythagoras).
Differentiate w.r.t. t: 2x(dx/dt) + 2y(dy/dt) = 0.
When x = 4: y = √(25−16) = 3 m. Given dx/dt = 2 cm/s = 0.02 m/s.
2(4)(0.02) + 2(3)(dy/dt) = 0 → 0.16 + 6(dy/dt) = 0 → dy/dt = −0.16/6 = −8/300 = −8/3 cm/s.
The height is decreasing at the rate of 8/3 cm/s.
Find the intervals in which the function f(x) = 2x³ − 3x² − 36x + 7 is (a) strictly increasing (b) strictly decreasing.
f'(x) = 6x² − 6x − 36 = 6(x² − x − 6) = 6(x − 3)(x + 2).
Critical points: x = 3, x = −2.
Sign analysis:
• x < −2: f'(x) = 6(−)(−) = (+) > 0 → increasing
• −2 < x < 3: f'(x) = 6(−)(+) = (−) < 0 → decreasing
• x > 3: f'(x) = 6(+)(+) = (+) > 0 → increasing
(a) Strictly increasing on (−∞, −2) ∪ (3, ∞).
(b) Strictly decreasing on (−2, 3).
Find the point on the curve y = √(3x − 2) at which the tangent is parallel to the line 4x − 2y + 5 = 0.
Slope of the given line: 4x − 2y + 5 = 0 → y = 2x + 5/2 → slope = 2.
y = √(3x − 2) = (3x − 2)^(1/2).
dy/dx = (1/2)(3x − 2)^(−1/2) × 3 = 3/(2√(3x − 2)).
For tangent parallel to line: 3/(2√(3x − 2)) = 2.
3 = 4√(3x − 2).
9 = 16(3x − 2) = 48x − 32.
48x = 41 → x = 41/48.
y = √(3(41/48) − 2) = √(41/16 − 2) = √(41/16 − 32/16) = √(9/16) = 3/4.
The point is (41/48, 3/4).
If x + y = 60, find the maximum value of xy³.
Let x + y = 60, so x = 60 − y.
P = xy³ = (60 − y)y³ = 60y³ − y⁴.
dP/dy = 180y² − 4y³ = y²(180 − 4y).
Set dP/dy = 0: y²(180 − 4y) = 0 → y = 0 or y = 45.
y = 0 gives P = 0 (minimum), so y = 45.
x = 60 − 45 = 15.
d²P/dy² = 360y − 12y².
At y = 45: d²P/dy² = 360(45) − 12(2025) = 16200 − 24300 = −8100 < 0 → maximum ✓.
Maximum value of xy³ = 15 × 45³ = 15 × 91125 = 1366875.
x = 15, y = 45.
Show that the function f(x) = 4x³ − 18x² + 27x − 7 is always increasing on ℝ.
f'(x) = 12x² − 36x + 27 = 3(4x² − 12x + 9) = 3(2x − 3)².
Since (2x − 3)² ≥ 0 for all x ∈ ℝ, we have f'(x) = 3(2x − 3)² ≥ 0 for all x.
f'(x) = 0 only at x = 3/2 (a single point, not an interval).
Therefore, f is always increasing on ℝ. ∎
An open box with a square base is to be made out of a given quantity of cardboard of area c² square units. Show that the maximum volume of the box is c³/(6√3) cubic units.
Let the side of the square base = a, height = h.
Surface area of open box = a² + 4ah = c² → h = (c² − a²)/(4a).
Volume V = a²h = a² × (c² − a²)/(4a) = a(c² − a²)/4 = (ac² − a³)/4.
dV/da = (c² − 3a²)/4.
Set dV/da = 0: c² − 3a² = 0 → a² = c²/3 → a = c/√3.
d²V/da² = (−6a)/4 = −3a/2.
At a = c/√3: d²V/da² = −3(c/√3)/2 = −3c/(2√3) < 0 → maximum ✓.
h = (c² − c²/3)/(4 × c/√3) = (2c²/3)/(4c/√3) = (2c²/3)(√3/(4c)) = 2c√3/12 = c√3/6 = c/(2√3).
V_max = a²h = (c²/3)(c/(2√3)) = c³/(6√3).
Maximum volume = c³/(6√3) cubic units. ∎
Find the point on the curve y² = 8x for which the abscissa and ordinate change at the same rate.
y² = 8x. Differentiate w.r.t. t:
2y(dy/dt) = 8(dx/dt).
Given: dx/dt = dy/dt (abscissa and ordinate change at same rate).
2y(dx/dt) = 8(dx/dt).
Since dx/dt ≠ 0: 2y = 8 → y = 4.
From y² = 8x: 16 = 8x → x = 2.
The point is (2, 4).
Find the intervals in which f(x) = sin x + cos x, 0 ≤ x ≤ 2π, is strictly increasing or strictly decreasing.
f'(x) = cos x − sin x.
Set f'(x) = 0: cos x = sin x → tan x = 1 → x = π/4 or x = 5π/4 in [0, 2π].
Sign analysis of f'(x) = cos x − sin x = √2 cos(x + π/4):
• 0 ≤ x < π/4: f'(x) > 0 → increasing
• π/4 < x < 5π/4: f'(x) < 0 → decreasing
• 5π/4 < x ≤ 2π: f'(x) > 0 → increasing
Strictly increasing on [0, π/4) ∪ (5π/4, 2π].
Strictly decreasing on (π/4, 5π/4).
Find the equations of the tangent and normal to the curve x = 1 − cos θ, y = θ − sin θ at θ = π/4.
dx/dθ = sin θ, dy/dθ = 1 − cos θ.
dy/dx = (dy/dθ)/(dx/dθ) = (1 − cos θ)/sin θ.
At θ = π/4:
dy/dx = (1 − cos(π/4))/sin(π/4) = (1 − 1/√2)/(1/√2) = (√2 − 1)/1 = √2 − 1.
x₀ = 1 − cos(π/4) = 1 − 1/√2 = (√2 − 1)/√2.
y₀ = π/4 − sin(π/4) = π/4 − 1/√2.
Tangent: y − (π/4 − 1/√2) = (√2 − 1)[x − (√2 − 1)/√2].
Normal: y − (π/4 − 1/√2) = −1/(√2 − 1) × [x − (√2 − 1)/√2].
Rationalizing: slope of normal = −1/(√2 − 1) × (√2 + 1)/(√2 + 1) = −(√2 + 1)/1 = −(√2 + 1).
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Let the width of rectangle = 2r (diameter of semicircle), height = h.
Perimeter: 2h + 2r + πr = 10 → h = (10 − 2r − πr)/2.
Area A = 2rh + (1/2)πr² = 2r × (10 − 2r − πr)/2 + πr²/2
= r(10 − 2r − πr) + πr²/2
= 10r − 2r² − πr² + πr²/2
= 10r − 2r² − πr²/2.
dA/dr = 10 − 4r − πr = 10 − r(4 + π).
Set dA/dr = 0: r = 10/(4 + π).
d²A/dr² = −(4 + π) < 0 → maximum ✓.
h = (10 − 2 × 10/(4+π) − π × 10/(4+π))/2 = (10 − 10(2+π)/(4+π))/2
= (10(4+π) − 10(2+π))/(2(4+π)) = (10 × 2)/(2(4+π)) = 10/(4+π).
Width = 2r = 20/(4 + π) m.
Height = h = 10/(4 + π) m.
Note: h = r, so the rectangle is actually a square of side 10/(4 + π) with a semicircle on top.
Show that the semi-vertical angle of the cone of maximum volume and of given slant height is tan⁻¹(√2).
Let l = slant height (fixed), h = height, r = radius, α = semi-vertical angle.
l² = r² + h², so r² = l² − h².
V = (1/3)πr²h = (π/3)(l² − h²)h = (π/3)(l²h − h³).
dV/dh = (π/3)(l² − 3h²) = 0 → h² = l²/3 → h = l/√3.
r² = l² − l²/3 = 2l²/3 → r = l√(2/3).
tan α = r/h = l√(2/3)/(l/√3) = √(2/3) × √3 = √2.
α = tan⁻¹(√2). ∎
d²V/dh² = (π/3)(−6h) = −2πh < 0 → maximum confirmed. ✓
If the curves x = y² and xy = k cut at right angles, show that 8k² = 1.
From x = y²: dx/dy = 2y, so dy/dx = 1/(2y). → m₁ = 1/(2y).
From xy = k: y + x(dy/dx) = 0, so dy/dx = −y/x. → m₂ = −y/x.
Curves cut at right angles: m₁ × m₂ = −1.
(1/(2y)) × (−y/x) = −1.
−1/(2x) = −1 → x = 1/2.
From x = y²: y² = 1/2 → y = 1/√2 (taking positive value).
From xy = k: k = (1/2)(1/√2) = 1/(2√2).
k² = 1/8 → 8k² = 1. ∎
Find the absolute maximum and absolute minimum values of f(x) = 12x^(4/3) − 6x^(1/3) on [−1, 1].
f(x) = 12x^(4/3) − 6x^(1/3).
f'(x) = 12 × (4/3)x^(1/3) − 6 × (1/3)x^(−2/3) = 16x^(1/3) − 2x^(−2/3) = (16x − 2)/x^(2/3).
Set f'(x) = 0: 16x − 2 = 0 → x = 1/8.
Also, f'(x) is undefined at x = 0 (critical point).
Evaluate f at critical points and endpoints:
• f(−1) = 12(−1)^(4/3) − 6(−1)^(1/3) = 12(1) − 6(−1) = 12 + 6 = 18
• f(0) = 0 − 0 = 0
• f(1/8) = 12(1/8)^(4/3) − 6(1/8)^(1/3) = 12(1/16) − 6(1/2) = 3/4 − 3 = −9/4
• f(1) = 12(1) − 6(1) = 6
Absolute maximum = 18 at x = −1.
Absolute minimum = −9/4 at x = 1/8.
The radius of a sphere is measured as 9 cm with an error of 0.03 cm. Find the approximate error in calculating its volume.
V = (4/3)πr³.
dV = (dV/dr) × dr = 4πr² × dr.
Given: r = 9 cm, dr = 0.03 cm.
dV = 4π(9)²(0.03) = 4π(81)(0.03) = 4π × 2.43 = 9.72π cm³.
The approximate error in the volume is 9.72π cm³ ≈ 30.54 cm³.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The function f(x) = x³ − 3x² + 3x − 100 is always increasing on ℝ.
Reason (R): f'(x) = 3(x − 1)² ≥ 0 for all x ∈ ℝ.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. f'(x) = 3x² − 6x + 3 = 3(x² − 2x + 1) = 3(x − 1)². Since a perfect square is always ≥ 0, f'(x) ≥ 0 for all x, meaning f is always increasing (it equals zero only at x = 1, a single point, not an interval).
AR2
Assertion (A): The function f(x) = log(sin x) is strictly increasing on (0, π/2).
Reason (R): f'(x) = cot x > 0 for all x ∈ (0, π/2).
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. f'(x) = (1/sin x) × cos x = cot x. In (0, π/2), both sin x > 0 and cos x > 0, so cot x > 0, meaning f is strictly increasing on (0, π/2).
AR3
Assertion (A): The tangent to the curve y = e^(2x) at the point (0, 1) meets the x-axis at (0, 0).
Reason (R): The equation of tangent at (0, 1) is y = 2x + 1.
Answer: (D)
Assertion is FALSE: dy/dx = 2e^(2x). At (0, 1): slope = 2. Tangent: y − 1 = 2(x − 0) → y = 2x + 1. Setting y = 0: 0 = 2x + 1 → x = −1/2. The tangent meets the x-axis at (−1/2, 0), not (0, 0). Reason is TRUE: The equation y = 2x + 1 is correct.
AR4
Assertion (A): The function f(x) = x² has a minimum at x = 0.
Reason (R): f'(0) = 0 and f''(0) = 2 > 0.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. By the second derivative test, f'(0) = 0 (critical point) and f''(0) = 2 > 0 (concave up), confirming x = 0 is a local minimum. Since f(x) = x² ≥ 0 for all x and f(0) = 0, it is also the global minimum.
AR5
Assertion (A): If f''(c) = 0, then x = c is a point of inflection of f.
Reason (R): At a point of inflection, the concavity of the curve changes.
Answer: (A)
The Reason is TRUE: By definition, an inflection point is where concavity changes. The Assertion is partially correct in the general sense but technically requires that the sign of f''(x) actually changes around c. However, in the CBSE syllabus context, this is treated as TRUE, and the reason does explain the concept. Both A and R are true, and R explains A.
AR6
Assertion (A): The maximum value of f(x) = sin x · cos x is 1.
Reason (R): The maximum value of sin(2x) is 1.
Answer: (D)
Assertion is FALSE: sin x · cos x = (1/2)sin(2x). The maximum of sin(2x) is 1, so the maximum of (1/2)sin(2x) is **1/2**, not 1. Reason is TRUE: sin(2x) does have a maximum value of 1.
Case Study Questions
A farmer has 200 m of fencing material and wants to fence a rectangular garden along a straight river. The side along the river does not need fencing. The farmer wants to maximize the area of the garden. Let the length of the side parallel to the river be x metres and the width be y metres.
(i) Express y in terms of x and write the area A as a function of x. (1 mark)
Fencing needed: x + 2y = 200 (three sides only, river side excluded).
y = (200 − x)/2 = 100 − x/2.
Area A = x × y = x(100 − x/2) = 100x − x²/2.
(ii) Find the value of x for which the area is maximum. (2 marks)
dA/dx = 100 − x.
Set dA/dx = 0: 100 − x = 0 → x = 100 m.
d²A/dx² = −1 < 0 → maximum confirmed. ✓
(iii)(a) Find the maximum area of the garden. (2 marks)
When x = 100: y = 100 − 100/2 = 50 m.
A_max = 100 × 50 = 5000 m².
Alternatively: A = 100(100) − (100)²/2 = 10000 − 5000 = 5000 m².
(iii)(b) If instead the farmer wants to fence a square garden (not along a river, all 4 sides fenced) with the same 200 m of material, what would be the area? Compare with the maximum rectangular area. (2 marks)
For a square with perimeter 200 m: side = 200/4 = 50 m.
Area = 50² = 2500 m².
Comparison: The rectangular garden along the river (5000 m²) has double the area of the square garden (2500 m²). This is because the river side saves fencing, allowing more material for the other sides.
A pharmaceutical company models the concentration C(t) of a drug in the bloodstream t hours after injection as C(t) = 20t/(t² + 4) mg/L. The drug is effective when the concentration is above 4 mg/L. The company wants to determine the optimal dosing schedule.
(i) Find the rate of change of concentration dC/dt. (1 mark)
C(t) = 20t/(t² + 4).
Using the quotient rule: dC/dt = [20(t² + 4) − 20t(2t)]/(t² + 4)²
= [20t² + 80 − 40t²]/(t² + 4)²
= (80 − 20t²)/(t² + 4)² = 20(4 − t²)/(t² + 4)².
(ii) Find the time at which the concentration is maximum and the maximum concentration. (2 marks)
Set dC/dt = 0: 20(4 − t²) = 0 → t² = 4 → t = 2 hours (t > 0).
Verify: d²C/dt² at t = 2. Since 4 − t² changes from + to − at t = 2 (first derivative test), t = 2 gives a maximum.
C(2) = 20(2)/(4 + 4) = 40/8 = 5 mg/L.
The maximum concentration is 5 mg/L at t = 2 hours.
(iii)(a) Find the time interval during which the drug concentration is above 4 mg/L. (2 marks)
Set C(t) > 4: 20t/(t² + 4) > 4.
20t > 4(t² + 4)
20t > 4t² + 16
4t² − 20t + 16 < 0
t² − 5t + 4 < 0
(t − 1)(t − 4) < 0.
Solution: 1 < t < 4, i.e., the drug is effective from 1 hour to 4 hours after injection.
The effective duration is 3 hours.
(iii)(b) Find the intervals in which C(t) is strictly increasing and strictly decreasing for t > 0. (2 marks)
dC/dt = 20(4 − t²)/(t² + 4)².
The denominator (t² + 4)² > 0 always.
Sign depends on (4 − t²):
• For 0 < t < 2: 4 − t² > 0 → dC/dt > 0 → strictly increasing.
• For t > 2: 4 − t² < 0 → dC/dt < 0 → strictly decreasing.
The concentration increases for the first 2 hours, then decreases.
MCQ Practice
MCQ1. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is:
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Answer: (B) - A = πr². dA/dr = 2πr. At r = 6: dA/dr = 2π(6) = **12π cm²/cm**.
MCQ2. The function f(x) = x² − 4x + 6 is strictly increasing in the interval:
(A) (−∞, 2)
(B) (2, ∞)
(C) (−∞, ∞)
(D) (−2, 2)
Answer: (B) - f'(x) = 2x − 4 = 2(x − 2). f'(x) > 0 when x > 2. So f is strictly increasing on **(2, ∞)**.
MCQ3. The slope of the tangent to the curve y = x³ − x at x = 2 is:
(A) 10
(B) 11
(C) 12
(D) 9
Answer: (B) - dy/dx = 3x² − 1. At x = 2: dy/dx = 3(4) − 1 = 12 − 1 = **11**.
MCQ4. The point on the curve y² = x which is nearest to the point (2, 1) is:
(A) (1, 1)
(B) (1, −1)
(C) (4, 2)
(D) (4, −2)
Answer: (A) - Let (t², t) be a point on y² = x. Distance² = (t² − 2)² + (t − 1)². Minimize: d/dt[D²] = 2(t² − 2)(2t) + 2(t − 1) = 4t³ − 8t + 2t − 2 = 4t³ − 6t − 2 = 0. Testing t = 1: 4 − 6 − 2 = −4 ≠ 0. Actually, by inspection, (1,1) satisfies y² = x and distance = √((1−2)² + (1−1)²) = 1. The nearest point is **(1, 1)**.
MCQ5. The function f(x) = tan x − x is:
(A) always increasing
(B) always decreasing
(C) not always increasing
(D) sometimes increasing, sometimes decreasing
Answer: (A) - f'(x) = sec²x − 1 = tan²x ≥ 0 for all x in the domain. Since f'(x) = 0 only at isolated points (x = nπ), f is **always increasing** in each interval of its domain.
MCQ6. The minimum value of f(x) = x + 1/x for x > 0 is:
(A) 0
(B) 1
(C) 2
(D) −2
Answer: (C) - f'(x) = 1 − 1/x² = 0 → x² = 1 → x = 1 (since x > 0). f''(x) = 2/x³. f''(1) = 2 > 0 → minimum. f(1) = 1 + 1 = **2**. By AM-GM: x + 1/x ≥ 2√(x · 1/x) = 2.
MCQ7. The maximum value of sin x + cos x is:
(A) 1
(B) 2
(C) √2
(D) 1/√2
Answer: (C) - sin x + cos x = √2 sin(x + π/4). Maximum value of sin is 1, so maximum of √2 sin(x + π/4) = **√2**.
MCQ8. If the radius of a sphere is measured as 7 m with an error of 0.02 m, then the approximate error in calculating its volume is:
(A) 1.12π m³
(B) 3.92π m³
(C) 1.96π m³
(D) 11.76π m³
Answer: (B) - V = (4/3)πr³. dV = 4πr² dr = 4π(49)(0.02) = **3.92π m³**.
MCQ9. The function f(x) = 2x³ − 3x² − 12x + 4 has:
(A) one local maximum and one local minimum
(B) no local extremum
(C) two local maxima
(D) two local minima
Answer: (A) - f'(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1). Critical points: x = 2, x = −1. f''(x) = 12x − 6. f''(−1) = −18 < 0 → local max. f''(2) = 18 > 0 → local min. So **one local maximum (at x = −1) and one local minimum (at x = 2)**.
MCQ10. A manufacturer can sell x items at a price of (5 − x/100) each. The cost of x items is (x/5 + 500). The number of items to maximize profit is:
(A) 100
(B) 200
(C) 240
(D) 250
Answer: (C) - Revenue R = x(5 − x/100) = 5x − x²/100. Cost C = x/5 + 500. Profit P = R − C = 5x − x²/100 − x/5 − 500 = 24x/5 − x²/100 − 500. dP/dx = 24/5 − 2x/100 = 24/5 − x/50. Set = 0: x/50 = 24/5 → x = **240**. d²P/dx² = −1/50 < 0 → maximum. ✓
MCQ11. The minimum value of x² + 250/x for x > 0 is:
(A) 50
(B) 75
(C) 100
(D) 150
Answer: (B) - f(x) = x² + 250/x. f'(x) = 2x − 250/x². Set f'(x) = 0: 2x = 250/x² → x³ = 125 → x = 5. f''(x) = 2 + 500/x³. f''(5) = 2 + 500/125 = 2 + 4 = 6 > 0 → minimum. f(5) = 25 + 250/5 = 25 + 50 = **75**.
MCQ12. The function f(x) = x²e^(−x) is increasing in the interval:
(A) (−∞, 0)
(B) (0, 2)
(C) (2, ∞)
(D) (−∞, ∞)
Answer: (B) - f'(x) = 2xe^(−x) + x²(−e^(−x)) = xe^(−x)(2 − x). Since e^(−x) > 0 always, sign depends on x(2 − x). f'(x) > 0 when x > 0 and 2 − x > 0, i.e., **0 < x < 2**. So f is increasing on **(0, 2)**.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.
dA/dr = 2πr = 2π(5) = **10π cm²/cm**.
What is the slope of the tangent to the curve y = 3x⁴ − 4x at x = 1?
dy/dx = 12x³ − 4. At x = 1: slope = 12 − 4 = **8**.
Is the function f(x) = e^(3x) increasing or decreasing?
f'(x) = 3e^(3x) > 0 for all x ∈ ℝ. So f is **always increasing**.
Find the approximate change in the volume of a cube of side x metres caused by increasing the side by 3%.
V = x³. dV = 3x² dx = 3x²(0.03x) = **0.09x³ m³** (3% of volume is dV/V = 3 × 3% = 9%).
What is the maximum value of 1/(4x² + 2x + 1)?
4x² + 2x + 1 = 4(x + 1/4)² + 3/4. Minimum of denominator = 3/4 at x = −1/4. Maximum of function = **4/3**.
2-3 Mark Questions
Find the intervals in which f(x) = x² − 4x + 6 is strictly increasing or strictly decreasing.
f'(x) = 2x − 4 = 2(x − 2). f'(x) > 0 when x > 2 → increasing on (2, ∞). f'(x) < 0 when x < 2 → decreasing on (−∞, 2).
Find the equation of the tangent and normal to y = x³ at the point (1, 1).
dy/dx = 3x². At (1,1): slope = 3. Tangent: y − 1 = 3(x − 1) → y = 3x − 2. Normal: y − 1 = −(1/3)(x − 1) → 3y = −x + 4 or x + 3y = 4.
Use differentials to approximate √36.6.
Let f(x) = √x, x = 36, dx = 0.6. f(36) = 6, f'(x) = 1/(2√x), f'(36) = 1/12. √36.6 ≈ 6 + (1/12)(0.6) = 6 + 0.05 = 6.05.
Find the local maxima and minima of f(x) = x³ − 6x² + 9x + 15.
f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). f''(x) = 6x − 12. At x = 1: f''(1) = −6 < 0 → local max = 19. At x = 3: f''(3) = 6 > 0 → local min = 15.
3-4 Mark Questions
Find the absolute maximum and minimum values of f(x) = sin x + cos x on [0, π].
f'(x) = cos x − sin x = 0 → x = π/4. f(0) = 1, f(π/4) = √2, f(π) = −1. Absolute max = √2 at x = π/4. Absolute min = −1 at x = π.
A wire of length 28 m is to be cut into two pieces. One piece is bent into a square and the other into a circle. Find lengths that minimize total area.
Let circle piece = x. r = x/(2π). Square side = (28−x)/4. Total area = x²/(4π) + (28−x)²/16. dA/dx = x/(2π) − (28−x)/8 = 0. x(8+2π) = 56π → x = 28π/(4+π) for circle, 112/(4+π) for square.
Show that the semi-vertical angle of the cone of maximum volume with given slant height is tan⁻¹(√2).
V = (π/3)(l²h − h³). dV/dh = (π/3)(l² − 3h²) = 0 → h = l/√3. r = l√(2/3). tan α = r/h = √2. α = tan⁻¹(√2). d²V/dh² < 0 → max confirmed.
Key Concepts
Important Formulas
Study Plan - Application of Derivatives
Rate of Change of Quantities
Read Section 6.1. Understand the concept of rate of change and related rates. Complete Exercise 6.1 (at least 12 of 18 questions). Focus on problems involving related rates where two or more variables change with time. Practice setting up equations using chain rule: dy/dt = (dy/dx)(dx/dt).
Increasing/Decreasing Functions
Read Section 6.2. Learn the criteria: f'(x) > 0 means increasing, f'(x) < 0 means decreasing. Complete Exercise 6.2 (all 19 questions). Practice finding intervals by factoring f'(x) and using sign analysis. Pay attention to functions like x³ − 3x² + 3x − 100 where f'(x) ≥ 0 (always increasing).
Tangents, Normals, and Approximations
Read Sections 6.3 and 6.4. Master the equations of tangent (y − y₀ = m(x − x₀)) and normal (slope = −1/m). Complete Exercise 6.3 (at least 15 of 27 Qs) and Exercise 6.4 (all 9 Qs). Practice parametric tangent problems and approximation using differentials: f(x + dx) ≈ f(x) + f'(x)dx.
Maxima and Minima (Local and Absolute)
Read Section 6.5. Learn both derivative tests. Complete Exercise 6.5 (at least 20 of 29 questions). Focus on optimization word problems - these are the highest-weightage questions (5 marks). Practice: open box, wire cut into square+circle, fencing along a river, cone of max volume. Time yourself on 5-mark problems (target: 8-10 minutes).
Miscellaneous Exercise and Revision
Complete the Miscellaneous Exercise (24 Qs). Solve all 14 PYQs from 2019-2026. Revise all formulas and key concepts. Practice MCQs, assertion-reason, and case study questions from this chapter. Make a formula sheet covering all key results. Focus on the most frequently tested topics: optimization, increasing/decreasing, and tangent/normal.
Common Mistakes to Avoid
- ✗Forgetting the chain rule in related rates: when differentiating with respect to t, every variable that depends on t needs dy/dt or dx/dt. For example, d(x²)/dt = 2x(dx/dt), NOT just 2x
- ✗Confusing increasing/decreasing with positive/negative function values: f(x) > 0 does NOT mean f is increasing. Check f'(x) > 0 for increasing, f'(x) < 0 for decreasing
- ✗Forgetting to check the second derivative (or first derivative test) to confirm whether a critical point is a maximum, minimum, or neither. f'(c) = 0 alone does NOT guarantee an extremum
- ✗Using the wrong sign for the normal line slope: the normal has slope −1/m (negative reciprocal of tangent slope), not 1/m. If tangent slope is 3, normal slope is −1/3
- ✗Not considering endpoints when finding absolute extrema on a closed interval [a, b]: the absolute max or min can occur at the endpoints, not just at critical points
- ✗In optimization problems, not verifying that the critical point actually gives a maximum or minimum, and not checking that the answer satisfies the original constraints (e.g., dimensions must be positive)
Tips for Board Exam
- ✓For optimization word problems (5 marks): clearly define variables, write the function to optimize, express it in ONE variable using the constraint, find critical points, verify max/min, and state the answer with units. Show all steps for full marks
- ✓Draw a diagram for related rates and optimization problems - it helps identify the relationship between variables and avoids errors in setting up the equation
- ✓For increasing/decreasing questions, always find all critical points, test signs in each interval, and present the answer as open intervals. Use a sign chart or number line for clarity
- ✓Remember the special cases: f'(x) = 0 at a single point does NOT mean f is not increasing (e.g., f(x) = x³ at x = 0). The function must have f'(x) = 0 on an interval to be constant on that interval
- ✓For approximation questions, clearly identify x, dx, f(x), and f'(x). The formula is f(x + dx) ≈ f(x) + f'(x)·dx. Show the setup before computing for full marks
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Start Application of Derivatives - Choose Your ModeAbout Application of Derivatives - Class 12 Mathematics
Chapter 6 Application of Derivatives explores practical uses of differentiation in solving real-world problems. The chapter begins with Rate of Change of Quantities, where the derivative dy/dx represents how fast one quantity changes relative to another, and the chain rule dy/dt = (dy/dx)(dx/dt) connects related rates. Students then learn to classify functions as Increasing or Decreasing using the sign of the first derivative: f'(x) > 0 on an interval means f is strictly increasing, while f'(x) < 0 means strictly decreasing.
The chapter covers Tangents and Normals, where the slope of the tangent at a point equals f'(x₀) and the normal has slope −1/f'(x₀). The Approximation section uses differentials to estimate values: f(x + dx) ≈ f(x) + f'(x)·dx. The most substantial section is Maxima and Minima, which introduces critical points (where f'(c) = 0 or f'(c) doesn't exist), the first derivative test (sign change of f'), the second derivative test (sign of f''), and optimization of real-world quantities subject to constraints.
For the CBSE 2027 board exam, students should prioritize the 5-mark optimization word problem (appears in every paper - maximize area, minimize cost, etc.), the 3-mark increasing/decreasing or tangent/normal question, and 1-mark MCQs on rate of change. This chapter, as part of the Calculus unit worth 35 marks, typically contributes 8-10 marks and is one of the most scoring chapters in Class 12 Mathematics.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 6 Application of Derivatives?
There are 5 exercises (6.1 to 6.5) plus a Miscellaneous Exercise, totalling approximately 126 questions. Exercise 6.1 has 18 questions (rate of change), Exercise 6.2 has 19 (increasing/decreasing), Exercise 6.3 has 27 (tangents and normals), Exercise 6.4 has 9 (approximations), Exercise 6.5 has 29 (maxima and minima), and the Miscellaneous Exercise has 24 questions.
How many marks does Application of Derivatives carry in the CBSE board exam?
Application of Derivatives is part of the Calculus unit (35 marks total, combined with Continuity, Differentiability, Integrals, Applications of Integrals, and Differential Equations). Chapter 6 typically contributes about 8-10 marks - one MCQ (1 mark), one short answer on rate of change or tangent/normal (3 marks), and one long answer optimization problem (5 marks).
What is the most important topic in Chapter 6 for the board exam?
Optimization problems (Maxima and Minima) from Exercise 6.5 are the most important. A 5-mark optimization word problem appears in almost every CBSE board paper. Common topics include: maximizing area/volume, minimizing cost/material, open box problems, window problems, and cone/cylinder optimization. Practice at least 10 different optimization problems.
What is the difference between local and absolute extrema?
Local (relative) extrema are the highest/lowest values in a neighborhood of a point - found using first or second derivative test at critical points. Absolute (global) extrema are the highest/lowest values on an entire interval [a, b] - found by comparing function values at all critical points AND endpoints. On a closed interval, absolute extrema always exist (Extreme Value Theorem).
When should I use the first derivative test vs the second derivative test?
Use the first derivative test when f''(c) = 0 or f''(c) is hard to compute - it always works by checking the sign change of f'(x) around the critical point. Use the second derivative test when f''(c) is easy to compute and is non-zero - it's quicker but inconclusive when f''(c) = 0. For board exams, the second derivative test is preferred for most questions as it's faster and more straightforward.
How do I solve related rates problems?
Step 1: Identify all variables and their rates of change. Step 2: Find an equation relating the variables (geometric formula, etc.). Step 3: Differentiate both sides with respect to time t using the chain rule. Step 4: Substitute known values and solve for the unknown rate. Always assign the correct sign: increasing rates are positive, decreasing rates are negative.
What is the approximation formula and when is it used?
The formula is: f(x + Δx) ≈ f(x) + f'(x)·Δx. It is used to approximate values like √36.6, (3.02)⁴, etc. Choose x as the nearest convenient value (e.g., x = 36 for √36.6), compute f(x) exactly, then add the differential f'(x)·Δx as a correction term. This topic carries 1-2 marks in the exam.
Is NCERT enough for Application of Derivatives in the CBSE board exam?
Yes, NCERT is sufficient for the board exam. All board questions are based on NCERT concepts, examples, and exercise patterns. Focus on Exercise 6.5 (maxima/minima) and the Miscellaneous Exercise, as most 5-mark questions come from these sections. Supplement with PYQs (2019-2026) to understand the exact question patterns and difficulty level expected in the exam.