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CBSE · Class 12 · Mathematics

Application of Derivatives

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The radius of a circle grows at 3 cm/s. When the radius is 3 cm, its area grows at:
  1. 18π cm2/s
  2. 36π cm2/s
  3. 27π cm2/s
  4. 9π cm2/s
2.
f(x) = 2x3 − 27x2 + 120x + 6 is strictly increasing on:
  1. (4, 5)
  2. (−∞, −5) ∪ (−4, ∞)
  3. (−∞, 4) ∪ (5, ∞)
  4. (−∞, 5)
3.
The slope of the tangent to y = −x3 − 5x2 + x − 9 at x = 3 is:
  1. −56
  2. −78
  3. 56
  4. −55
4.
The maximum value of f(x) = −2x2 + 2 is:
  1. 0
  2. 2
  3. −2
  4. 3
5.
f is strictly increasing on an interval if, on that interval:
  1. f(x) > 0
  2. f′(x) < 0
  3. f′(x) > 0
  4. f′(x) = 0
6.
f(x) = 2x3 − 9x2 − 5 has a local minimum at x =
  1. 4
  2. 0
  3. 3
  4. −3
7.
The slope of the normal to y = −2x2 − 5x at x = 4 is:
  1. −1/21
  2. −21
  3. 1/21
  4. 21
8.
Assertion (A): f(x) = 5x3 + 8x + 3 is strictly increasing on R.
Reason (R): A function with f′(x) > 0 at every point of an interval is strictly decreasing there.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Air is pumped into a spherical balloon at 25 cm3/s. How fast is the radius increasing when the radius is 15 cm?
10.
Find the intervals on which f(x) = 2x3 − 24x2 + 90x + 7 is strictly increasing and strictly decreasing.
11.
Find the equation of the tangent to y = 3x3 + 3x2 − 3x − 5 at the point where x = 3.
12.
Find the equation of the normal to y = x2 + 2x − 6 at the point where x = 2.
13.
Find the local maximum and local minimum values of f(x) = 2x3 − 3x2 − 36x + 6.
14.
Find the absolute maximum and absolute minimum values of f(x) = 2x3 + 12x2 + 18x + 5 on [−5, 0].
15.
The total revenue in rupees from selling x units is R(x) = 9x2 + 18x + 76. Find the marginal revenue when x = 15.
16.
Find the point on y = x2 + x + 8 where the tangent is parallel to the line y = 11x + 7.
Section C — Word problems4 × 4 = 16 marks
17.
An open box is made from a square sheet of cardboard of side 78 cm by cutting equal squares from each corner and folding up the sides. Find the side of the square to be cut off for the box to have the greatest volume, and that volume.
18.
Two positive numbers have sum 84. Find them if the product of the square of the first and the second is as large as possible, and the largest value.
19.
A farmer has 36 m of fencing to enclose a rectangular field along a straight river; no fence is needed along the river. Find the dimensions that give the largest area, and that area.
20.
A ladder 17 m long leans against a wall. Its foot is pulled away from the wall at 3 m/s. How fast is the top sliding down the wall when the foot is 8 m from the wall?
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Mathematics

Answer key

Application of Derivatives · AW-12M-AOD-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 18π cm2/sdifferentiate the formula with respect to t
2(c) (−∞, 4) ∪ (5, ∞)f′(x) = 6(x − 4)(x − 5)
3(a) −56dy/dx = −3x2 − 10x + 1
4(b) 2f′(x) = 0 at x = 0; f(0) = 2
5(c) f′(x) > 0—
6(c) 3f′(x) = 6x(x − 3); f″(3) > 0
7(c) 1/21slope of tangent = −21; normal slope = −1/(−21)
8(c) A is true, but R is false.A is true: f′(x) = 15x2 + 8 > 0 for every x. R is false.
91/(36π) cm/sdV/dt = 4πr2 dr/dt ⇒ dr/dt = 25/(4π × 225)
10Increasing on (−∞, 3) ∪ (5, ∞); decreasing on (3, 5)f′(x) = 6(x − 3)(x − 5)
11y = 96x − 194point (3, 94), slope 96
12x + 6y = 14point (2, 2); tangent slope 6; normal: (x − 2) + 6(y − 2) = 0
13local maximum 50 at x = −2; local minimum −75 at x = 3f′(x) = 6(x + 2)(x − 3); f″(x) = 12x − 6
14maximum 5 at x = −3 and 0; minimum −35 at x = −5f(−5) = −35, f(−3) = 5, f(−1) = −3, f(0) = 5
15Rs 288R′(x) = 18x + 18
16(5, 38)dy/dx = 2x + 1 = 11
1713 cm; 35,152 cm3V = x(78 − 2x)2; V′ = 0 at x = 13 (x = 39 gives no box); V″ < 0 there
1856 and 28; largest value 87,808write the product in one variable, set its derivative to zero, and check the second derivative
199 m by 18 m (the side along the river is 18 m); 162 m2A = x(36 − 2x); A′ = 36 − 4x = 0 at x = 9
208/5 m/sx2 + y2 = 289; when x = 8, y = 15; dy/dt = −(x/y) dx/dt = −(8/15) × 3