Class 12 Maths · Original EZT practice
Application of Derivatives (Class 12 Chapter 6): summary and practice
The derivative is a rateometer: it measures how fast one quantity changes with another, tells you where a function climbs or falls, and pinpoints its peaks and valleys. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Rate of change of quantities
dy/dx measures how fast y changes with x. In related rates, two changing quantities are linked by a formula, so differentiating with respect to time connects their rates through the chain rule.
2. Increasing and decreasing functions
Where f'(x) is positive the function increases; where it is negative the function decreases. The sign chart of f' splits the domain into climbing and falling stretches.
3. Maxima and minima
Critical points are where f'(c) = 0 or f' does not exist - the chapter stresses this is necessary, not sufficient. The first derivative test reads the sign change of f' across a critical point: plus to minus gives a local maximum, minus to plus a local minimum.
4. Absolute extrema on a closed interval
On a closed interval, the absolute maximum and minimum are found by evaluating f at every critical point inside and at both end points, then comparing the values.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find where a function increases or decreases
Choose the method: Factor f' and read its sign on each interval.
EZT practice question
For f(x) = 2x^3 - 9x^2 + 12x + 1, find where f is increasing and where decreasing.
Show the worked solution
- f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2).
- f' is positive for x < 1 and x > 2, negative for 1 < x < 2.
Answer: Increasing on (-inf, 1) and (2, inf); decreasing on (1, 2).
Check: The roots of f' split the line, and the upward parabola shape of 6(x-1)(x-2) fixes the signs.
Watch for this: Do not judge increase or decrease from f itself - only the sign of f' speaks.
Now try it yourself
Where does g(x) = x^3 - 3x increase?
Check your answer
On (-inf, -1) and (1, inf).
g'(x) = 3x^2 - 3 = 3(x-1)(x+1) is positive outside [-1, 1].
Classify critical points
Choose the method: Sign change of f': plus-to-minus is a max, minus-to-plus is a min.
EZT practice question
For f(x) = 2x^3 - 9x^2 + 12x + 1, find the local maximum and minimum values.
Show the worked solution
- Critical points: f'(x) = 6(x-1)(x-2) = 0 gives x = 1 and x = 2.
- f' goes plus to minus at x = 1 (local max) and minus to plus at x = 2 (local min).
- f(1) = 2 - 9 + 12 + 1 = 6 and f(2) = 16 - 36 + 24 + 1 = 5.
Answer: Local maximum value 6 at x = 1; local minimum value 5 at x = 2.
Check: A cubic with positive leading coefficient rises, dips, then rises - max first, min second.
Watch for this: A critical point is not automatically an extremum - the chapter warns the condition f'(c) = 0 is necessary, not sufficient.
Now try it yourself
Classify the critical points of g(x) = x^3 - 3x.
Check your answer
Local max 2 at x = -1; local min -2 at x = 1.
g'(x) = 3(x-1)(x+1): plus to minus at -1, minus to plus at 1.
Find absolute extrema on a closed interval
Choose the method: Evaluate f at critical points and both end points; compare.
EZT practice question
Find the absolute maximum and minimum of f(x) = 2x^3 - 9x^2 + 12x + 1 on [0, 3].
Show the worked solution
- Critical points inside: x = 1 and x = 2.
- f(0) = 1, f(1) = 6, f(2) = 5, f(3) = 54 - 81 + 36 + 1 = 10.
- Largest value 10, smallest value 1.
Answer: Absolute maximum 10 at x = 3; absolute minimum 1 at x = 0.
Check: Both extrema landed on end points here - which is exactly why end points must be checked.
Watch for this: Skipping the end points is the standard error; a closed-interval answer needs every candidate.
Now try it yourself
Absolute extrema of g(x) = x^3 - 3x on [0, 2]?
Check your answer
Max 2 at x = 2; min -2 at x = 1.
g(0) = 0, g(1) = -2, g(2) = 2 - compare all three.
Interpret dy/dx as a rate
Choose the method: The derivative is the instantaneous rate of change.
EZT practice question
The volume of a sphere is V = (4/3) pi r^3. If the radius grows at 1 cm/s, how fast does V grow when r = 3 cm?
Show the worked solution
- dV/dt = 4 pi r^2 dr/dt.
- Substitute r = 3, dr/dt = 1: dV/dt = 4 pi x 9 x 1.
Answer: dV/dt = 36 pi = 113.1 cm^3/s.
Check: The rate equals the surface area times dr/dt - a thin shell of volume added each second.
Watch for this: Differentiate with respect to time, not r alone - both V and r change with t.
Now try it yourself
Same sphere, r = 3 cm but dr/dt = 2 cm/s - dV/dt?
Check your answer
226.2 cm^3/s (72 pi).
The rate doubles with dr/dt.
Spot a critical point that is not an extremum
Choose the method: f'(c) = 0 with no sign change means no extremum.
EZT practice question
f(x) = x^3 has f'(0) = 0. Is x = 0 a local maximum or minimum?
Show the worked solution
- f'(x) = 3x^2 is positive on both sides of 0.
- No sign change across x = 0, so the first derivative test gives no extremum.
Answer: Neither - x = 0 is a critical point but not an extremum.
Check: The cubic x^3 keeps rising through the origin; it just flattens momentarily.
Watch for this: Never announce an extremum from f'(c) = 0 alone - always check the sign change.
Now try it yourself
Does g(x) = x^4 have an extremum at x = 0?
Check your answer
Yes, a local minimum.
g'(x) = 4x^3 changes minus to plus across 0.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Rate of change of A = pi r^2?
Show answer and reason
dA/dt = 2 pi r dr/dt.
Chain rule with respect to time.
2. f increasing where?
Show answer and reason
Where f'(x) > 0.
Positive derivative means climbing.
3. First derivative test for a max?
Show answer and reason
f' changes plus to minus.
The function stops rising and starts falling.
4. Closed-interval method?
Show answer and reason
Evaluate f at critical points and both end points, then compare.
Absolute extrema can sit on the boundary.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A cube's side is 4 cm and grows at 0.5 cm/s: find dV/dt.
- Classify the critical points of g(x) = x^3 - 3x.
- Find the absolute maximum and minimum of g(x) = x^3 - 3x on [0, 2].
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Questions about this practice page
- What is a critical point?
- A point where f'(c) = 0 or f' does not exist. The chapter stresses this condition is necessary for an extremum but not sufficient - x^3 at 0 is the standard counterexample.
- What is the difference between a local and an absolute maximum?
- A local maximum beats nearby points; the absolute maximum beats every point on the whole interval. On a closed interval the absolute value may sit at an end point.
- How do I test where a function increases?
- Compute f', find its roots, and check its sign on each interval between them: positive means increasing, negative means decreasing.
- Why do related-rates answers carry units?
- Because the derivative is a physical rate: dA/dt in cm^2/s, dV/dt in cm^3/s. Dropping the units usually means a factor went missing.
- Are these the NCERT exercise questions for Application of Derivatives (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.