Chapter 6 of 13

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CBSE Class 12 Mathematics★ Unit III · part of the 35-mark Calculus block, CBSE states no chapter-wise split

Applications of Derivatives

This chapter puts the derivative to work — rates of change, monotonicity, and optimisation via the first and second derivative tests. It does not cover tangents and normals, or approximations using differentials, despite the chapter's own introduction promising both; neither exists anywhere in the current book's actual sections. CBSE Class 12 Mathematics Chapter 6.

⏱ 9 min readHard difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Solve related-rates problems by differentiating a geometric or physical relationship with respect to time
2Determine the intervals on which a function is increasing or decreasing by analysing the sign of its derivative
3Find local maxima and minima using the first derivative test and, where it applies, the faster second derivative test
4Find the absolute maximum and minimum of a function on a closed interval by checking both critical points and endpoints
5Set up and solve optimisation word problems (maximum volume, minimum surface area, maximum area) by expressing the target quantity as a function of one variable first

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Why this chapter matters

This is where the differentiation machinery built in the previous chapter earns its keep — every optimisation problem in engineering, economics, and physics reduces to exactly the first/second derivative test technique this chapter drills, and it is one of the highest-weightage single chapters for JEE Main.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Continuity and Differentiability (Class 12): chain rule, implicit differentiation, and second-order derivatives, used throughout every related-rates and second-derivative-test question in this chapter

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Coordinate geometry and mensuration formulas (Class 9-10): area and volume formulas for circles, spheres, cylinders, and cones, needed to set up almost every optimisation word problem

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Continuity and Differentiability (Class 12 Ch 5): every technique here begins by differentiating, so the chain, product and quotient rules must be automatic

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Solving polynomial equations (Class 10-11): critical points come from setting f'(x) = 0, which is usually a quadratic or cubic to be factorised

Applications of Derivatives

1. Check this before you revise anything

The chapter's own introduction promises two topics it never delivers. Its opening paragraph lists three uses of the derivative it will cover: "(i) to determine rate of change of quantities, (ii) to find the equations of tangent and normal to a curve at a point, (iii) to find turning points" — and closes with "we use the derivative to find approximate value of certain quantities."

Only rate of change and turning points (maxima/minima, increasing/decreasing) are actually taught. There is no tangent/normal section, and no differentials/approximations section, anywhere in the current book's four actual sections (6.2 Rate of Change, 6.3 Increasing and Decreasing Functions, 6.4 Maxima and Minima).

The old stub taught tangents and normals as a full section with the standard slope formulas — content the current edition's introduction still name-drops but never teaches. Removed entirely from this rebuild, matching the syllabus line itself, which lists only "rate of change of quantities, increasing/decreasing functions, maxima and minima" with nothing about tangents, normals, or approximations.

The old stub also collapsed the chapter's three real exercises into one invented 40-question group with no solutions file behind it. The book has Exercise 6.1 (18 questions, rates of change), Exercise 6.2 (19 questions, increasing/decreasing functions), Exercise 6.3 (29 questions, maxima/minima and optimisation), and a Miscellaneous Exercise (16) — 82 questions in total.

2. What this chapter covers

Textbook section	Topic
6.2	Rate of change of quantities — related rates, marginal cost/revenue
6.3	Increasing and decreasing functions — sign of $f^{'} (x)$
6.4	Maxima and minima — first and second derivative tests, absolute extrema, optimisation

3. Rate of change and monotonicity

If $y = f (x)$ , then $\frac{d y}{d x}$ (or $f^{'} (x)$ ) is the rate of change of $y$ with respect to $x$ . When two quantities $x$ and $y$ both vary with a third variable $t$ (a related rates problem), the chain rule connects them: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ .

$f$ is increasing on an interval if $f^{'} (x) \geq 0$ throughout it, and decreasing if $f^{'} (x) \leq 0$ throughout it. The practical test is always the same: differentiate, find where $f^{'} (x) = 0$ or is undefined (the critical points), and check the sign of $f^{'} (x)$ on each interval between them.

Reading a rate correctly. $\frac{d y}{d x}$ evaluated at $x = a$ gives the rate of change of $y$ per unit change in $x$ at that instant, not an average. A positive value means $y$ increases as $x$ does; the magnitude says how fast.

The related-rates procedure. These word problems all follow one shape, and writing the steps out explicitly is what earns the method marks:

Name the variables and identify which rate is given and which is wanted.
Write a geometric relation connecting the variables (area, volume, Pythagoras).
Differentiate that relation with respect to $t$ , applying the chain rule to every variable.
Substitute the given values only at this final stage.

Substituting numbers before differentiating is the classic error — it freezes a variable that is still changing, and the derivative of the resulting constant is zero.

Strict versus non-strict. The book uses $f^{'} (x) > 0$ for strictly increasing and $f^{'} (x) \geq 0$ for increasing. A question asking for strict monotonicity wants the open condition, and intervals are normally reported as open intervals for that reason.

Why the sign test works. Between consecutive critical points $f^{'}$ cannot change sign without passing through zero or a break, so its sign is constant on each such interval. Testing one convenient point per interval therefore settles the whole interval — there is no need to test several.

A worked monotonicity case. For $f (x) = x^{3} - 3 x^{2} + 4 x$ , $f^{'} (x) = 3 x^{2} - 6 x + 4$ . Completing the square gives $3 (x - 1)^{2} + 1$ , which is strictly positive for every real $x$ . So $f$ is strictly increasing on all of $R$ , with no critical points at all — a reminder that "find the intervals" sometimes has the answer "everywhere".

4. Maxima and minima

A critical point is a point in the domain where $f^{'} (c) = 0$ or $f$ is not differentiable. The first derivative test: if $f^{'} (x)$ changes from positive to negative as $x$ increases through $c$ , then $c$ is a local maximum; negative to positive, a local minimum; no sign change, neither.

The second derivative test is faster when it applies: if $f^{'} (c) = 0$ and $f^{''} (c) < 0$ , $c$ is a local maximum; if $f^{''} (c) > 0$ , a local minimum; if $f^{''} (c) = 0$ , the test is inconclusive and you fall back to the first derivative test.

Local versus absolute. A local maximum is only the largest value in some neighbourhood of $c$ ; an absolute maximum is the largest over the entire interval under consideration. A local maximum can be smaller than a value the function reaches elsewhere, so the two questions have genuinely different answers and must not be conflated.

Critical points need not be extrema. For $f (x) = x^{3}$ , $f^{'} (0) = 0$ but $x = 0$ is neither a maximum nor a minimum — the derivative is positive on both sides, so there is no sign change. This is the "neither" branch of the first derivative test, and it is examined.

When the second derivative test fails. $f^{''} (c) = 0$ tells you nothing on its own. Both $f (x) = x^{4}$ (a minimum at $0$ ) and $f (x) = x^{3}$ (no extremum at $0$ ) have $f^{''} (0) = 0$ . In that situation you must return to the first derivative test rather than guess.

For absolute (global) extrema on a closed interval $[a, b]$ : evaluate $f$ at every critical point inside $(a, b)$ and at both endpoints $f (a)$ , $f (b)$ — the largest of all these values is the absolute maximum, the smallest is the absolute minimum. Endpoints are easy to forget and are exactly where these questions like to hide the actual answer.

Worked, mirroring the textbook's own optimisation technique. A square piece of tin of side 18 cm is made into an open box by cutting a square of side $x$ from each corner and folding up the flaps. Volume $V (x) = x (18 - 2 x)^{2}$ . Differentiating and setting $V^{'} (x) = 0$ gives $x = 3$ or $x = 9$ — but $x = 9$ collapses the box to zero volume, since the flaps meet in the middle.

So the real answer is $x = 3$ , giving maximum volume $V (3) = 432 cm^{3}$ . That second, algebraically valid but physically meaningless critical point is the recurring trap in every optimisation question of this type.

Establishing the domain first. In that example the physical constraint is $0 < x < 9$ , because the cut cannot be negative and two cuts cannot exceed the side. Writing this down before differentiating makes the rejection of $x = 9$ automatic rather than an afterthought.

The general optimisation recipe:

Draw the situation and name the variable to be optimised.
Write the quantity as a function of one variable, using a constraint to eliminate any others.
State the valid domain from the physical setup.
Differentiate, set to zero, and solve for the critical points.
Discard those outside the domain, then classify the rest with the first or second derivative test.
Answer the question that was actually asked — sometimes the dimensions, sometimes the maximum value itself.

Step 2 is where most marks are lost: leaving two variables in the expression makes the differentiation meaningless.

Summary

$\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ connects related rates through the chain rule.
$f^{'} (x) \geq 0$ on an interval means increasing there; $f^{'} (x) \leq 0$ means decreasing.
First derivative test: sign change of $f^{'}$ around a critical point determines local max/min/neither.
Second derivative test: $f^{''} (c) < 0$ is a local max, $f^{''} (c) > 0$ is a local min, $f^{''} (c) = 0$ is inconclusive.
Absolute extrema on $[a, b]$ : check every interior critical point and both endpoints.
In optimisation word problems, always discard critical points that fall outside the physically valid range (negative lengths, degenerate shapes) before picking the answer.
In related-rates problems, differentiate the geometric relation first and substitute the given values only at the end.
Between consecutive critical points the sign of $f^{'}$ is constant, so one test point settles each interval.
A critical point need not be an extremum — $f (x) = x^{3}$ at $x = 0$ has $f^{'} (0) = 0$ but no sign change.
$f^{''} (c) = 0$ is inconclusive: $x^{4}$ has a minimum there and $x^{3}$ has none, so fall back to the first derivative test.
Local and absolute extrema are different questions; a local maximum can be smaller than the function's value elsewhere.
In optimisation, reduce to a single variable and state the physical domain before differentiating.
Tangents, normals, and approximations using differentials are not part of the current edition of this chapter, even though the chapter's own introduction still mentions them — they don't appear in any of its four actual sections.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Related rates via the chain rule

dy/dt = (dy/dx).(dx/dt), whenever both x and y are functions of a shared variable t

The core technique for every rate-of-change word problem in Exercise 6.1

Marginal cost and marginal revenue

Marginal cost MC = dC/dx, marginal revenue MR = dR/dx, evaluated at the given production/sales level x

Just the derivative of the given cost or revenue function, evaluated at a point

Increasing / decreasing test

f is increasing on an interval if f'(x) >= 0 throughout it; decreasing if f'(x) <= 0 throughout it

Find critical points first, then test the sign of f' on each interval between them

First derivative test

If f'(x) changes from + to - as x increases through c, c is a local maximum. If - to +, a local minimum. No sign change, neither.

Always works, even when f''(c) doesn't exist or equals 0

Second derivative test

If f'(c)=0: f''(c)<0 means local maximum, f''(c)>0 means local minimum, f''(c)=0 means the test is inconclusive

Faster than the first derivative test when it applies, but falls back to the first derivative test when f''(c)=0

Absolute extrema on a closed interval

On [a,b], the absolute maximum/minimum of f is the largest/smallest value among f at every interior critical point and f(a), f(b)

Endpoints must always be checked — the extremum can occur there even if f'(x) is never 0 nearby

The related-rates procedure

Name the variables, write a geometric relation, differentiate it with respect to t, and substitute the given values ONLY at the final stage

Substituting numbers before differentiating freezes a variable that is still changing, and its derivative wrongly becomes zero

Strict versus non-strict monotonicity

f'(x) > 0 gives strictly increasing; f'(x) >= 0 gives increasing

A question asking for strict monotonicity wants the open condition, which is why intervals are normally reported as open

Why one test point settles an interval

Between consecutive critical points f' cannot change sign without passing through zero or a break, so its sign is constant there

Testing several points inside the same interval is wasted effort

Local versus absolute extrema

A local maximum is largest only in a neighbourhood; an absolute maximum is largest over the whole interval

A local maximum can be smaller than a value the function reaches elsewhere, so the two questions have different answers

A critical point need not be an extremum

For f(x) = x^3, f'(0) = 0 but the derivative is positive on both sides, so there is no extremum

This is the 'neither' branch of the first derivative test, and it is examined directly

When the second derivative test fails

f''(c) = 0 is inconclusive: x^4 has a minimum at 0 while x^3 has no extremum, and both have f''(0) = 0

Return to the first derivative test rather than guessing

The optimisation recipe

Draw the situation, express the quantity in ONE variable using a constraint, state the valid domain, differentiate, discard critical points outside the domain, then classify

Reducing to a single variable is where most marks are lost — two variables make the differentiation meaningless

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Forgetting to check the endpoints when finding absolute extrema on a closed interval

✓ Absolute maximum/minimum on [a,b] requires comparing f at every interior critical point AND at both f(a) and f(b) — the true extremum is often at an endpoint, not a critical point.

WATCH OUT

✗ Accepting every algebraically valid critical point in an optimisation word problem without checking it makes physical sense

✓ Discard any critical point that gives a negative length, a degenerate (zero-volume) shape, or a value outside the problem's stated range before picking the final answer.

WATCH OUT

✗ Using the second derivative test when f''(c)=0

✓ f''(c)=0 makes the second derivative test inconclusive — it does not mean 'no extremum.' Fall back to the first derivative test (or check higher derivatives) in that case.

WATCH OUT

✗ Setting up an optimisation problem with two independent variables instead of reducing to one

✓ Always use the given constraint (fixed perimeter, fixed volume, etc.) to eliminate one variable before differentiating — you can only apply the derivative tests to a function of a single variable.

WATCH OUT

✗ Trying to find the equation of a tangent or normal line for this chapter

✓ That topic is not taught anywhere in the current edition, despite being mentioned in the chapter's introduction — it does not appear in any of the four actual sections or in any exercise.

WATCH OUT

✗ Substituting the given numerical values before differentiating in a related-rates problem

✓ Differentiate the geometric relation with respect to t first. Substituting early turns a changing quantity into a constant, whose derivative is zero.

WATCH OUT

✗ Assuming that f'(c) = 0 makes c a maximum or minimum

✓ It makes c a critical point only. f(x) = x^3 at x = 0 has zero derivative but no extremum, because f' does not change sign.

WATCH OUT

✗ Treating a local maximum as the absolute maximum

✓ They are different questions. For an absolute extremum on [a,b] you must also evaluate the endpoints and compare all candidate values.

WATCH OUT

✗ Failing to state the physical domain before solving an optimisation problem

✓ Writing the domain first, such as 0 < x < 9 for a box cut from an 18 cm square, makes the rejection of the degenerate critical point automatic rather than an afterthought.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Determining intervals of increase/decrease for polynomial, trigonometric, and logarithmic functions

Questions

Ex 6.3

Exercise 6.3

Local and absolute maxima/minima, and optimisation word problems (boxes, cylinders, cones, wires)

Questions

Miscellaneous Ex

Miscellaneous Exercise

Mixed rate-of-change, monotonicity, and optimisation problems, including inscribed-shape proofs

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Applications of Derivatives?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~35 marks in CBSE exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•dy/dt = (dy/dx).(dx/dt): the core technique for every related-rates problem
•f is increasing where f'(x)>=0, decreasing where f'(x)<=0 — find critical points first, then test signs on each interval
•First derivative test: sign change of f' around a critical point (+ to - is a max, - to + is a min, no change is neither)
•Second derivative test: f''(c)<0 is a local max, f''(c)>0 is a local min, f''(c)=0 is inconclusive
•Absolute extrema on [a,b] require checking every interior critical point AND both endpoints
•Optimisation word problems: use the given constraint to reduce to one variable, then discard any critical point that isn't physically valid
•Tangents, normals, and approximations using differentials are not part of the current edition of this chapter, despite being mentioned in its introduction

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit III: part of the 35-mark Calculus block, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Rate of Change, Related Rates and Marginal Cost and Revenue	3-5	1	Setting up dy/dt=(dy/dx)(dx/dt) for circles, spheres, cubes, cones, and marginal cost/revenue
Increasing and Decreasing Functions	3-5	1	Finding critical points and analysing the sign of f'(x) on the resulting intervals
Maxima, Minima and Optimisation, including Absolute Extrema, the First Derivative Test and the Second Derivative Test	5-6	1-2	First/second derivative tests, absolute extrema on a closed interval, optimisation word problems

Prep strategy

For related-rates questions, write down every given rate and the relationship between the quantities before differentiating — most errors come from differentiating before identifying what's actually constant versus changing
For increasing/decreasing questions, always find every critical point first, then test the sign of f' on each resulting sub-interval systematically rather than guessing
For optimisation word problems, reduce to a single variable using the given constraint before differentiating, and always discard critical points that don't make physical sense
Do not attempt to find a tangent or normal line equation for this chapter — that topic is not part of the current syllabus here

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Manufacturing and packaging design

Companies use exactly the box/cylinder optimisation technique in this chapter to design packaging that minimises material cost (surface area) for a required volume, directly saving money at scale.

Economics: marginal cost and profit maximisation

Marginal cost and marginal revenue (the derivatives of cost and revenue functions) are core concepts in microeconomics; profit is maximised exactly where marginal revenue equals marginal cost, a direct application of the first derivative test.

Physics: velocity and acceleration from position

Rate-of-change problems generalise directly to kinematics, where velocity is the derivative of position and acceleration is the derivative of velocity — the same related-rates technique used for expanding circles and balloons.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

For related-rates problems, explicitly write 'Given: ... Find: ...' before setting up the relationship — this catches setup errors before they propagate through the differentiation

For increasing/decreasing and maxima/minima questions, present the critical points and the sign analysis as an explicit table or interval chart — examiners award marks for showing the sign check, not just the final classification

For absolute extrema, explicitly list every candidate (critical points and both endpoints) with its function value before stating the conclusion

For optimisation word problems, state which critical point was discarded and why (negative length, degenerate shape) — this demonstrates the physical reasoning step examiners look for

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Lagrange multipliers generalise this chapter's single-variable-constraint optimisation technique to problems with two or more variables and multiple constraints, a standard tool in multivariable calculus beyond this syllabus

STRETCH

Convexity and concavity (related to the sign of the second derivative) extend beyond just classifying individual critical points, characterising the global shape of a function's graph

STRETCH

The AM-GM inequality often provides a faster, calculus-free route to the same optimisation answers this chapter finds via derivatives — for example, the classic 'maximise xy given x+y is fixed' problems are solved instantly by AM-GM

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainRelated rates with a right-triangle constraintDifferentiating a Pythagorean relationship with respect to time

A man of height 2 m walks away from a lamp post of height 6 m at the rate of 1.5 m/s. Find the rate at which the length of his shadow is increasing.

Stuck? Show the approach

Set up similar triangles between the lamp post and the man to relate the shadow length to the man's distance from the post, then differentiate with respect to time.

Show the full solution

Let $x$ be the man's distance from the post and $s$ the shadow length. By similar triangles, $\frac{6}{x + s} = \frac{2}{s}$ , so $6 s = 2 x + 2 s$ , giving $4 s = 2 x$ , so $s = x /2$ . Differentiating: $\frac{d s}{d t} = \frac{1}{2} \frac{d x}{d t} = \frac{1}{2} (1.5) = 0.75$ .

Answer: The shadow length increases at 0.75 m/s

The trap

Confusing the rate of increase of the shadow's tip position (which would need $d (x + s) / d t$ ) with the rate of increase of the shadow's length ( $d s / d t$ ) — the question asks specifically for the length.

JEE MainAbsolute extrema requiring a domain restrictionOptimisation with an implicit constraint on the variable's range

Find the maximum value of $f (x) = x^{3} - 3 x^{2} + 6$ on the interval $[- 1, 2]$ .

Stuck? Show the approach

Find critical points via f'(x)=0, keep only those inside the interval, then compare with both endpoint values.

Show the full solution

$f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2)$ , zero at $x = 0$ and $x = 2$ . Both lie in $[- 1, 2]$ (with $x = 2$ being the right endpoint itself). Evaluating: $f (- 1) = - 1 - 3 + 6 = 2$ , $f (0) = 6$ , $f (2) = 8 - 12 + 6 = 2$ .

Answer: Maximum value is 6, at x=0

The trap

Since $x = 2$ is both a critical point and an endpoint, it's easy to only check it once and assume it must be extremal — always compare the actual numeric value against all other candidates rather than assuming a boundary critical point is automatically the max or min.

JEE MainOptimisation with a non-obvious constraint substitutionMinimising a sum of distances

Find the point on the curve $y^{2} = 8 x$ closest to the point $(4, 0)$ .

Stuck? Show the approach

Express the squared distance from a general point on the curve to (4,0) as a function of x alone (using the curve's own equation to eliminate y), then minimise.

Show the full solution

For a point $(x, y)$ on the curve, $y^{2} = 8 x$ , so the squared distance is $D (x) = (x - 4)^{2} + y^{2} = (x - 4)^{2} + 8 x = x^{2} - 8 x + 16 + 8 x = x^{2} + 16$ . $D^{'} (x) = 2 x = 0$ gives $x = 0$ . $D^{''} (x) = 2 > 0$ , confirming a minimum. At $x = 0$ , $y^{2} = 0$ , so $y = 0$ .

Answer: The closest point is (0,0), at distance 4 from (4,0)

The trap

Trying to minimise the distance formula directly (with the square root) instead of the squared distance — minimising $D (x)$ instead of $D (x)$ gives the same critical point with much less algebra, since square root is an increasing function.

JEE MainIncreasing/decreasing test on a rational functionSign analysis of a derivative with multiple factors

Find the intervals in which $f (x) = \frac{x}{x ^{2} + 1}$ is increasing.

Stuck? Show the approach

Differentiate using the quotient rule, factor the numerator, and determine the sign of the derivative on each interval it defines.

Show the full solution

$f^{'} (x) = \frac{( x ^{2} + 1 ) - x ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}}$ . Since the denominator is always positive, the sign of $f^{'} (x)$ matches the sign of $1 - x^{2} = (1 - x) (1 + x)$ , which is positive exactly when $- 1 < x < 1$ .

Answer: f is increasing on the interval (-1, 1)

The trap

Forgetting that the denominator $(x^{2} + 1)^{2}$ is always positive and never zero — this means the sign of the whole derivative is decided entirely by the numerator, simplifying the analysis considerably.

JEE AdvancedOptimisation of an inscribed shape with a non-trivial constraintMaximising area subject to a geometric constraint using a trigonometric parametrisation

A right circular cylinder is inscribed in a sphere of radius $R$ . Find the height of the cylinder (in terms of $R$ ) that maximises its volume, and find that maximum volume.

Stuck? Show the approach

Parametrise the cylinder's radius and half-height using the sphere's radius and a single angle (or express the cylinder radius in terms of its height via the sphere's equation), reducing volume to a function of one variable.

Show the full solution

If the cylinder has height $2 h$ (half-height $h$ ) and radius $ρ$ , the sphere's constraint gives $ρ^{2} + h^{2} = R^{2}$ , so $ρ^{2} = R^{2} - h^{2}$ . Volume $V = π ρ^{2} (2 h) = 2 π h (R^{2} - h^{2}) = 2 π (R^{2} h - h^{3})$ . $\frac{d V}{d h} = 2 π (R^{2} - 3 h^{2}) = 0$ gives $h = R / 3$ . $\frac{d ^{2} V}{d h ^{2}} = - 12 π h < 0$ at this point, confirming a maximum. Height $= 2 h = 2 R / 3$ . Maximum volume: $V = 2 π (R^{2} \cdot \frac{R}{3} - \frac{R ^{3}}{3 3}) = 2 π \cdot \frac{2 R ^{3}}{3 3} = \frac{4 π R ^{3}}{3 3}$ .

Answer: Height = 2R/sqrt(3); maximum volume = 4.pi.R^3/(3.sqrt(3))

The trap

Using the cylinder's full height as a single variable without recognising that expressing it as $2 h$ (symmetric about the sphere's centre) makes the constraint equation and the calculus noticeably cleaner than working with the asymmetric setup directly.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 BoardVery High

JEE MainVery High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No — even though the chapter's own introduction mentions it as an upcoming topic, none of the four actual sections (rate of change, increasing/decreasing, maxima/minima) ever teaches it. It simply isn't part of the current edition.

Use the second derivative test first since it's faster — compute f''(c) at each critical point. Only fall back to the first derivative test (checking the sign of f' on either side) when f''(c)=0, since the second derivative test gives no information in that case.

Differentiating a cubic or higher-degree volume/area function often produces more than one root. Physical constraints (a length can't be negative, a cut can't exceed half the original side) usually eliminate all but one — always check every critical point against the problem's real-world constraints before picking the final answer.

A local maximum only needs to be bigger than nearby points, found via the derivative tests. An absolute maximum on a closed interval must be the single largest value anywhere in that interval, which means you also have to check both endpoints, not just the interior critical points.

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Trig functions repeat, so they aren't one-one on their full domain — restrict each one to an interval where it is, and you get an inverse with a well-defined principal value branch. The current book teaches every simplification through one substitution technique (x=sinθ, x=cosθ, x=tanθ, x=secθ) rather than a list of identities to memorise. CBSE Class 12 Mathematics Chapter 2.

Matrices

This chapter builds matrices from notation up through addition, multiplication, transpose, symmetric/skew-symmetric decomposition, and invertibility — but stops at proving an inverse is unique, not computing one. The old "find the inverse using elementary row operations" method has been dropped from the current book entirely, and its exercise shows it: Exercise 3.4 now has exactly one leftover question. CBSE Class 12 Mathematics Chapter 3.

Determinants

This chapter computes determinants, uses them to test for collinearity and invertibility, and solves linear systems — but only by the matrix method X=A⁻¹B, never Cramer's Rule, which the current book never mentions. Its own Miscellaneous Exercise also still points to questions 11-15 and 17-19 that no longer exist after rationalisation trimmed it down to 9. CBSE Class 12 Mathematics Chapter 4.

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