Inequalities, Maxima–Minima & Mensuration of Solids
Inequalities and maxima–minima supply two to three of the twenty-two QA questions, and mensuration of solids one or two more, so this chapter is worth roughly a tenth of the section. The inequality questions are almost always counting questions in disguise — "how many integer values of x satisfy…" — and many of them appear as TITA items, where a wrong count costs nothing but a right one earns 3, so they should always be attempted. The maxima–minima questions rarely need calculus: AM ≥ GM with its equality condition, or completing the square, answers nearly all of them. Solid mensuration at CAT depth means melting, recasting, scaling and painted-cube counting rather than plugging into a formula. Read every worked example for the equality condition or the critical points it identifies, because that is where the mark is decided.
The Wavy-Curve Method and Modulus Inequalities
For a rational inequality, bring everything to one side so that it reads P(x)/Q(x) ≤ 0 (or <, >, ≥) with both P and Q factorised into linear factors with positive leading coefficients. Mark every root of P and Q on a number line — these are the critical points. The expression is positive to the right of the largest critical point, and its sign alternates as you cross each critical point of odd multiplicity; it does not change across a root of even multiplicity such as (x − 1)². Finally, include roots of P when the inequality is non-strict, and always exclude roots of Q, since the expression is undefined there.
- Worked example (basic wavy curve). Solve (x − 1)(x − 4)/(x − 2) ≤ 0. Critical points 1, 2, 4. For x > 4 all three factors are positive: sign +. Crossing 4: − on (2, 4). Crossing 2: + on (1, 2). Crossing 1: − on (−∞, 1). We want ≤ 0, so take the negative regions and add the roots of the numerator: x ≤ 1 or 2 < x ≤ 4. The point 2 is excluded because the denominator vanishes there; 1 and 4 are included because the numerator vanishes and the inequality allows equality.
- Worked example (even power). Solve (x − 1)²(x − 3)/(x + 2) > 0. Critical points −2, 1, 3. For x > 3: +. Crossing 3: − on (1, 3). Crossing 1, an even power: still − on (−2, 1). Crossing −2: + on (−∞, −2). Strict inequality, so no critical point is included, and x = 1 in particular gives 0, which is not > 0. Solution: x < −2 or x > 3. Treating (x − 1)² as an ordinary factor would wrongly add the interval (−2, 1).
- Modulus as distance. |x − a| is the distance of x from a on the number line. So |x − a| < r means a − r < x < a + r, and |x − a| > r means x < a − r or x > a + r. Example: |x − 3| < 5 gives −2 < x < 8, which contains the 9 integers −1, 0, …, 7. And |2x − 1| ≥ 3 splits as 2x − 1 ≥ 3 or 2x − 1 ≤ −3, giving x ≥ 2 or x ≤ −1.
- Two moduli — sum of distances. |x − 1| + |x − 3| is the total distance from x to the points 1 and 3. Anywhere between them that total is exactly 2, and outside it grows, so the minimum is 2, attained for every x in [1, 3], not at a single point. Similarly |x − 3| + |x − 7| ≤ 6: between 3 and 7 the sum is 4 (fine); for x < 3 it is 10 − 2x ≤ 6, so x ≥ 2; for x > 7 it is 2x − 10 ≤ 6, so x ≤ 8. Solution 2 ≤ x ≤ 8, holding 7 integers.
- Modulus with a variable on the right. |x − 2| < x has no solution for x ≤ 0 (a modulus is never less than a non-positive number). For x > 0 square both sides safely: (x − 2)² < x², so −4x + 4 < 0, x > 1. Solution x > 1.
🧠 Read the number line, not the algebra
For any inequality built from moduli of x minus constants, draw the constants on a line and ask "how far is x from each". The answer to |x − 3| + |x − 7| ≤ 6 is visible without algebra: the two points are 4 apart, so a total distance of 6 allows x to wander 1 unit beyond each end, giving [2, 8].AM ≥ GM ≥ HM and Maxima–Minima
For positive numbers, the arithmetic mean is at least the geometric mean, which is at least the harmonic mean: for two numbers, (a + b)/2 ≥ √(ab) ≥ 2ab/(a + b), with equality exactly when a = b. For 4 and 16 the three means are 10, 8 and 6.4. The inequality is the CAT tool for extremes without calculus, and the equality condition is what tells you where the extreme occurs — which is usually the actual question.
| Given | Result | Equality when |
|---|---|---|
| Fixed sum a + b = S | Product ab is at most S²/4 | a = b = S/2 |
| Fixed product ab = P | Sum a + b is at least 2√P | a = b = √P |
| x > 0 | x + 1/x ≥ 2 and x + k/x ≥ 2√k | x = 1, or x = √k |
| Any real a, b | a² + b² ≥ 2ab | a = b |
| Positive a, b | (a + b)(1/a + 1/b) ≥ 4 | a = b |
| Cauchy–Schwarz | (a² + b²)(c² + d²) ≥ (ac + bd)² | a/c = b/d |
- Worked example (weighted sum fixed). Positive x and y satisfy 2x + 3y = 12. Find the maximum of xy. Apply AM ≥ GM to the two terms 2x and 3y: (2x + 3y)/2 ≥ √(2x · 3y), so 6 ≥ √(6xy), 36 ≥ 6xy and xy ≤ 6. Equality needs 2x = 3y = 6, i.e. x = 3, y = 2, and indeed 3 × 2 = 6 ✓. The slip is to apply the sum-fixed rule to x and y directly, which gives 36, but x + y is not fixed here.
- Worked example (fixed product). Minimise x + 4/x for x > 0. The product of the two terms is x × 4/x = 4, fixed, so the sum is at least 2√4 = 4, with equality when x = 4/x, i.e. x = 2. Check x = 2: 2 + 2 = 4 ✓; x = 1: 1 + 4 = 5 > 4 ✓
- Worked example (Cauchy–Schwarz). Maximise 3x + 4y subject to x² + y² = 25. By Cauchy–Schwarz, (3x + 4y)² ≤ (3² + 4²)(x² + y²) = 25 × 25 = 625, so 3x + 4y ≤ 25, with equality when x/3 = y/4, giving the point (3, 4). Geometrically this is the farthest a line 3x + 4y = c can be pushed while still touching a circle of radius 5.
- Counting integer solutions. The number of positive integer solutions of x + y + z = 10 is C(10 − 1, 3 − 1) = C(9, 2) = 36; of non-negative solutions, C(10 + 2, 2) = C(12, 2) = 66. A lower bound is absorbed by substitution: for positive solutions with x ≥ 3, put x = x′ + 2 with x′ ≥ 1, so x′ + y + z = 8 and the count is C(7, 2) = 21. An upper bound such as x ≤ 4 is handled by counting the complement: total minus the solutions with x ≥ 5.
🎯 Why the equality condition is the answer
AM ≥ GM only gives a bound; it becomes the maximum or minimum only if the bound is actually reached, and it is reached exactly when the terms you averaged are equal. Every time you use the inequality, write the equality condition and check that it is compatible with the constraint. If the terms cannot be made equal — say, because x must be an integer — the true extreme is at the nearest feasible point, not at the bound.Mensuration of Solids — the Formula Set
| Solid | Volume | Surface |
|---|---|---|
| Cube, edge a | a³ | TSA 6a²; diagonal a√3 |
| Cuboid l × b × h | lbh | TSA 2(lb + bh + hl); diagonal √(l² + b² + h²) |
| Cylinder, radius r, height h | πr²h | CSA 2πrh; TSA 2πr(r + h) |
| Cone, radius r, height h, slant l = √(r² + h²) | ⅓πr²h | CSA πrl; TSA πr(l + r) |
| Sphere, radius r | (4/3)πr³ | 4πr² |
| Hemisphere, radius r | (2/3)πr³ | CSA 2πr²; TSA 3πr² |
| Frustum of a cone, radii R and r, height h, slant l = √(h² + (R − r)²) | ⅓πh(R² + Rr + r²) | CSA π(R + r)l; TSA adds πR² + πr² |
| Prism (any base), height h | Base area × h | Lateral area = base perimeter × h |
| Pyramid (any base), height h | ⅓ × base area × h | Lateral area = ½ × base perimeter × slant height |
- Worked example (cone). A cone has base radius 6 cm and height 8 cm. The slant height is √(36 + 64) = √100 = 10 cm — a 3-4-5 triple doubled. CSA = π × 6 × 10 = 60π cm²; TSA = π × 6 × (10 + 6) = 96π cm²; volume = ⅓ × π × 36 × 8 = 96π cm³. The volume and TSA coincide numerically here, which is a coincidence of these dimensions and not a rule. The common error is CSA = πrh = 48π, using the height instead of the slant height.
- Worked example (frustum). A bucket is a frustum with radii 6 cm and 3 cm and height 4 cm. Slant height l = √(4² + (6 − 3)²) = √(16 + 9) = 5 cm. CSA = π(6 + 3) × 5 = 45π cm². Volume = ⅓ × π × 4 × (36 + 18 + 9) = ⅓ × π × 4 × 63 = 84π cm³. Dropping the middle term Rr gives ⅓ × π × 4 × 45 = 60π, the most frequent frustum error.
- Worked example (longest rod in a room). The longest straight rod that fits in a room 12 m × 9 m × 8 m is the cuboid's diagonal, √(144 + 81 + 64) = √289 = 17 m. The longest rod on the floor is only the floor diagonal √(144 + 81) = 15 m — read which one the question wants.
- Worked example (square pyramid). A pyramid has a square base of side 6 and height 4. The slant height to the midpoint of a base edge is √(4² + 3²) = 5, so the lateral area is 4 × (½ × 6 × 5) = 60, and the volume is ⅓ × 36 × 4 = 48. Using the height 4 in place of the slant height 5 in the lateral area gives 48 — a wrong area that happens to equal the right volume, so the options can be genuinely confusing.
ℹ️ Slant height is a right triangle
Every slant height in this table is the hypotenuse of a right triangle whose legs you already know: r and h for a cone, (R − r) and h for a frustum, half the base edge and h for a square pyramid. Setters choose the legs to form 3-4-5, 5-12-13 or 8-15-17 triples, so a slant height that is not an integer is a signal to recheck.Scaling, Melting and Recasting, Painted Cubes
When every linear dimension of a solid is multiplied by k, every area is multiplied by k² and the volume by k³. Doubling the radius of a sphere makes its surface 4 times and its volume 8 times as large. Melting and recasting preserves volume and changes surface area, and CAT's favourite question is by how much. The lever is the ratio of the two: for a sphere, surface/volume = 4πr²/(4πr³/3) = 3/r, so smaller pieces have more surface per unit volume.
- Worked example (sphere into small spheres). A solid sphere of radius 6 cm is melted into spheres of radius 2 cm. Number of spheres = (6/2)³ = 27. Surface area before: 4π × 36 = 144π. After: 27 × 4π × 4 = 432π. The total surface has become 3 times as large — exactly the ratio of radii 6/2, as the 3/r argument predicts. The value 27 answers "how many", not "how much more surface".
- Worked example (cone into sphere). A solid cone of radius 6 cm and height 24 cm is melted into a single sphere. Volume = ⅓ × π × 36 × 24 = 288π. Set (4/3)πR³ = 288π, so R³ = 216 and R = 6 cm. Note that the sphere's radius equals the cone's radius only because h = 4r; in general R³ = r²h/4.
- Worked example (rise in water level). A sphere of radius 3 cm is dropped into a cylindrical vessel of radius 6 cm containing water. The displaced volume is (4/3)π × 27 = 36π, spread over a base of area π × 36 = 36π, so the water rises by 36π/36π = 1 cm. The rise depends only on the ratio of volumes to base area, never on how much water was there.
- Largest solid inside another. The largest sphere in a cube of edge a has diameter a, volume (π/6)a³, about 52% of the cube. The largest cube inside a sphere of radius r has diagonal 2r, so edge 2r/√3. The largest cylinder in a cube of edge a has radius a/2 and height a, volume (π/4)a³.
| Faces painted | Formula | n = 4 | n = 5 |
|---|---|---|---|
| Exactly 3 (corners) | 8 | 8 | 8 |
| Exactly 2 (edges) | 12(n − 2) | 24 | 36 |
| Exactly 1 (faces) | 6(n − 2)² | 24 | 54 |
| None (interior) | (n − 2)³ | 8 | 27 |
| Total check | n³ | 64 | 125 |
⚠️ n − 2, not n − 1
Each edge of the big cube has n small cubes, of which the two at the ends are corners, so n − 2 are edge cubes. Each face is an n × n grid whose border is painted on more than one side, leaving an (n − 2) × (n − 2) interior with exactly one face painted. Writing n − 1 anywhere breaks the total, so always confirm that the four counts add to n³. "At least one face painted" is n³ − (n − 2)³, which for n = 5 is 125 − 27 = 98.Key takeaways
- A quadratic with positive leading coefficient is negative strictly between its roots and positive outside; completing the square gives its extreme value and the x at which it occurs.
- For a rational inequality, factorise, mark the critical points, start positive on the far right and alternate at every odd-multiplicity root; never include a zero of the denominator, and never multiply through by an unknown.
- |x − a| is the distance from a: |x − a| < r is the open interval (a − r, a + r), and a sum of two moduli is minimised everywhere between the two points.
- AM ≥ GM ≥ HM with equality only when the terms are equal: a fixed sum maximises the product and a fixed product minimises the sum at the equal split, and the equality condition tells you where.
- Scaling every length by k scales areas by k² and volumes by k³; melting preserves volume, so recasting into smaller pieces raises the total surface by the ratio of radii, and a painted cube of edge n has 8, 12(n − 2), 6(n − 2)² and (n − 2)³ cubes with 3, 2, 1 and 0 painted faces.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The complete solution set of the inequality x² − 5x + 6 < 0 is:
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Answer: D — 2 < x < 3
Factorise: x² − 5x + 6 = (x − 2)(x − 3). The leading coefficient is positive, so the quadratic is negative strictly between its roots and positive outside them; the solution of < 0 is 2 < x < 3. Test x = 2.5: 6.25 − 12.5 + 6 = −0.25 < 0 ✓. The set described by "x < 2 or x > 3" is where the expression is positive, the answer to the reversed inequality; the interval from −3 to −2 comes from reading the roots with the wrong sign, as if the factors were (x + 2)(x + 3); and "x > 3 only" keeps just one of the two boundary conditions.The solution set of (x − 1)(x − 4)/(x − 2) ≤ 0 is:
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Answer: C — x ≤ 1 or 2 < x ≤ 4
The critical points are 1, 2 and 4. For x > 4 all three factors are positive, so the sign is +; it alternates at each point: − on (2, 4), + on (1, 2), − on (−∞, 1). We need ≤ 0, so take the negative regions and include the numerator's zeros 1 and 4, but exclude 2 where the denominator vanishes: x ≤ 1 or 2 < x ≤ 4. The set "1 ≤ x ≤ 4, x ≠ 2" is what you get by multiplying through by (x − 2) without regard to its sign; "x ≤ 1 or 2 ≤ x ≤ 4" wrongly includes x = 2, where the expression is undefined; and "1 ≤ x < 2 or x ≥ 4" is the set where the expression is positive or zero, the reversed inequality.If x and y are positive real numbers with 2x + 3y = 12, the maximum possible value of xy is:
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Answer: B — 6
Apply AM ≥ GM to the two terms whose sum is fixed, 2x and 3y: (2x + 3y)/2 ≥ √(2x · 3y), so 6 ≥ √(6xy), 36 ≥ 6xy and xy ≤ 6. Equality holds when 2x = 3y = 6, i.e. x = 3 and y = 2, which does give xy = 6 ✓. The value 36 applies the rule "product ≤ (sum/2)²" to x and y themselves, but x + y is not what is fixed; the value 9 treats 12/2 = 6 as a fixed sum of x and y and takes (6/2)²; and 12 is just the given constant.The number of integer values of x satisfying |x − 3| + |x − 7| ≤ 6 is:
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Answer: D — 7
The left side is the total distance from x to the points 3 and 7. For 3 ≤ x ≤ 7 that total is exactly 4, so every such x works. For x < 3 the sum is (3 − x) + (7 − x) = 10 − 2x ≤ 6, giving x ≥ 2; for x > 7 it is 2x − 10 ≤ 6, giving x ≤ 8. The solution is 2 ≤ x ≤ 8, containing the 7 integers 2, 3, 4, 5, 6, 7, 8. The value 5 counts only the integers from 3 to 7, missing the two extra units allowed on each side; the value 6 is the length of the interval 8 − 2, forgetting to add one for the endpoints; and 13 is the count for |x − 3| ≤ 6 alone, ignoring the second modulus.A right circular cone has base radius 6 cm and height 8 cm. Its curved surface area is:
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Answer: A — 60π cm²
The curved surface area of a cone is πrl, where l is the slant height, not the vertical height. Here l = √(6² + 8²) = √100 = 10 cm, so CSA = π × 6 × 10 = 60π cm². The value 48π is πrh, using the height 8 in place of the slant height 10 — the single most common cone error; 96π is the total surface area πr(l + r) = π × 6 × 16, which adds the base (and happens also to equal the volume ⅓π × 36 × 8 in this case); and 36π is the base area πr² alone.A solid metal sphere of radius 6 cm is melted and recast into solid spheres of radius 2 cm each, with no metal wasted. Compared with the original sphere, the total surface area of all the small spheres is:
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Answer: B — 3 times as large
Volume is conserved, so the number of small spheres is (6/2)³ = 27. Original surface: 4π × 6² = 144π. Total surface after: 27 × 4π × 2² = 27 × 16π = 432π, which is 432/144 = 3 times the original — exactly the ratio of the radii, because surface per unit volume of a sphere is 3/r. The answer "27 times" confuses the count of spheres with the surface ratio; "9 times" is the square of the radius ratio, which would be the ratio of a single large sphere's surface to a single small one's, not the total; and "unchanged" is what is true of volume, not surface.A frustum of a cone has circular ends of radii 6 cm and 3 cm and a height of 4 cm. Its volume is:
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Answer: A — 84π cm³
The volume of a frustum is ⅓πh(R² + Rr + r²) = ⅓ × π × 4 × (36 + 18 + 9) = ⅓ × π × 4 × 63 = 84π cm³. Check by subtraction: the full cone has height 8 (similar triangles, since the radius halves over 4 cm), volume ⅓π × 36 × 8 = 96π, minus the removed top cone ⅓π × 9 × 4 = 12π, leaving 84π ✓. The value 60π drops the middle term Rr, using R² + r² = 45 only; 252π forgets the factor ⅓; and 45π is the curved surface area π(R + r)l with slant height l = √(16 + 9) = 5, a different quantity altogether.A solid cube of edge 5 cm is painted on all six faces and then cut into 125 unit cubes. The number of unit cubes with exactly one face painted is:
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Answer: C — 54
On each face of the 5 × 5 × 5 cube, the unit cubes with exactly one painted face form the interior (5 − 2) × (5 − 2) = 3 × 3 = 9 square, since the border cubes touch an edge or corner and carry two or three painted faces. Six faces give 6 × 9 = 54. Check the full count: 8 corners (3 faces) + 12 × 3 = 36 edge cubes (2 faces) + 54 (1 face) + 3³ = 27 interior (0 faces) = 125 ✓. The value 27 is the number of unpainted interior cubes; 36 is the number with exactly two faces painted; and 98 = 125 − 27 is the number with at least one face painted, not exactly one.