Senior calculus and functions problems (ages 16 to 18)

Answer: −2

1. dy/dx = 3x² = 3 at x = 1, so the tangent is y − 1 = 3(x − 1), i.e. y = 3x − 2.
2. Intersections: x³ − 3x + 2 = 0. The tangent point gives a double root, so (x − 1)² is a factor.
3. x³ − 3x + 2 = (x − 1)²(x + 2), so the other point has x = −2.

Why this works: A tangent touches the curve at a double root of the intersection equation, so knowing that root lets you factor out a square.

Where it leads: For any cubic y = x³ the tangent at x = a meets the curve again at x = −2a (the sum of the three roots is 0, by Vieta).

Answer: 1

1. The graph of |x − 1| on [0, 2] is a V with vertex at (1, 0), reaching height 1 at x = 0 and x = 2.
2. The area is two triangles, each with base 1 and height 1: ½ + ½.
3. The integral is 1.

Why this works: With absolute values, split at the point where the inside changes sign; often geometry gives the area directly.

Where it leads: ∫₀¹∫₀¹ |x − y| dy dx = 1/3 is the average distance between two random points on [0, 1]. The same splitting idea computes it.

Answer: B, 1/2

1. 1 − cos x = 2 sin²(x/2).
2. So (1 − cos x)/x² = 2 sin²(x/2)/x² = ½ (sin(x/2)/(x/2))².
3. As x → 0, sin(u)/u → 1, so the limit is 1/2 (B).

Why this works: A double-angle identity rewrites the expression in terms of sin u / u, the one standard trigonometric limit.

Where it leads: It matches the Maclaurin series cos x = 1 − x²/2 + x⁴/24 − …: near 0, cos x ≈ 1 − x²/2.

Answer: 4/3

1. They meet where x² = 2x: x = 0 and x = 2. For 0 < x < 2, 2x > x².
2. Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3.
3. = 4/3.

Why this works: The area between two graphs is the integral of (top − bottom) between their crossing points.

Where it leads: Archimedes found this without calculus: a parabolic segment has 4/3 the area of the triangle with the same base and the apex at the point where the tangent is parallel to the chord.

Answer: 3

1. dy/dx = 4x³ − 12x² + 8x = 4x(x² − 3x + 2) = 4x(x − 1)(x − 2).
2. This is zero at x = 0, 1, 2.
3. So there are 3 stationary points (minima at x = 0 and 2, a maximum at x = 1).

Why this works: Factorising the derivative shows all the stationary points at once.

Where it leads: The curve is y = (x(x − 2))², a perfect square, so it touches the x-axis at both minima. Spotting such structure saves differentiating at all.

Answer: 48 cm²

1. Volume: x²h = 32, so h = 32/x². Area A = x² + 4xh = x² + 128/x.
2. dA/dx = 2x − 128/x² = 0 gives x³ = 64, x = 4 (and d²A/dx² > 0, a minimum). Then h = 2.
3. A = 16 + 32 = 48 cm².

Why this works: Using the volume constraint to eliminate one variable reduces the problem to minimising a function of one variable.

Where it leads: The optimal open box has height half the base side: it is half of a cube, which makes sense, since a closed cube is optimal and a mirror image would close the box.

Answer: B, 1

1. ln y = x ln x. Differentiating: (1/y) dy/dx = ln x + 1.
2. So dy/dx = x^x(ln x + 1).
3. At x = 1: 1 × (0 + 1) = 1 (B).

Why this works: Neither the power rule nor the exponential rule applies to x^x; logarithmic differentiation handles a variable in both base and exponent.

Where it leads: x^x has its minimum at x = 1/e, where ln x + 1 = 0. The minimum value is e^−1/e ≈ 0.692.