ESAT 問題集

解答: B · y = 3x + 1

Gradient = (10 − 4) / (3 − 1) = 6 / 2 = 3.

Using y − 4 = 3(x − 1) gives y = 3x + 1. Check the second point: 3(3) + 1 = 10. ✓

解答: D · k = ±6

A repeated root means the discriminant b² − 4ac = 0.

Here a = 1, b = k, c = 9, so k² − 4(1)(9) = 0, giving k² = 36.

Therefore k = ±6. The trap is to keep only the positive root.

解答: C · 15 / 36

Possible sums run from 2 to 12. The prime sums are 2, 3, 5, 7 and 11.

Count the ordered outcomes for each: 2 → 1 way, 3 → 2, 5 → 4, 7 → 6, 11 → 2. Total favourable = 1 + 2 + 4 + 6 + 2 = 15.

Probability = 15 / 36 (which is also 5 / 12, but option C is written in the unsimplified form the question matches).

解答: B · x = 4

Combine the logs: log₂ [x(x − 2)] = 3, so x(x − 2) = 2³ = 8.

Then x² − 2x − 8 = 0, which factorises as (x − 4)(x + 2) = 0.

x = 4 or x = −2. Reject x = −2 because log of a negative number is undefined. So x = 4.

解答: C · 4

Differentiate: dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).

Stationary points at x = 1 and x = 3. The cubic rises, falls, then rises, so the maximum is at x = 1 (a = 1) and the minimum at x = 3 (b = 3).

a + b = 1 + 3 = 4. You do not even need to test which is which, since the question only asks for the sum.

解答: B · 20 m

Use s = ut + ½at² with u = 0, a = 10 m s⁻², t = 2 s.

s = 0 + ½ × 10 × 2² = ½ × 10 × 4 = 20 m.

The common error is forgetting the factor of ½ and writing 40 m (option C).

解答: C · 2 Ω

For parallel resistors, 1/R = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2.

So R = 2 Ω. A useful check: the combined value of parallel resistors is always smaller than the smallest individual resistor, here 3 Ω.

解答: A · 1 m s⁻¹

Conserve momentum: total before = total after.

Before: 0.5 × 4 + 1.5 × 0 = 2 kg m s⁻¹.

After: combined mass = 0.5 + 1.5 = 2 kg, moving at v, so 2v = 2, giving v = 1 m s⁻¹.

解答: B · 2 mol

Molar mass of O₂ = 2 × 16 = 32 g mol⁻¹.

Moles = mass / molar mass = 64 / 32 = 2 mol.

The trap is using the atomic mass 16 (giving 4 mol) instead of the molecular mass 32.

解答: B · Water molecules hydrogen-bond to one another

Boiling point depends on the strength of intermolecular forces, not the strength of the covalent bonds inside the molecule.

Oxygen is highly electronegative and small, so O–H bonds allow strong hydrogen bonding between water molecules. H₂S cannot hydrogen-bond effectively, so it boils far lower despite having a higher molecular mass.

解答: D · Cellulose cell wall

Mitochondria, ribosomes, the cell-surface membrane and the nucleus are common to both plant and animal cells.

A cellulose cell wall is a feature of plant cells (and absent from animal cells), which is the distinguishing structure here.

解答: A · A or B only

The father (AB) passes on either I^A or I^B. The mother (O) is genotype ii, so she always passes on i.

Offspring genotypes are therefore I^Ai (group A) or I^Bi (group B).

So children can be group A or group B only. AB and O are not possible.