In the first quadrant, the region bounded by y=x, y=x²/k and y=x²/(2k) has area 1, with k>0. Find k.

01 / Standard#Your turn

Official paper · jm02-2024 · 2(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

The line meets the parabolas again at x=k and x=2k.

Hint 2

The upper boundary changes at x=k.

Worked solution

Split the area into two integrals.
$A=\int_0^k\left(\frac{x^2}k-\frac{x^2}{2k}\right)dx+\int_k^{2k}\left(x-\frac{x^2}{2k}\right)dx$
Evaluate and retain the positive parameter.
$A=\frac{k^2}6+\frac{k^2}3=\frac{k^2}2=1\implies k=\sqrt2$

k=√2.

Checks and common pitfalls: Integrating one fixed pair of curves over the whole range gives the wrong region.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Split the area into two integrals.
$A=\int_0^k\left(\frac{x^2}k-\frac{x^2}{2k}\right)dx+\int_k^{2k}\left(x-\frac{x^2}{2k}\right)dx$
Evaluate and retain the positive parameter.
$A=\frac{k^2}6+\frac{k^2}3=\frac{k^2}2=1\implies k=\sqrt2$

Think first. Reveal a hint when the class is ready.

2024 JM02