Question 3(d) Merit

Two exponential terms to integrate, a log-form differential equation, an upper limit that appears on both sides, a bisected cosine area, and Newton’s law of cooling built from scratch.

The graph shows y = 2 cos(x/2) from x = 0 to x = π, where it meets the x-axis. The region under it is split by the line x = k into Area A, from 0 to k, and Area B, from k to π.

Find the value of k so that the shaded Area A will be equal to the shaded Area B.

The 2024 paper, page 12 ↗ Marking schedule, page 6 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Find the whole shaded area first, from 0 to π.

   int 2 cos(x/2)dx = 4 sin(x/2), so the whole area is 4 sinπ/2 − 4 sin 0 = 4.

2. 2 If A and B are equal, how big is each?

   Half each: Area A = 2.

3. 3 Write Area A as an integral from 0 to k and set it equal to 2.

   4 sin(k/2) − 0 = 2, so sin(k/2) = 1/2.

4. 4 Solve in radians. Which angle fits on the graph?

   k/2 = π/6, so k = π/3. The other angle with sine 1/2, 5π/6, gives k = 5π/3, which is beyond π and off the graph.

What the marker wanted

• Achieved The integration correct.
• Merit The value of k, π/3 = 1.047, from correct integration. Setting the two areas equal directly works too. Any sensible rounding is accepted.

In our words, from NZQA’s 2024 marking schedule.

Common mistake

How did it go?

Got it moves it further out. Review brings it back tomorrow.