Question 1(d) Merit

A sec–tan integral, an area under a cubic-in-a-bracket, a kinematics chain that starts 360 m from the reference point, a trig differential equation, and an area between two trig curves.

Consider the differential equation dy/dx = 24 cos(3x) sin(x). Given that y = 6 when x = π/3, find the value(s) of y when x = π/2.

The 2024 paper, page 4 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 This is cos(3x), the cosine of 3x, times sin x: two different angles. Which identity turns the product into a sum?

   2 sin A cos B = sin(A + B) + sin(A − B). With A = x and B = 3x:
   2 sin x cos 3x = sin 4x + sin(−2x) = sin 4x − sin 2x
   so 24 cos 3x sin x = 12(sin 4x − sin 2x).

2. 2 Integrate each sine, dividing by the number in front of x.

   y = 12(−(1/4) cos 4x + (1/2) cos 2x) + c = −3 cos 4x + 6 cos 2x + c

3. 3 Use y = 6 when x = π/3. What are cos4π/3 and cos2π/3?

   Both are −1/2.
   6 = −3(−1/2) + 6(−1/2) + c = 1.5 − 3 + c, so c = 7.5.

4. 4 Now put x = π/2 in.

   cos 2π = 1 and cosπ = −1:
   y = −3(1) + 6(−1) + 7.5 = −1.5

What the marker wanted

• Achieved The integration correct.
• Merit The value of y, −1.5, from correct integration.

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

How did it go?

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