Question 3(c) 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.

Find the value of m, given that int₋₁^m ((4x + 5)/(2x + 3)) dx = 2m.

The 2024 paper, page 11 ↗ Marking schedule, page 5 ↗

Work it out

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

1. 1 The top has the same degree as the bottom. Divide first.

   4x + 5 = 2(2x + 3) − 1, so (4x + 5)/(2x + 3) = 2 − 1/(2x + 3).

2. 2 Integrate both pieces.

   int (2 − 1/(2x + 3))dx = 2x − (1/2) ln |2x + 3|

3. 3 Put in the limits −1 and m. What is 2x + 3 at x = −1?

   It is 1, and ln 1 = 0.
   Bigl(2m − (1/2) ln(2m + 3)Bigr) − Bigl(−2 − (1/2) ln 1Bigr) = 2m − (1/2) ln(2m + 3) + 2

4. 4 Set that equal to 2m and solve.

   2m − (1/2) ln(2m + 3) + 2 = 2m
   ln(2m + 3) = 4
   2m + 3 = e⁴, so m = (e⁴ − 3)/2 ≈ 25.80

What the marker wanted

• Achieved Reaching Bigl[2x − (1/2) ln |2x + 3|Bigr] with its limits. Dividing by long division or by inspection both count.
• Merit The value of m, 25.8, from correct integration. Any sensible rounding is accepted.

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

How did it go?

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