Question 3(b) Achieved

Opens with a numerical estimate from measured depths, then a log integral left in terms of a limit, a separable differential equation, an area that must be turned into a cubic, and a balance point found from a ratio of two integrals.

Find int₁^k (10/(2x − 1)) dx, giving your answer in terms of k, where k is a constant and k > 1.

The 2025 paper, page 10 ↗ Marking schedule, page 5 ↗

Work it out

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

1. 1 The top is a number and the bottom is linear. What does int (1/(ax + b)) dx give?

   A log, divided by the number in front of x: int (1/(2x − 1)) dx = (1/2) ln |2x − 1| + c.

2. 2 So what is the antiderivative of 10/(2x − 1)?

   10 × (1/2) ln |2x − 1| = 5 ln |2x − 1|

3. 3 Put in the limits. What is ln 1?

   5 ln(2k − 1) − 5 ln 1 = 5 ln(2k − 1), since ln 1 = 0. As k > 1, 2k − 1 > 1, so no modulus is needed.

What the marker wanted

• Achieved The correct answer, 5 ln(2k − 1).

In our words, from NZQA’s 2025 marking schedule.

How did it go?

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