Question 1(c) Merit

Five independent parts, no shared scenario: a standard integral, a differential equation with a boundary condition, a pair of definite integrals that pin down two constants, a trig area, and a differential equation you have to rearrange before you can separate it.

It is given that int₁² (a − 6k/x²)dx = 3 and int₁^k 6x dx = a, where a and k are constants. Determine the possible value(s) of k.

The 2025 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 Start with the first integral. Write 6k/x² as a power of x; k and a are just numbers.

   int₁² (a − 6kx⁻²)dx = Bigl[ax + (6k/x)Bigr]₁²

2. 2 Put in the limits and simplify.

   (2a + 3k) − (a + 6k) = a − 3k
   so a − 3k = 3, that is a = 3 + 3k.

3. 3 Now the second integral. Its upper limit is k itself.

   int₁^k 6x dx = [3x²]₁^k = 3k² − 3, so a = 3k² − 3.

4. 4 You now have two expressions for a. Set them equal and solve.

   3k² − 3 = 3 + 3k
   3k² − 3k − 6 = 0
   k² − k − 2 = 0
   (k − 2)(k + 1) = 0

5. 5 Does either value break anything in the question?

   Neither does. k = 2 gives a = 9, and k = −1 gives a = 0; each pair satisfies both equations. The question asks for every possible value, so keep both.

What the marker wanted

• Achieved Both integrals done correctly.
• Merit Both values of k, k = 2 and k = −1, from correct integration.

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

How did it go?

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