Question 2(e) Merit to Excellence

A parametric derivative, a velocity from a log displacement, a second-derivative verification, an inflection point on (ln x)/x, and a turning point that survives an unknown constant.

The graph of y = xe^3x/(2x + k), where k is a non-zero constant, has a single turning point at Q. Find the x-coordinate of the point Q.

The 2024 paper, page 8 ↗ Marking schedule, page 4 ↗

Work it out

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

1. 1 This is a product on top of a quotient. Set up the quotient rule: what is the top’s derivative?

   u = xe^3x, so by the product rule u' = e^3x + 3xe^3x = e^3x(1 + 3x).
   v = 2x + k, v' = 2.

2. 2 Write dy/dx and tidy the numerator.

   dy/dx = (e^3x(1 + 3x)(2x + k) − 2xe^3x)/(2x + k)² = (e^3x(6x² + 3kx + k))/(2x + k)²

3. 3 Where is dy/dx = 0?

   e^3x is never zero, so 6x² + 3kx + k = 0.

4. 4 There is only one turning point. What does that tell you about this quadratic?

   It has exactly one root, so its discriminant is zero:
   (3k)² − 4 × 6 × k = 9k² − 24k = 3k(3k − 8) = 0
   k is not zero, so k = 8/3.

5. 5 Put k back in and solve for x.

   6x² + 8x + 8/3 = 0, so 9x² + 12x + 4 = (3x + 2)² = 0 and x = −2/3.

What the marker wanted

• Achieved A correct dy/dx.
• Merit k = 8/3, found by calculus.
• Excellence x = −2/3, with the calculus set out in full (E8). A correct solution with one minor error is E7.

The paper calls Q a turning point, and the schedule marks exactly this method. Strictly, a repeated root means the gradient touches zero without changing sign, so the curve flattens and carries on rising: a stationary point of inflection. Follow the method; the answer is −2/3 either way.

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.