Question 1(c) Merit

A chain-rule root, a product-rule tangent, a decreasing-interval question mixing a square with a log, stationary points of an exponential product, and a tangent at a point of inflection.

For the function y = 3(2x − 7)² + 60 ln x + 12, where x > 0, find the range of values of x for which the function is decreasing.

The 2024 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 A function is decreasing where its derivative is what?

   Negative. So find dy/dx and solve dy/dx < 0.

2. 2 Differentiate. The squared bracket needs the chain rule.

   dy/dx = 6(2x − 7) × 2 + 60/x = 24x − 84 + 60/x

3. 3 Set up the inequality and clear the fraction. Why is it safe to multiply by x?

   24x − 84 + 60/x < 0. Because x > 0, multiplying by x keeps the inequality the same way round:
   24x² − 84x + 60 < 0, so 2x² − 7x + 5 < 0.

4. 4 Factorise. Where is this quadratic negative?

   (2x − 5)(x − 1) < 0. The critical values are x = 1 and x = 2.5, and a parabola that opens upwards is negative between its roots.

What the marker wanted

• Achieved The derivative set less than zero, or set equal to zero and solved for the critical values 1 and 2.5.
• Merit The range 1 < x < 2.5, worked from the derivative.

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

How did it go?

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