Question 1(e) Merit to Excellence

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.

A curve is defined by y = (2x² − 1 − 2x ln x)/x, where x > 0. The curve has a point of inflection at the point P. Find the equation of the tangent to the curve at the point P.

The 2024 paper, page 4 ↗ Marking schedule, page 2 ↗

Work it out

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

1. 1 Before reaching for the quotient rule, can you split the fraction into simpler terms?

   Divide each term by x: y = 2x − 1/x − 2 ln x = 2x − x⁻¹ − 2 ln x.

2. 2 Differentiate once, then again.

   dy/dx = 2 + x⁻² − 2/x
   (d²y)/dx² = −2x⁻³ + 2x⁻²

3. 3 Where is the point of inflection?

   −2/x³ + 2/x² = 0. Multiplying by x³ gives 2x − 2 = 0, so x = 1.
   (Writing it as 2x³ = 2x² also gives x = 0, which is ruled out by x > 0.)

4. 4 Find P, and the gradient there.

   y = 2 − 1 − 2 ln 1 = 1, so P is (1, 1).
   dy/dx = 2 + 1 − 2 = 1

5. 5 Write the tangent.

   y − 1 = 1(x − 1), so y = x.

What the marker wanted

• Achieved A correct dy/dx.
• Merit x = 1, from correct first and second derivatives.
• Excellence The tangent y = x (E8). A tangent that follows correctly from a wrong P, or a correct solution with one minor error, is E7.

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

How did it go?

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