AS 91578 · 2024 paper · sec, graph, cone, triangle area

Question 3(e) Merit to Excellence

A product with sec, three readings off a drawn graph (continuity, stationary points, a limit), stationary points of a rational function, a related-rates cone, and an optimisation that has to be proved to a given exact value.

The diagram shows part of the graph of f(x) = e^−x², where x ≥ 0. The point P lies on the curve and the point Q lies on the x-axis so that OP = PQ, where O is the origin.

Prove that the largest possible area of the triangle OPQ is 1/√2e. You do not need to show that the area you have found is a maximum.

f(x)	e^−x², x ≥ 0
OP	PQ

The 2024 paper, page 14 ↗ Marking schedule, page 6 ↗

Work it out

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

1 Let P be (x, y) on the curve. If OP = PQ, where is Q?
The triangle is isosceles with P at the top, so P sits above the middle of OQ: Q = (2x, 0).
2 Write the triangle’s area as a function of x.
Base 2x, height y = e^−x²:
A = 1/2 × 2x × e^−x² = xe^−x²
3 Differentiate. The exponential needs the chain rule.
dA/dx = e^−x² + x · (−2x)e^−x² = e^−x²(1 − 2x²)
4 Set it to zero. Which root do you keep?
e^−x² is never zero, so 1 − 2x² = 0 and x = ±1/√2. Keep x = 1/√2, because x ≥ 0.
5 Find the area, exactly.
A = (1/√2)e^−1/2 = 1/√2 × 1/√e = 1/√2e

Answer The largest area is 1/√2e (about 0.429), when x = 1/√2.

What the marker wanted

Achieved A correct dA/dx.
Merit x = 1/√2 found by calculus, with the negative root rejected.
Excellence The exact area 1/√2e proved (E8). A proof with one minor error is E7.

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

Common mistake

Getting x ≈ 0.707 from a graphics calculator and trying to reach an exact value from the decimal. NZQA’s 2024 report says leaning on the calculator put this exact proof beyond all but the strongest students. Keep 1/√2 exact to the end.

How did it go?

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