NCEA Calculus 2024 ONE(e - Solution & Explanation

Question

A curve is defined by the equation

y = \dfrac{2x^2 - 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.

*You must use calculus and show any derivatives that you need to find when solving this problem.*

Official Answer

\dfrac{dy}{dx}=2+x^{-2}-\dfrac{2}{x}

\dfrac{d^2y}{dx^2}=-2x^{-3}+2x^{-2}

For an inflection, solve

\dfrac{d^2y}{dx^2}=-2x^{-3}+2x^{-2}=0

\dfrac{2}{x^2}=\dfrac{2}{x^3}

2x^3=2x^2

2x^3-2x^2=0

2x^2(x-1)=0

Either

x=0

ignore as not valid
Or

x=1

x=1

gives

y=1

, i.e.

P=(1,1)

x=1

gives

\dfrac{dy}{dx}=1

Then equation of tangent is:

y-1=1(x-1)

y=x

Grading Criteria

Achievement (u)

Correct expression for $\dfrac{dy}{dx}$ .

Merit (r)

Finds $x=1$ , with correct $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ .

Excellence T1

E7 / T1 Consistent equation of tangent, from incorrect P. OR Correct solution but with one minor error.

Excellence T2

E8 / T2 Correct equation of tangent found.

Video Explanation

2024 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 12:41)

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2024 ONE(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2