NCEA Calculus 2021 THREE(d - Solution & Explanation

Question

A curve has the equation

y = \dfrac{4x + k}{4x - k}

, where

k

is a constant and

x \ne \dfrac{k}{4}

.

The point P lies on the curve and has an

x

-coordinate of 3.

The gradient of the tangent to the curve at P is

\dfrac{-8}{27}

.

Find the possible value(s) of

k

.

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

Official Answer

\dfrac{dy}{dx}=\dfrac{(4x-k)4-(4x+k)4}{(4x-k)^2}

=\dfrac{16x-4k-16x-4k}{(4x-k)^2}

=\dfrac{-8k}{(4x-k)^2}

When

x=3

,

\dfrac{dy}{dx}=\dfrac{-8}{27}

\dfrac{-8k}{(12-k)^2}=\dfrac{-8}{27}

\dfrac{k}{(12-k)^2}=\dfrac{1}{27}

27k=144-24k+k^2

k^2-51k+144=0

k=48

or

k=3

Grading Criteria

-

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2021 NZQA Exam - Worked Answers by infinityplusone(starts at 68:55)

2021 THREE(d)