2024 TWO(e)

Excellence
Question
The graph of the function y=xe3x2x+ky = \dfrac{xe^{3x}}{2x + k}, where kk is a non-zero constant, has a single turning point at Q.

Find the xx-coordinate of the point Q.

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
dydx=e3x(6x2+3kx+k)(2x+k)2\dfrac{dy}{dx}=\dfrac{e^{3x}(6x^2+3kx+k)}{(2x+k)^2}

For turning points
dydx=0\dfrac{dy}{dx}=0
e3x(6x2+3kx+k)=0e^{3x}(6x^2+3kx+k)=0
Either e3x=0e^{3x}=0 No solutions
Or 6x2+3kx+k=06x^2+3kx+k=0 (a)
But, as only a single turning point, using b24ac=0b^2-4ac=0
(3k)24×6×k=0(3k)^2-4\times6\times k=0
9k224k=09k^2-24k=0
3k(3k8)=03k(3k-8)=0
Either k=0k=0 Ignore as not valid
Or 3k8=03k-8=0 i.e. k=83k=\dfrac{8}{3}

Substituting k=83k=\dfrac{8}{3} into equation (a) gives

6x2+8x+83=06x^2+8x+\dfrac{8}{3}=0
9x2+12x+4=09x^2+12x+4=0
(3x+2)(3x+2)=0(3x+2)(3x+2)=0
x=23x=-\dfrac{2}{3}
Grading Criteria

Achievement (u)

  • Correct dydx\dfrac{dy}{dx}.

Merit (r)

  • Finds the correct value of kk, with evidence of calculus methods.

Excellence T1

  • E7 / T1 Correct solution but with one minor error.

Excellence T2

  • E8 / T2
    Calculates xx-co-ordinate of Q, with clear and full calculus justification.
Video Explanation
2024 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 32:52)
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