2015 A-level H2 Mathematics (9740) Paper 1 Question 11 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

\frac{dy}{d\theta} = 6\mathrm{sin}\theta \mathrm{cos^2}\theta - 3 \mathrm{sin^3} \theta

\frac{dx}{d\theta} = 3 \mathrm{sin^2}\theta \mathrm{cos}\theta

\frac{dy}{dx} = \frac{6\mathrm{sin}\theta \mathrm{cos^2}\theta3 - \mathrm{sin^3} \theta}{3 \mathrm{sin^2}\theta \mathrm{cos}\theta}

\frac{dy}{dx} = \frac{3\mathrm{cos^2}\theta - \mathrm{sin^2} \theta}{ \mathrm{sin^2}\theta \mathrm{cos}\theta}

\frac{dy}{dx} = 2 \frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} - \frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}

\frac{dy}{dx} = 2 \mathrm{cot} \theta - \mathrm{tan} \theta

Let \frac{dy}{dx} = 0,  \frac{2}{\mathrm{tan}\theta} - \mathrm{tan}\theta = 0

\mathrm{tan^2} \theta = 2

\mathrm{tan} \theta = \sqrt{2} or - \sqrt{2} (reject since \theta \in [0, \frac{\pi}{2}]

\therefore, k = 2

When \mathrm{tan} \theta = \sqrt{2}, using trigonometry identity, we find

\mathrm{cos} \theta = \frac{\sqrt{3}}{3} ~and~ \mathrm{sin} \theta = \frac{\sqrt{6}}{3}

From \frac{dy}{dx}, \frac{d^2y}{dx^2} = -2\mathrm{cosec^2} \theta \frac{d\theta}{dx} - \mathrm{sec^2} \theta \frac{d\theta}{dx}

For \theta \in [0, \frac{\pi}{2}], \frac{d^2y}{dx^2} \textless 0 since \mathrm{sin} \theta > 0 and \mathrm{cos} \theta >0 for principal values of \theta

We have that when \mathrm{tan} \theta = \sqrt{2}, its is a maximum point.

x = \mathrm{sin^3} \theta = \frac{2\sqrt{6}}{9}

y = 3 \mathrm{sin^2}\theta \mathrm{cos} \theta = \frac{2 \sqrt{3}}{3}

\therefore, \text{Required}~\text{coordinates}~ = (\frac{2\sqrt{6}}{9}, \frac{2\sqrt{3}}{3})


\int_0^1 y dx

= \int_0^{\frac{\pi}{2}} 3 \mathrm{sin^2}\theta \mathrm{cos} \theta 3 \mathrm{sin^2}\theta \mathrm{cos}\theta d\theta

= \int_0^{\frac{\pi}{2}} 9 \mathrm{sin^4}\theta \mathrm{cos^2} \theta d\theta

Using GC, Area = 0.88357 \approx 0.884 \mathrm{units^2}

At P, y = ax
\Rightarrow 3 \mathrm{sin^2}\theta \mathrm{cos} \theta  = a \mathrm{sin^3} \theta

\mathrm{tan} \theta = \frac{3}{a}

At maximum point, \mathrm{tan} \theta = \sqrt{2}

\Rightarrow \sqrt{2} = \frac{3}{a}

\therefore, a = \frac{3\sqrt{2}}{2}

Back to 2015 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

KS Comments:

To find the exact coordinates, students need to be able to either identify the correct trigonometry identify, or draw the right angled triangle to find \mathrm{sin} \theta and \mathrm{cos} \theta carefully. As for the show part for the parametric integration, students need to start from the definition of area then proceed carefully, not forgetting to change the limits.

Showing 12 comments
  • PleasegivemeAforMath

    Those who got A for 2013 paper around how much did they get for Paper 2 and paper 1 separately? 🙂 And also for the last part if i leave answer as tan tether = 3/a how many marks I get?

    • KS Teng

      From what I understand, mostly >75. I had one student who left 30 marks blanks in paper 1 due to time constraint, she was very confident of paper 2 getting > 95 though. And she still got an A. But she was really confident, like she new how she till manage. In any case, the past year distinctions should not be much of an indication, since it largely still depends on your cohort and how you stack against the rest. Like I mentioned here, Bell curve distributes the score of students. It will help the best student for sure, But the weaker students are subjected to how well the best students did. I probably just lose the 1 mark.

      • aaa

        ooh so >75 for P1 and >75 for P2?

      • aaa

        how about 2014?

        • KS Teng

          Its quite hard to suggest a sufficient mark. Necessary mark, 70/100? Safe and guaranteed mark 170/100. The definition need to be really precise here. :/

          • aaa

            thanks for answering 🙂 hmmm what was the lowest mark you know that got A in 2014?

  • PleasegivemeAforMath

    for 11ii) if I did everything including proving second derivative except find the coordinates how many marks do I get? Thank u!!

    • KS Teng

      I think if you left it in trigonometry form, probably lose 2 marks maximum. :/

  • Sad

    I most probably lose around 40 marks or more for paper 1 due to careless and panic. Is there still hope for a B? What is the mark range for B like?

    • KS Teng

      most of the mark range you observe in the JC, >60 is necessary mark for B. But the curves shifts rightwards. I can’t really advise a sufficient mark. And definitely can get B la, There is still a paper 2.. Dont give up 🙂

pingbacks / trackbacks

Leave a Comment

twelve − eight =

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt

Start typing and press Enter to search