2014 A-level H2 Mathematics (9740) Paper 2 Question 1 Suggested Solutions

JC Mathematics

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

(i)
\frac{dx}{dt}=6t

\frac{dy}{dt}=6

\frac{dy}{dx}={\frac{dy}{dt}}\times {\frac{dt}{dx}}

\frac{dy}{dx}=\frac{1}{t}=0.4

t=\frac{5}{2}

(ii)
At P, t=p, \frac{dy}{dx}=\frac{1}{p}
Equation of tangent: y-6p=\frac{1}{p}(x-3p^{2})
At D, x=0,  y=-3p+6p=3p
Therefore, D(0,3p).
Midpoint of PD = (\frac{0+3p^{2}}{2}, \frac{6p+3p}{2})=(\frac{3p^{2}}{2}, \frac{9p}{2})

As p varies, x=\frac{3}{2} p^{2}, y= \frac{9}{2}p
We have that p=\frac{2y}{9}
Then, x=\frac{3}{2}(\frac{2y}{9})^{2}
Thus, 27x=2y^{2}

Personal Comments:
This questions test students on their understanding of parametric equations and finding gradient. Find the cartesian equation can be slightly tricky if a student hasn’t been exposed to such questions before.

2014 A-level H2 Mathematics (9740) Paper 1 Question 11 Suggested Solutions

JC Mathematics

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

(i)
\mathrm{Vol}, V = \frac{2}{3} \pi r^{3} + \frac{1}{3} \pi r^{2}h

Using Pythagoras’ Theorem, we have h^{2}=4^{2}-r^{2}

V = \frac{2}{3} \pi r^{3} + \frac{1}{3} \pi r^{2} \sqrt{4^{2}-r^{2}}

\frac{dV}{dr} = 2 \pi r^{2} + \frac{1}{3} \pi [2r\sqrt{4^{2}-r^{2}}+(r^{2})(\frac{-r}{\sqrt{4^{2}-r^{2}}})]

When r=r_{1}, \frac{dV}{dr} = 0,

2 \pi r^{2} + \frac{1}{3} \pi [2r\sqrt{4^{2}-r^{2}}-(\frac{r^{3}}{\sqrt{4^{2}-r^{2}}})] = 0

2 \pi r^{2}\sqrt{4^{2}-r^{2}} + \frac{1}{3} \pi [2r(4^{2}-r^{2})-r^{3}] = 0

2 r^{2}\sqrt{16-r^{2}} + \frac{1}{3} [32r-2r^{3}-r^{3}] = 0

2 r^{2}\sqrt{16-r^{2}} + \frac{1}{3} (32r-3r^{3}) = 0

2 r^{2}\sqrt{16-r^{2}} = \frac{1}{3} (3r^{3}-32r)

6r\sqrt{16-r^{2}} = (3r^{2}-32)

[6r\sqrt{16-r^{2}}]^{2} = (3r^{2}-32)^{2}

36r^{2}(16-r^{2}) = 9r^{4}-192r^{2}+1024

45r^{4}-768r^{2}+1024=0

(ii)
Using GC, r=1.207 \text{~or~} r=3.951

(iii)
When r=1.207, using GC, \frac{dV}{dr} = 18.3 \neq 0

Thus, r=1.207 does not give a stationary value of V.

When r=3.951=r_1, ~ h=0.625

(iv) Note than r \textless 4,

Graph for 11(iv)
Graph for 11(iv)

Personal Comments:
It is rather disappointing to see some students unable to reach this question. Some panicked and differentiated wrongly, then forcefully copied what they were required to show. Students should also still continue with subsequent parts even if they cannot show (i). The interesting part of this question is that the last few parts require students to know how to use the GC effectively, in particularly setting the correct window for the graph in (iv) while (ii) and (iii) is entirely GC work as I’ve shown numerous times in classes.

2014 A-level H2 Mathematics (9740) Paper 1 Question 2 Suggested Solutions

JC Mathematics

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

x^{2}y+xy^{2}+54=0
differentiating with respect to x,
2xy+x^{2}(\frac{dy}{dx})+y^{2}+2xy(\frac{dy}{dx})=0
When \frac{dy}{dx}=-1
-x^{2}+2xy+y^{2}-2xy=0
y^{2}=x^{2}
y=\pm{x}

When y=x,
y^{3}+y^{3}+54=0
y^{3}=-27
y=-3
x=-3

When y=-x
-y^{3}+y^{3}+54=0
There is no solutions.

Therefore, required coordinates are (-3,-3)

Personal Comments:
This question tests students on their understanding of basic differentiations. First pitfall was when students forget to introduce \pm{ }. And to SHOW there is only one such point baffled many as they didn’t know how to go about. But the most direct way is to solve for them and find there is only one. Lastly, read the question carefully and you will see they require the answers to be COORDINATES.

Let’s try some calculus questions

JC Mathematics

Many students often overlook that the coefficient of x in the integration or differentiation formulas in MF15 is 1!!! When it is not 1, many things changes. I’ll let the examples do the talking. 🙂

Differentiation (recall chain rule)

\frac {d}{dx}(sin^{-1}(3x^2)) = \frac {1}{\sqrt{1-(3x^2)^2}}(6x)

Integration

\int \frac {1}{4+9x^2} dx = \frac {1}{2} {tan^{-1}(\frac {3x}{2})}\times\frac{1}{3}

For my careless students, I usually recommend they make the case of the coefficient of x be ONE instead. So \int \frac {1}{4+9x^2} dx = \frac{1}{9}\int \frac {1}{\frac{4}{9}+x^2} dx and after applying formula gives, (\frac{1}{9})(\frac{1}{\frac{2}{3}}){tan^{-1}(\frac {x}{\frac{2}{3}})} which will give the same answers after simplifications.

Practice

You may practice a few of the following questions!

\int\frac {1}{4-9x^2}dx

\int\frac {1}{9x^{2}-4}dx

\int\frac {1}{\sqrt{4-9x^2}}dx

\int\frac {1}{2x^{2}-2x-10}dx

Let me know if you have problems!