### Thoughts on H2 Mathematics (9758) 2017 Paper 1

This is a new syllabus and this is the first time it will be tested. Personally, I don’t think it will be easy and students should not underestimate this upcoming A’levels. And I’m referring to the A’levels, on the whole. We saw how the Science Paper 4 were… unexpected.

The new H2 Mathematics (9758) syllabus has several topics removed, and these were mostly topics that were “drill-able”, aside from complex numbers. The new syllabus added in mainly, new integration forms, focus on parametric Equations with Cartesian equations, and of course, Discrete R.V. But let us leave the statistics out.

Students should familiarise themselves with the trigonometry Formulae in MF26. There are several topics that can be linked up with trigonometry, makes me wonder why it isn’t a chapter by itself. Complex numbers has a trigonometry form too, so make sure students know how to manipulate it, given the trigonometry Formulae.

Next, students should understand the use of Maclaurin’s. What does it mean for $x$ to be small, and the implications when they say $h$ is small compared to $R$… And also finding the general term of a Maclaurin’s Expansion.

It won’t hurt to review how to find the Area using Shoe-lace method. And not forgetting our Sine Rule and Cosine Rule.

Do know how to prove a one-one function… Non-graphically. (i.e. not using the Horizontal Line Test)

Do know that the oblique asymptote of $f(x)$ becomes $y=0$ when we do the $y = \frac{1}{f(x)}$ transformation too.

Lastly, students must READ really carefully and discern every information. Having marked many scripts, many students do not read carefully and lose marks here and there. And they do add up… Be alert and read, take note of the forms that they want. Here are 10 little things to take note when you read the question.

1. Cartesian/ Polar/ Exponential for complex
2. Scalar/ Parametric/ Cartesian for vectors
3. Set/ range/ interval of values
4. Algebraically => show all the workings without a GC.. usually discriminant, completing the square or maybe some differentiation will be involved.
5. Without using a calculator => show your workings and check with a GC (secretly)
6. Decimal places, etc…
7. Rounding off when you’re dealing with an inequality
8. Units used in the questions, (ten thousands, etc)
9. Rate of change; leaking means the rate is negative…
10. All answers should be in 3 SF UNLESS OTHERWISE STATED. Degrees to 1 DP. RADIANS to 3 SF.

Have fun and all the best!

### Random Questions from 2017 Prelims #3

Here is another question that is from CJC H2 Mathematics 9758 Prelim Paper 1. Its a question on differentiation. I think it is simple enough and tests student on their thinking comprehension skills. This is question 6.

A straight line passes through the point with coordinates (4, 3) cuts the positive x-axis at point P and positive y-axis at point Q. It is given that $\angle PQO = \theta$, where $0 < \theta < \frac{\pi}{2}$ and O is the origin.

(i) Show that equation of line PQ is given by $y = (4-x) \text{cot} \theta +3$.

(ii) By finding an expression for $OP + OQ$, show that as $\theta$ varies, the stationary value of $OP + OQ$ is $a + b \sqrt{3}$, where $a$ and $b$ are constants to be determined.

### Differentiation Question #1

Given that $y = \frac{8}{x^3} - \frac{6}{x^2} + \frac{5}{2x}$, find the approximate percentage change in $y$ when $x$ increases from 2 by 2%.

### Random Sec 4 Differentiations

B6

$y = 3e^x + \frac{4}{e^x}$

$\frac{dy}{dx} = 3e^x - \frac{4}{e^x}$

$\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}$

let $\frac{dy}{dx} = 0$

$3e^x - \frac{4}{e^x} = 0$

$3e^{2x} = 4$

$2x = \mathrm{ln} \frac{4}{3}$

$x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$

Sub $x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$ to $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} > 0$ Thus, it is a min point.

C7

$y = \mathrm{ln} \frac{5-4x}{3+2x}$

$y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)$

$\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}$

let $\frac{dy}{dx} = 0$

$\frac{-4}{5-4x} - \frac{2}{3+2x} = 0$

$\frac{-4}{5-4x} = \frac{2}{3+2x}$

$-4(3+2x) = 2(5-4x)$

$-12 - 8x = 10 - 8x$

$-12 = 10$ (NA).

There are no stationary points for this curve.

C8

$x = \frac{1}{3}e^{y(2x+5)}$

$\mathrm{ln}(3x) = y(2x+5)$

$\frac{\mathrm{ln}(3x)}{2x+5} = y$

$y = \frac{\mathrm{ln}(3x)}{2x+5}$

$\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}$

Let $x = e^2$

$\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}$

Evaluate with a calculator…

### A little reminder to students doing Calculus now

When $\frac{dy}{dx} = 0$, it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test or the second derivative.

The first derivative test:

Students should write the actual values of $\alpha^-, \alpha, \alpha^+$ and $\frac{dy}{dx}$ in the table.

We use this under these two situations:
1. $\frac{d^2y}{dx^2}$ is difficult to solve for, that is, $\frac{dy}{dx}$ is tough to be differentiated
2. $\frac{d^2y}{dx^2} = 0$

The second derivative test:

Other things students should take note is concavity and drawing of the derivative graph.

### Random Questions from 2016 Prelims #4

VJC P1 Q6

A curve has equation $y^2=4x$ and a line $l$ has equation $2x-y+1=0$ as shown above.

$B(b, 2\sqrt{b})$ is a fixed point on C and A is an arbitrary point on $l$. State the geometrical relationship between the line segment AB and $l$ is the distance from B to A is the least.

Taking the coordinates of A as $(a, 2a+1)$, find an equation relating $a$ and $b$ for which AB is the least.

Deduce that when AB is the least, $(AB)^2 = m (2b - 2\sqrt{b} +1)^2$ where $m$ is a constant to be found. Hence or otherwise, find the coordinates of the point on C that is nearest on $l$.

### June Revision Exercise 3 Q7

(a)

$\int x(\mathrm{ln}x)^2 ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \int \frac{x^2}{2} 2(\mathrm{ln}x)\frac{1}{x} ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \int x \mathrm{ln}x ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \int \frac{x^2}{2}\frac{1}{x}~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \frac{x^2}{4} + C$

(b)

$\int \frac{1}{1-\mathrm{cos}2x} ~dx$

$= \int \frac{1}{2\mathrm{sin}^2x} ~dx$

$= \frac{2}{2} \int \mathrm{cosec}^2x~dx$

$= -\frac{1}{2} \mathrm{cot}x + C$

(c)
(i)

(ii)

$\int_{\mathrm{ln}2}^{\mathrm{ln}3} x ~dy$

$=\int_2^3 t^2 \frac{1}{t} ~dt$

$=\Big| \frac{t^2}{2} \Big|_2^3$

$= 2.5 \text{units}^2$

(iii)
$\pi \int_4^9 y^2 ~dx$

$=\pi \int_2^3 (\mathrm{ln}t)^2(2t) ~dt$

$=2\pi \int_2^3 t(\mathrm{ln}t)^2 ~dt$

$= 13.6$

Back to June Revision Exercise 3

### June Revision Exercise 3 Q6

(a)(i)

Consider $y=\sqrt{a^2 - x^2} \Rightarrow y^2+x^2=a^2$ is a circle with centre Origin and radius $a$.

Thus, required area is the area of a quadrant with radius $a$

$\text{Area} = \frac{1}{4} \pi a^2$

(ii)

$\int_0^a \frac{x^2}{\sqrt{a^2-x^2}} ~dx$

$= \int_0^2 x\frac{x}{\sqrt{a^2-x^2}} ~dx$

$= - x \sqrt{a^2-x^2} \Big|_0^a - \int_0^a -\sqrt{a^2-x^2} ~dx$

$= 0 + \frac{1}{4} \pi a^2$

$= \frac{1}{4} \pi a^2$

(b)

$u=3x+5 \Rightarrow \frac{du}{dx}=3$

$= \int_{-2}^{-\frac{5}{3}} (x-2)(3x+5)^3 ~dx$

$=\int_{-1}^0 (\frac{u-5}{3}-2)u^3 \frac{1}{4} ~du$

$=\frac{1}{9} \int_{-1}^0 u^4 - 11u^3 ~du$

$=\frac{1}{9} \Big| \frac{1}{5}u^5 - \frac{11}{4}u^4 \Big|_{-1}^0$

$= \frac{59}{180}$

Back to June Revision Exercise 3

### June Revision Exercise 4 Q7

(i)

$f(x) = \frac{4}{(1-3)\sqrt{x^2+4}}$

$= 4(1-3x)^{-1} (4)^{-\frac{1}{2}} (1 + \frac{x^2}{4})^{-\frac{1}{2}}$

$= 2[1 + 3x + (3x)^2 + (3x)^3 + \ldots][1 - \frac{1}{2} \frac{x^2}{4} + \ldots]$

$= 2[1 + 3x + (-\frac{1}{8}+9)x^2 + (-\frac{3}{8}+27)x^3 + \ldots]$

$= 2 + 6x + \frac{71}{4}x^2 + \frac{213}{4}x^2 + \ldots$

$|3x| \textless 1 \text{ and } |\frac{x^2}{4}| \textless 1$

$\Rightarrow \text{Range of values} = (-\frac{1}{3}, \frac{1}{3})$

$2 \mathrm{tan} y = f(x)$

$2 \mathrm{sec}^2 y \frac{dy}{dx} = f'(x)$

$2[2\mathrm{sec}^2 y \mathrm{tan}y (\frac{dy}{dx})^2 + \mathrm{sec}^2 y \frac{d^2y}{dx^2}] = f''(x)$

When $x = 0, f(0) = 2, f'(0)=6, f''(0)=\frac{71}{2}$

$2 \mathrm{tan}y=2 \Rightarrow y= \frac{\pi}{4}$

$2(2)\frac{dy}{dx}=6 \Rightarrow \frac{dy}{dx}=\frac{3}{2}$

$2[2(2)(1)(\frac{3}{2})^2 + 2 \frac{d^2y}{dx^2}]= \frac{71}{2} \Rightarrow = \frac{35}{8}$

Hence $y = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{8}\frac{x^2}{2!}+ \ldots = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{16}x^2 + \ldots$

### June Revision Exercise 3 Q5

(a)
$\int_0^{\mathrm{ln}2} \frac{e^3x}{e^x+2} ~dx$

$= \int_3^4 \frac{(u-2)^2}{u} ~du$

$= \int_3^4 (u-4+\frac{4}{u}) ~du$

$= \frac{1}{2}u^2 - 4u + 4 \mathrm{ln}u \Big|_3^4$

$= 4 \mathrm{ln}\frac{4}{3} - \frac{1}{2}$

(b)
$\int_0^2 |e^x - 2| dx = \int_e^{e^{\sqrt{k}}} \frac{1}{x \mathrm{ln}x} ~dx$

$- \int_0^{\mathrm{ln}2} (e^x - 2) dx + \int_{\mathrm{ln}2}^2 (e^x - 2) ~dx = \int_e^{e^{\sqrt{k}}} \frac{\frac{1}{x}}{\mathrm{ln}x} ~dx$

$-(e^x - 2x) \Big|_0^{\mathrm{ln}2} + (e^x - 2x) \Big|_{\mathrm{ln}2}^2 = \mathrm{ln}\sqrt{k}$

$\mathrm{ln}k = 2 (e^2 + 4 \mathrm{ln}2 -7)$

Back to June Revision Exercise 3