2011 A-level H2 Mathematics (9740) Paper 1 Question 5 Suggested Solutions

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

(i)

(ii)
$0 \le x \le 2$

(iii)
$\int_{-1}^1 f(|x|) dx = \int_1^a |f(x)| dx$

$2 \int_0^1 2-x dx = \int_1^2 2-x dx + \int_2^a -(2-x) dx$

$2[2x - \frac{x^2}{2}]\bigl|_0^1 = [2x - \frac{x^2}{2}]\bigl|_1^2 + [\frac{x^2}{2} - 2x]\bigl|_2^a$

$3 = \frac{1}{2} + \frac{a^2}{2} - 2a + 2$

$a = 2 \pm \sqrt{5}$

$\therefore a = 2 + \sqrt{5}$ since $a > 2$

There are other alternative methods to solving the integration like using the area of trapezoid.

2012 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.

(i)
Since $\frac{dx}{d\theta} = 1 - cos \theta$ and $\frac{dy}{d\theta} = sin\theta$

$\Rightarrow \frac{dy}{dx} = \frac{sin \theta}{1 - cos \theta} = \frac{2sin \frac{\theta}{2} cos \frac{\theta}{2}}{2 sin^2 \frac{\theta}{2}} = cot \frac{\theta}{2}$

When $\theta = \pi$, $tan\frac{\theta}{2} \rightarrow \infty$, $\text{cot}\theta \rightarrow 0$

$\therefore$ gradient of C $= 0$ when $\theta = \pi$.

As $\theta \rightarrow 0, \mathrm{~or~} \theta \rightarrow 2\pi, \frac{dy}{dx} \rightarrow \infty$

Thus the tangents are parallel to y-axis.

(ii)

(iii)
Area $= \int_0^{2\pi} (1-cos\theta)^2 d\theta$

$= \int_0^{2\pi} 1 - 2cos\theta + cos^2 \theta d\theta$

$= \int_0^{2\pi} 1 - 2cos\theta + \frac{1+cos2\theta}{2} d\theta$

$= \frac{3}{2}\theta - 2 sin\theta + \frac{sin2\theta}{4} \biggl|_0^{2\pi}$

$= 3 \pi$

(iv)
Normal: $y - (1-cos p) = - tan\frac{p}{2} [x- (p-sin p)]$

When $y=0, - 1+cos p = - tan\frac{p}{2} (x- p + sin p)]$

$sin p = x - p + sin p$

$x = p$

Thus, normal cross x-axis at $(p, o)$

For (i), some students have difficulties proving. They should note that they found the expression in terms of $\theta$ but they are trying to prove an expression in terms of $\frac{\theta}{2}$. This should prompt them to apply the double angle formulas that are found in the MF15. Next, students should learn how to describe the tangents and keep it short and sweet. Some describe tangents become straighter, which is awkward since the tangents are lines. Students can check their definite integral with the graphing calculator.

2012 A-level H2 Mathematics (9740) Paper 1 Question 2 Suggested Solutions

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

(i)
$\int \frac{x^{3}}{1+x^{4}}dx = \frac{1}{4} \mathrm{ln} (1 + x^{4}) + C$

(ii)
Given $u = x^{2}, dx = \frac{1}{2x} du$

$\int \frac{x}{1+x^{4}} dx$

$= \int \frac{x}{1 +u^{2}} \frac{1}{2x} du$

$= \frac{1}{2} \int \frac{1}{1+u^{2}} du$

$= \frac{1}{2} tan^{-1}u + C$

$= \frac{1}{2} tan^{-1} (x^{2}) + C$

(iii)
Using a Graphing Calculator, $\int_0^1 (\frac{x}{1+x^{4}})^{2} dx = 0.186$

Do remember to change the $dx$ when doing substitution. Students are reminded that the integration formula can be found in MF15. The last part can be solved using GC since there is no mention of exact value.

2014 A-level H1 Mathematics (8864) Question 5 Suggested Solutions

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

5.
(i)
$\frac{dy}{dx} = 3x^{2} + 2kx + 7$

When $x = 1, ~ \frac{dy}{dx} = 0$

$\Rightarrow 3 + 2k + 7 = 0$

$k = -5$

$\Rightarrow 1^{3} - 5 (1)^{2} + 7 + c = 2$

$\therefore, c = -1$

(ii)
$\frac{dy}{dx} = 3x^{2} - 10 x + 7 = 0$

$x= \frac{7}{3} \mathrm{~or~} x = 1$

When $x = \frac{7}{3}, y = \frac{22}{27}$

$\therefore, (\frac{7}{3}, \frac{22}{27})$

(iii)

(iv)
Area $= \int_1^2 x^{3} - 5 x^{2} + 7x - 1 dx$

$= \frac{x^{4}}{4} - \frac{5x^{3}}{3} + \frac{7x^{2}}{2} - x \biggl|_1^2$

$= \frac{19}{12}$

Students must be careful to leave answers in fractions and not just decimals. They should also check their answers with the Graphing Calculator.

2014 A-level H1 Mathematics (8864) Question 1 Suggested Solutions

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

$\int_{1}^{6} \frac{1}{\sqrt{1+4x}} dx$

$= \frac{1}{4} \int_{1}^{6} {4}{\sqrt{1+4x}}^{-0.5} dx$

$= \frac{1}{4} \frac{(1+4x)^{0.5}}{0.5}$

$= \frac{1}{2}[5-\sqrt{5}]$

Students should check the answers with the Graphing Calculator.

2013 A-level H2 Mathematics (9740) Paper 1 Question 5 Suggested Solutions

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

(i)

(ii)
Since $x=asin\theta$, we have $dx=acos\theta ~d\theta$
When $x= \frac{\sqrt{3}a}{2}, \theta = \frac{\pi}{3}$
When $x= \frac{a}{2}, \theta = \frac{\pi}{6}$

$\int_{\pi / 6}^{\pi / 3} \sqrt{1-sin^{2}\theta}~(acos\theta) ~d\theta$

$= a \int_{\pi / 6}^{\pi / 3} cos^{2} \theta ~d \theta$

$= a \int_{\pi / 6}^{\pi / 3} \frac{1+cos2\theta}{2} ~d \theta$

$= \frac{a}{2}[\theta + \frac{sin 2 \theta}{2}]\biggl|_{\pi / 6}^{\pi / 3}$

$= \frac{a \pi}{12}$

The function is a periodic curve and some students do have problem reading such functions. They should be alert of the domain that they are required to draw too. Some students were unsure how the hint that $f(x+3a)=f(x)$ should help. The parametric integration should be quite simple to handle as students just need to be cautious and introduce the double angle formula from MF15 effectively.