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

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

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

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

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### June Revision Exercise 3 Q4

(i)
$A = 2\{ \frac{1}{n} (1) + \frac{1}{n}e^{\frac{1}{4n}} + \frac{1}{n}e^{\frac{2}{4n}} + \ldots + \frac{1}{n}e^{\frac{n-1}{4n}}\}$

$= \frac{2}{n} \{e^{\frac{0}{4n}} + e^{\frac{1}{4n}} + e^{\frac{2}{4n}} + \ldots + e^{\frac{n-1}{4n}} \}$

$= \frac{2}{n}\sum_{r=0}^{n-1} e^{\frac{r}{4n}}$

(ii)
Area of R $2 \int_0^1 e^{\frac{y}{4}} ~dy = 8(e^{\frac{1}{4}}-1)$

As $n \rightarrow \infty, A \rightarrow \text{Actual Area} = 8(e^{\frac{1}{4}}-1)$

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### June Revision Exercise 3 Q3

(a)(i)

Let $AB=y=\frac{40-x}{2}$

$h = \sqrt{y^2-(\frac{x}{2})^2}=2\sqrt{100-5x}$

$z = \frac{1}{2}xh=\frac{1}{2}x(2\sqrt{100-5x}=x\sqrt{100-5x}$

(a)(ii)

$\frac{dz}{dx}=\sqrt{100-5x} + x\frac{1}{2}(-5)(100-5x)^{-\frac{1}{2}}$

$\frac{dz}{dx}=\frac{100-\frac{15}{2}x}{\sqrt{100-5x}}$

$\frac{dz}{dx}=0 \Rightarrow x=\frac{40}{3}$

*Students are expected to prove that $x = \frac{40}{3}$ gives the maximum area.

(b)(i)
$x^3 + 2y^3 +3xy=k$

$3x^2+6y^2\frac{dy}{dx}+3x\frac{dy}{dx}+3y=0$

$\frac{dy}{dx}=\frac{-y-x^2}{2y^2+x} (b)(ii) Tangent parallel to$latex x\$-axis $\Rightarrow \frac{dy}{dx}=0$

$y=-x^2$

Sub $y=-x^2$ into $x^3 + 2y^3 +3xy=k$

$x^3 + 2(-x^2)^3 +3x(-x^2)=k$

$2x^6 + 2x^3 + k=0$

(b)(iii)
When the line $y=-1$ is a tangent to C,

$-1=-x^2$

$x =\pm x$

When $x = 1, k=-4$

When $x = -1, k=0$

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### June Revision Exercise 3 Q2

(i)

$\frac{dx}{dt}=-\frac{3a}{t^4}$

$\frac{dy}{dt}=-\frac{a}{t^2}$

$\frac{dy}{dx}= \frac{dy}{dt} \times \frac{dt}{dx} = \frac{t^2}{3}$

When $t= \frac{1}{2}, \frac{dy}{dx}= \frac{1}{12}$

Equation of tangent: $y-2a = \frac{1}{12}(x-8a) \Rightarrow y=\frac{1}{12}x + \frac{4}{3}a$

Equation of normal: $y-2a = -12(x-8a) \Rightarrow y=-12x+98a$

(ii)
$\frac{a}{t}=\frac{1}{12}\frac{a}{t^3}+\frac{4}{3}a$

$16t^3 - 12t^2 + 1 =0$

$t=\frac{1}{2} \text{ (rej) or } t=-\frac{1}{4}$

$\text{When } t=-\frac{1}{4}, x= -64a, y=-4a$

Hence, tangent cuts curve again at $(-64a, -4a)$

(iii)
At Q, $y=0 \Rightarrow x=-16a$

At R, $y=0 \Rightarrow x=\frac{49}{6}a$

$\text{Area } = \frac{1}{2}[\frac{49}{6}a-(-16a)](2a) = \frac{145}{6}a^2$ $\text{units}^2$

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### June Revision Exercise 3 Q1

(i)

$2x^2 + 8xy + 5y^2 = -3$

$4x + (8x \frac{dy}{dx} + 8y) + 10y \frac{dy}{dx}=0$

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

For tangents to be parallel to the $x$-axis, $\frac{dy}{dx}=0$

$\Rightarrow x=-2y$

Sub $x=-2y$ into $2x^2 + 8xy + 5y^2 = -3$

$y= \pm 1$

Thus, equations of tangents which are parallel to $x$-axis are $y= 1 \text{ or } y=-1$

(ii)

$\frac{dy}{dx}=\frac{dy}{dx}\bullet \frac{dx}{dt} = \frac{2[3+2(-1.15)]}{4(3)+5(-1.15)} = -0.488$ units per second.

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### June Revision Exercise 3

Please do check through the solutions on your own, especially for questions that we did not have chance to properly discuss during class. You may whatsapp me too if you have a burning question.

Note: You should not spend more than 180mins on the entire exercise.

### Integration related articles

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### 2010 A-level H2 Mathematics (9740) Paper 1 Question 6 Suggested Solutions

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

(i)
Using GC, $\beta = 0.347, \gamma = 1.532$

(ii)
Using GC, Area $= |\int_0.347^1.532 x^3 - 3x + 1 dx| = 0.781$

(iii)
Area $= \int_{-\sqrt{3}}^0 x^3 -3x + 1 dx - \sqrt{3} \times 1$

$= \frac{9}{4}$

(iv)
$\frac{dy}{dx}=2x^2 - 3 = 0$

$x = \pm 1$

$y = 3 \mathrm{~or~}-1$

Thus, $-1 \textless k \textless 3$