AY2021 June Revision J1 Revision Paper 1 Solutions

1. (a) $\vec{BA} = \begin{pmatrix}-1\\ 3\\ \lambda \end{pmatrix}, \quad \vec{BC} = \begin{pmatrix}2\\2\\1 \end{pmatrix}$

$\vec{BA} \cdot \vec{BC} = |BA| |BC |\cos \angle ABC$

$\begin{pmatrix}-1\\ 3\\ \lambda \end{pmatrix} \cdot \begin{pmatrix}2\\2\\1 \end{pmatrix}= \vert \begin{pmatrix}-1\\ 3\\ \lambda \end{pmatrix} \vert \vert \begin{pmatrix}2\\2\\1 \end{pmatrix} \vert (\frac{1}{3})$

$\Rightarrow \lambda = -0.75$

1. (b) $\textbf{p} \times \textbf{q} = 3 \textbf{p} \times \textbf{r}$

$\textbf{p} \times ( \textbf{q} - 3 \textbf{r}) = \textbf{0}$

$\Rightarrow \textbf{p}$ is parallel to $(\textbf{q} - 3 \textbf{r})$

$\Rightarrow \textbf{q} - 3 \textbf{r} = \lambda \textbf{p}$ , where $\lambda$ is a scalar.

$( \textbf{q} - 3 \textbf{r}) \cdot (\textbf{q} - 3 \textbf{r}) = (\lambda \textbf{p})\cdot (\lambda \textbf{p})$

$\vert \textbf{q} \vert ^2 - 6 \textbf{q} \cdot \textbf{r} + 9 \vert \textbf{r} \vert ^2 = \lambda ^2 \vert \textbf{p} \vert ^2$

$\lambda = \pm \sqrt{11}$

2. Using partial fractions, we find that $\frac{r}{(r+1)(r+2)(r+3)} = \frac{-0.5}{r+1} + \frac{2}{r+2} + \frac{-1.5}{r+3}$

$\sum_{r=1}^n \frac{r}{(r+1)(r+2)(r+3)} = \sum_{r=1}^n \bigg( \frac{-0.5}{r+1} + \frac{2}{r+2} + \frac{-1.5}{r+3} \bigg)$

Using method of differences (DIY), we have that

$\sum_{r=1}^n \frac{r}{(r+1)(r+2)(r+3)} = - \frac{1}{4} + \frac{2}{3} - \frac{1}{6} + \frac{2}{n+2} - \frac{3}{2(n+2)} - \frac{3}{2(n+3)} = \frac{1}{4} + \frac{1}{2(n+2)} - \frac{3}{2(n+3)}$

$\therefore \sum_{r=1}^n \frac{r}{(r+1)(r+2)(r+3)} = \frac{1}{4} + \frac{1}{2(n+2)} - \frac{3}{2(n+3)}$

$\sum_{r=7}^{\infty} \frac{r}{(r+1)(r+2)(r+3)} = \sum_{r=1}^{\infty} \frac{r}{(r+1)(r+2)(r+3)} - \sum_{r=1}^{6} \frac{r}{(r+1)(r+2)(r+3)}$

$\sum_{r=7}^{\infty} \frac{r}{(r+1)(r+2)(r+3)} = \frac{1}{4} - [ \frac{1}{4} + \frac{1}{2(6+2)} - \frac{3}{2(6+3)}]$

$\sum_{r=7}^{\infty} \frac{r}{(r+1)(r+2)(r+3)} = \frac{5}{48}$

3. $f(x)$ differs from $g(x)$ by less than 1 $\Rightarrow \vert f(x) - g(x) \vert < 1$

Plot the graph of $y = | f(x) - g(x) |= |e^x \sin x - (x^2 + 2x - 6)|$ and $y = 1$ in the GC, sketch it out. * You can key in modulus by pressing (alpha)(window)(1).

Observe that the $x$ -coordinates of the intersections points between the graphs are $3.83, -3.45, 2.60, 2.74$ .

$\therefore -3.83 < x < -3.45 \quad \text{or} \quad 2.60 < x < 2.74$ .

4. $\frac{x^2 -2x - 5}{x(1-x)} \le \frac{1}{x}$

$\frac{x^2 -2x - 5}{x(1-x)} - \frac{1}{x} \le 0$

$\frac{x^2 - x - 6}{x(1-x)} \le 0$

$\frac{(x-3)(x+2)}{x(1-x)} \le 0$

$\therefore \{ x: x \in \mathbb{R}, x \le -2 \quad \text{or} \quad 0 < x < 1 \quad \text{or} \quad x \ge 3 \}$

Replace $x$ with $e^x$ .

$e^x \le -2 \quad \text{or} \quad 0 < e^x < 1 \quad \text{or} \quad e^x \ge 3$

$\Rightarrow x < 0 \quad \text{or} \quad x \ge \ln 3$

$\therefore \{ x: x \in \mathbb{R}, x < 0 \quad \text{or} \quad x \ge \ln 3\}$

5. (i) $R = \sqrt{1+3} = 2 \quad \text{and} \quad \alpha = \tan^{-1} \sqrt{3} = \frac{\pi}{3}$ .

$\Rightarrow \cos x - \sqrt{3} \sin x = 2 \cos ( x + \frac{\pi}{3} )$

$R_f = [-2, 2]$

5. (ii) Let $\text{f}(x) = y$ , then $\text{f}^{-1}(x) = \cos^{-1} \frac{x}{2} - \frac{\pi}{3}, -2 \le x \le 2$ .

5. (iii) $\text{fg}(x) = x - 3, 1 \le x \le 3$

$\text{f}^{-1} \text{fg}(x) = \text{f}^{-1}(x-3)$

$\text{g}(x) = \cos ^{-1} \frac{x-3}{2} - \frac{\pi}{3}, 1 \le x \le 3$

6. (a) $P = \frac{150}{2}[2a + (150-1)d] = 150a + 11175d$

Sum of first 40 even-numbered terms $= \frac{40}{2} [2(a+d) + (40-1)(2d)] = 40a + 1600d$

We have $40a + 1600d = \frac{4}{15}(150a + 11175d) - 2760$

$\Rightarrow d = 2$

6. (b) $a = 15000, \quad r = 0.95$

$S_{\infty} = \frac{15000}{1-0.95} = 300000$

$(15000)(0.95)^{n-1} < 800$

$0.95^{n-1} < \frac{8}{150}$

$\ln (0.95^{n-1}) < \ln \frac{8}{150}$

$(n-1) \ln 0.95 < \ln \frac{8}{150}$

$n - 1 > \frac{\ln \frac{8}{150}}{\ln 0.95}$

$n > 58.15$

$\therefore n=59$ years of extraction

8. We substitute the given points $(2, 5), (4, -5)$ and $(3, 6)$ into $x^2 + y^2 + ax + by + c = 0$ . We have

$2a + 5b + c = -29$ —(1)

$4a - 5b + c = -41$ —(2)

$3a + 6b + c = -45$ —(3)

Using GC, $a = - \frac{43}{3}, b = - \frac{5}{3}, c = 8$

We have $x^2 + y^2 - \frac{43}{3} x - \frac{5}{3}y + 8 = 0 \Rightarrow (x - \frac{43}{6})^2 + ( y - \frac{5}{6})^2 = \frac{793}{18}$ .

We observe we have a circle with centre $(\frac{43}{6}, \frac{5}{6} )$ and radius $= \sqrt{\frac{793}{18}}$

9. (i) Equations of asymptotes: $x = -1, x =2, y=0$ .

9. (ii) Max $(5, \frac{1}{9})$ and Min $(1, 1)$ . Axial intercepts: $(0, 1.5)$ and $(3, 0)$ .

9. (iii) $\frac{x-3}{x+1} - 2k + kx = 0 \Rightarrow \frac{x-3}{(x-2)(x+1)} = -k$

(a) From the sketch, the line $y=-k$ does not cut the curve when $\frac{1}{9} < -k < 1 \Rightarrow -1 < k < - \frac{1}{9}$ .

(b) From the sketch, two real roots when $-k \in (-\infty, 0) \cup (0, \frac{1}{9} ) \cup (1, \infty)$

$\therefore k < -1 \quad \text{or} \quad -\frac{1}{9} < k < 0 \quad \text{or} \quad k > 0$ .

10. $w = 4 + 2i \quad \text{or} \quad w = 2-2i$

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

AY2021 June Revision J1 Revision Paper 1 Solutions

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