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 = 1in 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

7.

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

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