Solutions to Set A

Hopefully, you guys have started on the Set A. You will find the following solutions useful. Click on the question. Please do attempt them during this December Holidays. 🙂

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Firstly, either consider $|a| \le b \Rightarrow -b \le a \le b$ or square both sides. Its easier to do the latter (so I will).

$\bigg| \frac{2x-3}{x+5} \bigg| \le \frac{2}{3}$

$\bigg[ \frac{2x-3}{x+5} \bigg]^2 \le \bigg[ \frac{2}{3} \bigg]^2$

$\frac{4x^2 -12x+9}{x^2 + 10x + 25} \le \frac{4}{9}$

$\frac{4x^2 -12x+9}{x^2 + 10x + 25} - \frac{4}{9} \le 0$

$\frac{9(4x^2-12x+9) - 4 (x^2 + 10x+25)}{x^2 + 10x+25} \le 0$

Since $(x+5)^2 \ge 0$ for all $x \in \mathbb{R}, x \neq -5$

$36x^2 - 108x + 81 -4x^2 -40x - 100 \le 0$

$32 x^2 -148x - 19 \le 0$

$(8x+1)(4x-19) \le 0$

$- \frac{1}{8} \le x \le \frac{19}{4}$

(i)

LHS

$= \frac{1}{(n-1)^2} - \frac{1}{n^2}$

$= \frac{n^2 - (n^2 -2n+1)}{n^2(n^2-1)}$

$= \frac{2n-1}{ n^2 (n-1)^2} =$ RHS.

(ii)

$\sum_{n=3}^N \frac{2n-1}{(2n)^2 (n-1)^2}$

$= \frac{1}{4} \sum_{n=3}^N \bigg( \frac{1}{(n-1)^2} - \frac{1}{n^2} \bigg)$

(Showing workings for Method of Difference)

$= \frac{1}{16} - \frac{1}{4N^2}$

(iii)

As $N \mapsto \infty, \frac{1}{N^2} \mapsto 0$ .

Hence, the series converges to $\frac{1}{16}$ .

(i)

Let $f(x) = ax^3 + bx^2 + cx$

$27a+9b+3c =6$ — (1)

$8a+4b+2c =6 \frac{2}{3}$ — (2)

$12a+4b+c =0$ — (3)

Using GC, $a = \frac{1}{3}, b = -3, c = 8$

(ii)

$V = \pi \int_0^3 \bigg( \frac{1}{3} x^3 - 3x^2 + 8x \bigg)^2 - (2x)^2 ~dx$

$= 176.92$

$\approx 177 \text{~units}^3$

$\frac{a+2d}{a+9d} = \frac{a}{a+2d}$

$a^2 + 9ad = a^2 + 4ad + 4d^2$

$d(4d-5a) = 0$

Since arithmetic series is increasing, $d \neq 0, d \textgreater 0$ ,

$\Rightarrow d = \frac{5a}{4}$

Common ratio $= \frac{a}{a+2d} = \frac{2}{7}$

Since $\bigg| \frac{2}{7} \bigg| \textless 1$ , the geometric series converges.

(ii)

First term $= \bigg( \frac{7}{2} \bigg)^5 a = \frac{16807}{32} a$

$S_{\infty} = \frac{ \frac{16807}{32} a}{ 1 - \frac{2}{7} }$

$S_{\infty} = \frac{117649}{160}a$

By Pythagoras’ Theorem,

$(2+h)^2 + (\frac{k}{2})^2 = 4^2$

$k^2 = 4(12 - 4h-h^2)$

$k = 2 \sqrt{12 - 4h -h^2}$ since $k \textgreater 0$

Area, $A = hk = 2h \sqrt{12 - 4h - h^2}$

$\frac{dA}{dh} = 2h \bigg( \frac{4+2h}{2\sqrt{12-4h-h^2} } \bigg) +2 \sqrt{12-4h-h^2}$

Let $\frac{dA}{dh} = 0$

$\Rightarrow h(2+h) = 12 - 4h -h^2$

$h = 1.37228$

Check with first derivative test.

Hence, $h = 1.37228$ maximises A.

$k = 4.302$

Thus, dimensions are $h = 1.37$ m and $k = 4.30$ m (to 3.s.f.).

$\frac{dx}{dt} = \frac{2}{3}x - \frac{x^2}{12} - 1$

$\frac{dx}{dt} = - \frac{1}{12}(x^2 - 8x + 12)$

(i)

$\frac{dx}{dt} = - \frac{1}{12} (x-6)(x-2)$

$\int \frac{1}{(x-6)(x-2)} ~dx = - \frac{1}{12} \int ~dt$

$\int \frac{1}{x-6} - \frac{1}{x-2} ~dx = - \frac{1}{3} \int ~dt$

$\text{ln} |x - 6 | - \text{ln}|x - 2| = - \frac{1}{3} t + C$

$\frac{x-6}{x-2} = Ae^{- \frac{1}{3}t}$ , where $A = \pm e^{C}$ is an arbitrary constant.

When $t = 0, x = 4, A = -1$ .

$\Rightarrow x = \frac{6 + 2 e^{- \frac{1}{3} t}}{1 + 2e^{- \frac{1}{3} t}}$

(ii)

As $t \mapsto \infty, x \mapsto 6$ .

Thus, the population of salmon in the fish farm will decrease towards 6,000 in the long run.

(i)

$f(x) = \frac{3x}{2x-5}, -5 \textless x \le 5, x \neq \frac{5}{2}$

Let $f(x) = y$

$y = \frac{3x}{2x-5}$

$x = \frac{5y}{2y-3}$

$f^{-1}(x) = \frac{5x}{2x-3}$

$D_{f^{-1}} = R_f = ( - \infty, 1 ) \cup [3, \infty )$

$R_{f^{-1}} = D_f = (-5, 2.5) \cup (2.5, 5]$

(ii)

$g(4) = \frac{5-4}{2} = 0.5$

(iii)

(iv)

Since $R_g = (-0.5, 1) \subset D_f = (-5, 2.5) \cup (2.5, 5], ~fg(x)$ exists.

$R_{fg} = (-1, 0.25)$

(a)

$\frac{d}{dx} e^{x^2 + 1} = 2x e^{x^2 + 1}$

$\int_0^n x^3 e^{x^2 + 1} ~dx$

$= \frac{1}{2} \int_0^n (x^2)(2xe^{x^2+1}) ~dx$

$= \frac{1}{2} \big[ x^2 e^{x^ +1} \big]_0^n - \frac{1}{2} \int_0^n 2xe^{x^2 + 1} ~dx$

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

(b)

$\int \frac{x^2 e^{x^3}}{ e^{x^3} + 1} ~dx$

$= \frac{1}{3} \int \frac{3x^2 e^{x^3}}{e^{x^3} - 1} ~dx$

$= \frac{1}{3} \text{ln} |e^{x^3} - 1| + C$

(c)

$u = e^x + 1 \Rightarrow \frac{du}{dx} = e^x$

$\int \frac{e^{3x}}{e^x + 1} ~dx$

$= \int \frac{(u-1)^3}{u} \cdot \frac{1}{u-1} ~du$

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

$= \frac{1}{2}u^2 - 2u + \text{ln}|u| + C$

$= \frac{1}{2} (e^x + 1)^2 - 2(e^x + 1) + \text{ln} (e^x + 1) + C$

$x + k + \frac{k+2}{x-k} = 0 \Rightarrow x^2 = k^2 - k - 2$

For curve to cut the x axis at 2 distinct points,

$k^2 -k -2 \textgreater 0$

$k \textless -1$ or $k \textgreater 2$

(i)

(ii)

$y = x + 4$ — (1)

$x = 4$ — (2)

Coordinate of $P = (4, 8)$

When $x = 4, y = m(4) + 4 - 2m = 8$

Observe that $y = mx + 8 -4m$ passes through $P$ for all real values of $m$ .

Thus, for line to not intersect C, $m \le 1$ .

(iii)

$x = t + 4 + \sqrt{6} \Rightarrow t = x - (4 + \sqrt{6})$

$\Rightarrow y = a - (x - 4 - \sqrt{6})^2$

$a \textless 8 + 2 \sqrt{6}$ or $a \textless 12.9$

Q10

(a)

$\frac{1}{\sqrt{2 + \sqrt{2} \text{sin} (2 \theta - \frac{\pi}{4}}}$

$= \frac{1}{ \sqrt{2 + \sqrt{2} (\text{sin}2 \theta \text{cos} \frac{\pi}{4} - \text{cos} 2 \theta \text{sin} \frac{\pi}{4} )}}$

$\approx \frac{1}{ \sqrt{ 2 + \sqrt{2} \frac{1}{\sqrt{2}} [ 2 \theta - (1 - \frac{(2 \theta )^2}{2!}]} }$

$= \frac{1}{\sqrt{1 + 2 \theta + 2 \theta ^2}}$

$= (1 + 2\theta + 2 \theta ^2)^{-\frac{1}{2}}$

$= 1 + (- \frac{1}{2}) (2 \theta + 2 \theta ^2 ) + \frac{(- \frac{1}{2})( - \frac{3}{2})}{2!} (2 \theta + 2 \theta ^2)^2 + ...$

$\approx 1 - \theta + \frac{1}{2} \theta ^2$

(b)

(i)

$y = e^x \text{sec}x$

$\frac{dy}{dx} = e^x \text{sec} x \text{tan} x + e^x \text{sec} x$

$\frac{dy}{dx} = y \text{tan}x + y$

$\frac{dy}{dx} = y(1 + \text{tan}x )$

(ii)

$\frac{d^2y}{dx^2} = y \text{sec}^2 x + \frac{dy}{dx} (1 + \text{tan}x)$

$\frac{d^3y}{dx^3} = y (2 \text{sec}^2 x \text{tan}x) + 2 \frac{dy}{dx} \text{sec}^2 x + \frac{d^2y}{dx^2} (1 + \text{tan} x)$

When $x = 0, y = 1, \frac{dy}{dx} = 1, \frac{d^2y}{dx^2} = 2, \frac{d^3y}{dx^3} = 4$

$e^x \text{sec}x = 1 + x + x^2 + \frac{2}{3} x^3 + ...$

Q11

(a)

Firstly, scale by factor $\frac{1}{3}$ parallel to the y – axis.

Secondly, translate by 3 units in the positive x – direction.

Curve D is a circle.

(b)

(i)

(ii)

Q12

(i)

$RS = \frac{1}{2} \times \text{length~of~diagonal~of~square}$

$RS = \frac{1}{2} \times \sqrt{2^2 + 2^2} = \sqrt{2}$

Since $\angle RSP = \frac{\pi}{4}, \angle SPR = \pi - (\frac{\pi}{2} - \theta) = \frac{\pi}{2} + \theta$

Using Sine Rule,

$\frac{RS}{ \text{sin} ( \frac{\pi}{2} + \theta )} = \frac{PR}{ \text{sin} \frac{\pi}{4} }$

$\Rightarrow PR = \sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{- \text{cos} \theta} = - \text{sec} \theta$

So y coordinate is given by $\sqrt{2} - \text{sec} \theta$

(ii)

Gradient of $QS = \frac{PM}{MS} = \text{tan} \theta$

Since QS is always tangential to hump, gradient of $QS = \frac{PS}{NS} = \text{tan} \theta$

Since $\frac{dy}{dx} = \frac{dy}{d \theta} \times \frac{d \theta }{dx}$

$\Rightarrow \frac{dx}{d \theta} = - \text{sec} \theta$

(iii)

Area $= \int_{\frac{\pi}{4}}^{- \frac{\pi}{4}} (\sqrt{2} - \text{sec} \theta)(- \text{sec} \theta) ~ d \theta$

$= 0.4929$ (4dp)

Relevant Materials: MF26

Solutions to Set A2017-11-302017-11-26http://theculture.sg/wp-content/uploads/2018/03/the-culture-logo.pngThe Culture SGhttp://theculture.sg/wp-content/uploads/2017/11/angelina-litvin-32188.jpg200px200px

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for almost two decades. 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.

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