Solutions to Set A

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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

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