# Q14 Solutions (1 Viewer)

#### Sy123

##### This too shall pass
$\bg_white a)$

$\bg_white i) \ \ \frac{1}{(k+1)^2} - \frac{1}{k} + \frac{1}{k+1} = \frac{1}{(k+1)^2} - \frac{1}{k(k+1)} = \frac{1}{k+1} \left( \frac{1}{k+1} - \frac{1}{k} \right) = \frac{-1}{k(k+1)^2} < 0 \ \ as \ \ k> 0$

$\bg_white ii)$

$\bg_white Step 1 \ \ n=2$

$\bg_white \frac{1}{1^2} + \frac{1}{2^2} < 2 - \frac{1}{2}$

$\bg_white \frac{5}{4} < \frac{3}{2}$

$\bg_white Therefore true for \ \ n=2$

$\bg_white Step 2 \ \ n=k$

$\bg_white \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{k^2} < 2 - \frac{1}{k}$

$\bg_white assume true for \ \ n=k$

$\bg_white Step 3 \ \ n=k+1$

$\bg_white Prove \ \ \sum_{m=1}^{k+1} \frac{1}{k^2} < 2 - \frac{1}{k+1}$

$\bg_white From Step 2$

$\bg_white \sum_{m=1}^{k} \frac{1}{m^2} + \frac{1}{(k+1)^2} < 2 - \frac{1}{k} + \frac{1}{(k+1)^2}$

$\bg_white \sum_{m=1}^{k+1} \frac{1}{m^2} < 2 + \frac{1}{(k+1)^2} - \frac{1}{k} < 2 - \frac{1}{k+1}$

$\bg_white Proof complete by mathematical induction$

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$\bg_white b)$

$\bg_white i) \binom{4n}{2n}$

$\bg_white ii)$

$\bg_white (1+x^2 + 2x)^{2n} = (x(x+2) + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (x(x+2))^{2n-k} 1^k = \sum_{k=0}^{2n} \binom{2n}{k} x^{2n-k} (x+2)^{2n-k}$

$\bg_white iii) Take co-efficient of \ \ x^{2n} \ \ both sides of equation in (ii)$

$\bg_white (1+x^2+2x)^{2n} = ((x+1)^2)^{2n} = (x+1)^{4n}$

$\bg_white \therefore \ \ \binom{4n}{2n} = co-efficient of \ x^{2n} \ in \ \ \sum_{k=0}^{2n} \binom{2n}{k} \left(\sum_{m=0}^{2n-k} \binom{2n-k}{m} 2^{2n-k-m} x^{2n-k + m} \right)$

$\bg_white Co-efficient of \ \ x^{2n} \ \ when$

$\bg_white k=0 \Rightarrow \ \ \binom{2n}{0} \binom{2n}{0} 2^{2n} x^{2n}$

$\bg_white k=1 \Rightarrow \ \ \binom{2n}{1} \binom{2n-1}{1}2^{2n-2} x^{2n}$

$\bg_white k=2 \Rightarrow \ \ \binom{2n}{2} \binom{2n-2}{2} 2^{2n-4}x^{2n}$

.
.
.

$\bg_white k=n \Rightarrow \ \ \binom{2n}{n} \binom{n}{n} 2^0 x^{2n}$

Since largest co-efficient is x^{4n-2k}, we cannot go beyond k=n in order to maintain x^{2n} co-efficient

$\bg_white \binom{4n}{2n} = \sum_{k=0}^n x^{2n-2k} \binom{2n}{k} \binom{2n-k}{k}$

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$\bg_white c) \ \ f(t) = e^t - \frac{1}{t}$

$\bg_white f'(t) = e^t + \frac{1}{t^2}$

$\bg_white t_1 = t_0 - \frac{e^{t_0} - \frac{1}{t_0}}{e^{t_0} + \frac{1}{t_0^2}} = 0.56$

$\bg_white ii) let intersection be \ x=a$

$\bg_white e^{ra} = \ln a \ \ \fbox{1}$

$\bg_white gradients of each curve at \ \ x=a \ \ must be equal$

$\bg_white re^{ra} = \frac{1}{a} \ \ \fbox{2}$

$\bg_white e^{ra} = \frac{1}{ra}$

$\bg_white \therefore \ \ ra = t_1 \ \ is an approximation$

$\bg_white From \ \ \fbox{1} \ \ \fbox{2}$

$\bg_white r\ln a = \frac{1}{a}$

$\bg_white \frac{t_1}{a} \ln a = \frac{1}{a}$

$\bg_white \therefore \ \ a=e^{1/t_1}$

$\bg_white \therefore \ \ r= \frac{t_1}{e^{1/t_1}}$

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Hopefully that is correct

#### Locus

##### Member
$\bg_white a)$

$\bg_white i) \ \ \frac{1}{(k+1)^2} - \frac{1}{k} + \frac{1}{k+1} = \frac{1}{(k+1)^2} - \frac{1}{k(k+1)} = \frac{1}{k+1} \left( \frac{1}{k+1} - \frac{1}{k} \right) = \frac{-1}{k(k+1)^2} < 0 \ \ as \ \ k> 0$

$\bg_white ii)$

$\bg_white Step 1 \ \ n=2$

$\bg_white \frac{1}{1^2} + \frac{1}{2^2} < 2 - \frac{1}{2}$

$\bg_white \frac{5}{4} < \frac{3}{2}$

$\bg_white Therefore true for \ \ n=2$

$\bg_white Step 2 \ \ n=k$

$\bg_white \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{k^2} < 2 - \frac{1}{k}$

$\bg_white assume true for \ \ n=k$

$\bg_white Step 3 \ \ n=k+1$

$\bg_white Prove \ \ \sum_{m=1}^{k+1} \frac{1}{k^2} < 2 - \frac{1}{k+1}$

$\bg_white From Step 2$

$\bg_white \sum_{m=1}^{k} \frac{1}{m^2} + \frac{1}{(k+1)^2} < 2 - \frac{1}{k} + \frac{1}{(k+1)^2}$

$\bg_white \sum_{m=1}^{k+1} \frac{1}{m^2} < 2 + \frac{1}{(k+1)^2} - \frac{1}{k} < 2 - \frac{1}{k+1}$

$\bg_white Proof complete by mathematical induction$

----

$\bg_white b)$

$\bg_white i) \binom{4n}{2n}$

$\bg_white ii)$

$\bg_white (1+x^2 + 2x)^{2n} = (x(x+2) + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (x(x+2))^{2n-k} 1^k = \sum_{k=0}^{2n} \binom{2n}{k} x^{2n-k} (x+2)^{2n-k}$

$\bg_white iii) Take co-efficient of \ \ x^{2n} \ \ both sides of equation in (ii)$

$\bg_white (1+x^2+2x)^{2n} = ((x+1)^2)^{2n} = (x+1)^{4n}$

$\bg_white \therefore \ \ \binom{4n}{2n} = co-efficient of \ x^{2n} \ in \ \ \sum_{k=0}^{2n} \binom{2n}{k} \left(\sum_{m=0}^{2n-k} \binom{2n-k}{m} 2^{2n-k-m} x^{2n-k + m} \right)$

$\bg_white Co-efficient of \ \ x^{2n} \ \ when$

$\bg_white k=0 \Rightarrow \ \ \binom{2n}{0} \binom{2n}{0} 2^{2n} x^{2n}$

$\bg_white k=1 \Rightarrow \ \ \binom{2n}{1} \binom{2n-1}{1}2^{2n-2} x^{2n}$

$\bg_white k=2 \Rightarrow \ \ \binom{2n}{2} \binom{2n-2}{2} 2^{2n-4}x^{2n}$

.
.
.

$\bg_white k=n \Rightarrow \ \ \binom{2n}{n} \binom{n}{n} 2^0 x^{2n}$

Since largest co-efficient is x^{4n-2k}, we cannot go beyond k=n in order to maintain x^{2n} co-efficient

$\bg_white \binom{4n}{2n} = \sum_{k=0}^n x^{2n-2k} \binom{2n}{k} \binom{2n-k}{k}$

-----

$\bg_white c) \ \ f(t) = e^t - \frac{1}{t}$

$\bg_white f'(t) = e^t + \frac{1}{t^2}$

$\bg_white t_1 = t_0 - \frac{e^{t_0} - \frac{1}{t_0}}{e^{t_0} + \frac{1}{t_0^2}} = 0.56$

$\bg_white ii) let intersection be \ x=a$

$\bg_white e^{ra} = \ln a \ \ \fbox{1}$

$\bg_white gradients of each curve at \ \ x=a \ \ must be equal$

$\bg_white re^{ra} = \frac{1}{a} \ \ \fbox{2}$

$\bg_white e^{ra} = \frac{1}{ra}$

$\bg_white \therefore \ \ ra = t_1 \ \ is an approximation$

$\bg_white From \ \ \fbox{1} \ \ \fbox{2}$

$\bg_white r\ln a = \frac{1}{a}$

$\bg_white \frac{t_1}{a} \ln a = \frac{1}{a}$

$\bg_white \therefore \ \ a=e^{1/t_1}$

$\bg_white \therefore \ \ r= \frac{t_1}{e^{1/t_1}}$

------

Hopefully that is correct
Don't we sub in t1 approximation from part i)?

#### Sy123

##### This too shall pass
Question 13

$\bg_white a)$

$\bg_white i)$

$\bg_white \frac{dr}{dt} = \frac{dV}{dt} \times \frac{dr}{dV}$

$\bg_white \frac{dV}{dr} = 4\pi r^2 \Rightarrow \ \frac{dr}{dV} = \frac{1}{4\pi r^2} = \frac{1}{A}$

$\bg_white \therefore \ \frac{dr}{dt} = -10^{-4} A \times \frac{1}{A} = -10^{-4} = constant$

$\bg_white ii)$

$\bg_white \frac{dr}{dt} = -10^{-4}$

$\bg_white r(t) = -10^{-4} t + c$

$\bg_white t= 0 \Right arrow \ V=10^{-6}$

$\bg_white 10^{-6} = \frac{4}{3} \pi (r(0))^3$

$\bg_white \therefore \ r(0) = \sqrt[3] {\frac{3}{4\pi} 10^{-6}} = k$

$\bg_white c= r(0) = k$

$\bg_white \therefore \ \ r(t) = -10^{-4} t + k$

$\bg_white V=0 \Rightarrow \ r=0$

$\bg_white t = 10^4 k$

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$\bg_white b)$

$\bg_white i)$

$\bg_white G\left(\frac{2ap - 0}{2-1} , \frac{0 - (2a + ap^2)}{2-1} \right) \Rightarrow \ (2ap, -2a-ap^2)$

$\bg_white ii)$

$\bg_white x=2ap$

$\bg_white y=-2a-ap^2$

$\bg_white y = -2a - \frac{x^2}{4a}$

$\bg_white 4a(y+2a) = -x^2$

$\bg_white Upside down parabola with focal length \ a \ with vertex \ \ (0,-2a) \ \ therefore directrix \ \ y=-a$

$\bg_white Same as original$

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$\bg_white c)$

$\bg_white i)$

$\bg_white \dot{y} = 0$

$\bg_white -gt + V\sin \theta = 0$

$\bg_white t = \frac{u\sin \alpha}{g}$

$\bg_white ii)$

$\bg_white Times taken to collide at max height are the same$

$\bg_white \therefore \ \frac{u\sin \alpha}{g} = \frac{w \sin \beta}{g} \Rightarrow \ u\sin \alpha = w\sin \beta$

$\bg_white iii)$

$\bg_white Horizontal distance to origin from max height is$

$\bg_white x = V\cos \theta \frac{Vt \sin \theta}{g} = \frac{v^2 \sin 2\theta}{2g}$

$\bg_white x_A = horizontal distance from projection of max height$

$\bg_white x_A + x_B = d$

$\bg_white d = \frac{u^2 \sin \alpha \cos \alpha}{g} + \frac{w^2 \sin \beta \cos \beta}{g}$

$\bg_white d = \frac{uw}{g} \left(\frac{u}{w} \sin \alpha \cos \alpha + \frac{w}{u} \sin \beta \cos \beta \right)$

$\bg_white \frac{u}{w} \sin \alpha = \sin \beta$

$\bg_white \frac{w}{u} \sin \beta = \sin \alpha$

$\bg_white d = \frac{uw}{g} (\sin \alpha \cos \beta + \sin beta \cos \alpha) = \frac{uw}{g} \sin(\alpha + \beta)$

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$\bg_white d)$

$\bg_white Draw tangent at \ T$

$\bg_white Construct points on this tangent \ X \ \ at the top \ \ Y \ \ at the bottom$

$\bg_white \angle QBT = \angle PAT \ alternate angles$

$\bg_white \angle YTB = \angle BQT$

$\bg_white \angle XTA = \angle APT$

$\bg_white The above two due to alternate segment$

$\bg_white \angle XTA = \angle YTB \ vertically opposite$

$\bg_white \therefore \ \ \angle TQB = \angle APT$

$\bg_white \therefore \ \ \angle QTB = \angle ATP \ \ angle sum of triangle$

$\bg_white \therefore \ \ P,Q,T \ \ collinear, vertically opposite$

Last edited:

#### Locus

##### Member
for the water droplet
Can you integrate the dV/dt -> and you get like v = (-10^-4)(4.pi.r^2)t + c
Sub in conditions
Sub in r (from subbing in V in the formula volume formula given) to get the answer

#### Sy123

##### This too shall pass
for the water droplet
Can you integrate the dV/dt -> and you get like v = (-10^-4)(4.pi.r^2)t + c
Sub in conditions
Sub in r (from subbing in V in the formula volume formula given) to get the answer
r is a variable though

#### johnnysk

##### New Member
for the water droplet
Can you integrate the dV/dt -> and you get like v = (-10^-4)(4.pi.r^2)t + c
Sub in conditions
Sub in r (from subbing in V in the formula volume formula given) to get the answer
Lol that's what I did. What was ur answer though ?

#### Locus

##### Member
Yeah ... i think i got it wrong ... obviously didn't use part a_

#### Locus

##### Member
Do you think i could get 1 mark for that question ....

#### Sy123

##### This too shall pass
Question 12

$\bg_white a) i)$

$\bg_white 2\cos(x + \alpha) = 2\cos x \cos \alpha - 2\sin x \sin \alpha$

$\bg_white 2\cos \alpha = \sqrt{3}$

$\bg_white 2\sin \alpha = 1$

$\bg_white \alpha = \frac{\pi}{6}$

$\bg_white \therefore \ \sqrt{3} \cos x - \sin x = 2\cos(x+ \frac{\pi}{6})$

$\bg_white a) ii)$

$\bg_white \sqrt{3} \cos x - \sin x = 1 \Rightarrow \ \cos(x + \pi/6) = \frac{1}{2}$

$\bg_white x+ \pi/6 = \frac{\pi}{3} , \frac{5\pi}{3}$

$\bg_white x= \frac{\pi}{6} , \frac{3\pi}{2}$

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$\bg_white b)$

$\bg_white V = \pi \int_0^{3\pi/2} 9\sin^{2} x/2 \ dx = \frac{9\pi}{2} \int_0^{3\pi/2} 1-\cos x \ dx$

$\bg_white = \frac{9\pi}{2} (x-\sin x)^{3\pi/2}_0 = \frac{9\pi}{2} \left( \frac{3\pi}{2} + 1 \right)$

--

$\bg_white c)$

$\bg_white T = A + Be^{-kt}$

$\bg_white t=0 \rightarrow T=80 \fbox{1}$

$\bg_white t=10 \rightarrow T=60 \fbox{2}$

$\bg_white t \to \infty \rightarrow T=22 \fbox{3}$

$\bg_white from \ \fbox{3} \rightarrow A = 22$

$\bg_white from \ \fbox{1} \rightarrow 80 = 22 + B \rightarrow \ B = 58$

$\bg_white from \ \fbox{2} \rightarrow 60 = 22 + 58e^{-k\times 10}$

$\bg_white k = -\frac{1}{10} \ln \frac{38}{58}$

$\bg_white Find \ t \ when \ T=40$

$\bg_white 40 = 22 + 58e^{-kt}$

$\bg_white t = \frac{-1}{k} \ln \frac{18}{58}$

---

$\bg_white d) i)$

$\bg_white (t,t^2+3) \Rightarrow \ 2x - y - 1 = 0$

$\bg_white D(t) = \frac{|2t - (t^2+3) -1|}{\sqrt{2^2+1}}$

$\bg_white for \ -t^2 + 2t -4 \ \ \triangle < 0$

$\bg_white \therefore \ |2t - t^2 -4| = t^2 -2t + 4$

$\bg_white \therefore \ D(t) = \frac{t^2 - 2t + 4}{\sqrt{5}}$

$\bg_white d) ii)$

$\bg_white D'(t) = \frac{1}{\sqrt{5}} (2t-2) \ \ D'(1) = 0 \ \ D''(t) > 0 \ \ therefore \ \ t=1 \ \ distance is minimised$

$\bg_white d) iii)$

$\bg_white When \ t=1 \ P(1,4)$

$\bg_white y=x^2+3$

$\bg_white y'=2x \ y'(1) = 2$

$\bg_white therefore gradient of tangent is parralell to \ l \ which as gradient \ 2$

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$\bg_white e)$

$\bg_white v^2 + 9x^2 = k$

$\bg_white \frac{1}{2}v^2 + \frac{9}{2}x^2 = k/2$

$\bg_white \left(1/2 v^2 \right) \frac{d}{dx} + 9x = 0$

$\bg_white a = -9x$

$\bg_white therefore particle in SHM$

$\bg_white n = 3$

$\bg_white period \ \frac{2\pi}{n} = \frac{2\pi}{3}$

#### panda15

##### Alligator Navigator
That feeling when you get all of Q14 out

#### panda15

##### Alligator Navigator
$\bg_white e)$

$\bg_white v^2 + 9x^2 = k$

$\bg_white \frac{1}{2}v^2 + \frac{9}{2}x^2 = k/2$

$\bg_white \left(1/2 v^2 \right) \frac{d}{dx} + 9x = 0$

$\bg_white a = -9x$

$\bg_white therefore particle in SHM$

$\bg_white n = 3$

$\bg_white period \ \frac{2\pi}{n} = \frac{2\pi}{3}$
I got to the point that a=-9x, but didn't write a concluding statement that the period would be 2pi/3
Do you think that I would still get full marks?

#### Sy123

##### This too shall pass
Question 11

$\bg_white a) \ \ \frac{-7}{2}$

$\bg_white b) \ \ \int \frac{dx}{\sqrt{49-4x^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{49/4 - x^2}} = \frac{1}{2} \sin{-1} \frac{2}{7}x + c$

$\bg_white c) \ \ \binom{10}{7} (0.25)^7 (0.75)^3$

$\bg_white d) i)$

$\bg_white f(x) = \frac{x}{4-x^2}$

$\bg_white f'(x) = \frac{4-x^2 - x(-2x)}{(4-x)^2} = \frac{x^2+4}{(4-x^2)^2} > 0$

$\bg_white as \ \ \triangle \ of \ x^2+4 \ \ >0$

$\bg_white ii)$

Second graph

http://www.wolframalpha.com/input/?i=y=(x)/(4-x^2)

$\bg_white e) \frac{1}{6}$

$\bg_white f) \ u=e^{3x} \rightarrow \ 1/3 du= e^{3x} \ dx$

$\bg_white \int_0^{1/3} \frac{e^{3x} \ dx}{e^{6x} + 1} = \frac{1}{3} \int_1^{e} \frac{du}{u^2+1} = \frac{1}{3} \left(\tan^{-1} e - \frac{\pi}{4} \right)$

$\bg_white g)$

$\bg_white \frac{d}{dx} x^2 \sin^{-1} (5x) = 2x \sin^{-1}(5x) + \frac{5x^2}{\sqrt{1-25x^2}}$

#### Parvee

##### (╯°□°）╯︵ ┻━┻
seems like sy is beating carrot to it for MX1

#### Locus

##### Member
for the solution the f)
Would you lose a mark if you didn't factorise out the 1/3?

1) c

2) d

3) c

4) d

5) a

6) b

7) a

8) d

9) b

10) c

#### Sy123

##### This too shall pass
for the solution the f)
Would you lose a mark if you didn't factorise out the 1/3?
If your final answer is equivalent then you won't lose the mark
If the answer is wrong then yes you will lose something

#### RealiseNothing

##### what is that?It is Cowpea
for the solution the f)
Would you lose a mark if you didn't factorise out the 1/3?
Of course not.

#### superSAIyan2

##### Member
for 11f) did you have to type the actual calculator value or can you leave it as 1/3 (arctane - pi/4).

#### Sy123

##### This too shall pass
for 11f) did you have to type the actual calculator value or can you leave it as 1/3 (arctane - pi/4).
You can leave it in exact form