Q14 Solutions (1 Viewer)

Sy123 · Oct 24, 2013

$a)$

$$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$

$ii)$

$$Step 1$ \ \ n=2$

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

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

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

$$Step 2$ \ \ n=k$

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

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

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

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

$From Step 2$

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

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

$Proof complete by mathematical induction$

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$b)$

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

$ii)$

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

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

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

$\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)$

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

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

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

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

.
.
.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Hopefully that is correct

Locus · Oct 24, 2013

Sy123 said:
$a)$

$$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$

$ii)$

$$Step 1$ \ \ n=2$

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

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

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

$$Step 2$ \ \ n=k$

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

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

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

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

$From Step 2$

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

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

$Proof complete by mathematical induction$

----

$b)$

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

$ii)$

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

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

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

$\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)$

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

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

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

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

.
.
.

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

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

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

-----

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

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

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

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

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

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

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

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

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

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

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

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

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

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

------

Hopefully that is correct

Don't we sub in t1 approximation from part i)?

Sy123 · Oct 24, 2013

Question 13

$a)$

$i)$

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

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

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

$ii)$

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

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

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

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

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

$c= r(0) = k$

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

$V=0 \Rightarrow \ r=0$

$t = 10^4 k$

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$b)$

$i)$

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

$ii)$

$x=2ap$

$y=-2a-ap^2$

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

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

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

$Same as original$

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$c)$

$i)$

$\dot{y} = 0$

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

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

$ii)$

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

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

$iii)$

$Horizontal distance to origin from max height is$

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

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

$x_A + x_B = d$

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

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

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

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

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

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$d)$

$$Draw tangent at$ \ T$

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

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

$\angle YTB = \angle BQT$

$\angle XTA = \angle APT$

$The above two due to alternate segment$

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

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

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

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

Sy123 · Oct 24, 2013

Locus said:
Don't we sub in t1 approximation from part i)?

Yep

Locus · Oct 24, 2013

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 · Oct 24, 2013

Locus said:
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 · Oct 24, 2013

Locus said:
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 · Oct 24, 2013

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

Locus · Oct 24, 2013

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

Sy123 · Oct 24, 2013

Question 12

$a) i)$

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

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

$2\sin \alpha = 1$

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

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

$a) ii)$

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

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

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

----

$b)$

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

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

--

$c)$

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

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

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

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

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

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

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

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

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

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

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

---

$d) i)$

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

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

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

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

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

$d) ii)$

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

$d) iii)$

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

$y=x^2+3$

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

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

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$e)$

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

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

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

$a = -9x$

$therefore particle in SHM$

$n = 3$

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

panda15 · Oct 24, 2013

That feeling when you get all of Q14 out

panda15 · Oct 24, 2013

Sy123 said:
$e)$

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

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

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

$a = -9x$

$therefore particle in SHM$

$n = 3$

$$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 · Oct 24, 2013

Question 11

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

$$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$

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

$d) i)$

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

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

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

$ii)$

Second graph

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

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

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

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

$g)$

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

Parvee · Oct 24, 2013

seems like sy is beating carrot to it for MX1

Locus · Oct 24, 2013

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

Sy123 · Oct 24, 2013

1) c

2) d

3) c

4) d

5) a

6) b

7) a

8) d

9) b

10) c

Sy123 · Oct 24, 2013

Locus said:
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 · Oct 24, 2013

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

Of course not.

superSAIyan2 · Oct 24, 2013

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

Sy123 · Oct 24, 2013

superSAIyan2 said:
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

Q14 Solutions (1 Viewer)

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