DSE Math 2016 Paper 2 ANS

Posted: April 12, 2016 in Mathematics

Cristiano Ronaldo’s HAT TRICK miracle saved the day tonight! So happy although I am not a Madrid fan. But I do like Ronaldo, Modric, and Sergio Ramos. Can’t sleep and a few more hours to get to work so I decided to post the solution to the exam earlier from this morning. This year’s DSE core math exam paper 2 again isn’t that difficult, although I had a little problem with the very last question (It’s a stat problem, I am not very good at stat and I didn’t do any review or studying before it). Anyhow I managed to finish the whole thing (45 questions) in about 70 minutes.

So let’s begin1

Q1) A

8^{222}\cdot5^{666}=2^{666}\cdot5^{666}=10^{666}

Q2) A

Multiple both sides by xy yields

ay+bx=3xy \Rightarrow x=\dfrac{ay}{3y-b}

Q3) D

16-(2x-3y)^2 = (4-2x+3y)(4+2x-3y)

2.jpg

Q4) C

Obviously!

Q5) A

Solve the system of

4\alpha+\beta=5 \quad \text{and} \quad 7\alpha+3\beta=5

Multiple the first equation by 3 and then by the method of elimination,

\alpha=2 \Rightarrow \beta=-3

Q6) B

f(-\frac12)=4(-\frac12)^3+k(-\frac12)+3=0

\therefore k=5

\therefore f(-1)=4(-1)^3-5+3=-6

Q7) A

x<-7 \text{ and } x<3 \Rightarrow x<-7

3.jpg

Q8) C

\Delta = k^2-4(8k+36) = k^2 - 32k - 4\cdot36 = 0

(k-36)(k+4)=0

\therefore k=36\text{ or } k=-4

Q9) D

Let’s say a=-\frac12, then the equation becomes

(\frac{x}{2}-1)^2-\frac12

So it’s a “happy face” or “open up” parabola with vertex (2, -\frac12). Then let x=0, we can find the y-int to be \frac12 which is positive.

Q10) C

D=1.25P \text{ and } P=0.75T

\therefore T=\dfrac{D}{1.25\cdot 0.75} = \dfrac{33360}{\frac54\cdot\frac34} = \dfrac{33360\cdot16}{15} =35584

4.jpg

Q11) D

\dfrac{3y-4x}{2x+y}=\dfrac{5}{6}

18y-24x=10x+5y

13y=34x

\dfrac{x}{y}=\dfrac{13}{34}

Q12) D

z=\dfrac{k\sqrt{x}}{y}\quad \text{for some constant } k

\dfrac{k\sqrt{0.64x}}{1.6y}=\dfrac{1.8k\sqrt{x}}{1.6y}=\dfrac{1}{2}z

Q13) A

3(42)+2y=36(5) \Rightarrow y=27

Q14) C

t_7=9+6\cdot5=39

5.jpg

Q15) B

Add a parallel line in the middle then it is obvious that

180-c+a+b=360 \Rightarrow a+b-c=180

and obviously I and III are wrong!

Q16) D

By inspection, notice that

AB=24=3\times8

BD=32=4\times8

AD=40=5\times8

which form a 3-4-5 right triangle and hence \angle DBC is a right angle as well. By Pythagoras, we have

BC=\sqrt{68^2-32^2}=4\sqrt{17^2-8^2}=4\sqrt{289-64}=60

Q17) A

\angle CBE = \angle BCE = 180-114=66

\therefore \angle ABE = 114 - 66 = 48

6.jpg

Q18) C

It’s obvious that there is a 5-12-13 right triangle, hence the volume is

\dfrac{(4+4+12)5}{2}\cdot6 = 300

Q19) A

Let \angle AOB = \theta

(39^2\pi-33^2\pi)\dfrac{\theta}{360}=72\pi

\dfrac{9(13^2-11^2)\theta}{360}=72

\theta = \dfrac{72\cdot40}{169-121} = 60

So, I is right, now we can ignore II and check III. If you are smart, you don’t need to calculate the perimeter to know III must be wrong, because the arc length must be an integer over 6 times \pi and the two radii add to an integer. The perimeter must be in the form a\pi+b where a\in\mathbb{Q}, b\in\mathbb{Z}.

Q20) C

By mid-point theorem, we can divide those two region as follow and the answer comes out clearly!

tmp1

7

Q21) B

Join BE such that point B is on AD and line BE is parallel to the line CD. then

AD = AE+ED = AB\cos a + BC \sin c

Q22) D

Since rhombus is a parallelogram, \angle C = 180-118 = 62. And then by inscribed angle is half the central angle, \angle E = 31. Therefore

DFE = 180-31-(180-118) =87

Q23) A

Obviously!

8

Q24) B

180(n-2)=3240 \Rightarrow n=20, so A cannot be right. 360\div20=18, hence B is right!

Q25) D

The two lines are perpendicular hence

\dfrac{h}{k}=-\dfrac{3}{4} \Rightarrow h=-\dfrac{3k}{4}

Intersection lies on x-axis means

4x=5 \Rightarrow x=\dfrac54

\therefore -\dfrac{3k}{4}\cdot\dfrac{5}{4}+15=0

\therefore k=16

Q26) B

A(9,-2), B(-1,8), C(x,y), \quad x=2y

(x-9)^2+(y+2)^2 = (x+1)^2+(y-8)^2

x^2-18x+81+y^2+4y+4=x^2+2x+1+y^2-16y+64

20y+20=20x

2y+2=2x

x+2=2x

x=2

Q27) C

By completing the square,

3(x^2-4x+4-4)+3(y^2+10y+25-25)+65=0

3(x-2)^2-12 + 3(y+5)^2-75 + 65 = 0

(x-2)^2 + (y+5)^2 = \dfrac{22}{3}

r=\sqrt{\dfrac{22}{3}} \approx 2.7

Hence C is a circle with radius about 2.7 centered at (2,-5).

So II and III are right!

9.jpg

Q28) C

Obviously!

Q29) B

0.1\times90 + 0.3\times 20 + 0.6\times 10 = 9+6+6=21

Q30) B

Since mode is 68 and there are two 98 already,

so the data are 32 , 68 , 68 , 68 , 79 , 86 , 88 , 98 , 98 , unknown.

Since the mean is 77 therefore the unknown is 770-685 = 85

Therefore the median is average of 79 and 85 which is 82.

10.jpg

Q31) C

Obviously

Q32) D

\log_9 y = \dfrac12 x - 2

y=9^{\frac{x}{2}-2} = \dfrac{3^x}{81}

\therefore b=3

Q33) A

11\times16^{12}+12\times16^{11}+13\times16^7+14\times16^6

= (11\cdot16+12)\times16^{11} + (13\cdot16+14)\times16^6

= 188\times16^{11} + 222\times16^6

11.jpg

Q34) B

uv = \dfrac{49}{a^2+1}

Since a can be any real number such as \pi. \dfrac{49}{a^2+1} may not be rational. II doesn’t need to be checked, just check III. III is obviouly wrong. So answer is B.

Q35) D

Let the value be

t=7y-5x+3 \Rightarrow y=\dfrac{5}{7}x + \dfrac{t-3}{7}

Draw a family of lines with slope \dfrac{5}{7} and the line that touches the feasible set with the maximum value of t is at S obviously.

Q36) B

r^4 = \dfrac{189}{21} = 9 \Rightarrow r=\pm\sqrt3

I is wrong. So the answer is either B or D, so we can ignore II and check III.

If r=\sqrt3, then

S_{99} = \dfrac{7((\sqrt3)^{99}-1)}{\sqrt3-1}\approx 3.96 \times 10^{24}

However if r=-\sqrt3, then

S_{99} = \dfrac{7((-\sqrt3)^{99}-1)}{-\sqrt3-1}\approx 1.06 \times 10^{24}

Therefore only II must be true.

12.jpg

Q37) A

\cos 2x means cosine curve is compressed horizontally by a half, so the period is 180 degrees and therefore b=90. Amplitude is obviously 2 and the graph is flipped vertically, so a must be negative and hence a=-2.

Q38) B

(5 \sin\theta - 4)(\sin \theta + 1)=0

Two solution for the first bracket and one solution for the second bracket.

Q39) A

By 3-4-5 right triangle,

AC=20 \Rightarrow PQ=\sqrt{181}

By 3-4-5 right triangle,

FQ=25

By 5-12-13 right triangle,

FP=26

By cosine law,

PQ^2 = FQ^2 + FP^2 - 2 FQ \cdot FP \cos \angle PFQ

\cos \angle PFQ = \dfrac{25^2+26^2-181}{2\cdot25\cdot26} = \dfrac{56}{65}

\therefore \sin\angle PFQ = \sqrt{1-\left(\dfrac{56}{65}\right)^2} = \dfrac{33}{65}

13.jpg

Q40) D

By inscribed angle of a semi-circle,

\angle ABC = 90^\circ

Let the center of the circle be O, then in quadrilateral OBPD,

\angle BOD = 360^\circ - 90^\circ - 90^\circ - 68^\circ = 112^\circ

By inscribed angle is half the central angle,

\angle BAD = 56^\circ

By angle sum of a triangle,

\therefore \angle AQB = 180^\circ - 90^\circ - 56^\circ = 34^\circ

Q41) C

Plug y=2x-6 into x^2+y^2-8y-14=0 gives

x^2 - 8x + 14=0

x=4\pm\sqrt2

So the midpoint is 4, then plug x=4 into 2x-y-6=0 gives

y=2

Q42) A

\displaystyle \frac{\binom{3}{2} \binom{9}{2}+\binom{3}{3}\binom{9}{1}}{\binom{12}{4}} = \frac{3\cdot\frac{9\cdot8}{2}+9}{\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot2}} = \frac{13}{55}

14

Q43) D

\displaystyle \binom{15}{0} \binom{20}{6} + \binom{15}{1}\binom{20}{5} + \binom{15}{2}\binom{20}{4} = 780045

Q44) B

I is obviously wrong, 55 is lower quartile. Only need to check III. Use a calculator to find the standard deviation is about 11.58. So II only.

Q45) A

Because I forgot \mathrm{Var}(aX+b) = a^2 \; \mathrm{Var}(X) (later I read the book). I still remember what standard deviation is. I spent like 10 minutes to do this. Notice that

\mathrm{Var}(4X+9) = \sigma^2 = \dfrac{\sum(4 x_i+9-(4\mu+9))^2}{n}

= 16\dfrac{\sum(x_i-\mu)^2}{n} = 16 \; \mathrm{Var}(X)

Therefore variance of the new set of number is 49\times16=784.

An engineer, a physicist and a mathematician were asked to hammer a nail into a wall.

The engineer went to build a Universal Automatic Nailer – a device able to hammer every possible nail into every possible wall.

The physicist conducted series of experiments on strength of hammers, nails, and walls and developed a revolutionary technology of ultra-sonic nail hammering at super-low temperature.

The mathematician generalized the problem to an N dimensional problem of penetration of a knotted one dimensional nail into an N-1 dimensional hyper-wall. Several fundamental theorems are proved. Of course, the problem is too rich to suggest a possibility of a simple solution, even the existence of a solution is far from obvious.

Problem:

Suppose that a regular tetrahedron with edge length of s is inscribed in a sphere, then find the radius of the sphere.

Solution:

To start with, let’s draw a cube with a tetrahedron inside it as shown in the diagram.

Screenshot - 02162016 - 04:13:34 AM

This setup will simplify our work to find the radius because the cube ABCDEFGH is also inscribed in the sphere. And obviously the green line segment AH is the diameter of the sphere. Finding AH is extremely easy.

AE=s

AF=EF=EH=\dfrac{s}{\sqrt2}

AH = \sqrt{3\cdot AF^2} = \sqrt{\dfrac{3s^2}{2}} = \dfrac{\sqrt6 s}{2}

\therefore \text{radius} = \dfrac{1}{2}AH = \dfrac{\sqrt6 s}{4}

Problem:

x^2 + xy + y^2 = 3

y^2 + yz + z^2 = 4

z^2 + zx + x^2 = 5

x + y + z = \; ?

Solution:

Although there is a 3-4-5 ratio, it has nothing to do with Pythagoras Theorem. First let’s try to play around with the variables.

(z-x)(x+y+z)=z^2+yz-x^2-xy=4-3=1

(x-y)(x+y+z)=x^2+zx-y^2-yz=5-4=1

Obviously x+y+z \neq 0, hence

z-x=\dfrac{1}{x+y+z}=x-y

\therefore y+z=2x

\therefore x+y+z=3x

Then we have the following,

z-x = \dfrac{1}{3x} = x-y

\therefore y=x-\dfrac{1}{3x} \quad \text{and} \quad z=x+\dfrac{1}{3x}

Substitute these equations into y^2+yz+z^2=4,

\left(x-\dfrac{1}{3x}\right)^2 + \left(x-\dfrac{1}{3x}\right)\left(x+\dfrac{1}{3x}\right) + \left(x+\dfrac{1}{3x}\right)^2 = 4

2x^2 + \dfrac{2}{9x^2} + x^2 - \dfrac{1}{9x^2} = 4

3x^2 + \dfrac{1}{9x^2} - 4 = 0

27x^4-36x^2+1=0

By quadratic equation,

x^2=\dfrac{36\pm\sqrt{36^2-4\cdot27}}{2\cdot27}

x^2 = \dfrac{36\pm\sqrt{36^2-4\cdot9\cdot3}}{2\cdot3\cdot9}

x^2=\dfrac{6\pm\sqrt{33}}{9}

\therefore x=\dfrac{\pm\sqrt{6\pm\sqrt{33}}}{3}

Therefore,

x+y+z=3x=\pm\sqrt{6\pm\sqrt{33}}

Four solution for x+y+z, they are

\sqrt{6+\sqrt{33}}\; ,\; \sqrt{6-\sqrt{33}}\; , \;-\sqrt{6+\sqrt{33}}\; ,\; -\sqrt{6-\sqrt{33}}

Blob Pythagorean Theorem

Posted: June 11, 2015 in Mathematics

2014 in review

Posted: December 31, 2014 in Mathematics

The WordPress.com stats helper monkeys prepared a 2014 annual report for this blog.

Here's an excerpt:

The concert hall at the Sydney Opera House holds 2,700 people. This blog was viewed about 19,000 times in 2014. If it were a concert at Sydney Opera House, it would take about 7 sold-out performances for that many people to see it.

Click here to see the complete report.

a solution to another integral

Posted: October 8, 2014 in Mathematics

This week’s integral:

\displaystyle \int_0^1 \frac{\log(x+1)}{x^2+1} \;dx

Solution:

At first glance the integral looks scary because there is no way of making the numerator being the derivative of anything in the denominator. So what can we do about it? How about trying to simplify the denominator or get rid of it completely. Let’s make use of the trig sub,

x=\tan\theta, \quad dx=\sec^2{\theta}\;d\theta, \quad \tan^2\theta+1=\sec^2\theta

then we have,

\displaystyle \int_0^{\pi/4}\log(\tan\theta+1)\;d\theta

Now apply the trick of

\displaystyle \int_0^a f(t) \;dt = \int_0^a f(a-t) \;dt

we have,

I=\displaystyle \int_0^{\pi/4}\log\left(\tan\theta+1\right)\;d\theta = \int_0^{\pi/4}\log\left(\tan\left(\frac{\pi}{4}-\theta\right)+1\right)\;d\theta

\displaystyle I = \int_0^{\pi/4}\log\left(\frac{1-\tan\theta}{1+\tan\theta}+1\right)\;d\theta = \int_0^{\pi/4}\log\frac{2}{1+\tan\theta}\;d\theta

\displaystyle I = \int_0^{\pi/4} \log 2 \;d\theta - \int_0^{\pi/4}\log(\tan\theta+1)\;d\theta

\displaystyle \therefore\;I = \frac{\log 2}{2}\cdot\frac{\pi}{4} = \frac{\pi\log2}{8}