## DSE Math 2016 Paper 1 ANS

Posted: March 27, 2017 in Mathematics

Exam week’s coming, so here it is.

###### Section A (35 marks)

(1)  Simplify $\displaystyle\frac{(x^8 y^7)^2}{x^5 y^{-6}}$ and express your answer with positive indices.  (3 marks)

Solution

$\displaystyle\frac{(x^8 y^7)^2}{x^5 y^{-6}} = \frac{x^{16}y^{14}}{x^5 y^{-6}} = x^{11}y^{20}$

(2)  Make $x$ the subject of the formula $Ax=(4x+B)C$.  (3 marks)

Solution

$Ax=4Cx+BC$

$Ax-4Cx=BC$

$\therefore x = \dfrac{BC}{A-4C}$

(3)  Simplify $\dfrac{2}{4x-5}+\dfrac{3}{1-6x}$.  (3 marks)

Solution

$\displaystyle \frac{2(1-6x)+3(4x-5)}{(4x-5)(1-6x)} = \frac{2-12x+12x-15}{(4x-5)(1-6x)} = \frac{13}{(4x-5)(6x-1)}$

(4)  Factorize

(a)  $5m-10n$.

(b)  $m^2+mn-6n^2$.

(c)  $m^2+mn-6n^2-5m+10n$.  (4 marks)

Solution

(a)  $5(m-2n)$

(b)  $(m+3n)(m-2n)$

(c)  $(m+3n)(m-2n)-5(m-2n) = (m-2n)(m+3n-5)$

(5)  In a recreation club, there are 180 members and the number of male members is 40% more than the number of female members. Find the difference of the number of male members and the number of female members.  (4 marks)

Solution

Let $f$ be the number of female members.

$1.4f + f = 180 \quad \Rightarrow \quad f = \dfrac{180}{2.4} = 75$

$\therefore \text{difference} = 1.4f-f=0.4f=0.4(75)=30$

(6)  Consider the compound inequality

$x+6 < 6(x+11) \text{ or } x\leq -5 \qquad (*)$

(a)  Solve (*).

(b)  Write down the greatest negative integer satisfying (*).  (4 marks)

Solution

(a)  $x+6<6x+66 \text{ or } x\leq -5 \\ \Leftrightarrow 5x>-60 \text{ or } x\leq -5 \\ \Leftrightarrow x>-12 \text{ or } x\leq -5 \\ \Leftrightarrow x\in\mathbb{R}$

(b)  $-1$

(7)  In a polar coordinate system, $O$ is the pole. The polar coordinate of the points $A$ and $B$ are $(12, 75^\circ)$ and $(12, 135^\circ)$ respectively.

(a)  Find $\angle AOB$.

(b)  Find the perimeter of $\triangle AOB$.

(c)  Write down the number of folds of rotational symmetry of $\triangle AOB$.  (4 marks)

Solution

(a)  $\angle AOB = 135^\circ - 75^\circ = 60^\circ$

(b)  Obviously the triangle is isosceles and hence $\alpha = \beta = 60^\circ$ and hence $\triangle AOB$ is equilateral. Therefore the perimeter is 36.

(c)  3

(8)  It is given that $f(x)$ is the sum of two parts, one part varies as $x$ and the other part varies as $x^2$. Suppose that $f(3)=48$ and $f(9)=198$.

(a)  Find $f(x)$.

(b)  Solve the equation $f(x)=90$.  (5 marks)

Solution

(a)  Let $f(x) = ax^2+bx$ for some non-zero real numbers $a$ and $b$.

$f(3) = 9a+3b = 48 \Leftrightarrow 3a+b=16 \qquad (1)$

$f(9) = 81a+9b = 198 \Leftrightarrow 9a+b=22 \qquad (2)$

(2) – (1) gives $6a=6 \Rightarrow a=1 \Rightarrow b=13$

$\therefore f(x) = x^2+13x$

(b)  $x^2+13x-90=0 \Rightarrow (x+18)(x-5)=0 \Rightarrow x=-18 \text{ or } x=5$

(9)  The frequency distribution table and the cumulative frequency distribution table below show the distribution of the heights of the plants in a garden.

(a)  Find $x$, $y$, and $z$.

(b)  If a plant is randomly selected from the garden, find the probability that the height of the selected plant is less than 1.25 m but not less than 0.65 m.

Solution

(a)  $a=2, \therefore x=2+4=6, y=37-15=22, z=37+3=40$

(b)  $\dfrac{22-6}{40}=\dfrac{16}{40}=\dfrac{2}{5}$

(10)  The coordinates of the points $A$ and $B$ are $(5, 7)$ and $(13,1)$ respectively. Let $P$ be a moving point in the rectangular coordinate plane such that $P$ is equidistant from $A$ and $B$. Denote the locus of $P$ by $\Gamma$.

(a)  Find the equation of $\Gamma$.  (2 marks)

(b)  $\Gamma$ intersects the x-axis and y-axis at $H$ and $K$ respectively. Denote the origin by $O$. Let $C$ be the circle which passes through $O$, $H$, and $K$. Someone claims that the circumference of $C$ exceeds 30. Is the claim correct? Explain your answer.  (3 marks)

Solution

(a)  $(x-5)^2+(y-7)^2=(x-13)^2+(y-1)^2$

$\Rightarrow -10x+25-14y+49=-26x+169-2y+1$

$\Rightarrow 16x=12y+96$

$\Rightarrow 4x-3y-24=0$

(b)  Let $y=0$, then $x=6$, hence $H(6, 0)$

Let $x=0$, then $y=-8$, hence $K(0, -8)$

Since $\triangle HKO$ is a right triangle where $\angle O = 90^\circ$, hence $HK$ is the diameter of the circle $C$.

$\text{Circumference} = \pi d = \pi\sqrt{6^2+8^2} = 10\pi > 30$

Therefore, the claim is correct.

(11)  An inverted right circular conical vessel contains some milk. The vessel is held vertically. The depth of milk in the vessel is $12 \text{ cm}$. Peter pours $444\pi\text{ cm}^2$ of milk into the vessel without overflowing. He now finds that the depth of milk in the vessel is $16\text{ cm}$.

(a)  Express the final volume of milk in the vessel in terms of $\pi$.  (3 marks)

(b)  Peter claims that the final area of the wet curved surface of the vessel is at least $800\text{ cm}^2$. Do you agree? Explain your answer.  (3 marks)

Solution

(a)  Let $x$ be the original volume of milk.

$\displaystyle\frac{444\pi+x}{x}=\left(\frac{16}{12}\right)^3=\left(\frac{4}{3}\right)^3$

$27(444\pi+x)=64x$

$27\cdot444\pi=37x$

$x=27\cdot4\cdot3\pi = 81\cdot4\pi=324\pi$

$\therefore \text{Final volume} = 444\pi+324\pi = 768\pi\text{ cm}^3$

(b)  $\dfrac{\pi r^2 (16)}{3} = 768 \pi \Rightarrow r^2=144 \Rightarrow r=12\text{ cm}$

$\text{Curved surface area} = \pi r s = \pi(12)\sqrt{12^2+16^2} = 240\pi \approx 754 \text{ cm}^2 < 800\text{ cm}^2$

Therefore, disagree.

(12)  The bar chart below shows the distribution of the ages of the children in a group, where $a>11$ and $4. The median of the ages of the children in the group is 7.5.

(a)  Find $a$ and $b$.  (3 marks)

(b)  Four more children now join the group. It is found that the ages of these four children are all different and the range of the ages of the children in the group remains unchanged. Find

(i)  the greatest possible median of the ages of the children in the group.

(ii)  the least possible mean of the ages of the children in the group.  (4 marks)

Solution

(a)  Since the median age is 7.5, hence $11+a=11+b+4 \Rightarrow a=b+4$

Now, $a>11 \text{ and } 47 \text{ and } 4.

Since $a,b\in\mathbb{Z}^+ \therefore b=8\text{ or } 9$

$\therefore (a,b) = (12,8)\text{ or }(13,9)$

(b) (i)  To maximize the median, we add age 7, 8, 9, 10. Therefore the new median will be 8.

(b) (ii)  To minimize the mean, we add age 6, 7, 8, 9.

Case $(a, b)=(12,8)$,

$\text{mean} = \dfrac{12\cdot6+13\cdot7+12\cdot8+9\cdot9+4\cdot10}{12+13+12+9+4} = \dfrac{380}{50}=\dfrac{38}{5}=7.6$

Case $(a, b)=(13,9)$,

$\text{mean} = \dfrac{12\cdot6+14\cdot7+12\cdot8+10\cdot9+4\cdot10}{12+14+12+10+4} = \dfrac{396}{52}=\dfrac{99}{13}=7.\overline{615384}$

(13)  In figure 1, $ABC$ is a triangle. $D$, $E$, and $M$ are points lying on $BC$ such that $BD=CE$, $\angle ADC = \angle AEB$ and $DM=EM$.

(a)  Prove that $\triangle ACD\cong\triangle ABE$.  (2 marks)

(b) Suppose that $AD=15\text{ cm}$, $BD=7\text{ cm}$ and $DE=18\text{ cm}$.

(i)  Find $AM$.

(ii)  Is $\triangle ABE$ a right-angled triangle? Explain your answer.  (5 marks)

Solution

(a)  $\angle ADC = \angle AEB \quad \text{(given)}$

$AD=AE \quad \text{(sides opp. equal }\angle\text{s)}$

$BD=CE \quad\text{(given)}$

$BE = BD+DE = CE+DE = CD\quad (BD=CE)$

$\therefore \triangle ACD\cong\triangle ABE \quad\text{(SAS)}$

(b) (i)  $AD=AE=15 cm\quad \text{(given)}$

Hence, $\triangle ADE$ is isosceles.

$DM=EM\quad\text{(given)}$

$\therefore AM \perp DE\quad\text{(property of isos }\triangle \text{)}$

$\therefore AM=\sqrt{15^2-9^2} = 12\text{ cm} \quad\text{(Pyth. thm)}$

(b) (ii) $DM=EM=9\text{ cm}$

By Pyth. thm, $AB=\sqrt{12^2+(7+9)^2} = 20\text{ cm}$

$AB^2+AE^2 = 20^2+15^2 = 25^2 = AE^2$

Therefore, $\triangle ABE$ is right-angled.

(14)  Let $p(x)=6x^4+7x^3+ax^2+bx+c$, where $a$, $b$, and $c$ are constants. When $p(x)$ is divided by $x+2$ and when $p(x)$ is divided by $x-2$, the two remainders are equal. It is given that $p(x)=(lx^2+5x+8)(2x^2+mx+n)$, where $l$, $m$, and $n$ are constants.

(a)  Find $l$, $m$, and $n$.  (5 marks)

(b)  How many real roots does the equation $p(x)=0$ have? Explain your answer.  (5 marks)

Solution

(a)  $p(-2)=p(2)$

$6(-2)^4+7(-2)^3+a(-2)^2+b(-2)+c = 6(2)^4+7(2)^3+a(2)^2+b(2)+c$

$-56-2b = 56+2b$

$4b = -112$

$b=-28$

$\therefore p(x)=6x^4+7x^3+ax^2-28x+c$

$(lx^2+5x+8)(2x^2+mx+n) = (2l)x^4+(lm+10)x^3+(ln+5m+16)x^2+(5n+8m)x+8n$

By matching the coefficient,

$2l=6 \Rightarrow l=3$

$lm+10=7 \Rightarrow m=-1$

$5n+8m=-28 \Rightarrow n=-4$

(b)  $p(x)=(lx^2+5x+8)(2x^2+mx+n)=(3x^2+5x+8)(2x^2-x-4)=0$

$\therefore (3x^2+5x+8)=0\text{ or }(2x^2-x-4)=0$

For $3x^2+5x+8=0, \quad \Delta=25-4\cdot3\cdot8 = -71\quad \therefore$ no real root.

For $2x^2-x-4=0, \quad \Delta=1+4\cdot2\cdot4 = 33\quad \therefore$ two real roots.

Therefore, $p(x)=0$ has two real roots.

###### Section B (35 marks)

(15)  If 4 boys and 5 girls randomly form a queue, find the probability that no boys are next to each other in the queue.  (3 marks)

Solution

Consider this : _ G _ G _ G _ G _ G _

There are 6 places to put 4 B’s so there are 6P4 ways to place the boys. Moreover, there are 5! ways to permute those 5 G’s. Therefore,

$\displaystyle \text{P(no boys are next to each other)}=\frac{_6P_4 \cdot 5!}{9!} = \frac{6\cdot5\cdot4\cdot3}{9\cdot8\cdot7\cdot6} = \frac{5}{42}$

(16)  In a test, the mean of the distribution of the scores of a class of students is 61 marks. The standard scores of Albert and Mary are -2.6 and 1.4 respectively. Albert gets 22 marks. A student claims that the range of the distribution is at most 59 marks. Is the claim correct? Explain your answer.  (3 marks)

Solution

$Z=\dfrac{X-\mu}{\sigma} \quad \Rightarrow\quad -2.6 = \dfrac{22-61}{\sigma} \quad\Rightarrow\quad \sigma = 15$

So, the standard deviation is 15. Now, let’s calculate what Mary gets on the test.

$X=Z\sigma+\mu = 1.4\cdot 15 + 61 = 82$

$82-22 = 60$, which is greater than 59. Therefore the claim is false.

(17)  The 1st term and the 38th term of an arithmetic sequence are 666 and 555 respectively. Find

(a)  the common difference of the sequence.  (2 marks)

(b)  the greatest value of $n$ such that the sum of the first $n$ term of the sequence is positive.  (3 marks)

Solution

(a)  $t_{38} =t_1 + 37d$

$\therefore d=\dfrac{t_{38}-t_1}{37} = \dfrac{555-666}{37} = \dfrac{-111}{37}=-3$

(b)  $S_n = \dfrac{(2a+(n-1)d)n}{2} > 0$

$\displaystyle \frac{(2(666)+(n-1)(-3))n}{2}>0$

Since $n$ is positive integer, we can divide both side by $n$ without switching the sign.

$2(666)-3(n-1)>0$

$n-1 < 2(222)$

$n < 445$

Therefore, the greatest such $n$ is 444.

(18)  Let $f(x) = \dfrac{-1}{3}x^2 + 12x - 121$.

(a)  Using the method of completing the square, find the coordinates of the vertex of the graph of $y = f(x)$.  (2 marks)

(b)  The graph of $y=g(x)$ is obtained by translating the graph of $y=f(x)$ vertically. If the graph of $y=g(x)$ touches the x-axis, find $g(x)$.  (2 marks)

(c)  Under a transformation, $f(x)$ is changed to $\dfrac{-1}{3}x^2 - 12x - 121$. Describe the geometric meaning of the transformation.  (2 marks)

Solution

(a)  $\displaystyle f(x) = \frac{-1}{3}x^2 + 12x - 121$

$= -\frac13 (x^2-36x+18^2-18^2) - 121$

$= -\frac13(x-18)^2+\frac{18^2}{3}-121$

$= -\frac13(x-18)^2 -13$

Therefore, the vertex is $(18, -13)$.

(b)  Basically translate the graph of $f(x)$ 13 units upward.

$\therefore g(x) = -\frac13(x-18)^2$

(c)  Since $\displaystyle\frac{-1}{3}x^2-12x-121=\frac{-1}{3}(-x)^2+12(-x)-121 = f(-x)$

Therefore, the transformation is reflecting the graph of f(x) horizontally along the y-axis.

(19)  Figure 2 shows a geometric model $ABCD$ in the form of tetrahedron. It is given that $\angle BAD=86^\circ, \angle CBD=43^\circ, AB=10\text{ cm}, AC=6\text{ cm}, BC=8\text{ cm} \text{ and } BD=15\text{ cm}$.

(a)  Find $\angle ABD$ and $CD$.  (4 marks)

(b)  A craftsman claims that the angle between $AB$ and the face $BCD$ is $\angle ABC$. Do you agree? Explain your answer.  (2 marks)

Solution

(a)  By Sine Law, $\displaystyle \frac{\sin\angle ADB}{AB}=\frac{\sin\angle BAD}{BD}$

$\displaystyle \frac{\sin\angle ADB}{10}=\frac{\sin 86^\circ}{15}$

$\therefore \angle ADB = \sin^{-1} \dfrac{10\sin 86^\circ}{15}$

$\therefore \angle ABD = 180^\circ - 86^\circ - \sin^{-1} \dfrac{2\sin 86^\circ}{3} = 52.31439868 \approx 52.3^\circ \text{ (3 s.f.)}$

By Cosine Law, $CD^2=AC^2+AD^2-2AC\cdot AD \cos\angle CBD$.

$\therefore CD = \sqrt{8^2+15^2-2\cdot8\cdot15\cdot\cos 43^\circ} = 10.65246974 \approx 10.7 \text{ cm (3 s.f.)}$

(b)  By Sine law, $\displaystyle \frac{AD}{\sin\angle ABD}=\frac{BD}{\sin\angle BAD}$

$\therefore AD=\dfrac{15\sin\angle ABD}{\sin 86^\circ} \approx 11.8996447\ldots$

$AC^2+CD^2=6^2+10.65246974^2=149.4751116\neq AD^2=141.6015440$

Therefore, do not agree.

(20)  $\triangle OPQ$ is an obtuse-angled triangle. Denote the in-centre and the circumcentre of $\triangle OPQ$ by $I$ and $J$ respectively. It is given that $P$, $I$, and $J$ are collinear.

(a)  Prove that $OP=PQ$.  (3 marks)

(b)  A rectangular coordinate system is introduced so that the coordinates of $O$ and $Q$ are $(0,0)$ and $(40,30)$ respectively while the y-coordinate of $P$ is 19. Let $C$ be the circle which passes through $O$, $P$, and $Q$.

(i)  Find the equation of $C$.

(ii)  Let $L_1$ and $L_2$ be two tangents to $C$ such that the slope of each tangent is $\dfrac34$ and the y-intercept of $L_1$ is greater than that of $L_2$. $L_1$ cuts the x-axis and y-axis at $S$ and $T$ respectively while $L_2$ cuts the x-axis and the y-axis at $U$ and $V$ respectively. Someone claims that the area of the trapezium $STUV$ exceeds 17000. Is the claim correct? Explain your answer.  (9 marks)

Solution

(a)  Since $P$, $I$, and $J$ are collinear, we can draw a line passing through all three points and meet the line segment $OQ$ at $K$. Since $I$ is incenter, $PI$ is angle bisector of $\angle OPQ$, hence $\angle OPK = \angle QPK$. Since $J$ is circumcenter, then $PK$ is perpendicular bisector of $OQ$ and so $\angle PKO = \angle PKQ = 90^\circ$. $PK = PK$ obviously by common sides. Hence, $\triangle PKO \cong \triangle PKQ$ by ASA. Therefore the corresponding sides of the two congruent triangles are congruent and so $OP=PQ$.

(b) (i)  Let $P=(x, 19)$.  Since $OP = PQ$,

$\sqrt{x^2+19^2} = \sqrt{(x-40)^2+(19-30)^2}$

$x^2+361=x^2-80x+1600+121$

$80x=1360$

$x=17$

Let circle $C: x^2+y^2+Dx+Ey+F=0$

Substitute $O(0,0)$ into circle C yields $F=0$. Hence,

$P(17, 19): 17^2+19^2+17D+19E=0 \Rightarrow 650+17D+19E \qquad (1)$

$Q(40, 30): 40^2+30^2+40D+30E=0 \Rightarrow 250+4D+3E=0 \qquad (2)$

3(1)-19(2):

$1950+51D-4750-76D=0$

$\therefore D=-112$

$\therefore E=66$

Therefore, the equation of circle C is $x^2+y^2-112x+66y=0$

(b) (ii)  Since $L_1$ and $L_2$ have slope of $\dfrac34$, They both have the equation of the form: $y=\dfrac34x+b$

They are both tangents to the circle hence:

$x^2+y^2-112x+66y =0 \quad\text{and}\quad y=\dfrac34x+b$

$x^2+\left(\frac34x+b\right)^2-112x+66\left(\frac34x+b\right)=0$

$x^2+\frac{9}{16}x^2+\frac{3b}{2}x+b^2-112x+\frac{99}{2}x+66b=0$

$16x^2+9x^2+24bx+16b^2-1792x+792x+1056b=0$

$25x^2+(24b-1000)x+16b^2+1056b=0$

Since they are tangents to the circle,

$\Delta = (24b-1000)^2-4(25)(16b^2+1056b)=0$

$576b^2-48000b+1000000-1600b^2-105600b=0$

$1024b^2+153600b-1000000=0$

$16b^2+2400b-15625=0$

$(4b+625)(4b-25)=0$

$b=-\dfrac{625}{4} \text{ or } b=\dfrac{25}{4}$

$\displaystyle\therefore \quad L_1: y=\frac34x+\frac{25}{4} \quad\text{and}\quad L_2: y =\frac34x-\frac{625}{4}$

$L_1:$ Let $y=0 \Rightarrow x=-\dfrac{25}{3} \Rightarrow S\left(-\dfrac{25}{3},0\right) \Rightarrow T\left(0,\dfrac{25}{4}\right)$

$L_2:$ Let $y=0 \Rightarrow x=\dfrac{625}{3} \Rightarrow U\left(\dfrac{625}{3},0\right) \Rightarrow V\left(0,-\dfrac{625}{4}\right)$

$\displaystyle \therefore ST=\sqrt{\left(\frac{25}{3}\right)^2+\left(\frac{25}{4}\right)^2}=25\sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{25}{12}$

$\displaystyle \therefore UV=\sqrt{\left(\frac{625}{3}\right)^2+\left(\frac{625}{4}\right)^2}=625\sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{3125}{12}$

$C: x^2+y^2-112x+66y=0$

$\Rightarrow x^2-112x+56^2+y^2+66y+33^2=56^2+33^2=4225$

$\Rightarrow (x-56)^2+(y+33)^2=65^2 \Rightarrow r=65 \Rightarrow d=130$

$\displaystyle\therefore \text{Area} = \frac{(ST+UV)d}{2} = \left(\frac{125}{12}+\frac{3125}{12}\right)65 =17604.1\overline{6}>17000$

Therefore, the claim is correct.