# ပညာရေး ဝန်ကြီးဌာန၏ သင်္ချာ နမူနာ - မေးခွန်း ပုံစံ(၃) - အဖြေ

2019 SAMPLE QUESTION (3)
MATRICULATION EXAMINATION
DEPARTMENT OF MYANMAR EXAMINATION
MATHEMATICS                                  Time Allowed: 3 hours

SECTION (A)

1.(a)      If $\displaystyle f: R\to R$ is defined by $\displaystyle f(x) = x^2 + 3,$ find the function $\displaystyle g$ such that $\displaystyle (g\circ f)(x)=2{{x}^{2}}+3.$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{2}}+3\\\\\ \ \ \ (g\circ f)(x)=2{{x}^{2}}+3\\\\\ \ \ \ g\left( {f(x)} \right)=2{{x}^{2}}+3\\\\\ \ \ \ g\left( {{{x}^{2}}+3} \right)=2({{x}^{2}}+3)-3\\\\\therefore \ \ g(x)=2x-3\end{array}$

1.(b)      The expression $\displaystyle 6x^2-2x+3$ leaves a remainder of $\displaystyle 3$ when divided by $\displaystyle x-p.$ Determine the values of $\displaystyle p.$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ f(x)=6{{x}^{2}}-2x+3\\\\\ \ \ \ f(x)\ \text{leaves a remainder of}\ \text{3}\ \\\\\ \ \ \ \text{when divided by}\ x-p.\\\\\therefore \ \ \ f(p)=3\\\\\therefore \ \ \ 6{{p}^{2}}-2p+3=3\\\\\therefore \ \ \ 3{{p}^{2}}-p=0\\\\\therefore \ \ \ p(3p-1)=0\\\\\therefore \ \ \ p=0\ (\text{or})\ p=\displaystyle \frac{1}{3}\\\ \ \ \end{array}$

2.(a)      Find the coefficient of $\displaystyle {x}^{-10}$ in the expansion of $\displaystyle {{\left( {2-\frac{1}{{{{x}^{2}}}}} \right)}^{8}}$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}{{\left( {2-\displaystyle \frac{1}{{{{x}^{2}}}}} \right)}^{8}}\\\\\,\ \ \ ={}^{8}{{C}_{r}}{{2}^{{8-r}}}{{(-1)}^{r}}{{x}^{{-2r}}}\\\\\ \ \ \ \ \ \text{For}\ {{x}^{{-10}}},\ -2r=-10\Rightarrow r=5\\\\\therefore \ \ \ \ \text{Coefficient of}\ {{x}^{{-10}}}={}^{8}{{C}_{5}}{{2}^{3}}{{(-1)}^{5}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={}^{8}{{C}_{3}}{{2}^{3}}{{(-1)}^{5}}\ \ \left[ {\because {}^{n}{{C}_{r}}={}^{n}{{C}_{{n-r}}}} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{8\times 7\times 6}}{{1\times 2\times 3}}(8)(-1)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-448\end{array}$

2.(b)      Which term of the A.P. $\displaystyle 6, 13, 20, 27, … is 111?$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ 6,13,20,27,...,111\ \text{is an A}\text{.P}\text{.}\\\\\therefore \ \ a=6,\ d=13-6=7,{{u}_{n}}=111\\\\\ \ \ \ {{u}_{n}}=a+(n-1)d\\\\\therefore \ \ 6+(n-1)7=111\\\\\,\ \ \ 7(n-1)=105\\\\\ \ \ \ n-1=15\\\\\therefore \ \ n=16\ \ \end{array}$

3.(a)      If $\displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right),B=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)$ find the value of $\displaystyle k$ such that $\displaystyle AB = BA.$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ A=\displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right),B=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)\ \\\\\ \ \ \ AB=BA\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)\ =\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & k \end{array}} \right)\ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 1 & 5 \end{array}} \right)\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} {2+0} & {0+0} \\ {1+10} & {0+5k} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {2+0} & {0+0} \\ {4+k} & {0+5k} \end{array}} \right)\ \\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ {11} & {5k} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} 2 & 0 \\ {4+k} & {5k} \end{array}} \right)\ \\\\\therefore \ \ 4+k=11\Rightarrow k=7\end{array}$

3.(b)      If a die is rolled $\displaystyle 60$ times, what is the expected frequency of a number divisible by $\displaystyle 3$ turns up?
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{*{20}{l}} {\ \ \ \ \ \text{When a die is rolled}\ \text{once,}} \\ {} \\ {\ \ \ \ \ \text{Set of all possible outcomes = }\left\{ {\text{1, 2, 3, 4, 5, 6}} \right\}\text{ }} \\ {} \\ {\ \ \ \ \ \text{Number of possible outcomes = 6}} \\ {} \\ {\ \ \ \ \ \text{Set of favourable outcomes for a number divisible by 3 = }\left\{ {\text{3, 6 }} \right\}} \\ {} \\ {\ \ \ \ \ \text{Number of favourable outcomes = 2}} \\ {} \\ {\ \ \ \ \ P\text{(a number divisible by 3)}=\displaystyle \frac{2}{6}=\displaystyle \frac{1}{3}} \\ {} \\ {\ \ \ \ \ \text{When a die is rolled}\ \text{60 times,}} \\ {} \\ {\ \ \ \ \ \text{Expected frequency for a number divisible by 3 turns up}} \\ {} \\ {\ \ \ \ \ \ =\text{probability}\ \times \text{number of trials}} \\ {} \\ {\ \ \ \ \ \ =\displaystyle \frac{1}{3}\times 60} \\ {} \\ {\ \ \ \ \ \ =20} \end{array}$

4.(a)      In the given figure, $\displaystyle AB = BC.$ Find $\displaystyle x$ and $\displaystyle y.$

(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ ABPC\ \text{is a cyclic quadrilateral}\text{.}\\\\\therefore \ \ \ \ y+36{}^\circ =180{}^\circ \\\\\therefore \ \ \ \ y=144{}^\circ \\\\\ \ \ \ \ \ \text{Since}\ AB=BC,\\\\\ \ \ \ \ \ \angle ACB=36{}^\circ \\\\\therefore \ \ \ \ \angle ABC=180{}^\circ -(36{}^\circ +36{}^\circ )=108{}^\circ \\\\\ \ \ \ \ \ \text{Since}\ ABCD\ \text{is a cyclic quadrilateral,}\\\\\ \ \ \ \ \ x+108{}^\circ =180{}^\circ \\\\\therefore \ \ \ \ x=72{}^\circ \end{array}$

4.(b)      It is given that $\displaystyle {\vec{a}}$ and $\displaystyle {\vec{b}}$ are non-zero and non-parallel vectors. If $\displaystyle 3\vec{a}\ +x\left( {\vec{b}-\vec{a}} \right)=y\left( {\vec{a}+2\vec{b}} \right)$ find the values of $\displaystyle x$ and $\displaystyle y.$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \vec{a}\ \text{and}\ \vec{b}\ \text{are non-zero and non-parallel vectors}\text{.}\\\\\ \ \ \ 3\vec{a}\ +x\left( {\vec{b}-\vec{a}} \right)=y\left( {\vec{a}+2\vec{b}} \right)\\\\\therefore \ \ \ 3\vec{a}\ +x\vec{b}-x\vec{a}=y\vec{a}+2y\vec{b}\\\\\therefore \ \ \ (3-x)\vec{a}\ +x\vec{b}=y\vec{a}+2y\vec{b}\\\\\therefore \ \ \ 3-x=y\ \text{and}\ x=2y\\\\\therefore \ \ \ 3-2y=y\Rightarrow 3y=3\Rightarrow y=1\\\\\therefore \ \ \ x=2(1)=2\end{array}$

5.(a)      Prove that $\displaystyle \operatorname{cosec}\theta =\frac{{\cos 2\theta }}{{\sin \theta }}+\frac{{\sin 2\theta }}{{\cos \theta }}.$
(3 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \text{LHS }=\operatorname{cosec}\theta \\\\\ \ \ \text{RHS }=\displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\cos (2\theta -\theta )}}{{\sin \theta \cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\cos \theta }}{{\sin \theta \cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sin \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\operatorname{cosec}\theta \\\\\therefore \ \ \ \operatorname{cosec}\theta =\displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\end{array}$

5.(b)      Evaluate $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-8}}{{{{x}^{2}}+3x-10}}\ \operatorname{and}\ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{{2x}}-\sqrt{a}}}{{\sqrt{{2x}}+\sqrt{a}}}.$

(3 marks)

Show/Hide Solution
$\displaystyle \ \ \ \ \ \ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,\displaystyle \frac{{{{x}^{3}}-8}}{{{{x}^{2}}+3x-10}}$

$\displaystyle \ \ \ =\ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,\displaystyle \frac{{(x-2)({{x}^{2}}+2x+4)}}{{(x-2)(x+5)}}$

$\displaystyle \ \ \ =\ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,\displaystyle \frac{{{{x}^{2}}+2x+4}}{{x+5}}$

$\displaystyle \ \ \ =\ \ \displaystyle \frac{{{{2}^{2}}+2(2)+4}}{{2+5}}$

$\displaystyle \ \ \ =\ \ \displaystyle \frac{{12}}{7}$

$\displaystyle \ \ \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{{2x}}-\sqrt{a}}}{{\sqrt{{2x}}+\sqrt{a}}}$

$\displaystyle \ \ \ =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{x}\left( {\sqrt{2}-\displaystyle \frac{{\sqrt{a}}}{{\sqrt{x}}}} \right)}}{{\sqrt{x}\left( {\sqrt{2}+\displaystyle \frac{{\sqrt{a}}}{{\sqrt{x}}}} \right)}}$

$\displaystyle \ \ \ =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{2}-\sqrt{{\displaystyle \frac{a}{x}}}}}{{\sqrt{2}+\sqrt{{\displaystyle \frac{a}{x}}}}}$

$\displaystyle \ \ \ =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\sqrt{2}-\sqrt{0}}}{{\sqrt{2}+\sqrt{0}}}$

$\displaystyle \ \ \ =\ \ 1$

SECTION (B)

6.(a)      A function f is defined by $\displaystyle f(x) = 3x - 1.$ Determine whether $\displaystyle {{(f\circ f)}^{{-1}}}(x)$ is the same as $\displaystyle ({{f}^{{-1}}}\circ {{f}^{{-1}}})(x).$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ f(x)=3x-1\\\\\ \ \ \ \ \text{Let}\ {{f}^{{-1}}}(x)=y,\text{then}\ f(y)=x\\\\\therefore \ \ \ 3y-1=x\\\\\therefore \ \ \ y= \displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \ {{f}^{{-1}}}(x)= \displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \ \left( {{{f}^{{-1}}}\circ {{f}^{{-1}}}} \right)(x)={{f}^{{-1}}}\left( {{{f}^{{-1}}}(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( { \displaystyle \frac{{x+1}}{3}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{ \displaystyle \frac{{x+1}}{3}+1}}{3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{x+4}}{9}\\\\\ \ \ \ \ \text{Let}\ {{(f\circ f)}^{{-1}}}(x)=z\ \text{then}\ (f\circ f)(z)=x\\\\\therefore \ \ \ f\left( {f(z)} \right)=x\\\\\therefore \ \ \ f(3z-1)=x\\\\\therefore \ \ \ 3(3z-1)-1=x\\\\\therefore \ \ \ 3z-1= \displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \ 3z= \displaystyle \frac{{x+4}}{3}\\\\\therefore \ \ \ z= \displaystyle \frac{{x+4}}{9}\\\\\therefore \ \ \ {{(f\circ f)}^{{-1}}}(x)= \displaystyle \frac{{x+4}}{9}\\\\\therefore \ \ \ {{(f\circ f)}^{{-1}}}(x)=\ \left( {{{f}^{{-1}}}\circ {{f}^{{-1}}}} \right)(x)\end{array}$

6.(b)      Given that $\displaystyle f(x) = x^3 + px^2 - 2x + 4\sqrt{3}$ has a factor $\displaystyle x + \sqrt{2},$ find the value of $\displaystyle p.$ Show that $\displaystyle x - 2\sqrt{3}$ is also a factor and solve the equation f(x) = 0.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{*{20}{l}} {\ \ \ \ f(x)={{x}^{3}}+p{{x}^{2}}-2x+4\sqrt{3}} \\ {} \\ {\ \ \ \ x+\sqrt{2}\ \text{is a factor of }f(x).} \\ {} \\ {\therefore \ \ f(-\sqrt{2})=0} \\ {} \\ {\ \ \ \ {{{(-\sqrt{2})}}^{3}}+p{{{(-\sqrt{2})}}^{2}}-2(-\sqrt{2})+4\sqrt{3}=0} \\ {} \\ {\ \ \ \ -2\sqrt{2}+2p+2\sqrt{2}+4\sqrt{3}=0} \\ {} \\ {\therefore \ \ p=-2\sqrt{3}} \\ {} \\ {\therefore \ \ f(x)={{x}^{3}}-2\sqrt{3}{{x}^{2}}-2x+4\sqrt{3}} \\ {} \\ {\therefore \ \ f(2\sqrt{3})={{{(2\sqrt{3})}}^{3}}-2\sqrt{3}{{{(2\sqrt{3})}}^{2}}-2(2\sqrt{3})+4\sqrt{3}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{{(2\sqrt{3})}}^{3}}-{{{(2\sqrt{3})}}^{3}}-4\sqrt{3}+4\sqrt{3}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0} \\ {} \\ {\therefore \ \ \ x-2\sqrt{3}\ \text{is also a factor of }f(x).} \\ {} \\ {\ \ \ \ \text{Let}\ f(x)=(x+\sqrt{2})(x-2\sqrt{3})(x-k)} \\ {} \\ {\therefore \ \ \ (x+\sqrt{2})(x-2\sqrt{3})(x-k)={{x}^{3}}-2\sqrt{3}{{x}^{2}}-2x+4\sqrt{3}} \\ {} \\ {\therefore \ \ \ \sqrt{2}(-2\sqrt{3})(-k)=4\sqrt{3}} \\ {} \\ {\therefore \ \ \ k=\sqrt{2}} \\ {} \\ \begin{array}{l}\therefore \ \ f(x)=(x+\sqrt{2})(x-2\sqrt{3})(x-\sqrt{2})\\\\\ \ \ f(x)=0\Rightarrow (x+\sqrt{2})(x-2\sqrt{3})(x-\sqrt{2})=0\\\\\therefore \ \ x=-\sqrt{2}\ (\text{or})\ x=2\sqrt{3}\ (\text{or})\ x=\sqrt{2}\end{array} \end{array}$

7.(a)      Let $\displaystyle J^{+}$ be the set of all positive integers. Is the function $\displaystyle \odot$ defined by $\displaystyle x\odot y = x + 3y$ a binary operation on $\displaystyle J^{+}?$ Why? If it is a binary operation, find $\displaystyle (2\odot 3) \odot 4.$ Solve the equation $\displaystyle (k\odot 5) - (3\odot k) = 2k + 8.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ {{J}^{+}}=\text{the set of all positive integers}\text{.}\\\\\ \ \ x\odot y=x+3y\ \ \\\\\ \ \ \text{Since}\ x,y\in {{J}^{+}},3y\in \ \ {{J}^{+}}.\\\\\therefore \ x+3y\in {{J}^{+}}\\\\\therefore \ x\odot y\in {{J}^{+}}\\\\\therefore \ \text{Closure property is satisfied}\text{.}\\\\\therefore \ \odot \ \text{is a binary operation}.\\\\\therefore \ 2\odot 3=2+3(3)=11\\\\\therefore \ \left( {2\odot 3} \right)\odot 4=11\odot 4\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =11+3(4)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =23\\\\\ \ \ k\odot 5=k+3(5)=15+k\\\\\ \ \ 3\odot k=3+3(k)=3+3k\\\\\ \ \ (k\odot 5)-(3\odot k)=2k+8\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ (15+k)-(3+3k)=2k+8\\\\\therefore \ 12-2k=2k+8\\\\\therefore \ 4k=4\Rightarrow k=1\end{array}$

7.(b)      The first three terms in the expansion of $\displaystyle (a + b)^n,$ in ascending powers of $\displaystyle b,$ are denoted by $\displaystyle p, q$ and $\displaystyle r$ respectively. Show that $\displaystyle \frac{{{{q}^{2}}}}{{pr}}=\frac{{2n}}{{n-1}}.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ {{(a+b)}^{n}}=p+q+r+...\\\\\ \ \ \ \ {}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{{n-1}}}b+{}^{n}{{C}_{2}}{{a}^{{n-2}}}{{b}^{2}}+...=p+q+r+...\\\\\ \ \ \ \ {{a}^{n}}+n{{a}^{{n-1}}}b+ \displaystyle \frac{{n(n-1)}}{2}{{a}^{{n-2}}}{{b}^{2}}+...=p+q+r+...\\\\\therefore \ \ \ p={{a}^{n}},\ \ q=n{{a}^{{n-1}}}b,\ \ r= \displaystyle \frac{{n(n-1)}}{2}{{a}^{{n-2}}}{{b}^{2}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}= \displaystyle \frac{{{{{\left( {n{{a}^{{n-1}}}b} \right)}}^{2}}}}{{{{a}^{n}}\cdot \displaystyle \frac{{n(n-1)}}{2}{{a}^{{n-2}}}{{b}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}= \displaystyle \frac{{{{n}^{2}}{{a}^{{2n-2}}}{{b}^{2}}}}{{ \displaystyle \frac{{n(n-1)}}{2}{{a}^{{2n-2}}}{{b}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}={{n}^{2}}\cdot \displaystyle \frac{2}{{n(n-1)}}\\\\\therefore \ \ \ \displaystyle \frac{{{{q}^{2}}}}{{pr}}= \displaystyle \frac{{2n}}{{n-1}}\end{array}$

8.(a)      Find the solution set of the inequation $\displaystyle 2 + 3x > 5x^2$ and illustrate it on the number line.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ 2+3x>5{{x}^{2}}\\\\\therefore \ \ 5{{x}^{2}}-3x-2<0\\\\\ \ \ \ \text{Let}\ y=5{{x}^{2}}-3x-2\\\\\ \ \ \ \text{When}\ y=0,\\\\\ \ \ \ 5{{x}^{2}}-3x-2=0\\\\\ \ \ \ (5x+2)(x-1)=0\\\\\therefore \ \ x=-\displaystyle \frac{2}{5}\ (\text{or})\ x=1\\\\\therefore \ \ \text{The graph cuts the x-axis at (}-\displaystyle \frac{2}{5},0)\ \text{and}\ \text{(1,0)}\text{.}\\\\\ \ \ \ \text{When}\ x=0,y=-2\\\\\therefore \ \ \text{The graph cuts the y-axis at (}0,-2)\text{.}\end{array}$

$\displaystyle \begin{array}{l}\therefore \ \ \text{Solution set}=\{x|-\displaystyle \frac{2}{5}<x<0\}\\\\\ \ \ \ \text{Number Line:}\end{array}$

8.(b)      The sum of four consecutive numbers in an A.P. is $\displaystyle 28.$ The product of the second and third numbers exceeds that of the first and last by $\displaystyle 18.$ Find the numbers.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the four consecutive numbers in A}\text{.P}\text{. be}\\\\\ \ \ \ \ a,a+d,a+2d\ \operatorname{and}\ a+3d.\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ a+a+d+a+2d+a+3d=28\\\\\ \ \ \ \ 4a+6d=28\\\\\therefore \ \ \ 2a+3d=14\ \ \ \ -----\ (1)\\\\\ \ \ \ \ (a+d)(a+2d)-a(a+3d)=18\\\\\therefore \ \ \ {{a}^{2}}+3ad+2{{d}^{2}}-{{a}^{2}}-3ad=18\\\\\therefore \ \ \ {{d}^{2}}=9\Rightarrow d=\pm 3\ -----\ (2)\\\\\ \ \ \ \ \text{When }d=-3,\\\\\ \ \ \ \ \ 2a-9=14\ \Rightarrow a= \displaystyle \frac{{23}}{2}\\\\\therefore \ \ \ \ \ {{\text{1}}^{{\text{st}}}}\text{ number}\ \text{=} \displaystyle \frac{{23}}{2}\\\\\ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\text{ number}\ \text{=} \displaystyle \frac{{17}}{2}\\\\\ \ \ \ \ \ {{\text{3}}^{{\text{rd}}}}\text{ number}\ \text{=} \displaystyle \frac{{11}}{2}\\\\\ \ \ \ \ \ {{\text{4}}^{{\text{th}}}}\text{ number}\ \text{=} \displaystyle \frac{5}{2}\\\\\ \ \ \ \ \text{When }d=3,\\\\\ \ \ \ \ \ 2a+9=14\ \Rightarrow a= \displaystyle \frac{5}{2}\\\\\therefore \ \ \ \ {{\text{1}}^{{\text{st}}}}\text{ number}\ \text{=} \displaystyle \frac{5}{2}\\\\\ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\text{ number}\ \text{=} \displaystyle \frac{{11}}{2}\\\\\ \ \ \ \ \ {{\text{3}}^{{\text{rd}}}}\text{ number}\ \text{=} \displaystyle \frac{{17}}{2}\\\\\ \ \ \ \ \ {{\text{4}}^{{\text{th}}}}\text{ number}\ \text{=} \displaystyle \frac{{23}}{2}\end{array}$

9.(a)      The product of the first $\displaystyle 3$ terms of a G.P. is $\displaystyle 1$ and the product of the third, fourth and fifth terms is $\displaystyle 11\frac{25}{64}.$ Find the fifth term of the G.P.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \text{Let}\ {{u}_{1}},{{u}_{2}},{{u}_{3}},...\ \text{be the given G}\text{.P}\text{.}\\\\\ \ \ \ \ \text{Let}\ a\text{ be the first term and }r\ \text{be }\\\ \ \ \ \ \text{the}\ \text{common ratio of given G}\text{.P}\text{.}\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ {{u}_{1}}\times {{u}_{2}}\times {{u}_{3}}=1\\\\\therefore \ \ \ a\times ar\times a{{r}^{2}}=1\\\\\ \ \ \ \ {{(ar)}^{3}}=1\Rightarrow a= \displaystyle \frac{1}{r}\\\\\ \ \ \ \ {{u}_{3}}\times {{u}_{4}}\times {{u}_{5}}=11 \displaystyle \frac{{25}}{{64}}\\\\\ \ \ \ \ a{{r}^{2}}\times a{{r}^{3}}\times a{{r}^{4}}=11 \displaystyle \frac{{25}}{{64}}\\\\\therefore \ \ \ {{a}^{3}}{{r}^{9}}= \displaystyle \frac{{729}}{{64}}\\\\\ \ \ \ \ \displaystyle \frac{1}{{{{r}^{3}}}}\times {{r}^{9}}= \displaystyle \frac{{729}}{{64}}\\\\\ \ \ \ \ {{r}^{6}}= \displaystyle \frac{{729}}{{64}}\Rightarrow r=\pm \displaystyle \frac{3}{2}\\\\\therefore \ \ \ {{u}_{5}}=a{{r}^{4}}= \displaystyle \frac{1}{r}\times {{r}^{4}}={{r}^{3}}=\pm \displaystyle \frac{{27}}{8}\end{array}$

9.(b)      Show that the matrix $\displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right)$ satisfies the equation $\displaystyle A^2 - 4A - 5I = O,$ where $\displaystyle I$ is the unit matrix of order $\displaystyle 2.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ A= \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right),I= \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \ {{A}^{2}}-4A-5I\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right)-4 \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 3 \\ 3 & 2 \end{array}} \right)-5 \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} {4+9} & {6+6} \\ {6+6} & {9+4} \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {-8} & {-12} \\ {-12} & {-8} \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & 0 \\ 0 & {-5} \end{array}} \right)\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} {4+9-8-5} & {6+6-12+0} \\ {6+6-12+0} & {9+4-8-5} \end{array}} \right)\\\\\ \ =\ \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\\ \ =\ O\\\\\therefore \ \ {{A}^{2}}-4A-5I=O\end{array}$

10.(a)    Find the inverse of the matrix $\displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)$ and use it to solve the system of equations $\displaystyle 7x + 8y = 10, 5x + 6y = 7.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \text{Let}\ A= \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)\\\\\therefore \ \ \ \det A=42-40=2\ne 0\\\\\therefore \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\\\\\therefore \ \ {{A}^{{-1}}}=\ \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} 6 & {-8} \\ {-5} & 7 \end{array}} \right)\\\\\ \ \ \ 7x+8y=10\\\\\ \ \ \ 5x+6y=7\\\\\ \ \ \ \text{Transforming into matrix form,}\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 7 \end{array}} \right)\\\\\therefore \ \ {{ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)}^{{-1}}} \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)={{ \displaystyle \left( {\begin{array}{*{20}{c}} 7 & 8 \\ 5 & 6 \end{array}} \right)}^{{-1}}} \displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 7 \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} 6 & {-8} \\ {-5} & 7 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 7 \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} {60-56} \\ {-50+49} \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \frac{1}{2} \displaystyle \left( {\begin{array}{*{20}{c}} 4 \\ {-1} \end{array}} \right)\ \\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} 2 \\ {- \displaystyle \frac{1}{2}} \end{array}} \right)\ \\\\\therefore \ \ x=2,y=- \displaystyle \frac{1}{2}\end{array}$

10.(b)    A coin is tossed three times. Head or tail is recorded each time. Drawing a tree diagram, find the probabilities of getting exactly one head, and getting no head.
(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\therefore \ \ \ \ \text{Number of possible outcomes}=8\\\\\ \ \ \ \ \ \text{Set of favourable outcomes for getting }\\\ \ \ \ \ \ \text{exactly one head}=\ \{HTT,THT,TTH\}\\\\\therefore \ \ \ \ \text{Number}\ \text{favourable outcomes for getting }\\\ \ \ \ \ \ \text{exactly one head}=3\\\\\therefore \ \ \ \ P(\text{getting exactly one head)}= \displaystyle \frac{3}{8}\\\\\ \ \ \ \ \ \text{Set of favourable outcomes for getting }\\\ \ \ \ \ \ \text{no head}=\ \{TTT\}\\\\\therefore \ \ \ \ \text{Number}\ \text{favourable outcomes for getting }\\\ \ \ \ \ \ \text{no head}=1\\\\\therefore \ \ \ \ P(\text{getting no head)}= \displaystyle \frac{1}{8}\ \ \ \ \ \ \end{array}$

SECTION (C)

11.(a)    Two circles intersect at $\displaystyle A, B.$ At $\displaystyle A$ a tangent is drawn to each circle meeting the circles again at $\displaystyle P$ and $\displaystyle Q$ respectively. Prove that $\displaystyle ∠ ABP = ∠ ABQ.$
(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\ \ \ \ \text{In}\odot ABP,\\\\\ \ \ \ \angle ABP=\alpha \ \ \ \ \ \ \text{(}\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment)}\\\\\ \ \ \ \text{Similarly,}\ \text{in}\odot ABQ,\\\\\ \ \ \ \angle ABQ=\beta \ \ \ \ \ \ \text{(}\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment)}\\\\\ \ \ \ \text{But}\ \alpha =\beta \ \ \ \ \ \ \ (\text{vertically opposite }\ \angle \text{s})\\\\\therefore \ \ \angle ABP=\angle ABQ\end{array}$

11.(b)    In $\displaystyle ∆ABC, AD$ and $\displaystyle BE$ are altitudes. If $\displaystyle ∠ACB = 45°,$ prove that $\displaystyle α(∆DEC) = α(ABDE).$
(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Since }\angle AEB\text{ }=\angle ADB=\text{90}{}^\circ \text{,}\\\\\ \ \ \ \ \ \ ABDE\text{ is cyclic}\text{.}\\\\\ \ \ \ \ \ \ \angle ABC\text{ }=\angle DEC\\\\\ \ \ \ \ \ \ \text{In }\vartriangle DEC\text{ and }\vartriangle ABC\text{,}\\\text{ }\\\ \ \ \ \ \ \ \angle ABC\text{ }=\angle DEC\ \ \text{(proved)}\\\\\ \ \ \ \ \ \ \angle C=\angle C\ \text{(common }\angle \text{)}\\\\\therefore \ \ \ \ \ \vartriangle DEC\sim \vartriangle ABC\ \ \ \text{(AA corollary)}\\\\\ \ \ \ \ \ \ \displaystyle \frac{{\alpha (\vartriangle DEC)}}{{\alpha (\vartriangle ABC)}}= \displaystyle \frac{{D{{C}^{2}}}}{{A{{C}^{2}}}}\\\\\ \ \ \ \ \ \ \ \text{In right }\vartriangle ACD,\\\\\ \ \ \ \ \ \ \ \displaystyle \frac{{DC}}{{AC}}=\cos (\angle ACB)=\cos 45{}^\circ = \displaystyle \frac{1}{{\sqrt{2}}}\\\\\therefore \ \ \ \ \ \ \displaystyle \frac{{D{{C}^{2}}}}{{A{{C}^{2}}}}= \displaystyle \frac{1}{2}\Rightarrow \displaystyle \frac{{\alpha (\vartriangle DEC)}}{{\alpha (\vartriangle ABC)}}= \displaystyle \frac{1}{2}\\\\\therefore \ \ \ \ \ \ 2\alpha (\vartriangle DEC)=\alpha (\vartriangle ABC)\\\\\therefore \ \ \ \ \ \ 2\alpha (\vartriangle DEC)=\alpha (\vartriangle DEC)+\alpha (ABDE)\\\\\therefore \ \ \ \ \ \ \alpha (\vartriangle DEC)=\alpha (ABDE)\end{array}$

12.(a)    In $\displaystyle ∆ABC, AB = AC.\ P$ is a point on $\displaystyle BC,$ and $\displaystyle Y$ is a point on $\displaystyle AP.$ The circles $\displaystyle BPY$ and $\displaystyle CPY$ cut $\displaystyle AB$ and $\displaystyle AC$ respectively at $\displaystyle X$ and $\displaystyle Z.$ Prove $\displaystyle XZ ∥ BC.$
(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{In}\odot BPY\text{,}\\\\\ \ \ \ \ \ \ AX\cdot AB=AY\cdot AP\\\\\ \ \ \ \ \ \ \text{Similarly, in}\odot CPY\text{,}\\\\\ \ \ \ \ \ \ AZ\cdot AC=AY\cdot AP\\\\\therefore \ \ \ \ \ AX\cdot AB=AZ\cdot AC\\\\\ \ \ \ \ \ \ BCZX\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ \ \ \theta =\gamma \ \ \ \text{(exterior }\angle \ \text{of cyclic quadrilateral}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\text{interior opposite }\angle \text{)}\\\\\ \ \ \ \ \ \ \text{Since}\ AB=AC,\beta =\gamma .\\\\\therefore \ \ \ \ \ \theta =\beta \\\\\ \ \ \ \ \ \ \text{Since }\theta \text{ and }\beta \text{ are corresponding }\\\ \ \ \ \ \ \ \text{angles, we can say}\ XZ\parallel BC.\end{array}$

12.(b)    Given that $\displaystyle \sin α = \frac{15}{17}$ and that $\displaystyle \cos β = -\frac{3}{5}$ and that $\displaystyle α$ and $\displaystyle β$ are in the same quadrant, find without using tables, the values of $\displaystyle \sin 2α , \cos \frac{α}{2}$ and $\displaystyle \cos 2β.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \sin \alpha = \displaystyle \frac{{15}}{{17}},\ \ \ \cos \beta =- \displaystyle \frac{3}{5}\\\ \ \ \ \text{and }\alpha \ \text{and}\ \beta \ \text{are in the same quadrant}\text{.}\\\\\ \ \ \ \text{Since}\sin \alpha \ \text{is positive and}\cos \beta \ \text{is negative,}\\\ \ \ \ \alpha \ \text{and}\ \beta \ \text{will be in }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\end{array}$

$\displaystyle \begin{array}{l}\therefore \ \ \sin \alpha =\displaystyle \frac{{15}}{{17}},\ \cos \alpha =-\displaystyle \frac{8}{{17}}\\\\\therefore \ \ \sin 2\alpha =2\sin \alpha \ \cos \alpha \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ \left( {\displaystyle \frac{{15}}{{17}}} \right)\left( {-\displaystyle \frac{8}{{17}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{240}}{{289}}\\\\\therefore \ \ \cos \displaystyle \frac{\alpha }{2}=\pm \sqrt{{\displaystyle \frac{{1+\cos \alpha }}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \sqrt{{\displaystyle \frac{{1-\displaystyle \frac{8}{{17}}}}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \displaystyle \frac{3}{{\sqrt{{34}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\pm \displaystyle \frac{{3\sqrt{{34}}}}{{34}}\\\\\ \ \ \text{Since}\ {{90}^{{}^\circ }}<\alpha <{{180}^{{}^\circ }},{{45}^{{}^\circ }}<\displaystyle \frac{\alpha }{2}<{{90}^{{}^\circ }},\\\\\therefore \ \ \cos \displaystyle \frac{\alpha }{2}=\displaystyle \frac{{3\sqrt{{34}}}}{{34}}\\\\\ \ \ \ \cos \beta =-\displaystyle \frac{3}{5}\ \ \ \ \ \left[ {\text{given}} \right]\\\\\ \ \ \ \cos 2\beta =2{{\cos }^{2}}\beta -1\\\\\therefore \ \ \cos 2\beta =2{{\left( {-\displaystyle \frac{3}{5}} \right)}^{2}}-1\\\\\therefore \ \ \cos 2\beta =-\displaystyle \frac{7}{{25}}\end{array}$

13.(a)    In A town $\displaystyle A$ is $\displaystyle 50$ miles away from a town $\displaystyle B$ in the direction $\displaystyle N 35° E$ and a town $\displaystyle C$ is $\displaystyle 68$ miles from $\displaystyle B$ in the direction $\displaystyle N 42° 1{2}' W.$ Calculate the distance and bearing of $\displaystyle A$ from $\displaystyle C.$
(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\ \ \ \ \ PQ=50\ \text{mi,}\ QR=68\ \text{mi}\\\\\ \ \ \ \ \angle PQR=35{}^\circ +42{}^\circ 1{2}'=77{}^\circ 1{2}'\\\\\ \ \ \ \ \text{By the law of cosines,}\\\\\,\ \ \ \ P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}-2\cdot PQ\cdot QR\cos Q\\\\\,\ \ \ \ P{{R}^{2}}={{50}^{2}}+{{68}^{2}}-2(50)(68)\cos 77{}^\circ 1{2}'\\\\\,\ \ \ \ P{{R}^{2}}=2500+4624-6800\cos 77{}^\circ 1{2}'\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 6800} & {3.8325} \\ { \cos 77 {}^\circ1{2}' } & {\overline{1}.3455}\\ \hline {1507} & {3.1780} \\ \hline \end{array}\end{array}\\\\\ \ \ \ \ P{{R}^{2}}=7124-1507\\\\\ \ \ \ \ P{{R}^{2}}=5617\\\\\ \ \ \ \ PR=74.95\ \text{mi}\\\\\ \ \ \ \ \cos \gamma = \displaystyle \frac{{P{{R}^{2}}+Q{{R}^{2}}-P{{Q}^{2}}}}{{2(PR)(QR)}}\\\\\ \ \ \ \ \cos \gamma = \displaystyle \frac{{5617+4624-2500}}{{2(74.95)(68)}}\\\\\ \ \ \ \ \cos \gamma = \displaystyle \frac{{7741}}{{74.95\times 136}}\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 7741} & {3.8888} \\ \hline { 74.95 } & {1.8748}\\ {136} & {2.1335} \\ \hline {} & {4.0083} \\ \hline {\cos 40 {}^\circ3{5}'} & {\overline{1}.8805} \\ \hline \end{array}\end{array} \\\\\ \ \ \ \ \cos \gamma =\cos 40{}^\circ 3{5}'\\\\\ \ \ \ \ \gamma =40{}^\circ 3{5}'\\\\\therefore \ \ \ \theta =42{}^\circ 1{2}'+40{}^\circ 3{5}'=82{}^\circ 4{7}'\\\\\therefore \ \ \ P\ \text{is 74}\text{.95 mi away from}\,R.\\\ \ \ \ \ \text{It is in the direction }S\text{ }82{}^\circ 4{7}'\text{ }E\text{ from }R\text{.}\\\ \ \ \ \ \end{array}$

13.(b)    If $\displaystyle y=x^2+2x+3$ show that $\displaystyle {{\left( {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right)}^{3}}+{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=4y.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ y={{x}^{2}}+2x+3\\\\\ \ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2x+2\\\\\ \ \ \ \ \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=2\\\\\therefore \ \ \ \ \ {{\left( {\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right)}^{3}}+{{\left( {\displaystyle \frac{{dy}}{{dx}}} \right)}^{2}}\\\\\ \ \ =\ \ {{2}^{3}}+{{(2x+2)}^{2}}\\\\\ \ \ =\ \ 8+4{{x}^{2}}+8x+4\\\\\ \ \ =\ \ 4{{x}^{2}}+8x+12\\\\\ \ \ =\ \ 4({{x}^{2}}+x+3)\\\\\ \ \ =\ \ 4y\end{array}$

14.(a)    The vector $\displaystyle \overrightarrow{{OP}}$ has a magnitude of $\displaystyle 26$ units and has the same direction as $\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right).$ The vector $\displaystyle \overrightarrow{{OQ}}$ has a magnitude of $\displaystyle 20$ units and has the same direction as $\displaystyle \left( {\begin{array}{*{20}{c}} {3} \\ {4} \end{array}} \right).$ Find the magnitude of $\displaystyle PQ.$
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let}\ \vec{a}=\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right).\\\\\therefore \ \ \ \ \ \displaystyle \left| {\vec{a}} \right|=\sqrt{{{{{(-5)}}^{2}}+{{{12}}^{2}}}}=\sqrt{{169}}=13\\\\\therefore \ \ \ \ \ \hat{a}\ =\displaystyle \frac{{\vec{a}}}{{\displaystyle \left| {\vec{a}} \right|}}=\displaystyle \frac{1}{{13}}\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right)\\\\\ \ \ \ \ \ \overrightarrow{{OP}}\ \text{has}\ \text{magnitude 26 units and the same direction as}\ \vec{a}.\\\\\therefore \ \ \ \ \ \overrightarrow{{OP}}=\text{26}\hat{a}=26\times \displaystyle \frac{1}{{13}}\displaystyle \left( {\begin{array}{*{20}{c}} {-5} \\ {12} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {-10} \\ {24} \end{array}} \right)\\\\\ \ \ \ \ \ \ \text{Let}\ \vec{b}=\displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right).\\\\\therefore \ \ \ \ \ \displaystyle \left| {\vec{b}} \right|=\sqrt{{{{3}^{2}}+{{4}^{2}}}}=\sqrt{{25}}=5\\\\\therefore \ \ \ \ \ \hat{b}\ =\displaystyle \frac{{\vec{b}}}{{\displaystyle \left| {\vec{b}} \right|}}=\displaystyle \frac{1}{5}\displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right)\\\\\ \ \ \ \ \ \overrightarrow{{OQ}}\ \text{has}\ \text{magnitude 20 units and the same direction as}\ \vec{b}.\\\\\therefore \ \ \ \ \ \overrightarrow{{OQ}}=\text{20}\hat{b}=20\times \displaystyle \frac{1}{5}\displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {12} \\ {16} \end{array}} \right)\\\\\therefore \ \ \ \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}=\displaystyle \left( {\begin{array}{*{20}{c}} {12} \\ {16} \end{array}} \right)-\displaystyle \left( {\begin{array}{*{20}{c}} {-10} \\ {24} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {22} \\ {-8} \end{array}} \right)\\\\\therefore \ \ \ \ \ \displaystyle \left| {\overrightarrow{{PQ}}} \right|=\sqrt{{{{{22}}^{2}}+{{{(-8)}}^{2}}}}=\sqrt{{548}}=2\sqrt{{137}}\ \text{units}.\end{array}$

14.(b)    Find the stationary points of the curve $\displaystyle y = x^2(3 - x)$ and determine their natures.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ y={{x}^{2}}(3-x)\\\\\therefore \ \ \ y=3{{x}^{2}}-{{x}^{3}}\\\\\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=6x-3{{x}^{2}}=3x(2-x)\\\\\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=0\ \text{when}\ 3x(2-x)=0\\\\\therefore \ \ \ \ x=0\ (\text{or)}\ x=2\\\\\ \ \ \ \ \text{When}\ x=0,y=0.\\\\\ \ \ \ \ \text{When}\ x=2,y={{2}^{2}}(3-2)=4.\\\\\therefore \ \ \ \text{The stationary points are (0,0) and (2,4)}\text{.}\\\\\ \ \ \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6-6x=6(1-x)\\\\\ \ \ \ \ \text{When}\ x=0,\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(1-0)=6>0\\\\\therefore \ \ \ \text{(0,0) is a minimum turning point}\text{.}\\\\\ \ \ \ \ \text{When}\ x=2,\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(1-2)=-6<0\\\\\therefore \ \ \ \text{(2,4) is a maximum turning point}\text{.}\end{array}$