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

2019 SAMPLE QUESTION (3)

MATRICULATION EXAMINATION
DEPARTMENT OF MYANMAR EXAMINATION
MATHEMATICS                                  Time Allowed: 3 hours

WRITE YOUR ANSWERS IN THE ANSWER BOOKLET.

SECTION (A)

(Answer ALL questions.)

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)

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

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)

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

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

(3 marks)

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$ \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)

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$ \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( {$$\begin{array}{cc}2& 0\\ 1& 5\end{array}$$} \right),B=\left( {$$\begin{array}{cc}1& 0\\ 2& k\end{array}$$} \right)$find the value of$ \displaystyle k$such that$ \displaystyle AB = BA.$

(3 marks)

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

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)

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

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

(3 marks)

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

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)

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

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

(3 marks)

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

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)

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

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

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

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

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


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

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

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

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

$\displaystyle \ \ \ =\ \ 1$

SECTION (B)

(Answer any FOUR questions.)

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)

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$ \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)

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$ \displaystyle $$\text{Missing end{array}}$$ \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)

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$ \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)

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$ \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)

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

$ \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)

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

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)

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$ \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( {$$\begin{array}{cc}2& 3\\ 3& 2\end{array}$$} \right)$satisfies the equation$ \displaystyle A^2 - 4A - 5I = O,$where$ \displaystyle I$is the unit matrix of order$ \displaystyle 2.$

(5 marks)

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

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

(5 marks)

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

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)

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$ \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)

(Answer any THREE questions.)

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)

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

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

(5 marks)

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

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)

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

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)

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$ \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)

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$ \displaystyle $$\text{Missing end{array}}$$\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}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$\text{Missing end{array}}$$\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)

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

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

(5 marks)

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

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

(5 marks)

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

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