# ပညာရေး ဝန်ကြီးဌာန၏ သင်္ချာ နမူနာ - မေးခွန်း ပုံစံ(၁) - အဖြေ

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

SECTION (A)

1.(a)      Given that $\displaystyle f(x) = 3x - 4, g(x) = x^2 - 1.$ Find the values of $\displaystyle x$ which satisfy the equation $\displaystyle (g∘ f )(x) = 9 - 3x.$
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ f(x)=3x-4,\ g(x)={{x}^{2}}-1.\\\\\ \ \ \ \ \ \ (g\circ f)(x)=9-3x\\\\\ \ \ \ \ \ \ g \displaystyle \left( {f(x)} \right)=9-3x\\\\\ \ \ \ \ \ \ g\displaystyle \left( {3x-4} \right)=9-3x\\\\\ \ \ \ \ \ \ {{\displaystyle \left( {3x-4} \right)}^{2}}-1=9-3x\\\\\ \ \ \ \ \ \ 9{{x}^{2}}-21x+6=0\\\\\ \ \ \ \ \ \ (3x-1)(x-2)=0\\\\\therefore \ \ \ \ \ x= \displaystyle \displaystyle \frac{1}{3}\ \text{(or)}\ x=2\end{array}$

1.(b)      When $\displaystyle f(x) = (x + 2)^3(x - 1) - px + 6$ is divided by $\displaystyle x + 3$, the remainder is $\displaystyle 28$. Find the value of $\displaystyle p$ and hence show that $\displaystyle x - 1$ is a factor of $\displaystyle f(x).$
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ f(x)={{(x+2)}^{3}}(x-1)-px+6\\\\\ \ \ \ \ \text{When}\ f(x)\ \text{is divided by}\ x+3\ \text{the remainder is}\ 28.\\\\\therefore \ \ \ f(-3)=28\\\\\therefore \ \ \ {{(-3+2)}^{3}}(-3-1)+3p+6=28\\\\\therefore \ \ \ 3p+10=28\\\\\therefore \ \ \ 3p=18\\\\\therefore \ \ \ p=6\\\\\therefore \ \ \ f(x)={{(x+2)}^{3}}(x-1)-6x+6\\\\\therefore \ \ \ f(1)={{(1+2)}^{3}}(1-1)-6(1)+6=0\\\\\ \ \ \ \ \text{When}\ x-1\ \text{is a factor of }f(x).\end{array}$

2.(a)      The coefficient of $\displaystyle x^3$ in the expansion of $\displaystyle (1 + \displaystyle \frac{ x}{2} )^n$ is $\displaystyle 7$, find the value of $\displaystyle n.$
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{term in the expansion of}{{\displaystyle \left( {1+\displaystyle \displaystyle \frac{x}{2}} \right)}^{n}}={}^{n}{{C}_{r}}{{\displaystyle \left( {\displaystyle \displaystyle \frac{1}{2}} \right)}^{r}}{{x}^{r}}\\\\\ \ \ \ \ \text{For }{{x}^{3}},\ r=3.\\\\\ \ \ \ \ \text{Coefficient of }{{x}^{3}}={}^{n}{{C}_{3}}{{\displaystyle \left( {\displaystyle \displaystyle \frac{1}{2}} \right)}^{3}}\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ {}^{n}{{C}_{3}}{{\displaystyle \left( {\displaystyle \displaystyle \frac{1}{2}} \right)}^{3}}=7\\\\\ \ \ \ \ \displaystyle \displaystyle \frac{{n(n-1)(n-2)}}{{1\times 2\times 3}}\displaystyle \left( {\displaystyle \displaystyle \frac{1}{8}} \right)=7\\\\\therefore \ \ \ n(n-1)(n-2)=8\times 7\times 6\\\\\therefore \ \ \ n(n-1)(n-2)=8(8-1)(8-2)\\\\\therefore \ \ \ n=8\end{array}$

2.(b)      Find n, if $\displaystyle 1 + 3 + 3^2 + 3^3 + ... + 3^n = 121.$
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ 1+3+{{3}^{2}}+{{3}^{3}}+...+{{3}^{n}}=121\\\\\therefore \ \ 3+{{3}^{2}}+{{3}^{3}}+...+{{3}^{n}}=120\\\\\ \ \ \ \text{Since }\displaystyle \displaystyle \frac{{{{3}^{2}}}}{3}=3\ \text{and }\displaystyle \displaystyle \frac{{{{3}^{3}}}}{{{{3}^{2}}}}=3,\\\\\ \ \ \ \text{Given terms are in G}\text{.P}\text{.}\\\\\therefore \ \ a=3,\ r=3\ \ \text{and}\ {{S}_{n}}=120\\\\\ \ \ \ \text{Since}\ {{S}_{n}}=\displaystyle \displaystyle \frac{{a({{r}^{n}}-1)}}{{r-1}},\\\\\ \ \ \ \displaystyle \displaystyle \frac{{3({{3}^{n}}-1)}}{{3-1}}=120\\\\\therefore \ \ {{3}^{n}}-1=80\\\\\therefore \ \ {{3}^{n}}=81\Rightarrow n=4\end{array}$

3.(a)      Find two matrices of the form $\displaystyle X=\displaystyle \left( {\begin{array}{*{20}{c}} x & 1 \\ 0 & y \end{array}} \right)$ such that $\displaystyle X^2 = I.$
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ X=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} x & 1 \\ 0 & y \end{array}} \right)\\\\\ \ \ \ {{X}^{2}}=I\ \ \ \displaystyle \displaystyle \left[ {\because \text{given}} \right]\\\\\therefore \ \ \ \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} x & 1 \\ 0 & y \end{array}} \right)\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} x & 1 \\ 0 & y \end{array}} \right)=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\therefore \ \ \ \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {{{x}^{2}}} & {x+y} \\ 0 & {{{y}^{2}}} \end{array}} \right)=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\therefore \ \ \ {{x}^{2}}=\pm 1\ \ \ \ ----(1)\\\\\ \ \ \ \ x+y=0\ \ ----(2)\\\\\ \ \ \ \ {{y}^{2}}=\pm 1\ \ \ \ ----(3)\\\\\ \ \ \ \ \text{By equation (2),}\\\\\ \ \ \ \ \text{When }x=-1\text{,}\ y=1\text{ and}\\\\\ \ \ \ \ \text{when }x=1\text{,}\ y=-1.\\\\\therefore \ \ \ X=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-1} & 1 \\ 0 & 1 \end{array}} \right)\ \ \text{(or)}\ X=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 1 \\ 0 & {-1} \end{array}} \right).\end{array}$

3.(b)      Two balls are drawn at random at the same time from a box containing $\displaystyle 3$ red balls and $\displaystyle 8$ white balls. Find the probability that both balls will be white.
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \text{Number of red balls}=3\\\\\ \ \ \text{Number of white balls}=8\\\\\ \ \ \text{Total number of balls}=11\\\\\ \ \ \text{When two balls are drawn at random at the same time,}\\\\\ \ \ P(\text{both balls}\ \text{are white)}\\\\\ \ \ =P({{\text{1}}^{{\text{st}}}}\text{ ball is white and }{{\text{2}}^{{\text{nd}}}}\text{ ball is also white)}\\\\\ \ \ =\displaystyle \displaystyle \frac{8}{{11}}\times \displaystyle \displaystyle \frac{7}{{10}}\\\\\ \ \ =\displaystyle \displaystyle \frac{{28}}{{55}}\end{array}$

4.(a)      In the figure $\displaystyle AT$ is a tangent segment; $\displaystyle ABEF$ and $\displaystyle BCD$ are straight lines. If $\displaystyle AT = 6\ \text{cm}, AB = BE = 2\ \text{cm}, BC = 3\ \text{cm},$ then find $\displaystyle EF$ and $\displaystyle CD.$
(3 marks)
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$\displaystyle \begin{array}{l}\ \ \ \text{Since}\ AE\cdot AF=A{{T}^{2}},\\\\\ \ \ 4(4+EF)=36\\\\\therefore \ 4+EF=9\\\\\therefore \ EF=5\ \text{cm}\\\\\ \ \ \ \text{Since}\ BC\cdot BD=BE\cdot BF,\\\\\ \ \ 3(3+CD)=2\times 7\\\\\therefore \ 3+CD=\displaystyle \displaystyle \frac{{14}}{3}\\\\\therefore \ EF=\displaystyle \displaystyle \frac{5}{3}\ \text{cm}\end{array}$

4.(b)      The coordinates of $\displaystyle A, B$ and $\displaystyle C$ are $\displaystyle (1, 2), (7, 1)$ and $\displaystyle (- 3, 7)$ respectively. If $\displaystyle O$ is the origin and $\displaystyle \overrightarrow{{OC}}=h\ \overrightarrow{{OA}}+k\ \overrightarrow{{OB}}$, where $\displaystyle h$ and $\displaystyle k$ are constants, find the value of $\displaystyle h$ and of $\displaystyle k.$
(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \overrightarrow{{OA}}= \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right),\ \overrightarrow{{OB}}= \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right),\ \overrightarrow{{OC}}= \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-3} \\ 7 \end{array}} \right).\\\\\ \ \ \ \overrightarrow{{OC}}=h\ \overrightarrow{{OA}}+k\ \overrightarrow{{OB}}\ \ \displaystyle \displaystyle \left[ {\text{given}} \right]\\\\\ \ \ \ \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-3} \\ 7 \end{array}} \right)= \displaystyle h\displaystyle \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right)+ \displaystyle k\displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right)\\\\\ \ \ \ \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-3} \\ 7 \end{array}} \right)= \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} h \\ {2h} \end{array}} \right)+ \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {7k} \\ k \end{array}} \right)\\\\\ \ \ \ \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-3} \\ 7 \end{array}} \right)= \displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {h+7k} \\ {2h+k} \end{array}} \right)\\\\\therefore \ \ \ h+7k=-3\ \ \ \ \ \ \ \ \ -----(1)\\\\\ \ \ \ \ 2h+k=7\ \ \ \ \ \ \ \ \ \ \ -----(2)\\\\\ \ \ \ \text{Solving equations (1) and (2),}\\\\\ \ \ \ h= \displaystyle \displaystyle \frac{{13}}{5}\ \text{and}\ k= \displaystyle -\displaystyle \frac{4}{5}\end{array}$

5.(a)      In $\displaystyle ΔABC, ∠A : ∠B : ∠C = 3 : 4 : 5$ and $\displaystyle AC = \sqrt{6}$, find $\displaystyle BC.$

(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \text{In}\ \vartriangle ABC\text{, }AC=\sqrt{6}\\\\\ \ \ \ \ \angle A\ \text{: }\angle B\text{ : }\angle C\text{ = 3 : 4 : 5}\\\\\ \ \ \ \ \text{Let}\ \angle A=3k,\text{ }\angle B=\text{4}k,\ \text{and }\angle C=5k\\\\\ \ \ \ \ \text{Since}\ \angle A\ \text{+ }\angle B\text{ + }\angle C=180{}^\circ ,\\\\\,\ \ \ \ 12k=180{}^\circ \Rightarrow k=15{}^\circ \\\\\therefore \ \ \ \angle A=45{}^\circ ,\text{ }\angle B=60{}^\circ ,\ \text{and }\angle C=75{}^\circ .\\\\\ \ \ \ \ \text{Since}\ \displaystyle \displaystyle \frac{{BC}}{{\sin A}}=\displaystyle \displaystyle \frac{{AC}}{{\sin B}},\\\\\ \ \ \ \ BC=\displaystyle \displaystyle \frac{{AC\sin A}}{{\sin B}}\\\\\ \ \ \ \ BC=\displaystyle \displaystyle \frac{{\sqrt{6}\sin 45{}^\circ }}{{\sin 60{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \displaystyle \frac{{\sqrt{6}\times \displaystyle \displaystyle \frac{{\sqrt{2}}}{2}}}{{\displaystyle \displaystyle \frac{{\sqrt{3}}}{2}}}\\\\\ \ \ \ \ \ \ \ \ \ =2\end{array}$

5.(b)      Given that the gradient of the curve $\displaystyle y = x^2 + ax + b$ at the point $\displaystyle (2, -1)$ is $\displaystyle 1.$ Find the values of $\displaystyle a$ and $\displaystyle b.$

(3 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \text{Curve:}\ y={{x}^{2}}+ax+b\\\\\ \ \ \ \ (2,-1)\ \text{lies on the curve}\text{.}\\\\\therefore \ \ \ 4+2a+b=-1\\\\\therefore \ \ \ b=-5-2a\\\\\ \ \ \ \displaystyle \displaystyle \frac{{dy}}{{dx}}=2x+a\\\\\ \ \ \ \text{At }(2,-1),\ \text{the gradient of curve is 1}\text{.}\\\\\therefore \ \ {{\displaystyle \left. {\displaystyle \displaystyle \frac{{dy}}{{dx}}} \right|}_{{(2,-1)}}}=1\\\\\therefore \ \ 4+a=1\Rightarrow a=-3\\\\\therefore \ \ b=-5-2(-3)=1\end{array}$

SECTION (B)

6.(a)      Functions $\displaystyle f : R\to R$ and $\displaystyle g: R\to R$ are defined by $\displaystyle f(x) = x + 7$ and $\displaystyle g(x) = 3x - 1.$ Find the value of $\displaystyle x$ for which $\displaystyle ({{g}^{{-1}}}\circ f)(x)=({{f}^{{-1}}}\circ g)(x)+8.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ f:R\to R,\ f(x)=x+7\\\\\ \ \ \ \ g:R\to R,\ g(x)=3x-1\\\\\ \ \ \ \ \text{Let }({{g}^{{-1}}}\circ f)(x)=y\\\\\therefore \ \ \ {{g}^{{-1}}}(f(x))=y\\\\\therefore \ \ \ f(x)=g(y)\\\\\therefore \ \ \ x+7=3y-1\\\\\therefore \ \ \ y=\displaystyle \displaystyle \frac{{x+8}}{3}\Rightarrow ({{g}^{{-1}}}\circ f)(x)=\displaystyle \displaystyle \frac{{x+8}}{3}\\\\\ \ \ \ \ \text{Let }({{f}^{{-1}}}\circ g)(x)=z\\\\\therefore \ \ \ {{f}^{{-1}}}(g(x))=z\\\\\therefore \ \ \ g(x)=f(z)\\\\\therefore \ \ \ 3x-1=z+7\\\\\therefore \ \ \ z=3x-8\Rightarrow ({{f}^{{-1}}}\circ g)(x)=3x-8\\\\\ \ \ \ \ ({{g}^{{-1}}}\circ f)(x)=({{f}^{{-1}}}\circ g)(x)+8\ \ \displaystyle \left[ {\text{given}} \right]\\\\\therefore \ \ \ \displaystyle \displaystyle \frac{{x+8}}{3}=3x-8+8\\\\\therefore \ \ \ 9x=x+8\\\\\therefore \ \ \ x=1\end{array}$

6.(b)      Given that $\displaystyle x^3 - 2 x^2 - 3 x - 11$ and $\displaystyle x^3 - x^2 - 9$ have the same remainder when divided by $\displaystyle x + a$, determine the values of $\displaystyle a$ and the corresponding remainders.
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Let}\ f(x)={{x}^{3}}-2{{x}^{2}}-3x-11\ \text{and}\\\\\ \ \ \ \ \ g(x)={{x}^{3}}-{{x}^{2}}-9\\\\\ \ \ \ \ \ f(x)\ \text{and}\ g(x)\ \text{have the same remainder }\\\ \ \ \ \ \ \text{when divided by }x\text{ + }a.\\\\\therefore \ \ \ \ f(-a)=g(-a)\\\\\therefore \ \ \ \ \ -{{a}^{3}}-2{{a}^{2}}+3a-11=-{{a}^{3}}-{{a}^{2}}-9\\\\\therefore \ \ \ \ \ {{a}^{2}}-3a+2=0\\\\\therefore \ \ \ \ \ (a-1)(a-2)=0\\\\\therefore \ \ \ \ \ a=1\ (\text{or)}\ a=2\\\\\therefore \ \ \ \ \ \text{When}\ a=1,\text{the remainder }=g(-1)={{(-1)}^{3}}-{{(-1)}^{2}}-9=-11\\\\\therefore \ \ \ \ \ \text{When}\ a=2,\text{the remainder }=g(-2)={{(-2)}^{3}}-{{(-2)}^{2}}-9=-21\end{array}$

7.(a)      Let $\displaystyle J^{+}$ be the set of all positive integers. A binary operation on the set $\displaystyle J^{+}$ is defined by $\displaystyle a⊙ b = a^2 + ab + b^2.$ Prove that the binary operation is commutative. Find the value of $\displaystyle x$ such that $\displaystyle 2⊙x = 12.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \ {{J}^{+}}\text{= the set of positive integers}\text{.}\ \\\\\ \ \ \ \ \ a\odot b={{a}^{2}}+ab+{{b}^{2}},\ a,b\in {{J}^{+}}\\\\\therefore \ \ \ \ b\odot a={{b}^{2}}+ba+{{a}^{2}}={{a}^{2}}+ab+{{b}^{2}}\\\\\therefore \ \ \ \ a\odot b=b\odot a\\\\\therefore \ \ \ \ \text{The binary operation is commutative}\text{.}\\\\\ \ \ \ \ \ 2\odot x=12\\\\\therefore \ \ \ \ {{2}^{2}}+2x+{{x}^{2}}=12\\\\\therefore \ \ \ \ {{x}^{2}}+2x-8=12\\\\\therefore \ \ \ \ (x-2)(x+4)=0\\\\\therefore \ \ \ \ x=2\ \text{or}\ x=-4\notin {{J}^{+}}\\\\\therefore \ \ \ \ x=2\end{array}$

7.(b)      If the coefficient of $\displaystyle x^2$ in the expansion of $\displaystyle (2x + k)^6$ is equal to the coefficient of $\displaystyle x^5$ in the expansion of $\displaystyle (2 + kx)^8,$ find $\displaystyle k.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \ \text{ }\ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of }{{(2x+k)}^{6}}\\\\\ \ \ \ \ \ ={}^{6}{{C}_{r}}\ {{(2x)}^{{6-r}}}{{k}^{r}}\\\\\ \ \ \ \ \ ={}^{6}{{C}_{r}}\ {{2}^{{6-r}}}{{k}^{r}}{{x}^{{6-r}}}\\\\\ \ \ \ \ \text{For }{{x}^{2}},\ 6-r=2\Rightarrow r=4\\\\\therefore \ \ \ \ \ \ \ \text{coefficient of }{{x}^{2}}\ \text{in the expansion of }{{(2x+k)}^{6}}\\\\\ \ \ \ \ \ ={}^{6}{{C}_{r}}\ {{2}^{2}}{{k}^{4}}\\\\\ \ \ \ \ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of }{{(2+kx)}^{8}}\\\\\ \ \ \ \ \ ={}^{8}{{C}_{r}}\ {{2}^{{8-r}}}{{(kx)}^{r}}\\\\\ \ \ \ \ \ ={}^{8}{{C}_{r}}\ {{2}^{{8-r}}}{{k}^{r}}{{x}^{r}}\\\\\ \ \ \ \ \text{For }{{x}^{5}},\ r=5\\\\\therefore \ \ \ \ \ \ \ \text{coefficient of }{{x}^{5}}\ \text{in the expansion of }{{(2+kx)}^{8}}\\\\\ \ \ \ \ \ ={}^{8}{{C}_{5}}\ {{2}^{3}}{{k}^{5}}\\\\\ \ \ \ \ \text{By the problem, }\\\\\ \ \ \ \ {}^{6}{{C}_{4}}\ {{2}^{2}}{{k}^{4}}={}^{8}{{C}_{5}}\ {{2}^{3}}{{k}^{5}}\\\\\therefore \ \ \ {}^{6}{{C}_{2}}\ ={}^{8}{{C}_{3}}\ 2k\ \ \displaystyle \left[ {\because {}^{n}{{C}_{r}}={}^{n}{{C}_{{n-r}}}} \right]\\\\\therefore \ \ \ \displaystyle \displaystyle \frac{{6\times 5}}{{1\times 2}}\ =\displaystyle \displaystyle \frac{{8\times 7\times 6}}{{1\times 2\times 3}}\ (2k)\\\\\therefore \ \ \ k=\displaystyle \displaystyle \frac{{15}}{{112}}\end{array}$

8.(a)      Find the solution set in R of the inequation $\displaystyle x^2 - 4x \le 0$ by algebraic method and illustrate it on the number line.
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ {{x}^{2}}-4x\le 0\\\ \\\ \ \ \ x(x-4)\le 0\\\\\ \ \ \ \text{Case I}\\\\\ \ \ \ x\le 0\ \text{(or)}\ x-4\ge 0\\\\\ \ \ \ x\le 0\ \text{(or)}\ x\ge 4\end{array}$

There is no point to satisfy both conditions.

$\displaystyle \begin{array}{l}\ \ \ \ \text{Case II}\\\\\ \ \ \ x\ge 0\ \text{(or)}\ x-4\le 0\\\\\ \ \ \ x\ge 0\ \text{(or)}\ x\le 4\end{array}$

$\displaystyle \therefore \ \ 0\le x\le 4$

$\displaystyle \therefore \ \ \text{Solution set}\ =\{x\in R|0\le x\le 4\}.$

Number line

8.(b)      The four angles of a quadrilateral are in $\displaystyle A.P.$ Given that the value of the largest angle is three times the value of the smallest angle, find the values of all four angles.
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \text{Let the measures of the four angles }\\\ \ \ \text{of the quadrilateral be}\,\alpha ,\beta ,\gamma \ \text{and}\ \delta \\\ \ \ \text{and }\alpha \ \text{be the smallest angle}\text{.}\\\\\ \ \ \text{By the problem,}\\\\\ \ \ \alpha ,\beta ,\gamma \text{,}\ \delta \ \text{is an }A.P.\\\\\ \ \ \text{Let the common difference be }d.\\\\\therefore \ \alpha =\alpha \\\\\ \ \ \beta =\alpha +d\\\\\ \ \ \gamma =\alpha +2d\\\\\ \ \ \delta =\alpha +3d\\\\\ \ \ \text{Since}\ \alpha +\beta +\gamma \text{+}\ \delta =360{}^\circ ,\\\\\ \ \ 4\alpha +6d=360{}^\circ \\\\\therefore \ 2\alpha +3d=180{}^\circ \ -------(1)\\\\\ \ \ \text{By the problem,}\\\\\ \ \ \delta =3\alpha \\\\\ \ \ \alpha +3d=3\alpha \\\\\therefore \ 2\alpha -3d=0{}^\circ \ \ \ -------(2)\\\\\ \ \ \text{Solving equations (1) and (2),}\\\\\ \ \ \alpha =45{}^\circ \ \text{and}\ d=30{}^\circ \\\\\therefore \ \alpha =45{}^\circ ,\beta =75{}^\circ ,\gamma =105{}^\circ \ \text{and}\ \delta =135{}^\circ \ \ \ \end{array}$

9.(a)      Given that $\displaystyle 8, p$ and $\displaystyle q$ are three consecutive terms of an $\displaystyle A.P.$ while $\displaystyle p, q$ and $\displaystyle 36$ are three consecutive terms of a $\displaystyle G.P.,$ find the possible values of $\displaystyle p$ and $\displaystyle q.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ 8,p,q\ \text{is an }A.P.\\\\\therefore \ \ p-8=q-p\\\\\therefore \ \ q=2p-8\\\\\ \ \ \ p,q,36\ \text{is a }G.P.\\\\\therefore \ \ \displaystyle \displaystyle \frac{q}{p}=\displaystyle \displaystyle \frac{{36}}{q}\\\\\therefore \ \ {{q}^{2}}=36p\\\\\therefore \ \ {{(2p-8)}^{2}}=36p\\\\\therefore \ \ {{p}^{2}}-17p+16=0\\\\\therefore \ \ (p-1)(p-16)=0\\\\\therefore \ \ p=1\,\ (\text{or)}\ p=16\\\\\ \ \ \text{When}\ p=1,\ q=2(1)-8=-6.\\\\\ \ \ \text{When}\ p=16,\ q=2(16)-8=24.\end{array}$

9.(b)      Given that $\displaystyle D=\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 2} & {-3} \\ {-2} & {\ \ 1} \end{array}} \right)$ and that $\displaystyle D^2 - 3D - kI = O,$ find the value of $\displaystyle k.$
(5 marks)

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$\displaystyle \begin{array}{l}D=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 2} & {-3} \\ {-2} & {\ \ 1} \end{array}} \right)\\\\{{D}^{2}}-3D-kI=O\\\\\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 2} & {-3} \\ {-2} & {\ \ 1} \end{array}} \right)\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 2} & {-3} \\ {-2} & {\ \ 1} \end{array}} \right)-3\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 2} & {-3} \\ {-2} & {\ \ 1} \end{array}} \right)-k\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 4+6} & {-6-3} \\ {-4-2} & {\ \ 6+1} \end{array}} \right)+\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-6} & {\ \ 9} \\ {\ \ 6} & {-3} \end{array}} \right)+\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {-k} & 0 \\ 0 & {-k} \end{array}} \right)=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 4+6-6-k} & {-6-3+9+0} \\ {-4-2+6+0} & {\ \ 6+1-3-k} \end{array}} \right)=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 4-k} & 0 \\ 0 & {\ \ 4-k} \end{array}} \right)=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\\therefore 4-k=0\\\\\therefore k=4\end{array}$

10.(a)    Given that $\displaystyle A=\displaystyle \left( {\begin{array}{*{20}{c}} 3 & 1 \\ 2 & 1 \end{array}} \right)$ and $\displaystyle B=\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ \ 2} & {\ \ 5} \\ {-1} & {-3} \end{array}} \right)$, write down the inverse matrix of $\displaystyle A.$ Use your result to find the matrix $\displaystyle Q$ such that $\displaystyle QA = B.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ A=\displaystyle \displaystyle \left( {\begin{array}{*{20}{c}} 3 & 1 \\ 2 & 1 \end{array}} \right),B=\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ \ 2} & {\ \ 5} \\ {-1} & {-3} \end{array}} \right)\\\\\ \ \ \det A=3-2=1\ne 0.\\\\\therefore \ {{A}^{{-1}}}\ \text{exists}\text{.}\\\\\therefore \ {{A}^{{-1}}}=\displaystyle \frac{1}{{\det A}}\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 1} & {-1} \\ {-2} & {\ \ 3} \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 1} & {-1} \\ {-2} & {\ \ 3} \end{array}} \right)\\\\\ \ \ QA=B\ \ \ \displaystyle \left[ {\text{given}} \right]\\\\\therefore \ QA{{A}^{{-1}}}=B{{A}^{{-1}}}\\\\\therefore \ QI=B{{A}^{{-1}}}\\\\\therefore \ Q=B{{A}^{{-1}}}\\\\\therefore \ Q=\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ \ 2} & {\ \ 5} \\ {-1} & {-3} \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} {\ \ 1} & {-1} \\ {-2} & {\ \ 3} \end{array}} \right)\\\\\therefore \ Q=\displaystyle \left( {\begin{array}{*{20}{c}} {2-10} & {-2+15} \\ {-1+6} & {\ \ \ 1-9} \end{array}} \right)\\\\\therefore \ Q=\displaystyle \left( {\begin{array}{*{20}{c}} {-8} & {13} \\ 5 & {-8} \end{array}} \right)\end{array}$

10.(b)    How many $\displaystyle 3$ digit numerals can you form from $\displaystyle 1, 5$ and $\displaystyle 7$, without repeating any digit ? Find the probability of a numeral which begins with $\displaystyle 1$.
(5 marks)

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$\displaystyle \begin{array}{l}\therefore \ \ \ \text{Number of possible outcomes}=6\\\\\ \ \ \ \ \text{Set of favourable outcomes for a numeral which begins with }1\\\\\ \ \ \ \ =\{157,175\}\\\\\therefore \ \ \ \text{Number of favourable outcomes}=2\\\\\therefore \ \ \ P(\text{a numeral which begins with}\ 1)=\displaystyle \displaystyle \frac{2}{6}=\displaystyle \frac{1}{3}\end{array}$

SECTION (C)

11.(a)    $\displaystyle OA$ and $\displaystyle OB$ are two radii of a circle meeting at right angle. From $\displaystyle A$ and $\displaystyle B$, two parallel chords $\displaystyle AX, BY$ are drawn. Prove $\displaystyle AY ⊥ BX.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ AY\ \text{cut}\ BX\ \text{at}\ D.\\\\\therefore \ \ \alpha =\gamma \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\angle \text{s in same arc)}\\\\\ \ \ \ \text{But }\alpha =\gamma =\displaystyle \frac{1}{2}\angle AOB\ \ \ (\text{inscribed }\angle =\displaystyle \frac{1}{2}\text{central }\angle \text{)}\\\\\therefore \ \ \alpha =\gamma =\displaystyle \frac{1}{2}(90{}^\circ )=45{}^\circ \\\\\ \ \ \ \text{Since}\ AX\parallel BY,\ \alpha =\beta \ \ (\text{alternating }\angle \text{s)}\\\\\therefore \ \ \beta =45{}^\circ \\\\\ \ \ \ \text{In}\ \vartriangle BDY,\ \\\\\ \ \ \ \angle ADB=\beta +\gamma \ \ \ \ (\text{ext: }\angle \ \text{of}\ \vartriangle \ \text{=}\ \text{sum of opp: int: }\angle \text{s})\\\\\therefore \ \ \angle ADB=90{}^\circ \\\\\therefore \ \ AY\bot BX.\end{array}$

11.(b)    Given $\displaystyle ABCD$ is a trapezium in which $\displaystyle AB ∥ DC$ and $\displaystyle ∠ADB = ∠C.$ Prove that $\displaystyle AD^2 : BC^2 = AB : CD.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ AB\parallel CD,\ \beta =\delta \ \ (\text{alternating }\angle \text{s)}\\\\\ \ \ \ \text{In}\ \vartriangle ABD\ \text{and}\ \vartriangle BDC,\\\\\ \ \ \ \angle ADB=\angle C\ \ \ \ (\text{given)}\\\\\ \ \ \ \beta =\delta \ \ \ (\text{proved)}\\\\\therefore \ \ \vartriangle ABD\ \sim \ \vartriangle BDC\ \ (\text{AA corollary)}\\\\\therefore \ \ \displaystyle \frac{{\alpha (\Delta ABD)}}{{\alpha (\Delta BDC)}}=\frac{{A{{D}^{2}}}}{{B{{C}^{2}}}}-----(1)\\\\\ \ \ \ \text{But }\vartriangle ABD\ \text{and}\ \vartriangle BDC\ \text{have the same altitude}\text{.}\\\\\therefore \ \ \displaystyle \frac{{\alpha (\Delta ABD)}}{{\alpha (\Delta BDC)}}=\frac{{AB}}{{CD}}-----(2)\\\\\therefore \ \ \text{From equations (1) and (2),}\\\\\ \ \ \ \displaystyle \frac{{A{{D}^{2}}}}{{B{{C}^{2}}}}=\frac{{AB}}{{CD}}\end{array}$

12.(a)    From any point $\displaystyle D$ on the base $\displaystyle BC$ of $\displaystyle ∆ ABC$ a line is drawn meeting $\displaystyle AB$ at $\displaystyle E$ and such that $\displaystyle ∠ BDE = ∠A.$ Prove that $\displaystyle BE · BA = BD · BC.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \angle BDE\text{ }+\angle CDE=180{}^\circ \ \ (\text{supplementry }\angle \text{s)}\\\\\ \ \ \ \text{Since}\ \angle BDE=\angle A\ \ \ \ \ (\text{given)}\\\\\ \ \ \ \angle A\text{ }+\angle CDE=180{}^\circ \\\\\therefore \ \ \ ACDE\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ BE\cdot BA=BD\cdot BC\end{array}$

12.(b)    Express $\displaystyle \cos 3x$ in terms of $\displaystyle \cos x.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \cos 3x=\cos (2x+x)\\\\\ \ \ \ \ \ \ \ \ \ \ =\cos 2x\cos x-\sin 2x\sin x\\\\\ \ \ \ \ \ \ \ \ \ \ =(2{{\cos }^{2}}x-1)\cos x-(2\sin x\cos x)\sin x\\\\\ \ \ \ \ \ \ \ \ \ \ =2{{\cos }^{3}}x-\cos x-2{{\sin }^{2}}x\cos x\\\\\ \ \ \ \ \ \ \ \ \ \ =2{{\cos }^{3}}x-\cos x-2\cos x(1-{{\cos }^{2}}x)\\\\\ \ \ \ \ \ \ \ \ \ \ =2{{\cos }^{3}}x-\cos x-2\cos x+2{{\cos }^{3}}x\\\\\ \ \ \ \ \ \ \ \ \ \ =4{{\cos }^{3}}x-3\cos x\end{array}$

ေအာက္ပါ ပံုေသးနည္းမ်ားကို အသံုးျပဳ၍ တြက္ပါသည္။

 $\displaystyle \begin{array}{l}\underline{{\text{Pythagorean Identity}}}\\\\{{\sin }^{2}}x+{{\cos }^{2}}x=1\\\\\underline{{\text{Sum and Difference Formula}}}\\\\\cos (x+y)=\cos \cos y-\sin x\sin y\\\\\underline{{\text{Double Angle Formulae}}}\\\\\sin 2x=2\sin x\cos x\\\\\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x\end{array}$

13.(a)    Solve $\displaystyle ∆ABC,$ with $\displaystyle BC = 3, AC = 4, AB = 6.$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \text{In}\ \vartriangle ABC,\ BC=3,AC=4,AB=6\\\\\ \ \ \ \text{To Find : }\alpha ,\ \beta ,\ \gamma \\\\\ \ \ \ \text{By the law of cosines,}\\\\\ \ \ \ \text{cos}\alpha =\displaystyle \frac{{A{{C}^{2}}+A{{B}^{2}}-B{{C}^{2}}}}{{2AC\cdot AB}}\\\\\therefore \ \ \text{cos}\alpha =\displaystyle \frac{{{{4}^{2}}+{{6}^{2}}-{{3}^{2}}}}{{2(4)(6)}}\\\\\therefore \ \ \text{cos}\alpha =\displaystyle \frac{{43}}{{48}}=0.8958\\\\\therefore \ \ \alpha =26{}^\circ 2{3}'\\\\\ \ \ \ \text{Similarly,}\\\\\ \ \ \ \text{cos}\beta =\displaystyle \frac{{B{{C}^{2}}+A{{B}^{2}}-C{{C}^{2}}}}{{2AC\cdot AB}}\\\\\therefore \ \ \text{cos}\beta =\displaystyle \frac{{{{3}^{2}}+{{6}^{2}}-{{4}^{2}}}}{{2(3)(6)}}\\\\\therefore \ \ \text{cos}\beta =\displaystyle \frac{{29}}{{36}}=0.8056\\\\\therefore \ \ \beta =36{}^\circ 2{0}'\\\\\therefore \ \ \gamma =180{}^\circ -(26{}^\circ 2{3}'+36{}^\circ 2{0}')\\\\\therefore \ \ \gamma =117{}^\circ 1{7}'\end{array}$

13.(b)    Given that $\displaystyle y = \cos^2 x,$ prove that $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4y=2$
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ y={{\cos }^{2}}x\\\\\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2\cos x(-\sin x)=-\sin 2x\\\\\ \ \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-\cos 2x(2)=-2\cos 2x\\\\\therefore \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4y=-2\cos 2x+4{{\cos }^{2}}x\\\\\therefore \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4y=-2(2{{\cos }^{2}}x-1)+4{{\cos }^{2}}x\\\\\therefore \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4y=-4{{\cos }^{2}}x+2+4{{\cos }^{2}}x\\\\\therefore \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4y=2\end{array}$

14.(a)    If apiece of string of fixed length is made to enclose a rectangle, show that the enclosed area is the greatest when the rectangle is a square.
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let the length of the string be}\ L.\\\\\ \ \ \ \ \ \ \text{Since the length of wire is fixed,}\ \\\ \ \ \ \ \ \ L\ \text{is a constant}\text{.}\\\\\ \ \ \ \ \ \ \text{Let the length and the breadth of }\\\ \ \ \ \ \ \ \text{the rectangle be }x\ \text{and }y\ \text{respectively}\text{.}\\\\\therefore \ \ \ \ \ 2x+2y=L\\\\\therefore \ \ \ \ \ y=\displaystyle \frac{L}{2}-x\\\\\ \ \ \ \ \ \ \text{Let the area of the rectangle be }A.\\\\\therefore \ \ \ \ \ A=xy=\displaystyle \frac{L}{2}x-{{x}^{2}}\\\\\therefore \ \ \ \ \ \displaystyle \frac{{dA}}{{dx}}=\displaystyle \frac{L}{2}-2x\\\\\therefore \ \ \ \ \ \displaystyle \frac{{dA}}{{dx}}=0\ \text{when}\ \displaystyle \frac{L}{2}-2x=0\\\\\therefore \ \ \ \ \,x=\displaystyle \frac{L}{4}\\\\\ \ \ \ \ \ \displaystyle \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=-2<0\\\\\therefore \ \ \ \ A\ \text{is maximum when }x=\displaystyle \frac{L}{4}.\\\\\therefore \ \ \ \ \text{When }x=\displaystyle \frac{L}{4},\ y=\displaystyle \frac{L}{2}-\displaystyle \frac{L}{4}=\displaystyle \frac{L}{4}\\\\\therefore \ \ \ \ x=y\ \text{and this means that the rectangle is a square}\text{.}\end{array}$

14.(b)    $\displaystyle OPRQ$ is a parallelogram and $\displaystyle OP$ is produced to $\displaystyle S$ such that $\displaystyle \overrightarrow{{OS}}=3\overrightarrow{{OP}}.$ If $\displaystyle X$ is a point on$\displaystyle PR$ such that $\displaystyle \overrightarrow{{PX}}=2\overrightarrow{{XR}},$ show that the points $\displaystyle Q, X$ and $\displaystyle S$ are collinear.
(5 marks)

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$\displaystyle \begin{array}{l}\ \ \ \ \text{Let }\overrightarrow{{OP}}=\vec{a}\ \text{and}\ \overrightarrow{{OQ}}=\vec{b}\\\\\therefore \ \ \ \overrightarrow{{OS}}=3\vec{a}\ \text{and}\ \overrightarrow{{PS}}=2\vec{a}\\\\\ \ \ \ \text{Since}OPRQ\ \text{is a parallelogram,}\\\\\ \ \ \ \overrightarrow{{OQ}}=\ \overrightarrow{{PR}}=\vec{b}\ \text{and}\ \overrightarrow{{OP}}=\overrightarrow{{QR}}=\vec{a}\\\\\therefore \ \ \overrightarrow{{PX}}=\displaystyle \frac{2}{3}\vec{b}\ \text{and}\ \overrightarrow{{XR}}=\displaystyle \frac{1}{3}\vec{b}.\\\\\therefore \ \ \overrightarrow{{QX}}=\overrightarrow{{QR}}+\overrightarrow{{RX}}\\\\\ \ \ \ \ \ \ \ \ \ =\overrightarrow{{QR}}-\overrightarrow{{XR}}\\\\\ \ \ \ \ \ \ \ \ \ =\vec{a}-\displaystyle \frac{1}{3}\vec{b}\\\\\therefore \ \ \overrightarrow{{XS}}=\overrightarrow{{PS}}-\overrightarrow{{PX}}\\\\\ \ \ \ \ \ \ \ \ \ =2\vec{a}-\displaystyle \frac{2}{3}\vec{b}\\\\\ \ \ \ \ \ \ \ \ \ =2\left( {\vec{a}-\displaystyle \frac{1}{3}\vec{b}} \right)\\\\\therefore \ \ \overrightarrow{{XS}}=2\overrightarrow{{QX}}\\\\\therefore \ \ Q,X\ \text{and}\ S\ \text{are collinear}\text{.}\end{array}$

ယခု sample question သည္ ေက်ာင္းသားတို႔၏ အေတြး ကို စစ္ေဆးေသာ ေမးခြန္း လံုး၀ မပါ၀င္သေလာက္ ျဖစ္ပါသည္။ ျပဌာန္းစာအုပ္ပါ Example ေမးခြန္း အမ်ားစု ႏွင့္ Exercise ကို တိုက္႐ိုက္ေမးထားေသာ ေမးခြန္းသာ ျဖစ္ပါသည္။