# Myanmar Matriculation (2019 FC) Solution

$\def\vecab{\overrightarrow}$ $\def\D{\displaystyle}$ $\def\frac{\dfrac}$ $\def\rbox#1#2{\begin{array}{|#1|}\hline #2\\ \hline\end{array}}$ $\let\rtable=\rbox$ $\def\matrix#1{\left(\begin{array}{ccc}#1\end{array}\right)}$ $\def\fbox#1{\begin{array}{|l|}\hline #1\\ \hline\end{array}}$ $\let\cvec=\matrix$ $\let\ruletable=\rbox$ $\def\tbox#1{\begin{array}{|l|}\hline \mbox{#1}\\ \hline\end{array}}$ $\def\dint{\displaystyle\int}\def\dlim{\displaystyle\lim}$ $\def\dsum{\displaystyle\sum}$ $\def\mathrm{}$ $\newcommand{\inside}[2]{\begin{array}{rcl}\leftarrow-- & \circ\!\frac{\qquad\qquad\qquad\qquad}{}\!\circ & --\rightarrow\\ & #1\qquad\qquad\qquad\qquad #2 & \end{array}}$ $\newcommand{\insideeq}[2]{\begin{array}{rcl}\leftarrow--&\bullet\!\frac{\qquad\qquad\qquad\qquad}{}\!\bullet & --\rightarrow\\[-5mm]& #1\qquad\qquad\qquad\qquad #2&\end{array}}$ $\newcommand{\outside}[2]{\begin{array}{rcl}\leftarrow\!\!\!\frac{\qquad\qquad}{}\!\circ & ---- &\circ\!\frac{\qquad\qquad}{}\!\!\!\rightarrow\\#1& & #2\end{array}}$ $\newcommand{\outsideeq}[2]{\begin{array}{rcl}\leftarrow\!\!\!\frac{\qquad\qquad}{}\!\bullet & ---- &\bullet\!\frac{\qquad\qquad}{}\!\!\!\rightarrow\\#1& & #2\end{array}}$

## Solution of Question 1(a)

\begin{aligned} &f(x)=x^{2}-1, g(x)=3 x+1 \\ &(g \circ f)(x)=7 x-4 \\ &g(f(x))=7 x-4^{-} \\ &g\left(x^{2}-1\right)=7 x-4 \\ &3\left(x^{2}-1\right)+1=7 x-4 \\ &3 x^{2}-3+1=7 x-4 \\ &3 x^{2}-2=7 x-4 \\ &3 x^{2}-7 x+2=0 \\ &(3 x-1)(x-2)=0 \\ &3 x-1=0 \text { or } x-2=0 \\ &x=\frac{1}{3} \text { or } x=2 \end{aligned}

## Solution of Question 1(b)

$f(x)=(x-1)^{3}+6(p x+4)^{2}$
When $\mathrm{f}(\mathrm{x})$ is divided by $\mathrm{x}+2$ remainder $=-3$. $$\begin{array}{l}\therefore \mathrm{f}(-2)=-3\\(-2-1)^{3}+6(-2 \mathrm{p}+4)^{2}=-3\\-27+6(-2 \mathrm{p}+4)^{2}=-3\\6(-2 \mathrm{p}+4)^{2}=24\\(-2 \mathrm{p}+4)^{2}=4\\-2 \mathrm{p}+4=\pm 2\\ -2 \mathrm{p}+4=2\text{ or }-2 \mathrm{p}+4=-2\\-2 \mathrm{p}=-2\text{ or }-2 \mathrm{p}=-6\\\mathrm{p}=1\text{ or }\mathrm{p}=3 \end{array}$$

## Solution of Question 2(a)

In the expansion of $\left(x-\frac{3}{x}\right)^{14}$ \begin{aligned} (\mathrm{r}+1) \text {-th term } &={ }^{14} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{14-\mathrm{r}}\left(-\frac{3}{\mathrm{x}}\right)^{\mathrm{r}} \\ &={ }^{14} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{14-\mathrm{r}}(-3)^{\mathrm{r}} \mathrm{x}^{-\mathrm{r}} \\ &={ }^{14} \mathrm{C}_{\mathrm{r}}(-3)^{\mathrm{r}} \mathrm{x}^{14-2 \mathrm{r}} \end{aligned} For the term containing $\mathrm{x}^{6}$, \begin{aligned} & 14-2 \mathrm{r}=6 \\ &\begin{aligned} -2 \mathrm{r} &=-8 \\ \mathrm{r}=& 4 \\ \text { coefficient of } \mathrm{x}^{6} &={ }^{14} \mathrm{C}_{\mathrm{r}}(-3)^{4} \\ &=\frac{14 \cdot 13 \cdot 12 \cdot 11}{1 \cdot 2 \cdot 3 \cdot 4} \cdot 81 \\ &=81081 \end{aligned} \end{aligned}

## Solution of Question 2(b)

In the A.P. $9,7,5, \ldots$ $\qquad a=9,d=7-9=-2,S_n=24.$, $$\begin{array}{rll} \therefore \frac{\mathrm{n}}{2}\{2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}\}&=24\\ \frac{\mathrm{n}}{2}\{2 \times 9+(\mathrm{n}-1)(-2)\}&=24\\ \frac{\mathrm{n}}{2}\{18-2 \mathrm{n}+2\}&=24\\ \frac{\mathrm{n}}{2}\{20-2 \mathrm{n}\}&=24\\ 10 \mathrm{n}-\mathrm{n}^{2}&=24\\ 0&=\mathrm{n}^{2}-10 \mathrm{n}+24\\ 0&=(\mathrm{n}-4)(\mathrm{n}-6)\\ \mathrm{n}-4&=0\text{ or }\mathrm{n}-6=0\\ \mathrm{n}=4\text{ or }\mathrm{n}=6 \end{array}$$ $\therefore 4$ terms or 6 terms of the given A.P. add up to $24 .$

## Solution of Question 3(a)

$$\begin{array}{rl} \left(\begin{array}{ll}a & b \\ 0 & c\end{array}\right) &=2\left(\begin{array}{rr}1 & 4 \\ -6 & 3\end{array}\right)+\left(\begin{array}{rr}1 & 2 \\ -1 & 3\end{array}\right)\left(\begin{array}{rr}3 & 4 \\ 5 & -4\end{array}\right)\\ &=\left(\begin{array}{rr}2 & 8 \\ -12 & 6\end{array}\right)+\left(\begin{array}{cc}13 & -4 \\ 12 & -16\end{array}\right)\\ &=\left(\begin{array}{cc}15 & 4 \\ 0 & -10\end{array}\right)\\ \end{array}$$ $\therefore a=15, b=4, c=-10$

## Solution of Question 3(b)

Solution: When a die is thrown,
the set of possible outcomes $=\{1,2,3,4,5,6\}$
number of possible outcomes $=6$.
Let the number of favourable outcomes=$k$.
Then, $\mathrm{P}($ the number turned up $< \mathrm{x})=\frac k6=\frac{2}{3}$
$\therefore k=4$
$\therefore$ There are exactly 4 numbers in the set $\{1,2,3,4,5,6\}$ which are less than $x.$
$\therefore$ the favourable outcome= $\{1,2,3,4\}$ and $4\le x < 5.$

## Solution of Question 4(a)

$\begin{array}{llrl} \text{Given:}& XY \text{ is tangent at }C\\ \text{To Prove:}&XY//DE\\ \text{Proof:}& \text{Draw }AB\end{array}$ $$\begin{array}{rll} \alpha&=\beta &(XY\text{ is a tangent at C and CA is a chord)}\\ \beta&=\gamma &ABED \text{ ia cyclic}\\ \therefore \alpha&=\gamma\\ \therefore &XY// DE&\text{ (alternate angles are equal)} \end{array}$$

## Solution of Question 4(b)

\begin{aligned} &\mathrm{A}=(1,0), \mathrm{B}=(4,2), \mathrm{C}=(5,4) \\ &\mathrm{ACBD} \text { is a parallelogram. } \end{aligned}
$\therefore \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{CB}}$
$\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OC}}$
$\overrightarrow{\mathrm{OD}}-\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}4 \\ 2\end{array}\right)-\left(\begin{array}{l}5 \\ 4\end{array}\right)$
$\overrightarrow{\mathrm{OD}}-\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}-1 \\ -2\end{array}\right)$ $\overrightarrow{\mathrm{OD}}=\left(\begin{array}{l}-1 \\ -2\end{array}\right)+\left(\begin{array}{l}1 \\ 0\end{array}\right)$
$\overrightarrow{\mathrm{OD}} \cdot=\left(\begin{array}{c}0 \\ -2\end{array}\right)$
$\therefore \quad \mathrm{D}=(0,-2)$

## Solution of Question 5(a)

\begin{aligned} &\cos ^{2} x=2+\cos x \\ &\cos ^{2} x-\cos x-2=0 \\ &(\cos x-2)(\cos x+1)=0 \\ &\cos x-2=0 \text { or } \cos x+1=0 \\ &\cos x=2 \text { or } \cos x=-1 \\ &\cos x=2 \text { is impossible. } \quad(\because-1 \leq \cos x \leq 1) \\ &\cos x=-1 \\ &\therefore x=180^{\circ} \end{aligned}

## Solution of Question 5(b)

$\quad$ Let $f(x)=x^{2}-3 x$ \begin{aligned} f^{\prime}(x) &=\lim _{x \rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x} \\ &=\lim _{x \rightarrow 0} \frac{\left[(x+\delta x)^{2}-3(x+6 x)\right]-\left[x^{2}-3 x\right]}{\delta x} \\ &=\lim _{x \rightarrow 0} \frac{x^{2}+2 x \delta x+(8 x)^{2}-3 x-38 x-x^{2}+3 x}{8 x} \\ &=\lim _{\alpha \rightarrow 0} \frac{2 x \delta x+(8 x)^{2}-38 x}{8 x} \\ &=\lim _{\alpha \rightarrow 0}[2 x+\delta x-3] \\ &=2 x+0-3 \\ &=2 x-3 \end{aligned}

## Solution of Question 6(a)

$$\begin{array}{ll} \text{Let }\left(f^{-1} \circ g\right)(x)=y\\ \text{Then, } f^{-1}(g(x))=y\\ g(x)=f(y)\\ x+2=3y-1\\y=\frac{x+3}{3}\\ \therefore \left(f^{-1} \circ g\right)(x)=\frac{x+3}{3}&\cdots (1)\\ \text{Let }\left(g^{-1} \cdot f\right)(x)=z\\ \text{Then, }g^{-1}(f(x))=z\\ f(x)=g(z)\\ 3x-1=z+2\\ z=3x-3\\ \therefore \left(g^{-1} \cdot f\right)(x)=3x-3&\cdots (2) \end{array}$$
Since $\left(f^{-1} \circ g\right)(x)=\left(g^{-1} \cdot f\right)(x)-4,$ \begin{eqnarray*} \frac{x+3}{3}&=& 3x-3-4\qquad\text{(by (1) and (2))}\\ x+3&=& 9x-21\\ 8x&=& 24\\x&=&3 \end{eqnarray*}

## Solution of Question 6(b)

Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{bx}^{2}-5 \mathrm{x}+2 \mathrm{a}$
$f(x)$ is divisible by $x^{2}-3 x-4$
$x^{2}-3 x-4$ is a factor of $f(x)$
$x^{2}-3 x-4=(x+1)(x-4)$
$\therefore x-4$ and $x+1$ are factors of $f(x)$
Thus $f(4)=0$. $$\begin{array}{rll} a(64)+b(16)-5(4)+2 a&=0\\ 64 a+16 b-20+2 a&=0\\ 66 a+16 b&=20\\ 33 a+8 b&=10&\cdots (1)\\ f(-1)&=0\\ a(-1)+b(1)-5(-1)+2 a&=0\\ -a+b+5+2 a&=0\\ a+b&=-5&\cdots (2)\\ (2) \times 8: 8 a+8 b&=-40&\cdots (3) \end{array}$$
By subtracting Eq.(1) and Eq.(3), $$\begin{array}{c}25 \mathrm{a}=50 \\ \mathrm{a} =2 \text {. } \\ \text { By substituting } \mathrm{a}=2 \text { in Eq. }(2) \text {, }\end{array}$$ $$\begin{array}{r} 2+b=-5 \\ b=-7 \\ f(x)=-2 x^{3}-7 x^{2}-5 x+4 \end{array}$$ When $f(x)$ is divided by $(x+2)$ \begin{aligned} \text { remainder } &=f(-2) \\ &=-16-28+10+4 \\ &=-30 . \quad \end{aligned}

## Solution of Question 7(a)

\begin{aligned}x\odot y&=x^2+y^2\\1\odot 3 &=1^2+3^2=10\\ (1\odot 3)\odot 2&=10 \odot 2 \\ &=10^{2}+2^{2} \\&=104\end{aligned} $$\begin{array}{rll} 3 \odot 2&=3^{2}+2^{2}=13\\ 1 \odot(3 \odot 2)&=1 \odot 13\\ &=1^{2}+13^{2}\\ &=170\end{array}$$ $$[(1 \odot 3) \odot 2]+[1 \odot(3,2)]=104+170=274$$ $$\begin{array}{rll} x\odot(y\odot x)&=x\odot (y^2+x^2)\\ &=x^{2}+\left(y^{2}+x^{2}\right)^{2}\\ &=x^{2}+y^{4}+2 y^{2} x^{2}+x^{4}&\cdots (1)\\ (x \odot y) \odot x&=\left(x^{2}+y^{2}\right) \odot x\\ &=\left(x^{2}+y^{2}\right)^{2}+x^{2}\\ &=x^{4}+2 x^{2} y^{2}+y^{4}+x^{2}&\cdots(2) \end{array}$$ By Eq.(1) and (2) $$x \odot(y \odot x)=(x \odot y) \odot x . \quad$$

## Solution of Question 7(b)

In the expansion of $(1+x)^{2 n},$
$(\mathrm{k}+1)$ - th term $={ }^{2 n} \mathrm{C}_{\mathrm{k}} 1^{2 \mathrm{n}-\mathrm{k}} \mathrm{x}^{\mathrm{k}}$ $={ }^{2 n} C_{k} x^{k}$
$\therefore$ coefficient of $\mathrm{x}^{\mathrm{k}}={ }^{2 n} \mathrm{C}_{\mathrm{k}}$
coefficient of $\mathrm{x}^{r}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}$
coefficient of $\mathrm{x}^{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{r+2}$
By the given condition} \begin{aligned} &{ }^{2 n} \mathrm{C}_{\mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+2} \\ \therefore \quad & \mathrm{r}+{2}=\mathrm{r} \text { or } \mathrm{r}+2=2 \mathrm{n}-\mathrm{r} \\ & \mathrm{r}+2=\mathrm{r} \text { is impossible } \\ \therefore & \mathrm{r}+2=2 \mathrm{n}-\mathrm{r} \\ & 2 \mathrm{r}=2 \mathrm{n}-2 \\ & \mathrm{r}=\mathrm{n}-1 \end{aligned}

## Solution of Question 8(a)

\begin{aligned} & x^{2}-2 x \leq 0 \\ & x(x-2)\leq 0 \\ &(x \geq 0 \text { and } x-2 \leq 0) \text { or }(x \leq 0 \text { and }x-2\geq 0) \\ &(x \geq 0 \text { and } x \leq 2) \text { or }\{x \leq 0 \text { and } x \geq 2) \end{aligned}There is no real number which satifies the second case. $$\begin{array}{r} \therefore x \geq 0 \text { and } x \leq 2 \\ 0 \leq x \leq 2 \end{array}$$ solution set $\{x \mid 0 \leq x \leq 2 \}$
Illustration of the solution set
$\insideeq 02$

## Solution of Question 8(b)

\begin{aligned} S_6&=96\\ \frac{6}{2}(2 a+5 d)&=96 \\ 3(2 a+5 d )&=96 \\ 2 a+5 d&=32&\cdots (1) \end{aligned} \begin{aligned} \mathrm{S}_{10}&=\frac{1}{3} \mathrm{~S}_{20}\\ 3S_{10}&=S_{20}\\ 3\times \frac{10}{2}\{2 \mathrm{a}+9 \mathrm{~d}\}&=\cdot \frac{20}{2}\{2 \mathrm{a}+19 \mathrm{~d}\}\\ 15\{2 \mathrm{a}+9 \mathrm{~d}\}&=10\{2 \mathrm{a}+19 \mathrm{~d}\}\\ 15\{2 \mathrm{a}+9 \mathrm{~d}\}&=10\{2 \mathrm{a}+19 \mathrm{~d}\}\\ 3\{2 \mathrm{a}+9 \mathrm{~d}\}&=2\{2 \mathrm{a}+19 \mathrm{~d}\}\\ 2a-11d&=0&\cdots (2)\\ (1)-(2): & 16 \mathrm{~d}=32\\ &\mathrm{d}=2\\ \text { By substituting } \mathrm{d}=2 \text { in (1) } \\ & 2 \mathrm{a}+5 \times 2=32 \\ 2 \mathrm{a} &=22 \\ \mathrm{a} &=11 \\ \mathrm{u}_{10} &=\mathrm{a}+9 \mathrm{~d} \\ &=11+9 \times 2 \\ &=29 \end{aligned}

## Solution of Question 9(a)

\begin{aligned} S_{n} &=\frac{1}{2}\left(3^{n}-1\right) \\ S_{1} &=\frac{1}{2}\left(3^{1}-1\right)=1 \\ S_{2} &=\frac{1}{2}\left(3^{2}-1\right)=4 \\ S_{3} &=\frac{1}{2}\left(3^{3}-1\right)=13 \\ u_{1} &=S_{1}=1 \\ u_{2} &=S_{2}-S_{1}=4-1=3 \\ u_{3} &=S_{3}-S_{2}=13-4=9 \\ u_{n} &=S_{n}-S_{n-1} \\ &=\frac{1}{2}\left(3^{n}-1\right)-\frac{1}{2}\left(3^{n-1}-1\right) \end{aligned} \begin{aligned} &=\frac{1}{2}\left(3^{n}-1-3^{n-1}+1\right) \\ &=\frac{1}{2}\left(3^{n}-3^{n-1}\right) \\ &=\frac{1}{2} \cdot 3^{n-1}\left(3^{1}-1\right) \\ &=3^{n-1} \end{aligned}

## Solution of Question 9(b)

Let $\mathrm{A}=\left(\begin{array}{ll}3 & 5 \\ 2 & 2\end{array}\right)$ and $\mathrm{A}^{-1}=\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right)$ By the definition of inverse of a matrix, $$\begin{array}{rll} A A^{-1} &=I \\ \therefore\left(\begin{array}{ll} 3 & 5 \\ 2 & 2 \end{array}\right)\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) &=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ \left(\begin{array}{ll} 3 a+5 c & 3 b+5 d \\ 2 a+2 c & b+2 d \end{array}\right) &=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ \therefore 3 a+5 c &=1 &\cdots (1)\\ 3 b+5 d &=0&\cdots (2)\\ 2 a+2 c&=0&\cdots (3) \\ 2 b+2 d&=1&\cdots (4)\\ (1)\times 2:6a+10c&=2&\cdots (5)\\ (3)\times 3:6a+6c&=0&\cdots (6)\\ (5)-(6):4c&=2\\ c&=\frac{1}{2} \end{array}$$ By substituting $\mathrm{c}=\frac{1}{2}$ in Eq (3), $$2 a-2 \times \frac{1}{2}=0\Rightarrow a=-\frac{1}{2}$$
$\mathrm{Eq}(2) \times 2: 6 \mathrm{~b}+10 \mathrm{~d}=0^{2}$
Eq. (4) $\times 3: 6 b+6 d=3$
By subtracting these equations, we get \begin{aligned} 4 d &=-3 \\ d &=-\frac{3}{4} \end{aligned} By substituting $\mathrm{d}=-\frac{3}{4}$ in Eq. (4), $$\begin{gathered} 2 \mathrm{~b}+2 \times\left(-\frac{3}{4}\right)=1 \\ \qquad b=\frac{5}{4} \\ \therefore \mathrm{A}^{-1}=\left(\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right)=\left(\begin{array}{rr} -\frac{1}{2} & \frac{5}{4} \\ \frac{1}{2} & -\frac{3}{4} \end{array}\right) \end{gathered}$$

## Solution of 10(a)

\begin{aligned} &\text { Let } \mathrm{A}=\left(\begin{array}{cc} 3 & 4 \\ 2 & 6 \end{array}\right) \cdot \\ &\operatorname{det} \mathrm{A}=18-8=10 \\ &\mathrm{~A}^{-1}=\frac{1}{10}\left(\begin{array}{cc} 6 & -4 \\ -2 & 3 \end{array}\right) \\ &=\left(\begin{array}{rr} \frac{3}{5} & -\frac{2}{5} \\ \frac{-1}{5} & \frac{3}{10} \end{array}\right) \end{aligned} Next, we will solve the system of equations \begin{aligned} & 3 x+4 y=18 \\ & 2 x+6 y=22 \end{aligned} In matrix form, we get \begin{aligned} &\qquad\left(\begin{array}{ll} 3 & 4 \\ 2 & 6 \end{array}\right)\left(\begin{array}{l} \mathrm{x} \\ \mathrm{y} \end{array}\right)=\left(\begin{array}{l} 18 \\ 22 \end{array}\right) \\ &\text { Let } \mathrm{X}=\left(\begin{array}{l} \mathrm{x} \\ \mathrm{y} \end{array}\right), \mathrm{B}=\left(\begin{array}{l} 18 \\ 22 \end{array}\right) \end{aligned} Then we have \begin{aligned} &\mathrm{AX}=\mathrm{B} \\ &\mathrm{A}^{-1} \mathrm{AX}=\mathrm{A}^{-1} \mathrm{~B} \\ &\mathrm{I} \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \end{aligned} $$\begin{array}{rll} \left(\begin{array}{l}x \\ y\end{array}\right) &=\frac{1}{10}\left(\begin{array}{ll}6 & -4 \\ -2 & 3\end{array}\right)\left(\begin{array}{l}18 \\ 22\end{array}\right)\\ &=\frac{1}{10}\left(\begin{array}{ll}6\times 18+(-4)\times 22 \\ -2\times 18+3\times 22 \end{array}\right)\\ &=\frac{1}{10}\left(\begin{array}{l}20 \\ 30\end{array}\right)\\ &=\left(\frac{2}{3}\right)\\ x=2, &y=3 .\end{array}$$ $\therefore$ The point of intersection of the given two lines is $(2,3)$.

## Solution of Question 10(b)

Blue Die $\begin{array}{c}\text{Black Die}\\ \begin{array}{|c||c|c|c|c|c|c|} \hline &1&2&3&4&5&6\\ \hline 1&(1,1)& (1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline 2&(2,1)& (2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline 3&(3,1)& (3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline 4&(4,1)& (4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline 5&(5,1)& (5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline 6&(6,1)& (6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline \end{array}\\ \end{array}$

number of possible outcomes $=36$
$\mathrm{P}$ (score on the blue die is less than that on the black) $=$ ?
set of favourable outcomes $=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5)$ $(2,6),(3,4)(3,5),(3,6),(4,5),(4,6),(5,6)\}$
number of favourable outcomes $=15$
$P$ (score on the blue die is less than that on the black) $=\frac{15}{36}=\frac{5}{12}$
P (score on the blue die is prime and that on the black die is even) $=$ ?
set of favourable outcomes $=\{(2,2),(2,4),(2,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}$
number of favourable outcomes $=9$
P (score on the blue die is
prime and that on the black is even $)=\frac{9}{36}=\frac{1}{4}$

## Solution of Question 11(a)

Given: In the figure $\mathrm{O}$ is the center of the circle and $\mathrm{OD} \perp \mathrm{BC}$.
To Prove: $\angle B O D=\angle B A C$
Proof: Draw $OC$. Since $OD\perp BC,\angle ODB=\angle ODC$. $\triangle BOD$ and $\triangle COD,$ $$\begin{array}{rll} OD&=OD&\text{(Common side)}\\ OB&=OC&\text{(Radius)}\\ \angle BDO&=\angle CDO=90^{\circ}\\ \triangle BOD &\cong \triangle COD &(\text{HL}) \end{array}$$ Thus $\alpha=\gamma.$ $$\begin{array}{rll} \angle BOC&=\alpha+\gamma\\ &=2\alpha\\&=2\beta&(\because \angle BOC=2\angle BAC)\\ \therefore\alpha&=\beta\\ \therefore\angle BOD&=\angle BAC\end{array}$$

## Solution of Question 11(b)

Given : $\triangle \mathrm{PQR}$ with two medians $\mathrm{PM}$ and $\mathrm{QN}$ intersecting at $\mathrm{K}$.
To Prove : $\alpha(\Delta \mathrm{PNK})=\alpha(\Delta \mathrm{QMK})$
Proof : Since $QM$ and $QR$ are the bases of $\triangle P Q M$ and $\triangle P Q R,$ these triangles have the same attitude. \begin{aligned} \therefore \frac{\alpha(\Delta \mathrm{PQM})}{\alpha(\Delta \mathrm{PQR})} &=\frac{\mathrm{QM}}{\mathrm{QR}} \\ &=\frac{\mathrm{QM}}{2 \mathrm{QM}}&(\because \mathrm{PM} \text { is a median of } \triangle \mathrm{PQR})\\ &=\frac{1}{2}\\ \therefore \alpha(\Delta \mathrm{PQM})&=\frac{1}{2} \alpha(\Delta \mathrm{PQR})\qquad (1) \end{aligned} Similarly we can show that $$\begin{array}{rll} \alpha(\Delta \mathrm{PQN})&=\frac{1}{2} \alpha(\Delta \mathrm{PQR})&(2)\\ \alpha(\triangle \mathrm{PQN})&=\alpha(\triangle \mathrm{PQM})&(\because \text{ by (1) and (2) })\\ \therefore \alpha(\Delta \mathrm{PQN})-\alpha(\Delta \mathrm{PQK})&=\alpha(\Delta \mathrm{PQM})-\alpha(\Delta \mathrm{PQK})\\ \alpha(\Delta \mathrm{PNK})&=\alpha(\Delta \mathrm{QMK}) \end{array}$$

## Solution of Question 12(a)

 Given: Two circles cut at $\mathrm{C}, \mathrm{D}$ and through $\mathrm{C}$ any line $\mathrm{ACB}$ is drawn to meet the circles at $A, B \cdot A D$ and $B D$ are joined and produced to meet the circles again at $E, F$. AF, BE produced meet at $G$.
To Prove: $\quad D, F, G, E$ are concyclic Proof : Draw $DC$ $$\begin{array}{rll} \alpha&=\theta&(ACDF\text{ is cyclic.)}\\ \beta&=\phi&(CBED\text{ is cyclic.)}\\ \therefore \alpha+\beta&=\theta+\phi\\ \alpha+\beta&=180^{\circ}&(ACB\text{ is a line segment})\\ \therefore \theta+\phi&=180^{\circ}\\ \therefore D,F,G,E &\text{ are concyclic.} \end{array}$$

## Solution of Question 12(b)

Proof:
$\begin{array}{lll}\quad& 4 \sin \alpha \sin \beta \sin \gamma \\&=(2 \sin \alpha \sin \beta)(2 \sin r) \\&=(\cos (\alpha-\beta)-\cos (\alpha+\beta))(2 \sin \gamma) &\mbox{(Product to Sum)}\\&=2 \cos (\alpha-\beta) \sin \gamma-2 \cos (\alpha+\beta) \sin \gamma \\&=\sin (\alpha-\beta+\gamma)-\sin (\alpha-\beta-\gamma)\\&\quad -(\sin (\alpha+\beta+\gamma)-\sin (\alpha+\beta-\gamma)) &\mbox{(Product to Sum)}\\&=\sin \left(180^{\circ}-2 \beta\right)-\sin \left(2 \alpha-180^{\circ}\right)\\&\quad -\sin \left(180^{\circ}\right)+\sin (180-2 \gamma) &\mbox{ (Trigular Properties)}\\&=\sin 2 \beta+\sin 2 \alpha+\sin 2 \gamma &\mbox{(Basic Trigonometric Identities)}\end{array}$

## Solution of Question 13(a)

\begin{aligned} &\angle \mathrm{A}: \angle \mathrm{B}: \angle \mathrm{C}=1: 3: 8 \\ &\therefore \angle \mathrm{A}=\mathrm{x}^{\circ}, \angle \mathrm{B}=3 \mathrm{x}^{\circ}, \angle \mathrm{C}=8 \mathrm{x}^{\circ} \text { for some } \mathrm{x}>0 . \\ &\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\ &\therefore \quad \mathrm{x}+3 \mathrm{x}+8 \mathrm{x}=180 \\ &\quad 12 \mathrm{x}=180 \\ &\quad \mathrm{x}=15 \\ &\therefore \quad \angle \mathrm{A}=15^{\circ} \\ &\angle \mathrm{B}=3 \times 15^{\circ}=45^{\circ} \\ &\angle \mathrm{C}=8 \times 15^{\circ}=120^{\circ} \end{aligned} By the law of sines \begin{aligned} \frac{A C}{\sin B} &=\frac{A B}{\sin C} \\ \frac{A C}{\sin 45^{\circ}} &=\frac{9}{\sin 120^{\circ}} \\ A C &=\frac{9 \sin 45^{\circ}}{\sin 120^{\circ}} \\ &=\frac{9 \sin 45^{\circ}}{\sin 60^{\circ}}\\ &= \frac{9 \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} \\ & =-\frac{9 \sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ & =\frac{9 \sqrt{6}}{3} \\ & =3 \sqrt{6} \end{aligned}

## Solution of Question 13(b)

\begin{aligned} &y=\ln \left(\sin ^{3} 2 x\right) \\ &\therefore y=3 \ln (\sin 2 x) \\ &\frac{d y}{d x}=3 \cdot \frac{1}{\sin 2 x} \cos 2 x \cdot 2 \\ &=6 \cot 2 x \\ &\begin{aligned} \frac{d^{2} y}{d x^{2}} &=6\left(-\operatorname{cosec}^{2} 2 x\right) 2 \\ &=-12 \operatorname{cosec}^{2} 2 x \\ 3 \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}+36 &=3\left(-12 \operatorname{cosec}^{2} 2 x\right)+(6 \cot 2 x)^{2}+36 \\ &=-36 \operatorname{cosec}^{2} 2 x+36 \cot ^{2} 2 x+36 \\ &=-36 \operatorname{cosec}^{2} 2 x+36\left(\cot ^{2} 2 x+1\right) \\ &=-36 \operatorname{cosec}^{2} 2 x+36 \operatorname{cosec}^{2} 2 x \\ &=6 \end{aligned} \end{aligned}

## Solution of Question 14(a)

$\def\Vec{\overrightarrow}\def\mathrm{}$
Given : $ABCD$ is a parallelogram.
To Prove: diagonals $\mathrm{AG}$ and $\mathrm{BD}$ bisect each other.
Proof: Let $AC$ and $BD$ meet at $O$, $AO=h\,OC,BO=k\,OD$. $$\begin{array}{rll} \Vec{DC}&=\Vec{AB}&(\because ABCD \text{ is a parallelogram}\\ \therefore \Vec{DO}+\Vec{OC}&= \Vec{AO}+\Vec{OB}\\ &=h\,\Vec{OC}+k\,\Vec{DO}\end{array}$$
Since $\overrightarrow{\mathrm{OC}} \neq \overline{0}, \overrightarrow{\mathrm{DO}} \neq \overline{0}$ and $\overrightarrow{\mathrm{OC}} \nparallel\overrightarrow{\mathrm{DO}}$, we have $$\begin{gathered} \mathrm{h}=1, \mathrm{k}=1 \\ \therefore \quad \mathrm{AO}=\mathrm{hOC}=\mathrm{OC}, \mathrm{BO}=\mathrm{kOD}=\mathrm{OD} \end{gathered}$$ $\therefore \mathrm{A} \mathrm{C}$ and $\mathrm{BD}$ bisect each other.

## Solution of Question 14(b)

Equation of the given curve is $$2 x y+\pi \sin y=2 \pi\qquad\cdots (1)$$ At the point $\left(1, \frac{\pi}{2}\right)$, \begin{aligned} 2 x y+\pi \sin y &=2 \times 1 \times \frac{\pi}{2}+\pi \sin \frac{\pi}{2} \\ &=\pi+\pi \times 1 \\ &=2 \pi \end{aligned} $\therefore$ The point $\left(1, \frac{\pi}{2}\right)$ lies on the given curve.
From Eq.(1) we get
$\frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{xy}+\pi \sin \mathrm{y})=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(2 \pi)$
$2 y+2 x \frac{d y}{d x}+\pi \cos y \frac{d y}{d x}=0$
At the point $\left(1, \frac{\pi}{2}\right)$, $$\begin{array}{rll} 2\left(\frac{\pi}{2}\right)+2(1)\frac{dy}{dx}+\pi\cos\left(\frac{\pi}{2}\right)\frac{dy}{dx}&=0\\ \pi+2\frac{dy}{dx}&=0& (\because \cos\left(\frac{\pi}{2}\right)=0)\\ \frac{dy}{dx}&=-\frac{\pi}{2}&\ldots(2)\end{array}$$
$\therefore$ gradient of the tangent at the point $\left(1, \frac{\pi}{2}\right)=-\frac{\pi}{2}$
$\therefore$ equation of the tangent at $\left(1, \frac{\pi}{2}\right)$ is \begin{aligned} &y-\frac{\pi}{2}=-\frac{\pi}{2}(x-1) \\ &y-\frac{\pi}{2}=-\frac{\pi}{2} x+\frac{\pi}{2} \\ &y=-\frac{\pi}{2} x+\pi \end{aligned} By Eq (2), gradient of the normal at the point $\left(1, \frac{\pi}{2}\right)=\frac{2}{\pi}$.
$\therefore$ equation of the normal at the point $\left(1, \frac{\pi}{2}\right)$ is \begin{aligned} &y-\frac{\pi}{2}=\frac{2}{\pi}(x-1) \\ &y-\frac{\pi}{2}=\frac{2}{\pi} x-\frac{2}{\pi} \\ &y=\frac{2}{\pi} x+\left(\frac{\pi}{2}-\frac{2}{\pi}\right) \end{aligned}