Myanmar Matriculation 2019 D Solution

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



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Solution 1


(a) Solution: $\quad f(x)=x+1, g(x)=2 x^{2}-x+3$

$(f \circ g) (x)=4 x+1$

$f(g(x))=4 x+1$

$f(2x^2-x+3)=4x+1$

$2x^2-x+3+1=4 x+1$

$2 x^{2}-5 x+3=0$

$(2 x-3)(x-1)=0$

$2 x-3=0$ or $x-1=0$

$x=\frac{3}{2}$ or $x=1 \square$


(b) Solution : Let $f(x)=2 x^{2}+5 x-3$. When $\mathrm{f}(\mathrm{x})$ is divided by $2 \mathrm{x}-\mathrm{p},$ $\begin{array}{rrlll} &\text { remainder }&=2 \mathrm{p}^{2}-3 \mathrm{p} \\ &\therefore \mathrm{f}\left(\frac{\mathrm{p}}{2}\right)&=2 \mathrm{p}^{2}-3 \mathrm{p} \\ &2 \cdot \frac{\mathrm{p}^{2}}{4}+5 \cdot \frac{\mathrm{p}}{2}-3&=2 \mathrm{p}^{2}-3 \mathrm{p} \\ &\frac{\mathrm{p}^{2}}{2}+\frac{5 \mathrm{p}}{2}-3-2 \mathrm{p}^{2}+3 \mathrm{p}&=0 \\ &\mathrm{p}^{2}+5 \mathrm{p}-6-4 \mathrm{p}^{2}+6 \mathrm{p}&=0 \\ &-3 \mathrm{p}^{2}+11 \mathrm{p}-6&=0 \\ &3 \mathrm{p}^{2}-11 \mathrm{p}+6&=0 \\ &(3 \mathrm{p}-2)(\mathrm{p}-3)&=0 \\ &3 \mathrm{p}-2&=0 \text { or } \mathrm{p}-3=0 \\ &\mathrm{p}=\frac{2}{3} \text { or } \mathrm{p}&=3\square \end{array}$




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Solution 2


(a) Solution : In the expansion of $\left(x^{2}+\frac{2}{x}\right)^{8}$, $$\begin{array}[t]{ll}(\mathrm{r}+1)^{\text{th}} \text { term } &={ }^{8} \mathrm{C}_{r}\left(\mathrm{x}^{2}\right)^{8-r}\left(\frac{2}{\mathrm{x}}\right)^{r} \\ &={ }^{8} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{16-2 r} 2^{\mathrm{r}} \mathrm{x}^{-r} \\ &={ }^{8} \mathrm{C}, 2^{\mathrm{r}} \mathrm{x}^{16-3 \mathrm{r}} \end{array}$$
The term containing $x^7$, we set $16-3 r=7$ $$\begin{aligned}-3 r &=-9 \\ r &=3 . \\ \therefore \text { coefficient of } x^{7} &={ }^{8} C_{3} 2^{3} \\ &=\frac{8 \times 7 \times 6}{1 \times 2 \times 3} \times 8 \\ &=448 \square \end{aligned}$$

(b) Solution: the required sum $=70+72+74+\ldots+146+148$
Since $72-70=74-72=\ldots .=148-146=2,$
the above series is an A. P. with $\mathrm{a}=70$,$d=2$ ,$u_n=148 .$
$\qquad$$\begin{array}{l}u_{n}=148 \\ a+(n-1) d=148 \\ 70+(n-1) 2=148 \\ (n-1) 2=78 \\ n-1=39 \\ n=40\end{array}$
The required sum $=\frac{40}{2}(a+l)=20(70+148)=4360 \square$


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Solution 3


(a) Solution : $$\begin{aligned}&A=\left(\begin{array}{ll}2 & 0 \\0 & 5\end{array}\right), B=\left(\begin{array}{ll} x & y \\ 0 & z \end{array}\right) \\ &A B=A+B \\ &\therefore\left(\begin{array}{ll} 2 & 0 \\ 0 & 5 \end{array}\right)\left(\begin{array}{ll} x & y \\ 0 & z \end{array}\right)=\left(\begin{array}{ll} 2 & 0 \\ 0 & 5 \end{array}\right)+\left(\begin{array}{ll} x & y \\ 0 & z \end{array}\right) \\ &\quad\left(\begin{array}{ll} 2 x & 2 y \\ 0 & 5 z \end{array}\right)=\left(\begin{array}{cc} 2+x & y \\ 0 & 5+z \end{array}\right) \\ &\therefore 2 x=2+x,2 y=y,5 z=5+z \end{aligned}$$ Thus, $x=2,y=0,z=\frac{5}{4}\square$
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(b) Solution: When a die is thrown,
the set of possible outcomes $=\{1,2,3,4,5,6\}$
number of possible outcomes $=6$
$P$ (getting a number not less than $ x )=\frac{2}{3}\cdots (1)$
Case 1: If $x>3$, the possible sets of favourable outcomes are $\{4,5,6\},$ $\{5,6\},$ $\{6\},$ or $\emptyset$. Thus $P$ (getting a number not less than $ x )\le \frac 36$, which is impossible by (1).
Case 2:If $x\le 2$, the set of favourable outcomes are $=\{2,3,4,5,6\}$ or $\{1,2,3,4,5,6\}$. Therefore $P$ (getting a number not less than $ x )\ge \frac 56$, contradicts (1).
Hence required number is $\{x|2 < x\le 3\}$.
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Solution 4



(a) Given : $A T$ and $B T$ are tangents to the circle $A B C$ at $A$ and $B$.
To Prove : $\angle \mathrm{BTX}=2 \angle \mathrm{ACB}$
Proof: $\begin{array}[t]{lll} \alpha&=\beta+\gamma&(\alpha \mbox{ is an exterior angle of triangle }ABT) \\ \beta&=\gamma=\theta&(TA,TB \mbox{ are tangents and } AB \mbox{ is chord})\\ \therefore\alpha&=\theta+\theta=2\theta\\ \angle BTX&=2\angle ACB  \end{array}$


Click to Question Paper 2019 (D) (b) Solution :
$$\begin{array}{l}\mathrm{A}=(1,0), \mathrm{B}=(4,2), \mathrm{C}=\mathrm{(5,4)}\\ \quad \mathrm{ABCD} \text{is a parallelogram.}\\ \therefore \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BC}}\\ \overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}\\ \overrightarrow{\mathrm{OD}}-\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}5 \\ 4\end{array}\right)-\left(\begin{array}{l}4 \\ 2\end{array}\right)\\ \overrightarrow{\mathrm{OD}}-\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2\end{array}\right)\\ \overrightarrow{\mathrm{OD}}=\left(\begin{array}{l}1 \\ 2\end{array}\right)+\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l} 2 \\ 2 \end{array}\right) \\ \therefore \mathrm{D}=(2,2) \end{array}$$
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Solution 5


(a) Solution :
$ \begin{aligned} &2 \cos x \sin x=\sin x \\ &2 \cos x \sin x-\sin x=0 \\ &\sin x(2 \cos x-1)=0 \\ &\sin x=0 \text { or } 2 \cos x-1=0 \\ &\sin x=0 \text { or } \cos x=\frac{1}{2} \\ &\text { If } \sin x=0 \text {, then } \\ &\qquad x=0^{\circ} \text { or } 180^{\circ} \text { or } 360^{\circ} \\ &\text { If } \cos x=\frac{1}{2} \text {, then } \end{aligned}$
basic acute angle of $\mathrm{x}=60^{\circ} .$
Since $\cos x=\frac{1}{2}>0, x$ lies in first or fourth quadrant,
$\therefore \mathrm{x}=60^{\circ}$ or $300^{\circ}$
$\therefore x=0^{\circ},60^{\circ},180^{\circ},300^{\circ},360^{\circ}.$

(b) Solution : Let $f(x)=x^{3}+2 x$ \begin{eqnarray*} f^{\prime}(x)&=&\displaystyle\lim _{\delta x \rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x}\\ \qquad &=&\displaystyle\lim _{\delta x \rightarrow 0} \dfrac{(x+\delta x)^{3}+2(x+\delta x)-\left(x^{3}+2 x\right)}{\delta x}\\ \qquad &=&\displaystyle\lim _{\delta x \rightarrow 0} \dfrac{x^{3}+3 x^{2} \delta x+3 x(\delta x)^{2}+(\delta x)^{3}+2 x+2 \delta x-x^{3}-2 x}{\delta x}\\ \qquad &=&\displaystyle\lim _{\delta x \rightarrow 0} \dfrac{3 x^{2} \delta x+3 x(\delta x)^{2}+(\delta x)^{3}+2 \delta x}{\delta x}\\ \qquad &=&\displaystyle\lim _{\delta x \rightarrow 0}\left(3 x^{2}+3 x \delta x+(\delta x)^{2}+2\right)\\ \qquad &=&3 \mathrm{x}^{2}+3 \mathrm{x} \times 0+0^{2}+2\\ \qquad &=&3 x^{2}+2 \end{eqnarray*}
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Solution 6

(a) Solution:
(1)$$ \begin{aligned} &f(x)=2 x-1, g(x)=4 x+3 \\ &(g \circ f)(x)=g(f(x))=g(2 x-1) \\ &=4(2 x-1)+3 \\ &=8 x-1 \end{aligned} $$ (2) Let $g^{-1}(x)=y$ Then $g(y)=x$ $$ \begin{gathered} 4 y+3=x \\ y=\frac{x-3}{4} \end{gathered} $$ $$ \begin{aligned} &\therefore \text { By Eq. }(2) \\ &g^{-1}(x)=\frac{x-3}{4} \end{aligned} $$ (3) Let $(g \circ f)^{-1}(x)=z$Then $(g \circ f)(z)=x$ $$ \begin{aligned} &8 z-1=x \\ &z=\frac{x+1}{8} \\ &\therefore \text { By Eq. (3)} \\ &(g \circ f)^{-1}(x)=\frac{x+1}{8} \end{aligned} $$ (4) Let $f^{-1}(x)=w.$ Then $f(w)=x$ $$ \begin{aligned} &2 w-1=x \\ &w=\frac{x+1}{2} \\ &\therefore \text { By Eq. (4)} \\ &f^{-1}(x)=\frac{x+1}{2} \end{aligned} $$ (5) Now $\left(f^{-1} \circ g^{-1}\right)(x)=f^{-1}\left(g^{-1}(x)\right)$ $$ \begin{aligned} &=f^{-1}\left(\frac{x-3}{4}\right) \\ &=\frac{\frac{x-3}{4}+1}{2}=\frac{\frac{x+1}{4}}{2}=\frac{x+1}{8} \end{aligned} $$ By Eq (3) and $(5)(g \circ f)^{-1}(x)=\left(f^{-1} \circ g^{-1}\right)(x)$

(b) Solution : Let $f(x)=a x^{3}-x^{2}+b x-1$ $$ \begin{aligned} &\text { When } \mathrm{f}(\mathrm{x}) \text { is divided by } \mathrm{x}+2 \text {, remainder }=-33 \text {. } \\ &\therefore \quad \mathrm{f}(-2)=-33 \\ &\quad-8 \mathrm{a}-4-2 \mathrm{~b}-1=-33 \\ &\quad-8 \mathrm{a}-2 \mathrm{~b}=-28 \\ &4 \mathrm{a}+\mathrm{b}=14 &(1) \end{aligned} $$ When $\mathrm{f}(\mathrm{x})$ is divided by $\mathrm{x}-3$, remainder $=77$
$\therefore \quad f(3)=77$
 $27 a-9+3 b-1=77$
$27 a+3 b=87$
$9 a+b=29\qquad (2)$ 
By subtracting Eq(1) and $(2)$
 $ \begin{aligned} &5 a=15 \\ &a=3 \end{aligned} $
By substituting $\mathrm{a}=3$ in Eq (1),
$ \begin{array}{r} 12+b=14 \\ \quad b=2 \\ \therefore \quad f(x)=3 x^{3}-x^{2}+2 x-1 \end{array} $
 When $f(x)$ is divided by $x-2$
$ \begin{aligned} \text { remainder } &=f(2) \\ &=24-4+4-1 \\ &=23 \end{aligned} $
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Solution 7


(a) Solution
$$ \begin{aligned} &\text { For } x, y \in R \\ &x \odot y=(3 y-x)^{2}-8 y^{2} \\ &\therefore x \odot y=9 y^{2}-6 x y+x^{2}-8 y^{2} \\ &=x^{2}-6 x y+y^{2} \\ &\therefore y \odot x=y^{2}-6 y x+x^{2} \\ &=x^{2}-6 x y+y^{2} \\ &\therefore x \odot y=y \odot x \text { for all } x, y \in R \end{aligned} $$ $\therefore$ The binary operation $Q$ is commutative. $$\begin{array}{l}2\odot k=-31 \\ \therefore 2^{2}-6 \times 2 \times k+k^{2}=-31 \\ 4-12 k+k^{2}=-31 \\ k^{2}-12 k+35=0 \\ (k-7)(k-5)=0 \end{array}$$ $\therefore k=7$ or $k=3$
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(b) Solution
In the expansion of $(1+x)^{2 n}$ $$ \begin{aligned} (k+1) \text {-th term } &={ }^{2 n} C_{k} 1^{2 n-k} x^{k} \\ &={ }^{2 n} C_{k} x^{k} \end{aligned} $$ $\therefore$ coefficient of $\mathrm{x}^{k}={ }^{\mathrm{2n}} \mathrm{C}_{\mathrm{k}}$
coefficient of $\mathrm{x}^{\mathrm{r}}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}}^{ }$

coefficient of $\mathrm{x}^{\mathrm{r}+2}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+2}$
By the given condition $$ \begin{aligned} &{ }^{2 n} C_{r}={ }^{2 n} C_{r+2} \\ \therefore & r+2=r \text { or } r+2=2 n-r \\ & r+2=r \text { is impossible } \\ \therefore & r+2=2 n-r \\ & 2 r=2 n-2 \\ & r=n-1\qquad\square \end{aligned} $$
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Solution 8


(a) Solution:
$$ \begin{aligned} &x^{2}-3 x+2 \leq 0 \\ &(x-1)(x-2) \leq 0 \\ &(x-1 \geq 0 \text { and } x-2 \leq 0) \text { or }(x-1 \leq 0 \text { and } x-2 \geq 0) \\ &(x \geq 1 \text { and } x \leq 2) \text { or }(x \leq 1 \text { and } x \geq 2) \end{aligned} $$ Here the second case is impossible. $$ \begin{aligned} &\therefore \quad x \geq 1 \text { and } x \leq 2 \\ &\quad 1 \leq x \leq 2 \end{aligned} $$ $\therefore$ solution set $=\{x \mid 1 \leq x \leq 2\}\qquad\square$
illustration of the solution set:
$\insideeq{1}{2}$

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(b) Solution:
$$ \begin{aligned} &\text { For the given A. P. } 44,40,36, \text {. } \\ &\qquad \begin{aligned} {a} &=44, {~d}=40-44=-4 \\ \therefore \quad {S}_{12} &=\frac{12}{2}\{2 {a}+11 {~d}\} \\ &=6\{2 \times 44+11 \times(-4)\} \\ &=6\{88-44\} \\ &=264 \qquad\square \end{aligned} \end{aligned} $$ sum of the terms between ${u}_{12}$ and ${u}_{26}$\\ $$ \begin{aligned} &={u}_{13}+{u}_{14}+{u}_{15}+\ldots .+{u}_{25} \\ &={S}_{25}-{S}_{12} \\ &=\frac{25}{2}\{2 {a}+24 {~d}\}-264 \\ &=\frac{25}{2}\{2 \times 44+24 \times(-4)\}-264 \\ &=\frac{25}{2}\{88-96\}-264 \\ &=\frac{25}{2} \times(-8)-264 \\ &=-100-264 \\ &=-364 \qquad\square \end{aligned} $$
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Solution 9


(a) Solution:
$$ \begin{aligned} &{S}_{{n}}=2^{{n}}-1 \\ &{~S}_{1}=2^{1}-1=1 \\ &{~S}_{2}=2^{2}-1=3 \end{aligned} $$ $$ \begin{aligned} S_{3} &=2^{3}-1=7 \\ u_{1} &=S_{1}=1 \\ u_{2} &=S_{2}-S_{1}=3-1=2 \\ u_{3} &=S_{3}-S_{2}=7-3=4 \\ u_{n} &=S_{n}-S_{n-1} \\ &=\left(2^{n}-1\right)-\left(2^{n-1}-1\right) \\ &=2^{n}-2^{n-1} \\ &=2^{n-1} 2^{1}-2^{n-1} \\ &=2^{n-1}(2-1) \\ &=2^{n-1}\qquad\square \end{aligned} $$
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(b) Solution :
Let ${A}=\left(\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right)$ and ${A}^{-1}=\left(\begin{array}{ll}{a} & {b} \\ c & {~d}\end{array}\right)$ By the definition of inverse of a matrix, $$ {AA}^{-1}={I} $$ $$ \begin{aligned} \therefore\left(\begin{array}{ll} 3 & 1 \\ 2 & 1 \end{array}\right)\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) &=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ \left(\begin{array}{ll} 3 a+c & 3 b+d \\ 2 a+c & 2 b+d \end{array}\right) &=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ \therefore 3 a+c &=1 &(1)\\ 3 b+d &=0&(2) \\ 2 a+c &=0 &(3)\\ 2 b+d &=1&(4) \end{aligned} $$ By solving equations ( 1 ) and (3), we get $a=1,c=-2.$
By solving equations ( 2 ) and (4), we get $b=-1,d=3.$
$\therefore A^{-1}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right)\qquad\square$
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Solution 10


(a) Solution
$$ \begin{aligned} &\text { Let } A=\left(\begin{array}{ll} 5 & 6 \\ 7 & 8 \end{array}\right) . \\ &\operatorname{det} A=40-42=-2 \\ &\therefore A^{-1}=-\frac{1}{2}\left(\begin{array}{cc} 8&-6 \\ -7 & 5 \end{array}\right) \\ &=\left(\begin{array}{cc} -4 & 3 \\ 3.5 & -2.5 \end{array}\right) \end{aligned} $$ Next, we will solve the system of equations $$ \begin{aligned} &5 x+6 y=7 \\ &7 x+8 y=10 \end{aligned} $$ In matrix form $$ \left(\begin{array}{ll} 5 & 6 \\ 7 & 8 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 7 \\ 10 \end{array}\right) $$ Let $X=\left(\begin{array}{l}x \\ y\end{array}\right), \quad B=\left(\begin{array}{l}7 \\ 10\end{array}\right)$ Then we get $$ \begin{aligned} &{AX}={B} \text {. } \\ &\therefore {A}^{-1}({AX})={A}^{-1} {~B} \\ &\left({~A}^{-1} {~A}\right) {X}={A}^{-1} {~B} \\ &{IX}={A}^{-1} {~B} \\ &{X}={A}^{-1} {~B} \\ &\left(\begin{array}{l} {x} \\ {y} \end{array}\right)=-\frac{1}{2}\left(\begin{array}{cc} 8-6 \\ -7 & 5 \end{array}\right)\left(\begin{array}{c} 7 \\ 10 \end{array}\right) =-\frac{1}{2}\left(\begin{array}{r} -4 \\ 1 \end{array}\right) =\left(\begin{array}{c} 2 \\ -0.5 \end{array}\right) \\ &\therefore {x}=2, {y}=-0.5 \end{aligned} $$ $\therefore$ The point of intersection of the given two lines is $(2,-0.5).\qquad\square$
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(b) Solution :
Second die
First Die $\begin{array}{|c||c|c|c|c|c|c|} \hline &1&2&3&4&5&6\\ \hline 1&(1,1)& (1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline 1&(2,1)& (2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline 1&(3,1)& (3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline 1&(4,1)& (4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline 1&(5,1)& (5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline 1&(6,1)& (6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline \end{array}$

number of possible outcomes $=36$
set of favourable outcomes
$=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5)$, $\quad (2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\}$
number of favourable outcomes $=15$
${P}$ (score on the first die is less than that on the second) $=\frac{15}{36}=\frac{5}{12}\qquad\square$
${P}$ (score on the first die is prime and that on the second 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 first die is prime and that on the second is even) $=\frac{9}{36}=\frac{1}{4}\qquad\square$
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Solution 11


(a) Solution
Given : $\qquad$$PT$ is a tangent and $PQR$ is a secant to a circle. A circle with center $T$ and radius $TQ$ meets $QR$ at $S$.
To Prove :$\angle RTS=\angle RPT$
Proof:
$$\begin{array}[t]{rll} TS&=TQ&\mbox{(radii of $\odot T$)}\\ \therefore\theta&=\phi\\ \theta&=\alpha+\gamma&\mbox{(exterior angle of $\Delta TSR$)}\\ \phi&=\beta+\delta&\mbox{(exterior angle of $\Delta TQP)$}\\ \therefore\alpha+\gamma&=\beta+\delta\\ \mbox{But }\gamma&=\delta &(\delta\mbox{ is an angle between a tangent and chord})\\ \therefore\alpha&=\beta\\ \therefore\angle RTS&=\angle RPT\qquad\square \end{array}$$
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(b) Solution
Given : $\qquad$ ${P}$ is the point on ${AC}$ such that ${AP}=3 {PC}, {R}$ is the point on BP such that ${BR}=2 {RP}$ and ${QR} / / {AC}$. Given that $\alpha(\triangle {BPA})=36 {~cm}^{2}$
To Find : $\alpha(\Delta {BPC})$ and $\alpha(\Delta {BRQ})$
Solution:
$AP=3PC,BR=2RP, QR\parallel AC$ $$\begin{array}{lll} \alpha (\Delta BPA)=36\text{ cm}^2\\\text{Since }\Delta BPA\ \operatorname{and}\ \Delta BPC\text{have the same altitude,}\\\displaystyle \frac{{\alpha \left( {\Delta BPC} \right)}}{{\alpha \left( {\Delta BPA} \right)}}=\displaystyle \frac{{PC}}{{AP}}\\\therefore \displaystyle \frac{{\alpha \left( {\Delta BPC} \right)}}{{36}}=\displaystyle \frac{{PC}}{{3PC}}\\\therefore \alpha \left( {\Delta BPC} \right)=12\ \text{c}{{\text{m}}^{2}}\qquad\square \end{array}$$ In $\triangle B R Q$ and $\triangle B P A$ $$ \begin{aligned} \alpha=\alpha & \text { (common angle) } \\ \theta=\phi &({QR} / / {AC}) \\ \therefore \triangle {BRQ} \sim \triangle {BPA} &({AA}) \\ \therefore \quad \frac{\alpha(\Delta {BRQ})}{\alpha(\Delta {BPA})} &=\left(\frac{{BR}}{{BP}}\right)^{2} \\ \frac{\alpha(\Delta {BRQ})}{36} &=\left(\frac{{BR}}{{BR}+{RP}}\right)^{2} \\ \alpha(\Delta {BRQ}) &=36\left(\frac{2 {RP}}{2 {RP}+{RP}}\right)^{2} \\ &=36\left(\frac{2 {RP}}{3 {RP}}\right)^{2} \\ &=36 \times \frac{4}{9} \\ &=16 {~cm}^{2}\qquad\square \end{aligned} $$
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Solution 12


(a) Solution
Given : $\qquad$Bisectors of the interior angles of the quadrilateral $ABCD$ form the quadrilateral $PQRS.$
To Prove : $ PQRS$ is cyclic.
Proof:
Since $2\alpha+2\beta+2\theta+2\phi=360^{\circ},$ $\alpha+\beta+\theta+\phi=180^{\circ}.$
In $\triangle {SAB}, \angle {S}+\alpha+\beta=180^{\circ}\cdots (1)$
In $\Delta QDC, \angle Q+\theta+\phi=180^{\circ}\cdots (2)$
By adding Eq. (1) and (2), $$\begin{array}{rl} \angle {S}+\angle {Q}+\alpha+\beta+\theta+\phi &=360^{\circ} \\ \text{Since, } \alpha+\beta+\theta+\phi &=180^{\circ},\\ \angle {S}+\angle {Q}&=180^{\circ}\end{array}$$ $\therefore {PQRS}$ is cyclic. $\qquad\square$

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(b) Solution :
$$ \begin{aligned} &\alpha+\beta+\gamma =180^{\circ} \\ &\therefore \frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=90^{\circ} \\ &\frac{\alpha}{2}+\frac{\beta}{2}=90^{\circ}-\frac{\gamma}{2} \\ &\tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\tan \left(90^{\circ}-\frac{\gamma}{2}\right) =\cot\left(\frac{\gamma}{2}\right) \\ &\frac{\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}}{1-\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}=\frac{1}{\tan \frac{\gamma}{2}} \\ &\tan \frac{\alpha}{2} \tan \frac{\gamma}{2}+\tan \frac{\beta}{2} \tan \frac{\gamma}{2}=1-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \\ &\tan \frac{\alpha}{2} \tan \frac{\beta}{2}+\tan \frac{\beta}{2} \tan \frac{\gamma}{2}+\tan \frac{\alpha}{2} \tan \frac{\gamma}{2}=1 \end{aligned}\qquad\square $$
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Solution 13


(a) Solution:
$$ \begin{aligned} &\angle B=\angle A+15^{\circ},\\ & \angle C=\angle B+15^{\circ} =\angle A+15^{\circ}+15^{\circ}=\angle A+30^{\circ}\\ &B C=6, A C=? \end{aligned}$$ $$\begin{array}{rll} \angle A+\angle B+\angle C&=180^{\circ}\\ \angle A+(\angle A+15^{\circ})+\left(\angle A+30^{\circ}\right)&=180^{\circ}\\ 3\angle A+45^{\circ}&=180^{\circ}\\ 3 \angle A&=135^{\circ}\\ \angle A&=45^{\circ} \\ \angle B&=\angle A+15^{\circ}\\ &=45^{\circ}+15^{\circ}=60^{\circ} \end{array}$$ By the law of sines $$ \begin{aligned} &\frac{{AC}}{\sin {B}}=\frac{{BC}}{\sin {A}} \\ &\frac{{AC}}{\sin 60^{\circ}}=\frac{6}{\sin 45^{\circ}} \\ &{AC}=\frac{6 \sin 60^{\circ}}{\sin 45^{\circ}}=\frac{6 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} \\ &=\frac{6 \sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{6 \sqrt{6}}{2} =3 \sqrt{6}\qquad\square \end{aligned}$$
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(b) Solution
$$\begin{array}{rlll} {y}&=&\ln \left(\sin ^{3} 2 {x}\right) \\ \therefore {y}&=& 3 \ln (\sin 2 {x}) \\ \dfrac{{d} y}{{dx}} &=&3 \cdot \dfrac{1}{\sin 2 {x}} \cos 2 {x} \cdot 2 \\ &=&6 \cot 2 {x} \\ \dfrac{{d}^{2} y}{{~d} {x}^{2}} &=&6\left(-\operatorname{cosec}^{2} 2 {x}\right) 2 =-12 \operatorname{cosec}^{2} 2 {x}\\ 3 \dfrac{{d}^{2} {y}}{{dx}^{2}}+\left(\frac{{dy}}{{dx}}\right)^{2}+36 &=&3\left(-12 \operatorname{cosec}^{2} 2 {x}\right)+(6 \cot 2 {x})^{2}+36 \\ &=&-36 \operatorname{cosec}^{2} 2 {x}+36 \cot ^{2} 2 {x}+36 \\ &=&-36 \operatorname{cosec}^{2} 2 {x}+36\left(\cot ^{2} 2 {x}+1\right) \\ &=&-36 \operatorname{cosec}^{2} 2 {x}+36 \operatorname{cosec}^{2} 2 {x}\\&=&0 \end{array} $$
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Solution 14


(a) Solution
Given: $\qquad$ In the quadrilateral ${ABCD}, {M}$ and ${N}$ are the midpoints of ${AC}$ and ${BD}$ respectively.
To Prove: $\overrightarrow{{AB}}+\overrightarrow{{CB}}+\overrightarrow{{AD}}+\overrightarrow{{CD}}=4 \overrightarrow{{MN}}$
Proof:
$$ \begin{aligned} &\overrightarrow{{AB}}=\overrightarrow{{AM}}+\overrightarrow{{MN}}+\overrightarrow{{NB}} \\ &\overrightarrow{{CB}}=\overrightarrow{{CM}}+\overrightarrow{{MN}}+\overrightarrow{{NB}} \\ &\overrightarrow{{AD}}=\overrightarrow{{AM}}+\overrightarrow{{MN}}+\overrightarrow{{ND}} \\ &\overrightarrow{{CD}}=\overrightarrow{{CM}}+\overrightarrow{{MN}}+\overrightarrow{{ND}} \end{aligned} $$ By adding the above equations we obtain $$ \begin{aligned} \overrightarrow{{AB}}+\overrightarrow{{CB}}+\overrightarrow{{AD}}+\overrightarrow{{CD}} &=2 \overrightarrow{{AM}}+2 \overrightarrow{{CM}}+4 \overrightarrow{{MN}}+2 \overrightarrow{{NB}}+2 \overrightarrow{{ND}} \\ &=2 \overrightarrow{{AM}}-2 \overrightarrow{{AM}}+4 \overrightarrow{{MN}}+2 \overrightarrow{{NB}}-2 \overrightarrow{{NB}} \\ &=4 \overrightarrow{{MN}} \qquad\square \end{aligned} $$
Click to Question Paper 2019 (D)
(b) Solution:
\begin{eqnarray*}x y+2 x-y&=&0 \\y(x-1) &=&-2 x \\y &=&\dfrac{-2 x}{x-1} =-2-\frac{2}{x-1}=-2-2(x-1)^{-1}\\\dfrac{d y}{d x} &=&0+2(x-1)^{-2} =\dfrac{2}{(x-1)^{2}}\end{eqnarray*}
Since gradient of normal is $-2,\dfrac{d y}{d x}=\dfrac{1}{2}.$
Thus, \begin{eqnarray*}\frac{2}{(x-1)^{2}} &=&\frac{1}{2} \\(x-1)^{2} &=&4 \\x-1 &=&\pm 2 \\x &=&-1 \text { or } x=3 . \end{eqnarray*} If $x=-1, y=-1.$
At $(-1,-1)$ the normal equation is $ y+1=-2(x+1) .\qquad\square$
If $x=3, y=-3.$
At $(3,-3)$ the normal equation is $y+3=-2(x-3).\qquad\square$

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