Myanmar Exam Board (2018D 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


$$\begin{array}{|rl|rl|rl|rl|rl|} \hline1&C& 2&D& 3&E& 4&A& 5&E\\ \hline6 &B& 7&E& 8&-& 9 &C& 10&B\\ \hline11&C&12&D&13&C&14&B&15&E\\ \hline16&-&17&B&18&A&19&D&20&C\\ \hline21&C&22&E&23&C&24&B&25&C\\ \hline \end{array}$$

Solution 2


$$ \begin{aligned} f(x)=y & \Leftrightarrow f^{-1}(y)=x \\ \left(f \circ f^{-1}\right)(y) &=f\left(f^{-1}(y)\right) \\ &=f(x) \\ &=y \\ \left(f^{-1} \circ f\right)(x) &=f^{-1}(f(x)) \\ &=f^{-1}(y) \\ &=x \end{aligned} $$ $$ \therefore\left(f \circ f^{-1}\right)(y)=y \text { and }\left(f^{-1} \circ f\right)(x)=x \text {. } $$
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Solution 2 (OR)


$$ \begin{aligned} \text { Let } f(x)=2 x^{3} &+k x^{2}+7 \\ f(2) &=\frac{1}{2} f\left(\frac{1}{2}\right) \\ 16+4 k+7 &=\frac{1}{2}\left(2 \times \frac{1}{8}+\frac{k}{4}+7\right) \\ 8 k+46 &=\frac{1+k+28}{4} \\ 32 k+184 &=k+29 \\ 31 k &=-155 \\ \therefore k &=-5 \end{aligned} $$

Solution 3


$\begin{aligned} \text { In an A.P, } & a=u_{1}=-27 \\ u_{10} &=S_{9} \\ a+9 d &=\frac{9}{2}\{2 a+8 d\} \\-27+9 d &=\frac{9}{2}\{-54+8 d\} \\ 9(-3+d) &=\frac{9}{2} \times 2(-27+4 d) \\-3+d &=-27+4 d \\ 3 d &=24 \\ \therefore \quad d &=8 \end{aligned}$
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Solution 3(OR)


In a GP, $a=12r,S=4.$ $$\begin{array}{rl}\frac{a}{1-r}&=4\\12r&=4-4r\\16r&=4\\r&=\frac 14\\a&=12\times \frac 14=3\\\therefore u_3&=ar^2=3\times \frac{1}{16}=\frac{3}{16}\end{array}$$
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Solution 4


Given: $\angle B C D=125^{\circ}$
To find: $x, y$
Solution: $y+125^{\circ}=180^{\circ}(A B C D$ is cyclic) $$ \begin{aligned} \therefore y &=55^{\circ} \\ \angle A B D &=90^{\circ}(A D \text { is a diameter }) \\ x+y+\angle A B D &=180^{\circ} \\ x+55^{\circ}+90^{\circ} &=180^{\circ} \\ \therefore \quad x &=180^{\circ}-145^{\circ}=35^{\circ} \end{aligned} $$ Click to Question Paper 2018 (D)

Solution 5


$$ \begin{aligned} \mathrm{A}+\mathrm{B} &=45^{\circ} \\ \tan (\mathrm{A}+\mathrm{B}) &=\tan 45^{\circ} \\ \frac{\tan A+\tan B}{1-\tan A \tan B} &=1 \\ \tan A+\tan B &=1-\tan A \tan B \\ \therefore \tan A+\tan B+\tan A \tan B=1 & \end{aligned} $$
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Solution 6


$$ \begin{aligned} f(x) &=1-2 x^{2} \\ f^{\prime}(2) &=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1-2(2+h)^{2}-\left(1-2(2)^{2}\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1-2\left(4+4 h+h^{2}\right)-(1-8)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1-8-8 h-2 h^{2}+7}{h} \\ &=\lim _{h \rightarrow 0} \frac{h(-8-2 h)}{h} \\ &=\lim _{4 \rightarrow 0}(-8-2 h) \\ &=-8-0 \\ f^{\prime}(2) &=-8 \end{aligned} $$
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Solution 7(a)


$$ \begin{aligned} f(x)=2 x &-1, g(x)=2 x+3 \\ \text { Let } f^{-1}(x) &=y \\ f(y) &=x \\ 2 y-1 &=x \\ y &=\frac{x+1}{2} \\ \therefore f^{-1}(x) &=\frac{x+1}{2} \end{aligned}$$ Let $g^{-1}(x)=z$ $$ \begin{aligned} g(z) &=x \\ 2 z+3 &=x \\ z &=\frac{x-3}{2} \\ g^{-1}(x) &=\frac{x-3}{2} \\ \left(g^{-1} \cdot f^{-1}\right)(2) &=g^{-1}\left(f^{-1}(2)\right) \\ &=g^{-1}\left(\frac{3}{2}\right) \\ &=\frac{\frac{3}{2}-3}{2}=\frac{-\frac{3}{2}}{2}=-\frac{3}{4} \end{aligned} $$ Click to Question Paper 2018 (D)

Solution 7(b)


$$\begin{array}{lll} x \odot y&=\frac{x^{2}+y^{2}}{2}-x y, \text { for all } x, y \in R\\ y \odot x&=\frac{y^{2}+x^{2}}{2}-y x=\frac{x^{2}+y^{2}}{2}-x y\end{array}$$ $x \odot y=y \odot x$ for all $x y \in R$
$\therefore \odot$ is commutative. $$ \begin{aligned} \frac{a^{2}+2^{2}}{2}-2 a &=a+2 \\ a^{2}+4-4 a &=2 a+4 \\ a^{2}-6 a &=0 \end{aligned} $$ $$\quad \therefore \begin{aligned} &a(a-6)=0 \\ &\therefore a=0 \text { (or) } a=6 \end{aligned} $$
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Solution 8(a)


$\operatorname{Let} f(x)=p x^{3}-5 x^{2}+q x+10$. Since $(2 x-1)$ is a factor of $f(x),$ $$ \begin{gathered} f\left(\frac{1}{2}\right)=0 \\ p\left(\frac{1}{2}\right)^{3}-5\left(\frac{1}{2}\right)^{2}+q\left(\frac{1}{2}\right)+10=0 \\ \frac{p}{8}-\frac{5}{4}+\frac{q}{2}+10=0 \end{gathered} $$ $$\begin{array}{rl} p-10+4 q+80 &=0 \\ p+4 q &=-70\qquad (1) \end{array}$$ When $f(x)$ is divided by $x+2$, the remainder $=f(-2)$ $$\therefore f(-2)=-20$$ $$ \begin{array}{r} p(-2)^{3}-5(-2)^{2}+q(-2)+10=-20 \\ -8 p-20-2 q+10=-20 \\ -8 p-2 q=-10 \qquad (2)\\ p+4 q=-70 \\ \text { Eq(2) } \times 2 \rightarrow-16 p-4 q=-20 \\ -15 p=-90 \\ \therefore p=6 \end{array} $$ $$ \begin{aligned} &\text { Substitute } p=6 \text { in } \mathrm{Eq}(2) \\ &\qquad \begin{array}{l} -48-2 q=-10 \\ \therefore q=-19 \end{array} \end{aligned} $$ $$\begin{array}{rl} &\begin{array}{l}\qquad\qquad 3x^2-x-10\end{array}\\ 2x-1&\begin{array}[t]{|l}\hline \quad 6x^3-5x^2-19x+10\\-(6x^3-3x^2)\\ \hline\end{array}\\ &\qquad \quad\begin{array}{l}-2x^2-19x+10\\-(-2x^2+x)\\ \hline\end{array}\\ &\qquad \qquad\qquad\begin{array}{l}-20x+10\\-20x+10\\ \hline\end{array}\\ \end{array}$$ $\therefore f(x)=(2x-1)(3x^2-x-10)=(2x-1)(x-2)(3x+5)$
Click to Question Paper 2018 (D)

Solution 8(b)


$\begin{aligned}(1-2 x)^{n} &=1+{ }^{n} \mathrm{C}_{1}(-2 x)+{ }^{n} \mathrm{C}_{2}(-2 x)^{2}+{ }^{n} \mathrm{C}_{3}(-2 x)^{3}+\ldots \\ &=1-2 \times{ }^{n} \mathrm{C}_{1} x+4 \times{ }^{n} \mathrm{C}_{2} x^{2}-8 \times{ }^{n} \mathrm{C}_{3} x^{3}+\ldots \end{aligned}$ $$\begin{aligned}\qquad-2 \times{ }^{n} \mathrm{C}_{1}+4 \times{ }^{n} \mathrm{C}_{2}=16 \\-2 n+4 \times \frac{n(n-1)}{2}=16 \\-2 n+2 n^{2}-2 n-16=0 \\ 2 n^{2}-4 n-16=0 \\ n^{2}-2 n-8=0 \\(n-4)(n+2)=0 \\ n=4(\text { or }) n=-2 \end{aligned}$$
Since $n>0, n=4$. The coefficient of $x^{3}$ $$ =-8 \times^{n} C_{3}=-8 \times{ }^{4} C_{3}=-8 \times{ }^{4} C_{1}=-8 \times 4=-32 $$ Click to Question Paper 2018 (D)

Solution 9(a)


$$ \begin{aligned} &2 x(x-1) < 3-x \\ &2 x^{2}-2 x+x-3 < 0 \\ &2 x^{2}-x-3 < 0 \\ & \text { let } y=2 x^{2}-x-3 \\ &\text{If } y=0,2 x^{2}-x-3=0 \end{aligned} $$ $$ \begin{array}{r} (2 x-3)(x+1)=0 \\ 2 x-3=0(\text { or }) x+1=0 \\ x=\frac{3}{2}(\text { or }) x=-1 \end{array} $$ The parabola cuts the $X$-axis at $\left(\frac{3}{2}, 0\right)$ and $(-1,0)$. If $x=0, y=-3$ The parabola cuts the $Y$-axis at $(0,-3)$.
The solution set is $\left\{x \mid-1 < x < \frac{3}{2}\right\}$

$\inside{-1}{\frac 32}$
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Solution 9(b)


In an A.P, $$\begin{aligned}u_{5} &=3 {u}_{2} \\ a+4 d &=3(a+d) \\ a+4 d &=3 a+3 d \\ 2 a-d &=0 \\ u_{5}+u_{6}+u_{7}+u_{8}&=240 \\ a+4 d+a+5 d+a+6 d+a+7 d&=240 \\ 4 a+22 d&=240 \\ 2 a+11 d&=120 \\ 2 a-d&=0 \\ 12 d&=120 \\ \therefore d&=10 \\ \text{Substitute }d&=10 \text{ in Eq (1) }\\ 2 a-10&=0 \\ 2 a&=10 \\ \therefore \quad a&=5 \end{aligned}$$ Click to Question Paper 2018 (D)

Solution 10(a)


In a G.P., first term $=a$, common ratio $=r$. In an A.P., first term $=a$, common difference $=d$. $$ \begin{aligned} a+a&=1\\\therefore a&=\frac 12. a r+a+d &=\frac{1}{2} \\ \frac{1}{2} r+\frac{1}{2}+d &=\frac{1}{2} \\ \frac{1}{2} r+d &=0 \\ \therefore d &=-\frac{1}{2} r \\ a r^{2}+a+2 d &=2 \\ \frac{1}{2} r^{2}+\frac{1}{2}+2\left(-\frac{1}{2} r\right) &=2 \\ \frac{1}{2} r^{2}-r &=\frac{3}{2} \\ r^{2}-2 r &=3 \\ r^{2}-2 r-3 &=0 \\ (r-3)(r+1) &=0 \\ r &=3(\text { or }) r=-1 \end{aligned} $$ Since the terms of G.Pare positive, $r=3$. $$ \begin{aligned} \text { The required sum } &=a r^{3}+a+3 d \\ &=\frac{1}{2}(27)+\frac{1}{2}+3\left(-\frac{1}{2} r\right) \\ &=\frac{27}{2}+\frac{1}{2}-\frac{9}{2} \\ &=\frac{19}{2} \end{aligned} $$
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Solution 10(b)


$$\begin{aligned} &\text { Solution }\\ &A=\left(\begin{array}{ll} 5 & 1 \\ a+1 & a \end{array}\right), \operatorname{det} A=7\\ &\begin{aligned} &\operatorname{det} A=5 a-(a+1)=5 a-a-1=4 a-1 \\ &4 a-1=7 \end{aligned}\\ &\therefore a=2\\ &A=\left(\begin{array}{ll} 5 & 1 \\ 3 & 2 \end{array}\right)\\ &A^{2}=A A=\left(\begin{array}{ll} 5 & 1 \\ 3 & 2 \end{array}\right)\left(\begin{array}{ll} 5 & 1 \\ 3 & 2 \end{array}\right)=\left(\begin{array}{cc} 25+3 & 5+2 \\ 15+6 & 3+4 \end{array}\right)=\left(\begin{array}{ll} 28 & 7 \\ 21 & 7 \end{array}\right)\\ &A^{-1}=\frac{1}{\operatorname{det} A}\left(\begin{array}{cc} 2 & -1 \\ -3 & 5 \end{array}\right)=\frac{1}{7}\left(\begin{array}{cc} 2 & -1 \\ -3 & 5 \end{array}\right)\\ &A^{2}-x A^{-1}-y=0\\ &\left(\begin{array}{ll} 28 & 7 \\ 21 & 7 \end{array}\right)-x \left(\frac{1}{7}\right)\left(\begin{array}{cc} 2 & -1 \\ -3 & 5 \end{array}\right)-y\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=O\\ &7\left(\begin{array}{ll} 28 & 7 \\ 21 & 7 \end{array}\right)-x\left(\begin{array}{cc} 2 & -1 \\ -3 & 5 \end{array}\right)-7\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)=O\\ &\left(\begin{array}{ll} 196 & 49 \\ 147 & 49 \end{array}\right)-\left(\begin{array}{cc} 2 x & -x \\ -3 x & 5 x \end{array}\right)-\left(\begin{array}{cc} 7 y & 0 \\ 0 & 7 y \end{array}\right)=O\\ &\left(\begin{array}{cc} 196-2 x-7 y & 49+x \\ 147+3 x & 49-5 x-7 y \end{array}\right)=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right)\\ & 196-2 x-7 y=0.49+x=0.147+3 x=0.49-5 x-7 y=9\\ &\text { Solving these equations, we have } x=-49, y=42 \text {. }\\ \end{aligned}$$ Click to Question Paper 2018 (D)

Solution 11(a)


Let $A=\left(\begin{array}{lr} 7 & -4 \\ -3 & 2 \end{array}\right)$. Then $\det A=14-12=2.$ $$\begin{aligned} &A^{-1}=\frac{1}{\det A}\left(\begin{array}{ll} 2 & 4 \\ 3 & 7 \end{array}\right)=\frac{1}{2}\left(\begin{array}{ll} 2 & 4 \\ 3 & 7 \end{array}\right)\\ & 7 x-4 y=13\\ & 2 y-3 x=-5\\ & -3 x+2 y=-5\\ &\left(\begin{array}{lr} 7 & -4 \\ -3 & 2 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 13 \\ -5 \end{array}\right)\\ &\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{lr} 7 & -4 \\ -3 & 2 \end{array}\right)^{-1}\left(\begin{array}{l} 13 \\ -5 \end{array}\right)\\ &=\frac{1}{2}\left(\begin{array}{ll} 2 & 4 \\ 3 & 7 \end{array}\right)\left(\begin{array}{l} 13 \\ -5 \end{array}\right)\\ &=\frac{1}{2}\left(\begin{array}{l} 26-20 \\ 39-35 \end{array}\right)=\frac{1}{2}\left(\begin{array}{l} 6 \\ 4 \end{array}\right)\\ &\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 3 \\ 2 \end{array}\right)\\ &\therefore x=3, y=2 \text {. } \end{aligned}$$ Click to Question Paper 2018 (D)

Solution 11(b)


Number of Possible outcomes=18
The set of favourable outcome for a number which is divisible by 3 but not divisible by $4=\{123,213,231,234,243,321,342\}$
Nurnber of fivearable cutemes $=7$
The sct of favourable cutoume for ansaberwtich is sat Givide $\mathrm{ly}$ $=(124,134,142,143,214,241,314,341)$
Number of favourable cutcomes $=8$.
$\therefore$ P(anumeral which is not divisible by 3 ) $=\frac{8}{18}=\frac{4}{9}$
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Solution 12(a)


Prove: $AC//BD$
Proof: Draw $PQ.$ $$ \begin{aligned} &\alpha+\gamma=180^{\circ}(A P Q C \text { is cyclic }) \\ &\theta=\gamma^{\circ}(C O r 1.4) \\ &\alpha+\theta=180^{\circ} \\ &\therefore A C / / B D \end{aligned} $$ Click to Question Paper 2018 (D)

Solution 12(b)


Given : $D E$ is tangent at $A, D E // F G$
Prove : $B F G C$ is a cyclic quadrileteral $$ \begin{aligned} \theta &=\beta \\ \beta &=\alpha \\ \therefore \quad \theta &=\alpha\\ \theta+\gamma&=180^{\circ}\qquad \text{ (Supplementary angles)}\\ \therefore \gamma+\alpha&=180^{\circ} \end{aligned} $$ Hence $BFGC$ is a cyclic quadrilateral. Click to Question Paper 2018 (D)

Solution 13(a)


Given: $\triangle A B C$ is a right triangle, $\triangle A C D$ and $\triangle B C E$ are equilateral, $\alpha(\triangle B C E)=2 \alpha(\triangle A C D)$
Prove: $\triangle A B C$ is an isosceles right triangle.
Proof: $\triangle A C D$ and $\triangle B C E$ are equilateral
$$ \begin{aligned} \therefore \quad \triangle A C D &\sim \triangle B C E \\ \frac{\alpha(\triangle A C D)}{\alpha(\triangle B C E)}&=\frac{A C^{2}}{B C^{2}} \\ \frac{\alpha(\triangle A C D)}{2 \alpha(\triangle A C D)}&=\frac{A C^{2}}{B C^{2}}\\ \frac 12&=\frac{AC^2}{BC^2}\\BC^2&=2AC^2 \end{aligned} $$ But $B C^{2}=A C^{2}+A B^{2}$ (Pythagoras Theorem) $$ \begin{aligned} 2 A C^{2} &=A C^{2}+A B^{2} \\ A C^{2} &=A B^{2} \\ A C &=A B \end{aligned} $$ $\therefore \triangle 4 B C$ is an isosceles right triangle. Click to Question Paper 2018 (D)

Solution 13(b)


$\begin{aligned} \text { Solution } & \\ \overrightarrow{O P}=3 \vec{i}+j &=\left(\begin{array}{l}3 \\ 1\end{array}\right) ; \quad \overrightarrow{O Q}=7 i-15 \hat{j}=\left(\begin{array}{r}7 \\ -15\end{array}\right) \\ 3 \overrightarrow{P R} &=\overrightarrow{R Q} \\ 3(\overrightarrow{O R}-\overrightarrow{O P}) &=\overline{O Q}-\overrightarrow{O R} \\ 3 \overrightarrow{O R}-3 \overrightarrow{O P} &=\overrightarrow{O Q}-\overrightarrow{O R} \\ 4 \overrightarrow{O R} &=\overrightarrow{O Q}+3 \overrightarrow{O P} \\ 4 \overrightarrow{O R} &=\left(\begin{array}{r}7 \\ -15\end{array}\right)+3\left(\begin{array}{l}3 \\ 1\end{array}\right) \\ 4 \overrightarrow{O R} &=\left(\begin{array}{r}7 \\ -15\end{array}\right)+\left(\begin{array}{l}9 \\ 3\end{array}\right) \\ 4 \overrightarrow{O R} &=\left(\begin{array}{r}16 \\ -12\end{array}\right) \\ \overrightarrow{O R} &=\left(\begin{array}{r}4 \\ -3\end{array}\right) \\|\overrightarrow{O R}| &=\sqrt{4^{2}+(-3)^{2}}=\sqrt{25}=5 \end{aligned}$
$\therefore$ the unit vector in the direction of $\overrightarrow{O R}$ $$ =\frac{\overrightarrow{O R}}{|\overrightarrow{O R}|}=\frac{1}{5}\left(\begin{array}{c} 4 \\ -3 \end{array}\right)=\frac{1}{5}(4 i-3 \hat{j})=\frac{4}{5} \hat{i}-\frac{3}{5}\hat{ j} $$ Click to Question Paper 2018 (D)

Solution 14(a)


$$ \tan \alpha=\frac{4}{3}, \alpha \text { and } \beta \text { are acute angles. } $$ $$ \begin{aligned} \tan (\alpha+\beta) &=-1 \\ \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} &=-1 \\ \tan \alpha+\tan \beta &=-1+\tan \alpha \tan \beta \\ \frac{4}{3}+\tan \beta &=-1+\frac{4}{3} \tan \beta \\ \frac{4}{3}+1 &=\frac{4}{3} \tan \beta-\tan \beta \\ \frac{7}{3} &=\frac{1}{3} \tan \beta \\ \therefore \quad \tan \beta &=7 \end{aligned} $$ $$ \begin{aligned} \text{cosec}^2\alpha&= 1+\cot^2\alpha\\&=1+\frac{9}{16}=\frac{25}{16}\\ \sin^2\alpha&=\frac{16}{25}\\ &\sin \alpha=\frac{4}{5} \\ \text{cosec}^2\beta&=1+\cot^2\beta\\&=1+\frac{1}{49}=\frac{59}{49}\\ \sin^2\beta&=\frac{49}{50}\\ &\sin \beta=\frac{7}{5 \sqrt{2}}=\frac{7 \sqrt{2}}{10} \end{aligned} $$ Click to Question Paper 2018 (D)

Solution 14(b)


$$ B\hat AC=180^{\circ}-\left(53^{\circ}+37^{\circ}\right)=90^{\circ} $$ In r. $\triangle B A C$, by the Pythagoras Theorem, $$ \begin{aligned} B C^{2} &=A B^{2}+A C^{2} \\ &=5^{2}+12^{2} \\ &=25+144 \\ &=169 \\ \therefore \quad BC &=13 \mathrm{~km} \end{aligned} $$ $$ \begin{aligned} \cot \gamma &=\frac{A C}{A B}=\frac{12}{5}=2.4000 \\ \cot \gamma &=\cot 22^{\prime \prime} 37^{\prime} \\ \gamma &=22^{*} 37^{\prime} \\ \theta+\gamma &=53^{*} \\ \theta &=53^{*}-22^{*} 37^{\prime} \\ \theta &=30^{\prime} 23^{\prime} \end{aligned} $$ $\therefore$ The direction of the ship from the lighthouse is $N 30^{\circ} 23^{\prime} E$ Click to Question Paper 2018 (D)

Solution 15(a)


$$ x y=\sin x $$ Differentiate with repect to $x$, $$ \begin{gathered} x \frac{d y}{d x}+y \frac{d x}{d x}=\cos x \\ x \frac{d y}{d x}+y-\cos x=0 \end{gathered} $$ Differentiate with respect to $x$, $$ \begin{gathered} x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \frac{d x}{d x}+\frac{d y}{d x}+\sin x=0 \\ x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+\sin x=0 \\ \frac{d^{2} y}{d x^{2}}+\frac{2}{x} \frac{d y}{d x}+\frac{\sin x}{x}=0 \\ \therefore \quad \frac{d^{2} y}{d x^{2}}+\frac{2}{x} \frac{d y}{d x}+y=0 \end{gathered} $$ Click to Question Paper 2018 (D)

Solution 15(b)


Let the volume of a sphere with radius $r \mathrm{~cm}$ be $\mathrm{Vem}^{3}$. $$ \begin{gathered} r=2 \mathrm{~cm}, r+\delta r=2.05 \mathrm{~cm}: \\ \delta r=0.05 \mathrm{~cm} \\ V=\frac{4}{3} \pi r^{3} \\ \frac{d V}{d r}=4 \pi r^{2} \end{gathered} $$ $$ \begin{aligned} &\text { When } r=2, \frac{d V}{d r}=16 \pi \\ &\qquad \begin{aligned} \delta V \approx & \frac{d V}{d r} \delta r \\ =&16 \pi \times 0.05 \\ =&0.8 \pi \end{aligned} \end{aligned} $$ The approximate change in the volume of a sphere is $0.8 \pi \mathrm{cm}^{3}$. Click to Question Paper 2018 (D)

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