2019 SAMPLE QUESTION (2)
MATRICULATION EXAMINATION
DEPARTMENT OF MYANMAR EXAMINATION
MATHEMATICS Time Allowed: 3 hours
DEPARTMENT OF MYANMAR EXAMINATION
MATHEMATICS Time Allowed: 3 hours
WRITE YOUR ANSWERS IN THE ANSWER BOOKLET.
SECTION (A)
(Answer ALL questions.)
1.(a) Functions $ \displaystyle f : R \to R$ and $ \displaystyle g : R \to R$ are defined by $ \displaystyle f(x) = 2x - 14$ and $ \displaystyle g(x) = x^2 + 3.$ Find the values of $ \displaystyle x$ if $ \displaystyle (f∘g)(x) = 42.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ f:R\to R,f(x)=2x-14\\\\\ \ \ \ g:R\to R,\ g(x)={{x}^{2}}+3\\\\\ \ \ \ (f\circ g)(x)=42\\\\\ \ \ \ f(g(x))=42\\\\\ \ \ \ f({{x}^{2}}+3)=42\\\\\ \ \ \ 2({{x}^{2}}+3)-14=42\\\\\ \ \ \ 2({{x}^{2}}+3)=56\\\\\ \ \ \ {{x}^{2}}+3=28\\\\\ \ \ \ {{x}^{2}}=\ 25\ \\\\\therefore \ \ x=\pm 5\ \ \end{array}$
1.(b) When the polynomial $ \displaystyle x^3-3x^2+ kx+7$ is divided by $ \displaystyle x + 3,$ the remainder is $ \displaystyle 1.$ Find the value of $ \displaystyle k.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ f(x)={{x}^{3}}-3{{x}^{2}}+kx+7.\\\\\ \ \ \ \text{When }f(x)\ \text{is divided by }x+3\text{ , the reamiader is}\ \text{1}\text{.}\\\\\therefore \ \ f(-3)=1\\\\\therefore \ \ {{(-3)}^{3}}-3{{(-3)}^{2}}+k(-3)+7=1\\\\\therefore \ \ -27-27-3k+7=1\\\\\therefore \ \ -3k=48\\\\\therefore \ \ k=-16\end{array}$
2.(a) Find the coefficient of $ \displaystyle x^4$ in the expansion of $ \displaystyle {{\left( {\frac{x}{3}-3} \right)}^{7}}.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}{{\left( {\displaystyle \frac{x}{3}-3} \right)}^{7}}\\\\\ \ \ \ \ =\ {}^{7}{{C}_{r}}{{\left( {\displaystyle \frac{x}{3}} \right)}^{{7-r}}}{{(-3)}^{r}}\\\\\ \ \ \ \ =\ {}^{7}{{C}_{r}}{{(-1)}^{r}}{{3}^{{2r-7}}}{{x}^{{7-r}}}\\\\\ \ \ \ \ \text{For}\ {{x}^{4}},\ 7-r=4\Rightarrow r=3\\\\\therefore \ \ \ \ \text{Coefficient of}\ {{x}^{4}}={}^{7}{{C}_{3}}{{(-1)}^{3}}{{3}^{{6-7}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{7\times 6\times 5}}{{1\times 2\times 3}}\cdot \displaystyle \frac{1}{3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{{35}}{3}\end{array}$
2.(b) The first term of an $ \displaystyle A.P.$ is $ \displaystyle 2$ and its $ \displaystyle {{n}^{{\text{th}}}}$ term is $ \displaystyle 20.$ If the sum of the first $ \displaystyle n$ terms of that $ \displaystyle A.P.$ is $ \displaystyle 110,$ find the value of $ \displaystyle n.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{In an A}\text{.P}\text{.,}\ a=2,\\\\\ \ \ \ \text{Let the common difference be}\ d.\ \\\ \ \\\ \ \ \ {{u}_{n}}=20\ \ \ \ \text{ }\!\![\!\!\text{ given }\!\!]\!\!\text{ }\\\\\therefore \ \ a+(n-1)d=20\\\\\therefore \ \ 2+(n-1)d=20\\\\\therefore \ \ (n-1)d=18\\\\\ \ \ \ {{S}_{n}}=110\ \ \ \ \text{ }\!\![\!\!\text{ given }\!\!]\!\!\text{ }\\\\\therefore \ \ \displaystyle \frac{n}{2}\left\{ {2a+(n-1)d} \right\}=110\\\\\therefore \ \ \displaystyle \frac{n}{2}\left\{ {4+18} \right\}=110\\\\\therefore \ \ n=10\end{array}$
3.(a) Let $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 4 \\ 6 & a \end{array}} \right)$. If $ \displaystyle \det (2A) = 2 \det A,$ find $ \displaystyle a.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ A=\displaystyle \left( {\begin{array}{*{20}{c}} 2 & 4 \\ 6 & a \end{array}} \right)\\\\\therefore \ \ 2A=\displaystyle \left( {\begin{array}{*{20}{c}} 4 & 8 \\ {12} & {2a} \end{array}} \right)\\\\\therefore \ \ \det A=2a-10=2(a-5)\\\\\ \ \ \ \det (2A)=8a-96=8(a-12)\\\\\ \ \ \ \text{By the problem},\\\\\ \ \ \ \det (2A)=2\det A\\\\\ \ \ \ 8(a-12)=4(a-5)\\\\\therefore \ \ 2a-24=a-5\\\\\therefore \ \ a=19\\\ \ \ \ \end{array}$
3.(b) When a die is rolled $ \displaystyle k$ times, the expected frequency of getting a factor of $ \displaystyle 4$ is $ \displaystyle 240.$ Find $ \displaystyle k.$
(3 marks)
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$ \displaystyle \begin{array}{*{20}{l}} {\ \ \ \ \ \text{When a die is rolled once,}} \\ {} \\ {\ \ \ \ \ \text{Set of all possible outcomes = }\left\{ {\text{1,2,3,4,5,6}} \right\}\text{ }} \\ {} \\ {\ \ \ \ \ \text{Number of possible outcomes }=\text{6}} \\ {} \\ {\ \ \ \ \ \text{Set of favourable outcomes for getting}} \\ {\ \ \ \ \ \text{a number which is a factor of 4 }=\text{ }\left\{ {\text{1,2,4}} \right\}\text{ }} \\ {} \\ {\ \ \ \ \ \text{Number of favourable outcomes }=\text{3 }} \\ {} \\ {\ \ \ \ P(\text{getting a factor of}\ 4)=\displaystyle \frac{3}{6}=\displaystyle \frac{1}{2}} \\ {} \\ {\ \ \ \ \text{When a die is rolled }k\ \text{times,}} \\ {} \\ {\ \ \ \ \text{Expected frequency of gettin a factor of 4}\ } \\ {} \\ {\ \ \ =P(\text{getting a factor of}\ 4)\times k} \\ {} \\ {\ \ \ =\displaystyle \frac{k}{2}} \\ {} \\ {\ \ \ \text{By the problem,}} \\ {} \\ {\ \ \ \displaystyle \frac{k}{2}=240\Rightarrow k=480} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ } \end{array}$
4.(a) Given $ \displaystyle ⊙O$ with diameter $ \displaystyle CI, CM ∥ ON,$ prove that arc $ \displaystyle MN =$ arc $ \displaystyle NI.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Draw }OM.\\\\\ \ \ \ \ \text{Since}\ \text{ }CM\parallel ON,\\\\\ \ \ \ \ \alpha =\gamma \ \ \ \ \text{(corresponding }\angle \text{s)}\\\\\ \ \ \ \ \beta =\delta \ \ \ \ \text{(alternating }\angle \text{s)}\\\\\ \ \ \ \ \text{Since}\ \text{ }OM=OC\ \ \ \text{(radii)}\\\\\ \ \ \ \ \gamma =\delta \ \\\\\therefore \ \ \ \beta=\alpha \\\\\therefore \ \ \ \text{arc }MN=\text{arc }NI\\\ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$
4.(b) Referred to a fixed origin $ \displaystyle O,$ the position vectors of the points $ \displaystyle A$ and $ \displaystyle B$ are $ \displaystyle 4\hat{\text{i}}-6\hat{\text{j}}$ and $ \displaystyle 8\hat{\text{i}}-p\hat{\text{j}}$ respectively. Given that $ \displaystyle O, A$ and $ \displaystyle B$ are collinear, find the value of $ \displaystyle p.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \overrightarrow{{OA}}=4\widehat{\text{i}}-6\widehat{\text{j}}\\\\\ \ \ \overrightarrow{{OB}}=8\widehat{\text{i}}-p\widehat{\text{j}}\\\\\ \ \ \text{Since }O,A\ \text{and }B\ \text{are collinear,}\\\\\ \ \ \text{Let }\ k\overrightarrow{{OA}}=\overrightarrow{{OB}}.\\\\\therefore \ 4k\widehat{\text{i}}-6k\widehat{\text{j}}=8\widehat{\text{i}}-p\widehat{\text{j}}\\\\\therefore \ 4k=8\Rightarrow k=2\\\\\therefore \ p=6k=12\end{array}$
5.(a) Show that $ \displaystyle \frac{{\sin 2\theta }}{{1+\cos 2\theta }}=\tan \theta. $
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{\sin 2\theta }}{{1+\cos 2\theta }}=\displaystyle \frac{{2\sin \theta \cos \theta }}{{1+2{{{\cos }}^{2}}\theta -1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2\sin \theta \cos \theta }}{{2{{{\cos }}^{2}}\theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\sin \theta }}{{\cos \theta }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan \theta \end{array}$
5.(b) Find the gradient of the curve $ \displaystyle y = 3x^2 - 4x + 3$ at the point where $ \displaystyle x = 2.$
(3 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Curve}:y=3{{x}^{2}}-4x+3\\\\\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=6x-4\\\\\ \ \ \ \ \text{When }x=2,\displaystyle \frac{{dy}}{{dx}}=6(2)-4=8\\\\\therefore \ \ \ \text{The gradient of tangent at }x=2\text{ is 8}\text{.}\end{array}$
SECTION (B)
(Answer any FOUR questions.)
6.(a) Let $ \displaystyle f : R\to R$ and $ \displaystyle g : R \to R$ be defined by $ \displaystyle f(x) = x + 7$ and $ \displaystyle g(x) = 3x –1.$ Find $ \displaystyle ({{f}^{{-1}}}\circ g)(x)$ and what is the value of $ \displaystyle b ∈ R$ for which $ \displaystyle ({{f}^{{-1}}}\circ g)(b)=4$ .
(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}\ \left( {{{f}^{{-1}}}\circ g} \right)(x)=y\\\\\therefore \ \ {{f}^{{-1}}}\left( {g(x)} \right)=y\\\\\therefore \ \ g(x)=f(y)\\\\\therefore \ \ 3x-1=y+7\\\\\therefore \ \ y=3x-8\\\\\therefore \ \ \left( {{{f}^{{-1}}}\circ g} \right)(x)=3x-8\\\\\therefore \ \ \left( {{{f}^{{-1}}}\circ g} \right)(b)=3b-8\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ \left( {{{f}^{{-1}}}\circ g} \right)(b)=4\\\\\ \ \ \ 3b-8=4\\\\\therefore \ \ \ b=4\end{array}$
6.(b) Given that $ \displaystyle f(x) = x^{2n}-(p+1)x^2+p,$ where $ \displaystyle n$ and $ \displaystyle p$ are positive integers. Show that $ \displaystyle x - 1$ is a factor of $ \displaystyle f(x),$ for all values of $ \displaystyle p.$ When $ \displaystyle p = 4,$ find the value of $ \displaystyle n$ for which $ \displaystyle x - 2$ is a factor of $ \displaystyle f(x)$ and hence factorize $ \displaystyle f(x)$ completely.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)={{x}^{{2n}}}-(p+1){{x}^{2}}+p\\\\\ \ \ \ \ f(1)={{(1)}^{{2n}}}-(p+1){{(1)}^{2}}+p\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =1-p+1+p\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =0\\\\\therefore \ \ \ x-1\ \text{is a factor of}\ f(x)\ \text{for all value of}\ p.\\\\\ \ \ \ \ \text{When}\ p=4,\ x-2\ \text{is a factor of}\ f(x).\\\\\therefore \ \ \ f(2)=0\ \text{when}\ p=4.\\\\\therefore \ \ \ {{(2)}^{{2n}}}-(4+1){{(2)}^{2}}+4=0\\\\\therefore \ \ \ {{4}^{n}}=16\\\\\therefore \ \ \ {{4}^{n}}={{4}^{2}}\\\\\therefore \ \ \ n=2\\\\\therefore \ \ \ f(x)={{x}^{4}}-5{{x}^{2}}+4\\\\\therefore \ \ \ f(x)=({{x}^{2}}-4)({{x}^{2}}-1)\\\\\therefore \ \ \ f(x)=(x-2)(x+2)(x-1)(x+1)\\\\\therefore \ \ \ \text{The factors of }f(x)\text{ are }x-1,x+1,x-2\ \text{and }x+2.\end{array}$
7.(a) Let R be the set of real numbers and a binary operation ⊙ on R be defined by $ \displaystyle x ⊙ y = xy - x - y$ for $ \displaystyle x, y ∈ R.$ Find the value of $ \displaystyle 3 ⊙ 4$ and $ \displaystyle (3 ⊙ 4) ⊙ 5.$ Find the values of $ \displaystyle x$ if $ \displaystyle x⊙ (x ⊙ 3) = 23.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ x\odot y=xy-x-y,\ \ x,y\in R.\\\\\ \ \ \ \ \ 3\odot 4=3(4)-3-4=5\\\\\ \ \ \ \ \ (3\odot 4)\odot 5=5\odot 5\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =5(5)-5-5\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =15\\\\\ \ \ \ \ \ x\odot 3=3x-x-3=2x-3\\\\\ \ \ \ \ \ x\odot (x\odot 3)=x\odot (2x-3)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x(2x-3)-x-(2x-3)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{x}^{2}}-3x-x-2x+3\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{x}^{2}}-6x+3\\\\\ \ \ \ \ \ \text{By the problem,}\ \ \\\\\ \ \ \ \ \ x\odot (x\odot 3)=23\\\\\therefore \ \ \ \ 2{{x}^{2}}-6x+3=23\\\\\therefore \ \ \ \ 2{{x}^{2}}-6x-20=0\\\\\therefore \ \ \ \ {{x}^{2}}-3x-10=0\\\\\therefore \ \ \ \ (x+2)(x-5)=0\\\\\therefore \ \ \ \ x=-2\ \text{(or)}\ x=5\end{array}$
7.(b) Find, in ascending powers of $ \displaystyle x,$ the first three terms of $ \displaystyle (1 + kx )^5( 1 - 4x ).$ If the coefficient of $ \displaystyle x$ is $ \displaystyle 16,$ find the value of $ \displaystyle k$ and the coefficient of $ \displaystyle x^2.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ {{(1+kx)}^{5}}(1-4x)=\left( {1+5(kx)+10{{{(kx)}}^{2}}+...} \right)(1-4x)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(1+5kx+10{{k}^{2}}{{x}^{2}}+...)(1-4x)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+5kx+10{{k}^{2}}{{x}^{2}}-4x-4x(5kx)-4x(10{{k}^{2}}{{x}^{2}})+...\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+(5k-4)x+(10{{k}^{2}}-20k){{x}^{2}}-40{{k}^{2}}{{x}^{3}}+...\\\\\ \ \ \ \ \ \text{coefficient of }x=16\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ \ \ 5k-4=16\Rightarrow k=4\\\\\therefore \ \ \ \ \text{coefficient of }{{x}^{2}}=10{{k}^{2}}-20k=80\end{array}$
8.(a) Find the solution set in R of the inequation $ \displaystyle (2x - 1)^2 ≥ 9$ by graphical method and illustrate it on the number line.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ {{(2x-1)}^{2}}\ge 9\\\\\ \ \ \ \ 4{{x}^{2}}-4x+1\ge 9\\\\\ \ \ \ \ {{x}^{2}}-x-2\ge 0\\\\\ \ \ \ \ \text{Let}\ y={{x}^{2}}-x-2.\\\\\ \ \ \ \ \text{When}\ y=0,\ {{x}^{2}}-x-2=0\\\\\therefore \ \ \ (x+1)(x-2)=0\\\\\therefore \ \ \ x=-1\ \text{(or)}\ x=2\\\\\therefore \ \ \ \text{The graph cuts the x-axis at (}-\text{1,0) and (2,0)}\text{.}\\\\\ \ \ \ \ \text{When}\ x=0,y=-2\\\\\therefore \ \ \ \text{The graph cuts the y-axis at (0,}-\text{2)}\text{.}\end{array}$
$ \displaystyle \therefore \ \ \ \ \text{Solution set}=\{x|x\le -1\ \text{(or)}\ x\ge 2\}$
$ \displaystyle \ \ \ \ \ \ \ \text{Number line}$
$ \displaystyle \therefore \ \ \ \ \text{Solution set}=\{x|x\le -1\ \text{(or)}\ x\ge 2\}$
$ \displaystyle \ \ \ \ \ \ \ \text{Number line}$
8.(b) Find the sum of all two-digit natural numbers which are divisible by $ \displaystyle 8.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \text{All two-digit natural numbers which are }\\\ \ \ \text{divisible by 8 are}\ \\\\\ \ \ 16,24,32,40,48,56,64,72,80,88,96.\\\\\ \ \ \text{It is an A}\text{.P}\text{. with }a=16,\ d=8\ \text{and}\ n=11\\\\\ \ \ {{S}_{n}}=\displaystyle \frac{n}{2}(a+l)\\\\\ \ \ {{S}_{{12}}}=\displaystyle \frac{{11}}{2}(16+96)=616\\\ \ \ \end{array}$
| Alternative Method As the problem does not state any restriction, we can find the sum simply as 16+24+32+40+48+56+64+72+80+88+96=616. ျပဌာန္းစာအုပ္တြင္ 3 ျဖင့္စာ၍ ျပတ္ေသာ သဘာ၀ကိန္းမ်ားေပါင္းလဒ္ကို ရွာခိုင္းသည္။ 12 မွ 99 အထိ ကိန္းလံုးအေရအတြက္ 30 ရွိရာ အထက္ပါအတိုင္း အလြယ္တကူ ေပါင္း၍ မရႏိုင္ပါ။ ေမးခြန္းထုတ္သူသည္ ေက်ာင္းသားမ်ားကို အလြန္သက္ညႇာေသာ သေဘာထားရွိသည္။ |
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9.(a) If $ \displaystyle m$ is a positive integer, show that the sum of the A.P. $ \displaystyle 2m + 1, 2m + 3, 2m + 5,$ … , $ \displaystyle 4m - 1$ is divisible by $ \displaystyle 3.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ 2m+1,2m+3,2m+5,...,4m-1\ \text{is an A}\text{.P}\text{.}\\\\\therefore \ \ \ a=2m+1\ \operatorname{and}\ \\\\\ \ \ \ \ d=2m+3-2m-1=2\\\\\ \ \ \ \ {{u}_{n}}=4m-1\\\\\ \ \ \ \ a+(n-1)d=4m-1\\\\\ \ \ \ \ 2m+1+(n-1)2=4m-1\\\\\ \ \ \ \ 2n-2=2m-2\\\\\therefore \ \ \ n=m\\\\\ \ \ \ \ \text{Since }m\ \text{is a positive integer, }n=m\ \text{is acceptable}\text{.}\\\\\ \ \ \ \ {{S}_{n}}= \displaystyle \frac{n}{2}\{2a+(n-1)d\}\\\\\ \ \ \ \ {{S}_{m}}= \displaystyle \frac{m}{2}\{2(2m+1)+(m-1)2\}\\\\\therefore \ \ \ {{S}_{m}}=3{{m}^{2}}\\\\\ \ \ \ \ \text{Since }m\ \text{is a positive integer, }3{{m}^{2}}\ \text{is divisible by 3}\text{.}\\\\\therefore \ \ \ \text{The sum of the A}\text{.P}\text{. is divisible by 3}\text{.}\end{array}$
9.(b) Solve the matrix equation $ \displaystyle P\left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)+2\left( {\begin{array}{*{20}{c}} {-6} & 0 \\ 5 & 7 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {-17} & {10} \\ 5 & {44} \end{array}} \right)$ for the $ \displaystyle 2×2$ matrix $ \displaystyle P.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ P \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)+2 \displaystyle \left( {\begin{array}{*{20}{c}} {-6} & 0 \\ 5 & 7 \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {-17} & {10} \\ 5 & {44} \end{array}} \right)\\\\\therefore \ \ \ P \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {-12} & 0 \\ {10} & {14} \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {-17} & {10} \\ 5 & {44} \end{array}} \right)\\\\\therefore \ \ \ P \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {-12} & 0 \\ {10} & {14} \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {12} & 0 \\ {-10} & {-14} \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {-17} & {10} \\ 5 & {44} \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} {12} & 0 \\ {-10} & {-14} \end{array}} \right)\\\\\therefore \ \ \ P \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)+ \displaystyle \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & {10} \\ {-5} & {30} \end{array}} \right)\\\\\therefore \ \ \ P \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)+O= \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & {10} \\ {-5} & {30} \end{array}} \right)\\\\\therefore \ \ \ P \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & {10} \\ {-5} & {30} \end{array}} \right)\\\\\ \ \ \ \ \text{Let}\ A= \displaystyle \left( {\begin{array}{*{20}{c}} {-3} & 2 \\ 1 & 6 \end{array}} \right)\ \text{and}\ B= \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & {10} \\ {-5} & {30} \end{array}} \right).\\\\\therefore \ \ \ PA=B\\\\\therefore \ \ \ PA{{A}^{{-1}}}=B{{A}^{{-1}}}\\\\\therefore \ \ \ PI=B{{A}^{{-1}}}\\\\\therefore \ \ \ P=B{{A}^{{-1}}}\ \text{if }{{A}^{{-1}}}\ \text{exists}\text{.}\\\\\ \ \ \ \ \det A=-18-2=-20\ne 0\\\\\therefore \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\\\\\therefore \ \ \ {{A}^{{-1}}}=- \displaystyle \frac{1}{{20}} \displaystyle \left( {\begin{array}{*{20}{c}} 6 & {-2} \\ {-1} & {-3} \end{array}} \right)\\\\\therefore \ \ \ P=- \displaystyle \frac{1}{{20}} \displaystyle \left( {\begin{array}{*{20}{c}} {-5} & {10} \\ {-5} & {30} \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} 6 & {-2} \\ {-1} & {-3} \end{array}} \right)\\\\\therefore \ \ \ P=- \displaystyle \frac{1}{{20}} \displaystyle \left( {\begin{array}{*{20}{c}} {-30-10} & {10-30} \\ {-30-30} & {10-90} \end{array}} \right)\\\\\therefore \ \ \ P=- \displaystyle \frac{1}{{20}} \displaystyle \left( {\begin{array}{*{20}{c}} {-40} & {-20} \\ {-60} & {-80} \end{array}} \right)\\\\\therefore \ \ \ P= \displaystyle \left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 4 \end{array}} \right)\end{array}$
10.(a) Find the solution set of the system of equations
$ \displaystyle \begin{array}{l}3x+2y=7\\5x-y=3\end{array}$ by matrix method.
$ \displaystyle \begin{array}{l}3x+2y=7\\5x-y=3\end{array}$ by matrix method.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \left. \begin{array}{l}3x+2y=7\\5x-y=3\end{array} \right\}-------(1)\\\\\ \ \ \ \text{Transforming into matrix equation, we have}\\\\\ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} 3 & 2 \\ 5 & {-1} \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)= \displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right)\ ---(2)\\\\\ \ \ \ \text{Let }A= \displaystyle \left( {\begin{array}{*{20}{c}} 3 & 2 \\ 5 & {-1} \end{array}} \right),X= \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ \text{and}\ B= \displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right).\\\\\ \ \ \ \text{Then equation (2) becomes }AX=B.\\\\\therefore \ \ {{A}^{{-1}}}AX={{A}^{{-1}}}B\\\\\ \ \ \ IX={{A}^{{-1}}}B\\\\\ \ \ \ X={{A}^{{-1}}}B\ \text{if }{{A}^{{-1}}}\ \text{exists}\text{.}\\\\\ \ \ \ \det A=-3-10=-13\ne 0.\\\\\therefore \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\\\\\therefore \ \ X=- \displaystyle \frac{1}{{13}} \displaystyle \left( {\begin{array}{*{20}{c}} {-1} & {-2} \\ {-5} & 3 \end{array}} \right) \displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ =- \displaystyle \frac{1}{{13}} \displaystyle \left( {\begin{array}{*{20}{c}} {-7-6} \\ {-35+9} \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ =- \displaystyle \frac{1}{{13}} \displaystyle \left( {\begin{array}{*{20}{c}} {-13} \\ {-26} \end{array}} \right)\\\\\therefore \ \ \displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ = \displaystyle \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right)\\\\\therefore \ \ x=1\ \text{and}\ y=2.\end{array}$
10.(b) Construct the table of outcomes for rolling two dice. Find the probability that the score on the second die is greater than that on the first. Find also the probability that the score on one die is prime and the score on the other is even.
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\ \text{die}\\ \ {{\text{1}}^{{\text{st}}}}\ \text{die}\ \ \ \begin{array}{|r||l|l|l|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}$
$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =36$ $ \displaystyle \begin{array}{l}\ \ \ \ \text{Set of favourable outcomes for the score on }\\\ \ \ \ \text{the second die is greater than that on the first}\\\\\ \ \ \ \ =(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)\}\\\\\therefore \ \ \text{Number of favourable outcomes}\ =15\\\\\therefore \ \ P(\text{score on}\ {{2}^{{\text{nd}}}}\text{ die}>\text{score on}\ {{1}^{{\text{st}}}}\text{ die})= \displaystyle \frac{{15}}{{36}}= \displaystyle \frac{5}{{12}}\\\\\ \ \ \ \text{Set of favourable outcomes for the score on }\\\ \ \ \ \text{one die is prime and the score on other is even}\\\\\ \ \ \ =\{(2,2)(2,3)(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,2),\\\ \ \ \ \ \ \ \ \ (4,3),(4,5),(5,2),(5,4),(5,6),(6,2),(6,3),(6,5)\}\\\\\therefore \ \ \text{Number of favourable outcomes}\ =17\\\\\therefore \ \ P(\text{the score one die is prime and }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{the score on other is even})= \displaystyle \frac{{17}}{{36}}\end{array}$
SECTION (C)
(Answer any THREE questions.)
11.(a) $ \displaystyle PQR$ is a triangle inscribed in a circle. The tangent at $ \displaystyle P$ meets $ \displaystyle RQ$ produced at $ \displaystyle T,$ and $ \displaystyle PC$ bisecting $ \displaystyle ∠ RPQ$ meets $ \displaystyle RQ$ at $ \displaystyle C.$ Prove that $ \displaystyle ∆TPC$ is isosceles.
(5 marks)
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11.(b) In the acute $ \displaystyle ∆ABC , AD$ and $ \displaystyle BE$ are altitudes. If $ \displaystyle α(∆DEC) =\frac{1}{3}α(∆ABDE)$ prove that $ \displaystyle ∠C =60°.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \alpha (\vartriangle DEC)= \displaystyle \frac{1}{3}\alpha (ABDE)\ \ \ \text{(given)}\\\\\therefore \ \ \alpha (ABDE)=3\alpha (\vartriangle DEC)\\\\\ \ \ \ \text{Since }AD\bot BC\ \text{and}\ BE\bot AC,\\\\\ \ \ \ \angle ADB=\angle AEB=90{}^\circ \\\\\therefore \ \ ABDE\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ \angle ABC=\angle CED\ \ \text{(exterior }\angle \ \text{of cyclic quadrilateral }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \text{interior opposite}\ \angle \text{)}\\\\\ \ \ \ \ \text{In}\ \vartriangle CDE\ \text{ and }\vartriangle CAB,\\\\\ \ \ \ \ \angle C=\angle C\ \ \ \ \ \ \ \ \ \ \ \ \ \text{(common}\ \angle )\\\\\ \ \ \ \ \angle ABC=\angle CED\ \ \ \ \ \text{(proved)}\\\\\therefore \ \ \ \vartriangle CDE\ \sim \vartriangle CAB\ \ \ \ \text{(AA corollary)}\\\\\therefore \ \ \ \displaystyle \frac{{\alpha (\vartriangle CDE)}}{{\alpha (\vartriangle CAB)}}= \displaystyle \frac{{C{{D}^{2}}}}{{A{{C}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{\alpha (\vartriangle CDE)}}{{\alpha (\vartriangle CDE)+\alpha (ABDE)}}= \displaystyle \frac{{C{{D}^{2}}}}{{A{{C}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{\alpha (\vartriangle CDE)}}{{\alpha (\vartriangle CDE)+3\alpha (\vartriangle CDE)}}= \displaystyle \frac{{C{{D}^{2}}}}{{A{{C}^{2}}}}\\\\\therefore \ \ \ \displaystyle \frac{{\alpha (\vartriangle CDE)}}{{4\alpha (\vartriangle CDE)}}= \displaystyle \frac{{C{{D}^{2}}}}{{A{{C}^{2}}}}\\\\\therefore \ \ \ \ \displaystyle \frac{{C{{D}^{2}}}}{{A{{C}^{2}}}}= \displaystyle \frac{1}{4}\Rightarrow \displaystyle \frac{{CD}}{{AC}}= \displaystyle \frac{1}{2}\\\\\ \ \ \ \ \ \text{In right}\ \vartriangle ACD, \displaystyle \frac{{CD}}{{AC}}=\cos C\\\\\therefore \ \ \ \ \cos C= \displaystyle \frac{1}{2}\\\\\therefore \ \ \ \ \angle C=60{}^\circ \end{array}$
12.(a) $ \displaystyle ABC$ is a triangle in which $ \displaystyle AB = AC. P$ is a point inside the triangle such that $ \displaystyle ∠ PAB = ∠ PBC.$ Given that $ \displaystyle Q$ is a point on $ \displaystyle BP$ produced such that $ \displaystyle ABCQ$ is a cyclic quadrilateral, prove that $ \displaystyle AQ = AP.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ ABCQ\text{ is cyclic,}\\\\\ \ \ \ \angle Q=\angle C\ \ \ \ (\angle \text{s in same segment)}\\\\\ \ \ \ \text{Since}\ AB=AC,\ \angle C=\angle ABC.\\\\\therefore \ \ \angle C=\theta +\beta \\\\\therefore \ \ \angle Q=\theta +\beta \\\\\ \ \ \ \text{Since}\ \angle PAB=\angle PBC,\ \ \ (\text{given)}\ \ \\\\\ \ \ \ \alpha =\beta \\\\\therefore \ \ \angle Q=\theta +\alpha \\\\\ \ \ \ \text{In}\ \vartriangle ABP,\\\\\ \ \ \ \angle APQ=\theta +\alpha \ (\text{exterior }\angle \ \text{of }\vartriangle \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\text{sum of interior opposite}\ \angle \text{s)}\\\\\therefore \ \ \angle Q=\ \angle APQ\\\\\therefore \ \ AQ=AP.\end{array}$
12.(b) Without using tables calculate the exact values of $ \displaystyle \sin 75°, \cos 195°$ and $ \displaystyle \tan 255°.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \sin 75{}^\circ =\sin (30{}^\circ +45{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin 30{}^\circ \cos 45{}^\circ +\cos 30{}^\circ \sin 45{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{1}{2}\times \displaystyle \frac{{\sqrt{2}}}{2}+ \displaystyle \frac{{\sqrt{3}}}{2}\times \displaystyle \frac{{\sqrt{2}}}{2}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{\sqrt{2}+\sqrt{6}}}{2}\\\\\ \ \ \cos 195{}^\circ =\cos (180{}^\circ +15{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 15{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos (45{}^\circ -30{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =-(\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\left( { \displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{{\sqrt{3}}}{2}+ \displaystyle \frac{{\sqrt{2}}}{2}\times \displaystyle \frac{1}{2}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =- \displaystyle \frac{{\sqrt{6}+\sqrt{2}}}{2}\\\\\ \tan 255{}^\circ =\tan (180{}^\circ +75{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan 75{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan (45{}^\circ +30{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{\tan 45{}^\circ +\tan 30{}^\circ }}{{1-\tan 45{}^\circ \tan 30{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{1+ \displaystyle \frac{1}{{\sqrt{3}}}}}{{1- \displaystyle \frac{1}{{\sqrt{3}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{ \displaystyle \frac{{\sqrt{3}+1}}{{\sqrt{3}}}}}{{ \displaystyle \frac{{\sqrt{3}-1}}{{\sqrt{3}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{\sqrt{3}+1}}{{\sqrt{3}-1}}\times \displaystyle \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \frac{{3+2\sqrt{3}+1}}{{3-1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =2+\sqrt{3}\end{array}$
13.(a) Solve $ \displaystyle ∆ ABC$ if $ \displaystyle BC = 9, AB = 13$ and $ \displaystyle ∠B = 108°$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \text{In}\ \vartriangle ABC,BC=9,AB=13,\angle B=108{}^\circ \\\\\ \ \ \text{To Solve}\ :AC,\ \angle A\ \operatorname{and}\angle C.\\\\\ \ \ \text{By the law of cosines,}\\\\\ \ \ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2AB\cdot BC\cos B\\\\\ \ \ \ \ \ \ \ \ \ \ =169+81-2(13)(9)\cos \ 108{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ =250+234\cos \ 72{}^\circ \ \ \ \left[ {\because \cos 108{}^\circ =\cos (180{}^\circ -72{}^\circ )} \right] \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 234} & {2.3692} \\ { \cos 72 {}^\circ } & {\overline{1}.4900}\\ \hline {72.31} & {1.8592} \\ \hline \end{array}\end{array}\\\\\ \ \ \ \ \ \ \ \ \ \ =\ 250+72.31\ \\\\\ \ \ \ \ \ \ \ \ \ \ =\ 322.31\\\\\ \ \ \ \ AC=17.95\\\ \ \ \\\ \ \ \ \end{array}$ $ \displaystyle \begin{array}{l}\ \ \ \text{By the law of sines,}\\\\\ \ \ \displaystyle \frac{{\sin A}}{{BC}}=\displaystyle \frac{{\sin B}}{{AC}}\\\\ \therefore \ \ \sin A=\displaystyle \frac{{BC\sin B}}{{AC}}\\\\ \therefore \ \ \sin A=\displaystyle \frac{{9\sin 108{}^\circ }}{{17.95}}\\\\ \therefore \ \ \sin A=\displaystyle \frac{{9\sin 72{}^\circ }}{{17.95}}\ \ \ \left[ {\because \sin 108{}^\circ =\sin (180{}^\circ -72{}^\circ )} \right]\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\text{No}} & {\text{Log}} \\ \hline { 9} & {0.9542} \\ { \sin 72 {}^\circ } & {\overline{1}.9782}\\ \hline {} & {0.9324} \\ \hline {17.95} & {1.2541} \\ \hline {\sin 28 {}^\circ{28}'} & {\overline{1}.6783} \\ \hline \end{array}\end{array}\\\\ \therefore \ \ \ \angle A=28{}^\circ 2{8}'\\\\ \therefore \ \ \angle C=180{}^\circ -(108{}^\circ +28{}^\circ 2{8}')=43{}^\circ 3{2}'\end{array}$
13.(b) If $ \displaystyle y \cos x = e^x,$ show that $x \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-2\tan x\frac{{dy}}{{dx}}-2y=0.$
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ y\cos x={{e}^{x}}\\\\\ \ \ \text{Differentiate with respect to}\ x.\\\\\ \ \ y\displaystyle \frac{d}{{dx}}(\cos x)+\cos x\displaystyle \frac{{dy}}{{dx}}={{e}^{x}}\\\\\ \ \ -y\sin x+\cos x\displaystyle \frac{{dy}}{{dx}}={{e}^{x}}\\\\\ \ \ \text{Differentiate again with respect to}\ x.\\\\\ \ \ -\left[ {y\displaystyle \frac{d}{{dx}}(\sin x)+\sin x\displaystyle \frac{{dy}}{{dx}}} \right]+\left[ {\cos x\displaystyle \frac{d}{{dx}}\left( {\displaystyle \frac{{dy}}{{dx}}} \right)+\displaystyle \frac{{dy}}{{dx}}\displaystyle \frac{d}{{dx}}(\cos x)} \right]={{e}^{x}}\\\\\ \ \ -y\cos x-\sin x\displaystyle \frac{{dy}}{{dx}}+\cos x\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\sin x\displaystyle \frac{{dy}}{{dx}}-{{e}^{x}}=0\\\\\ \ \ \cos x\cdot \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-2\cdot \sin x\cdot \displaystyle \frac{{dy}}{{dx}}-{{e}^{x}}-y\cos x=0\\\\\ \ \ \text{Dividing both sides with}\ \cos x,\ \text{we get}\\\\\ \ \ \displaystyle \frac{{\cos x}}{{\cos x}}\cdot \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-2\cdot \displaystyle \frac{{\sin x}}{{\cos x}}\cdot \displaystyle \frac{{dy}}{{dx}}-\displaystyle \frac{{{{e}^{x}}}}{{\cos x}}-\displaystyle \frac{{y\cos x}}{{\cos x}}=0\\\\\ \ \ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-2\tan x\displaystyle \frac{{dy}}{{dx}}-y-y=0\ \ \ \left[ {y\cos x={{e}^{x}}\Rightarrow y=\displaystyle \frac{{{{e}^{x}}}}{{\cos x}}} \right]\\\\\therefore \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-2\tan x\displaystyle \frac{{dy}}{{dx}}-2y=0\end{array}$
14.(a) What is the smallest perimeter possible for a rectangle of area $ \displaystyle 81\ {\text{in}}^2$ ?
(5 marks)
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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{ Let the length and breadth of }\\\\ \ \ \ \ \ \ \text{the}\ \text{rectangle be and y respectively}\text{.}\ \ \ \ \ \ \\\\ \ \ \ \ \ \ \text{By the problem, }\\\\ \ \ \ \ \ \ \text{area of rectangle = 81 i}{{\text{n}}^{\text{2}}}\\\\\therefore \,\,\ \ \ \ xy=81\Rightarrow y=\displaystyle \frac{{81}}{x}\\\\\ \ \ \ \ \ \text{ Let the perimeter of the rectangle be }P.\\\\\therefore \ \ \ \ \ P=2(x+y)=2\left( {x+\displaystyle \frac{{81}}{x}} \right)\\\\\therefore \ \ \ \ \ \ \displaystyle \frac{{dP}}{{dx}}=2\left( {1-\displaystyle \frac{{81}}{{{{x}^{2}}}}} \right)\\\\\ \ \ \ \ \ \ \ \displaystyle \frac{{dP}}{{dx}}=0\ \text{when}\ 2\left( {1-\displaystyle \frac{{81}}{{{{x}^{2}}}}} \right)=0\\\\\therefore \ \ \ \ \ \ {{x}^{2}}=81\Rightarrow x=9\ \ \ \left[ {\because x>0} \right]\\\\\ \ \ \ \ \ \ \ \displaystyle \frac{{{{d}^{2}}P}}{{d{{x}^{2}}}}=2\left( {1+\displaystyle \frac{{162}}{{{{x}^{3}}}}} \right)\\\\\ \ \ \ \ \ \ \ \text{When}\ x=9,\displaystyle \frac{{{{d}^{2}}P}}{{d{{x}^{2}}}}=2\left( {1+\displaystyle \frac{{162}}{{{{9}^{3}}}}} \right)>0\\\\\therefore \ \ \ \ \ \ P\ \text{is minimum when }x=9.\\\\\therefore \ \ \ \ \ \ \text{minimum value of }P=2\left( {9+\displaystyle \frac{{81}}{9}} \right)=36\ \text{in}\ \end{array}$
14.(b) Find the matrix which will reflect in the line $ \displaystyle OY$ followed by a rotation through $ \displaystyle 60°.$ What is the map of the point $ \displaystyle (−1, 0)?$
(5 marks)
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$ \displaystyle \begin{array}{l}\text{ Reflection matrix in the line }OY\ \text{is}\\\\\ \ \ \ \ \ \ F=\displaystyle \left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 0 & 1 \end{array}} \right)\\\\\text{ Reflection matrix about the origin through }60{}^\circ \ \text{is }\\\\\ \ \ \ \ \ \ R=\displaystyle \left( {\begin{array}{*{20}{c}} {\cos 60{}^\circ } & {-\sin 60{}^\circ } \\ {\sin 60{}^\circ } & {\cos 60{}^\circ } \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {\displaystyle \frac{1}{2}} & {-\displaystyle \frac{{\sqrt{3}}}{2}} \\ {\displaystyle \frac{{\sqrt{3}}}{2}} & {\displaystyle \frac{1}{2}} \end{array}} \right)\\\\\text{ Let the matrix which will reflect in the line }OY\\\ \ \ \ \ \ \ \text{followed by a rotation through 60}{}^\circ \ \text{be }T.\\\\\therefore \ \ \ \ \ \ T=RF\\\\\therefore \ \ \ \ \ \ T=\displaystyle \left( {\begin{array}{*{20}{c}} {\displaystyle \frac{1}{2}} & {-\displaystyle \frac{{\sqrt{3}}}{2}} \\ {\displaystyle \frac{{\sqrt{3}}}{2}} & {\displaystyle \frac{1}{2}} \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 0 & 1 \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {-\displaystyle \frac{1}{2}} & {-\displaystyle \frac{{\sqrt{3}}}{2}} \\ {-\displaystyle \frac{{\sqrt{3}}}{2}} & {\displaystyle \frac{1}{2}} \end{array}} \right)\\\\\text{ Let }\displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {-1} \\ 0 \end{array}} \right)\\\\\therefore \ \ \ \ \ \displaystyle \left( {\begin{array}{*{20}{c}} {{x}'} \\ {{y}'} \end{array}} \right)=T\displaystyle \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \left( {\begin{array}{*{20}{c}} {-\displaystyle \frac{1}{2}} & {-\displaystyle \frac{{\sqrt{3}}}{2}} \\ {-\displaystyle \frac{{\sqrt{3}}}{2}} & {\displaystyle \frac{1}{2}} \end{array}} \right)\displaystyle \left( {\begin{array}{*{20}{c}} {-1} \\ 0 \end{array}} \right)\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \left( {\begin{array}{*{20}{c}} {\displaystyle \frac{1}{2}} \\ {\displaystyle \frac{{\sqrt{3}}}{2}} \end{array}} \right)\ \\\\\therefore \ \ \ \ \text{The mapped point is}\ ({x}',{y}')=\displaystyle \left( {\displaystyle \frac{1}{2},\displaystyle \frac{{\sqrt{3}}}{2}} \right)\end{array}$
| ပုစာၦ၏ အေျဖမွန္ မမွန္ကို ေအာက္ပါအတိုင္း စဥ္းစားနိုင္ပါသည္။ reflection in the line OY ဆိုသည္မွာ y-axis အတိုင္း ေခါက္ခ်ိဳးညီ ေနရာကို ရွာယူလိုက္ျခင္း ျဖစ္သည္။ ထို႕ေၾကာင့္$ \displaystyle (-1,0)$ ၏ y-axis အတိုင္း ေခါက္ခ်ိဳးညီ ေနရာမွာ $ \displaystyle (1,0)$ ျဖစ္သည္။ ၎ $ \displaystyle (1,0)$ ကို 60° ေထာင့္ျဖင့္ ေရႊ႕လိုက္လွ်င္ $ \displaystyle x=\cos 60{}^\circ =\frac{1}{2}$ ႏွင့္ $ \displaystyle y=\sin 60{}^\circ =\frac{{\sqrt{3}}}{2}$ ျဖစ္မည္။ |
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