Sample Math Paper (2) - Section (A) Solution

๐Ÿ‘‰ แ€’ီေแ€”แ€›ာแ€™ွာ แ€แ€„္ေแ€•းแ€œိုแ€€္แ€ဲ့ ေแ€™းแ€ြแ€”္းแ€›ဲ့ section (A) แ€กေျแ€–ျแ€–แ€…္แ€•ါแ€แ€š္။ แ€’ီေแ€™းแ€ြแ€”္းแ€€ แ€‘ူးแ€แฝြแ€”္ ေแ€€်ာแ€„္းแ€žားแ€™်ား แ€กแ€ြแ€€္ แ€›แ€Š္แ€›ြแ€š္แ€•ါแ€แ€š္။ ေแ€กာแ€„္แ€™ွแ€္ แ€กแ€ြแ€€္แ€žာ แ€œုแ€•္ေแ€”แ€›แ€ဲ့ ေแ€€်ာแ€„္းแ€žား แ€™်ားแ€กแ€ြแ€€္ แ€กแ€†แ€„္แ€™ေျแ€•ႏိုแ€„္แ€•ါแ€˜ူး။ แ€’ါ့ေแพแ€€ာแ€„့္ แ€žာแ€™แ€”္แ€กแ€†แ€„့္ ေแ€€်ာแ€„္းแ€žားแ€™်ားแ€€ို ေแ€œ့แ€€်แ€„့္ေแ€•းแ€›แ€”္ แ€™แ€žแ€„့္ေแ€œ်ာ္ေแพแ€€ာแ€„္း แ€กแ‚€แ€€ံျแ€•ဳ แ€กแ€•္แ€•ါแ€แ€š္။ แ€›แ€Š္แ€™ွแ€”္းแ€‘ားแ€žแ€Š့္ แ€กแ€ိုแ€„္း ေแ€•ါแ€€္ေျแ€™ာแ€€္ ေแ€กာแ€„္ျแ€™แ€„္ ႏိုแ€„္แพแ€€แ€•ါေแ€…။...


Section (A)

1.  (a) The function $ \displaystyle g : N \to N$ is defined as $ \displaystyle g : x\mapsto$ smallest prime factor of $ \displaystyle x.$ (i) Find values for $ \displaystyle g(10), g (20)$ and $ \displaystyle g (81).$ (ii) Does $ \displaystyle g$ have an inverse? Give reasons for your answer.
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ g:N\to N\\\\\ \ \ \ g(x)=\text{smallest prime factor of}\ x.\\\\\ \ \ \ 10=2\times 5\\\\\therefore \ \ g(10)=2\\\\\ \ \ \ 20=2\times 2\times 5\\\\\therefore \ \ g(20)=2\\\\\ \ \ \ 81=3\times 3\times 3\times 3\\\\\therefore \ \ g(81)=3\\\\\ \ \ \ \text{Since}\ g(10)=g(20),\\\\\ \ \ \ g\ \text{is not one to one correspondence}\text{.}\\\\\therefore \ \ {{g}^{{-1}}}\ \text{does not}\ \text{exists}\text{.}\end{array}$


(1)  (b) If $ \displaystyle 2x-1$ is a factor of $ \displaystyle 2x^3-x^2-8x+k,$ find $ \displaystyle k$ and the other factors.
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \text{Let }f(x)=2{{x}^{3}}-{{x}^{2}}-8x+k\\\\\ \ \ \ \ \ \ 2x-1\ \text{is a factor of }f(x).\\\\\therefore \ \ \ \ \ f\left( {\displaystyle \frac{1}{2}} \right)=0\\\\\therefore \ \ \ \ \ 2{{\left( {\displaystyle \frac{1}{2}} \right)}^{3}}-{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-8\left( {\displaystyle \frac{1}{2}} \right)+k=0\\\\\therefore \ \ \ \ \ \displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}-4+k=0\\\\\therefore \ \ \ \ \ k=4\\\\\therefore \ \ \ \ \ f(x)=2{{x}^{3}}-{{x}^{2}}-8x+4\\\\\ \ \ \ \ \ \ \text{Let }f(x)=(2x-1)({{x}^{2}}+ax+b)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{x}^{3}}+2a{{x}^{2}}+2bx-{{x}^{2}}-ax-b\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{x}^{3}}+(2a-1){{x}^{2}}+(2b-a)x-b\\\\\therefore \ \ \ \ \ 2a-1=-1\ \operatorname{and}\,b=-4\\\\\therefore \ \ \ \ \ a=0\ \operatorname{and}\ b=-4\\\\\therefore \ \ \ \ \ f(x)=(2x-1)({{x}^{2}}-4)=(2x-1)(x-2)(x+2)\\\\\therefore \ \ \ \ \text{The other factors are }x-2\ \text{and }x+2.\end{array}$


2.  (a) Find the term independent of $ \displaystyle x$ in the expansion of $ \displaystyle {{\left( {x-\frac{2}{{{{x}^{2}}}}} \right)}^{9}}.$
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of }\ {{\left( {x-\displaystyle \frac{2}{{{{x}^{2}}}}} \right)}^{9}}\\\\\ \ \ \ ={}^{9}{{C}_{r}}{{x}^{{9-r}}}{{\left( {-\displaystyle \frac{2}{{{{x}^{2}}}}} \right)}^{r}}\\\\\ \ \ \ ={}^{9}{{C}_{r}}{{(-2)}^{r}}{{x}^{{9-3r}}}\\\\\ \ \ \ \ \ \ \text{For the term independent of }x,\ 9-3r=0\\\\\therefore \ \ \ \ \ r=3\\\\\therefore \ \ \ \ \ \text{The term independent of }x={}^{9}{{C}_{3}}{{(-2)}^{3}}=\displaystyle \frac{{9\times 8\times 7}}{{1\times 2\times 3}}(-8)=-672\end{array}$


(2)  (b) If the sum of n terms of a certain sequence is $ \displaystyle 2n + 3n^2,$ find the $ \displaystyle {n}^{\text{th}}$ term.
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ {{S}_{n}}=2n+3{{n}^{2}}\\\\\ \ \ \ \ \ {{u}_{n}}={{S}_{n}}-{{S}_{{n-1}}}\\\\\therefore \ \ \ \ {{u}_{n}}=2n+3{{n}^{2}}-\left[ {2(n-1)+3{{{(n-1)}}^{2}}} \right]\\\\\therefore \ \ \ \ {{u}_{n}}=2n+3{{n}^{2}}-2n+2+3{{n}^{2}}+6n-3\\\\\therefore \ \ \ \ {{u}_{n}}=4n-1\end{array}$


3.  (a) If $ \displaystyle X=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)$ and $ \displaystyle X-kI$ is singular, where $ \displaystyle I$ is a unit matrix of order $ \displaystyle 2,$ find $ \displaystyle k.$
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ X=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)\\\\\ \ \ \ \ X-kI=\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 2 & 3 \end{array}} \right)-k\displaystyle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)=\displaystyle \left( {\begin{array}{*{20}{c}} {1-k} & 0 \\ 2 & {3-k} \end{array}} \right)\\\\\ \ \ \ \ X-kI\ \text{is singular}.\\\\\therefore \ \ \ \det (X-kI)=0\\\\\therefore \ \ \ (1-k)(3-k)=0\\\\\therefore \ \ \ k=1\ (\text{or})\ k=3\end{array}$


(3)  (b) A number $ \displaystyle x$ is chosen at random from the numbers $ \displaystyle -4, -3, -2, -1, 0, 1, 2, 3, 4.$ What is the probability that $ \displaystyle |x| \le 2?$
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Set of possible outcomes}=\left\{ {-4,-3,-2,-1,0,1,2,3,4} \right\}\\\\\therefore \ \ \ \ \text{Number of possible outcomes}=9\\\\\ \ \ \ \ \ |x|\le 2\Leftrightarrow -2\le x\le 2\\\\\therefore \ \ \ \ \text{Set of favourable outcomes}=\left\{ {-2,-1,0,1,2} \right\}\\\\\therefore \ \ \ \ \text{Number of favourable outcomes}=5\\\\\therefore \ \ \ \ P\left( {|x|\le 2} \right)=\displaystyle \frac{5}{9}\end{array}$


4.  (a) $ \displaystyle TA$ is the tangent to the circle at$ \displaystyle A, AB = BC, ∠BAC = 41°$ and $ \displaystyle ∠ACT = 46°.$ Find $ \displaystyle ∠ATC.$
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \angle BAC=41{}^\circ ,\angle ACT=46{}^\circ (\text{given})\\\\\ \ \ \ \text{Since}\ AB=BC,\angle BCA=\angle BAC\\\\\therefore \ \ \angle BCA=41{}^\circ \\\\\therefore \ \ \angle ABC=180{}^\circ -(41{}^\circ +41{}^\circ )=98{}^\circ \\\\\ \ \ \ \text{Since}\ \angle CAT=\angle ABC,\angle CAT=98{}^\circ \\\\\ \ \ \ \text{In}\ \vartriangle CAT,\\\\\ \ \ \ \angle ATC=180{}^\circ -(\angle CAT+\angle ACT)\\\\\therefore \ \ \angle ATC=180{}^\circ -(98{}^\circ +46{}^\circ )=36{}^\circ \end{array}$


4.  (b) If $ \displaystyle 3\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}=\vec{0},$ show that the points $ \displaystyle A, B$ and $ \displaystyle C$ are collinear.
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 3\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}=\vec{0}\\\\\therefore \ \ \ 2\overrightarrow{{OA}}-2\overrightarrow{{OB}}+\overrightarrow{{OA}}-\overrightarrow{{OC}}=\vec{0}\\\\\therefore \ \ \ 2\left( {\overrightarrow{{OA}}-\overrightarrow{{OB}}} \right)+\left( {\overrightarrow{{OA}}-\overrightarrow{{OC}}} \right)=\vec{0}\\\\\therefore \ \ \ 2\overrightarrow{{BA}}+\overrightarrow{{CA}}=\vec{0}\\\\\therefore \ \ \ 2\overrightarrow{{BA}}=-\overrightarrow{{CA}}\\\\\therefore \ \ \ 2\overrightarrow{{BA}}=\overrightarrow{{AC}}\\\\\therefore \ \ \ A,B\ \operatorname{and}\ C\ \text{are collinear}\text{.}\end{array}$


5.  (a) If $\displaystyle \tan \alpha =x+1$ and $ \displaystyle \tan \beta =x-1$, find $ \displaystyle \cot (\alpha -\beta )$ in terms of $ \displaystyle x.$
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \tan \ \alpha =x+1,\ \tan \beta =x-1\\\\\therefore \ \ \ \ \tan (\alpha -\beta )=\displaystyle \frac{{\tan \ \alpha -\tan \beta }}{{1+\tan \ \alpha \tan \beta }}\\\\\therefore \ \ \ \ \tan (\alpha -\beta )=\displaystyle \frac{{x+1-x+1}}{{1+(x+1)(x-1)}}\\\\\therefore \ \ \ \ \tan (\alpha -\beta )=\displaystyle \frac{2}{{1+({{x}^{2}}-1)}}=\displaystyle \frac{2}{{{{x}^{2}}}}\\\\\therefore \ \ \ \ \cot (\alpha -\beta )=\displaystyle \frac{1}{{\tan (\alpha -\beta )}}=\displaystyle \frac{{{{x}^{2}}}}{2}\end{array}$


5.  (b) Evaluate $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(2x-3)(\sqrt{x}-1)}}{{2{{x}^{2}}+x-3}}$ and $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{3{{{\sin }}^{2}}x-2\sin {{x}^{2}}}}{{3{{x}^{2}}}}.$
(3 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)(\sqrt{x}-1)}}{{2{{x}^{2}}+x-3}}\\\\\ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)(\sqrt{x}-1)}}{{(2x+3)(x-1)}}\\\\\ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)(\sqrt{x}-1)}}{{(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)}}\\\\\ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\displaystyle \frac{{(2x-3)}}{{(2x+3)(\sqrt{x}+1)}}\\\\\ \ \ \ =\displaystyle \frac{{2-3}}{{(2+3)(1+1)}}\\\\\ \ \ \ =-\displaystyle \frac{1}{{10}}\\\\\\\ \ \ \ \ \ \ \underset{{x\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{3{{{\sin }}^{2}}x-2\sin {{x}^{2}}}}{{3{{x}^{2}}}}\\\\\ \ \ \ =\underset{{x\to 0}}{\mathop{{\lim }}}\,\left[ {\displaystyle \frac{{{{{\sin }}^{2}}x}}{{{{x}^{2}}}}-\displaystyle \frac{2}{3}\cdot \displaystyle \frac{{\sin {{x}^{2}}}}{{{{x}^{2}}}}} \right]\\\\\ \ \ \ =\underset{{x\to 0}}{\mathop{{\lim }}}\,\left[ {{{{\left( {\displaystyle \frac{{\sin x}}{x}} \right)}}^{2}}-\displaystyle \frac{2}{3}\cdot \displaystyle \frac{{\sin {{x}^{2}}}}{{{{x}^{2}}}}} \right]\\\\\ \ \ \ =1-\displaystyle \frac{2}{3}\\\\\ \ \ \ =\displaystyle \frac{1}{3}\end{array}$


แ€€်แ€”္ေแ€žာ แ€กေျแ€–แ€™်ားแ€€ို แ€†แ€€္แ€œแ€€္ แ€แ€„္ေแ€•းแ€žြားแ€•ါ့แ€™แ€š္။