👉 ဒီေနရာမွာ တင္ေပးလိုက္တဲ့ ေမးခြန္းရဲ့ section (A) အေျဖျဖစ္ပါတယ္။ ဒီေမးခြန္းက ထူးခၽြန္ ေက်ာင္းသားမ်ား အတြက္ ရည္ရြယ္ပါတယ္။ ေအာင္မွတ္ အတြက္သာ လုပ္ေနရတဲ့ ေက်ာင္းသား မ်ားအတြက္ အဆင္မေျပႏိုင္ပါဘူး။ ဒါ့ေၾကာင့္ သာမန္အဆင့္ ေက်ာင္းသားမ်ားကို ေလ့က်င့္ေပးရန္ မသင့္ေလ်ာ္ေၾကာင္း အႀကံျပဳ အပ္ပါတယ္။ ရည္မွန္းထားသည့္ အတိုင္း ေပါက္ေျမာက္ ေအာင္ျမင္ ႏိုင္ၾကပါေစ။...
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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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)
Show/Hide Solution
$ \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}$
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!