Target Mathematics

2020

MATRICULATION EXAMINATION

Sample Question Set (3)

MATHEMATICS Time allowed: 3hours
WRITE YOUR ANSWERS IN THE ANSWER BOOKLET.
SECTION (A)
(Answer ALL questions.)

1 (a). Let $ \displaystyle f:R\backslash \{\pm 2\}\to R$ be a function defined by $ \displaystyle f(x)=\frac{{3x}}{{{{x}^{2}}-4}}$.Find the positive value of $z$ such that $f(z) = 1$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}f(x)=\displaystyle \frac{{3x}}{{{{x}^{2}}-4}}\\[2ex]f(z)=1\\[2ex]\displaystyle \frac{{3z}}{{{{z}^{2}}-4}}=1\\[2ex]{{z}^{2}}-4=3z\\[2ex]{{z}^{2}}-3z-4=0\\[2ex](z+1)(z-4)=0\\[2ex]z=-1\ \text{or}\ z=4\\[2ex]\text{Since }z>0,\ z=4\end{array}$

1(b). If the polynomial $x^3 - 3x^2 + ax - b$ is divided by $(x - 2 )$ and $(x + 2)$, the remainders are $21$ and $1$ respectively. Find the values of $a$ and $b$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Let}\ f(x)={{x}^{3}}-3{{x}^{2}}+ax-b\\[2ex]\text{When}\ f(x)\ \text{is divided by }(x-2),\ \\[2ex]\text{The remainder is }21.\\[2ex]f(2)=21\\[2ex]{{(2)}^{3}}-3{{(2)}^{2}}+a(2)-b=21\\[2ex]2a-b=25\ --------(1)\\[2ex]\text{When}\ f(x)\ \text{is divided by }(x+2),\ \\[2ex]\text{The remainder is }1.\\[2ex]f(-2)=1\\[2ex]{{(-2)}^{3}}-3{{(-2)}^{2}}+a(-2)-b=1\\[2ex]2a+b=-21\ --------(2)\\[2ex](1)+(2)\Rightarrow 4a=4\Rightarrow a=1\\[2ex](1)-(2)\Rightarrow -2b=46\Rightarrow b=-23\end{array}$

2(a). Find the middle term in the expansion of $(x^2 - 2y)^{10}$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}{{(r+1)}^{{\text{th}}}}\ \text{term in the expansion of }{{({{x}^{2}}-2y)}^{{10}}}={}^{{10}}{{C}_{r}}{{\left( {{{x}^{2}}} \right)}^{{10-r}}}{{\left( {-2y} \right)}^{r}}\text{ }\\[2ex]\text{middle term in the expansion of }{{({{x}^{2}}-2y)}^{{10}}}={{6}^{{\text{th}}}}\ \text{term}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(5+1)}^{{\text{th}}}}\ \text{term}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={}^{{10}}{{C}_{5}}{{\left( {{{x}^{2}}} \right)}^{{10-5}}}{{\left( {-2y} \right)}^{5}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{10\times 9\times 8\times 7\times 6}}{{1\times 2\times 3\times 4\times 5}}{{x}^{{10}}}\left( {-32{{y}^{5}}} \right)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-8064{{x}^{{10}}}{{y}^{5}}\end{array}$

2(b). In a sequence if $u_1=1$ and $u_{n+1}=u_n+3(n+1)$, find $u_5$ .

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}{{u}_{1}}=1,\ {{u}_{{n+1}}}={{u}_{n}}+3(n+1)\\[2ex]\therefore {{u}_{2}}={{u}_{{1+1}}}={{u}_{1}}+3(1+1)=1+6=7\\[2ex]\ \ \ {{u}_{3}}={{u}_{{2+1}}}={{u}_{2}}+3(2+1)=7+9=16\\[2ex]\ \ \ {{u}_{4}}={{u}_{{3+1}}}={{u}_{3}}+3(3+1)=16+12=28\\[2ex]\ \ \ {{u}_{5}}={{u}_{{4+1}}}={{u}_{4}}+3(4+1)=28+15=43\end{array}$

3(a). If $ \displaystyle P=\left( {\begin{array}{*{20}{c}} x & {-4} \\ {8-y} & {-9} \end{array}} \right)$ and $ \displaystyle {{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-3x} & 4 \\ {-7y} & 3 \end{array}} \right)$, find the values of $x$ and $y$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}P=\left( {\begin{array}{*{20}{c}} x & {-4} \\[2ex] {8-y} & {-9} \end{array}} \right),\ {{P}^{{-1}}}=\left( {\begin{array}{*{20}{c}} {-3x} & 4 \\[2ex] {-7y} & 3 \end{array}} \right)\\[2ex]\text{Since}\ P{{P}^{{-1}}}=I,\\[2ex]\left( {\begin{array}{*{20}{c}} x & {-4} \\[2ex] {8-y} & {-9} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {-3x} & 4 \\[2ex] {-7y} & 3 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 & 0 \\[2ex] 0 & 1 \end{array}} \right)\\[2ex]\left( {\begin{array}{*{20}{c}} {-3{{x}^{2}}+28y} & {4x-12} \\[2ex] {-24x+3xy+63y} & {5-4y} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 & 0 \\[2ex] 0 & 1 \end{array}} \right)\\[2ex]\therefore -3{{x}^{2}}+28y=1\Rightarrow y=\displaystyle \frac{{3{{x}^{2}}+1}}{{28}}\\[2ex]\ \ \ 4x-12=0\Rightarrow x=3\\[2ex]\ \ -24x+3xy+63y=0\\[2ex]\ \ \ 5-4y=1\Rightarrow y=1\\[2ex]\therefore \ x=3\ \text{and}\ y=1.\end{array}$

3(b). A bag contains tickets, numbered $11, 12, 13, ...., 30$. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket is

(i) a multiple of $7$

(ii) greater than $15$ and a multiple of $5$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Set of possible outcomes}\ =\{11,\ 12,\ 13,\ ...,\ 30\}\\[2ex]\text{Number of possible outcomes}\ =20\\[2ex]\text{Set of favourable outcomes}\ \text{for a number }\\[2ex]\text{multiple of 7= }\!\!\{\!\!\text{ 14,}\ \text{21,}\ \text{28 }\!\!\}\!\!\text{ }\\[2ex]\text{Number of favourable outcomes}\ =3\\[2ex]P(\text{a number multiple of 7)}=\displaystyle \frac{3}{{20}}\\[2ex]\text{Set of favourable outcomes}\ \text{for a number }\\[2ex]\text{greater than 15 and a multiple of 5 = }\!\!\{\!\!\text{ }\ \text{20,}\ \text{25,}\ \text{30 }\!\!\}\!\!\text{ }\\[2ex]\text{Number of favourable outcomes}\ =3\\[2ex]P(\text{a number greater than 15 and amultiple of 5)}=\displaystyle \frac{3}{{20}}\end{array}$

4(a). Draw a circle and a tangent $TAS$ meeting it at $A$. Draw a chord $AB$ making $ \displaystyle \angle TAB=\text{ }60{}^\circ $ and another chord $BC \parallel TS$. Prove that $\triangle ABC$ is equilateral.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Since}\ BC\parallel TS,\\[2ex]\ \ \ \beta =\angle TAB\ \ \ [\text{alternating }\angle \text{s }\!\!]\!\!\text{ }\\[2ex]\therefore \beta =60{}^\circ \\[2ex]\ \ \ \gamma =\angle TAB\ \ \ [\angle \text{ between tangent and chord}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ }\text{=}\angle \ \text{in alternate segment }\!\!]\!\!\text{ }\\[2ex]\therefore \gamma =60{}^\circ \\[2ex]\text{Since}\ \alpha +\beta +\gamma =180{}^\circ ,\\[2ex]\alpha =180{}^\circ -(\beta +\gamma )\\[2ex]\alpha =180{}^\circ -(60{}^\circ +60{}^\circ )\\[2ex]\therefore \alpha =60{}^\circ \\[2ex]\therefore \alpha =\beta =\gamma \\[2ex]\therefore \ \triangle ABC\ \text{is equilateral}\text{.}\end{array}$

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

(3 marks)

Show/Hide Solution

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

5(a). Solve the equation $2 \sin x \cos x -\cos x + 2\sin x - 1 = 0$ for $ \displaystyle 0{}^\circ \le x\le \text{ }360{}^\circ $.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{For}\ 0{}^\circ \le x\le \text{ }360{}^\circ ,\\[2ex]2\sin x\cos x-\cos x+2\sin x-1=0\\[2ex]\cos x\left( {2\sin x-1} \right)+\left( {2\sin x-1} \right)=0\\[2ex]\left( {2\sin x-1} \right)\left( {\cos x+1} \right)=0\\[2ex]\therefore \sin x=\displaystyle \frac{1}{2}\ \text{or}\ \cos x=-1\\[2ex](\text{i})\ \sin x=\displaystyle \frac{1}{2}\\[2ex]\ \ \ \ x=30{}^\circ \ \text{or}\ x=150{}^\circ \\[2ex](\text{ii})\ \cos x=-1\\[2ex]\ \ \ \ \ x=180{}^\circ \\[2ex]\therefore x=30{}^\circ \ \text{or}\ x=150{}^\circ \ \text{or}\ x=180{}^\circ \end{array}$

5(b). Differentiate $ \displaystyle y=\frac{1}{{\sqrt[3]{x}}}$ from the first principles.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}y=\displaystyle \frac{1}{{\sqrt[3]{x}}}\\[2.5ex]y+\delta y=\displaystyle \frac{1}{{\sqrt[3]{{x+\delta x}}}}\\[2.5ex]\delta y=\displaystyle \frac{1}{{\sqrt[3]{{x+\delta x}}}}-\displaystyle \frac{1}{{\sqrt[3]{x}}}\\[2.5ex]\delta y=\displaystyle \frac{{\sqrt[3]{x}-\sqrt[3]{{x+\delta x}}}}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\\[2.5ex]\displaystyle \frac{{\delta y}}{{\delta x}}=\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{\sqrt[3]{x}-\sqrt[3]{{x+\delta x}}}}{{\delta x}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{\sqrt[3]{x}-\sqrt[3]{{x+\delta x}}}}{{x+\delta x-x}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{\sqrt[3]{x}-\sqrt[3]{{x+\delta x}}}}{{{{{\left( {\sqrt[3]{{x+\delta x}}} \right)}}^{3}}-{{{\left( {\sqrt[3]{x}} \right)}}^{3}}}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{-\left( {\sqrt[3]{{x+\delta x}}-\sqrt[3]{x}} \right)}}{{{{{\left( {\sqrt[3]{{x+\delta x}}} \right)}}^{3}}-{{{\left( {\sqrt[3]{x}} \right)}}^{3}}}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{-\left( {\sqrt[3]{{x+\delta x}}-\sqrt[3]{x}} \right)}}{{\left( {\sqrt[3]{{x+\delta x}}-\sqrt[3]{x}} \right)\left[ {{{{\left( {\sqrt[3]{{x+\delta x}}} \right)}}^{2}}+\left( {\sqrt[3]{{x+\delta x}}} \right)\left( {\sqrt[3]{x}} \right)+{{{\left( {\sqrt[3]{x}} \right)}}^{2}}} \right]}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{-1}}{{{{{\left( {\sqrt[3]{{x+\delta x}}} \right)}}^{2}}+\left( {\sqrt[3]{{x+\delta x}}} \right)\left( {\sqrt[3]{x}} \right)+{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}\\[2.5ex]\displaystyle \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{\delta y}}{{\delta x}}\\[2.5ex]\ \ \ \ \ =\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\left[ {\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{{x+\delta x}}}}\cdot \displaystyle \frac{{-1}}{{{{{\left( {\sqrt[3]{{x+\delta x}}} \right)}}^{2}}+\left( {\sqrt[3]{{x+\delta x}}} \right)\left( {\sqrt[3]{x}} \right)+{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}} \right]\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{x}\sqrt[3]{x}}}\cdot \displaystyle \frac{{-1}}{{{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\left( {\sqrt[3]{x}} \right)\left( {\sqrt[3]{x}} \right)+{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{1}{{{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}\cdot \displaystyle \frac{{-1}}{{{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{{-1}}{{3\cdot {{{\left( {\sqrt[3]{x}} \right)}}^{2}}{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}\\[2.5ex]\ \ \ \ \ =\displaystyle \frac{{-1}}{{3\cdot {{{\left( {\sqrt[3]{x}} \right)}}^{4}}}}\\[2.5ex]\ \ \ \ \ \end{array}$

SECTION (B)
(Answer Any FOUR questions.)

6 (a). Given that Given $ \displaystyle A=\{x\in R|\ x\ne -\frac{1}{2},x\ne \frac{3}{2}\}$. If $f:A\to A$ and $g:A\to A$ are defined by $f(x)=\displaystyle \frac{3x-5}{2x+1}$ and $g(x)=\displaystyle \frac{x+5}{3-2x}$, show that $f$ and $g$ are inverse of each other.

(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\begin{array}{c|c} { \displaystyle \begin{array}{l}A=\{x\in R|\ x\ne -\displaystyle \frac{1}{2},x\ne \displaystyle \frac{3}{2}\}\\[2ex]f:A\to A,f(x)=\displaystyle \frac{{3x-5}}{{2x+1}}\\[2ex]g:A\to A,g(x)=\displaystyle \frac{{x+5}}{{3-2x}}\\[2ex](f\cdot g)(x)=f\left( {g(x)} \right)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\displaystyle \frac{{x+5}}{{3-2x}}} \right)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{3\left( {\displaystyle \frac{{x+5}}{{3-2x}}} \right)-5}}{{2\left( {\displaystyle \frac{{x+5}}{{3-2x}}} \right)+1}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{3x+15-15+10x}}{{3-2x}}}}{{\displaystyle \frac{{2x+10+3-2x}}{{3-2x}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{13x}}{{3-2x}}}}{{\displaystyle \frac{{13}}{{3-2x}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =x\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =I(x)\end{array}} & { \displaystyle \begin{array}{l}(g\cdot f)(x)=g\left( {f(x)} \right)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( {\displaystyle \frac{{3x-5}}{{2x+1}}} \right)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\left( {\displaystyle \frac{{3x-5}}{{2x+1}}} \right)+5}}{{3-2\left( {\displaystyle \frac{{3x-5}}{{2x+1}}} \right)}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{3x-5+10x+5}}{{2x+1}}}}{{\displaystyle \frac{{6x+3-6x+10}}{{2x+1}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{13x}}{{2x+1}}}}{{\displaystyle \frac{{13}}{{2x+1}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =x\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ =I(x)\\[2ex]\therefore (f\cdot g)(x)=(g\cdot f)(x)=I(x)\\[2ex]\therefore f={{g}^{{-1}}}\ \text{and}\ g={{f}^{{-1}}}\end{array}}\end{array}\end{array}$

6 (b). Given that $x^5 + ax^3 + bx^2 - 3 = (x^2 - 1) Q(x) - x - 2$, where $Q(x)$ is a polynomial. State the degree of $Q (x)$ and find the values of $a$ and $b$. Find also the remainder when $Q (x)$ is divided by $x + 2$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}{{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2\\[2ex]{{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)\\[2ex]\text{degree}\ \text{of }Q\left( x \right)=3\\[2ex]{{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)\\[2ex]{{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}+x-1=(x-1)(x+1)Q(x)\\[2ex]\text{When}\ x=1\\[2ex]1+a+b+1-1=(1-1)(1+1)Q(x)\\[2ex]1+a+b=0\\[2ex]a+b=-1\ -------(1)\\[2ex]\text{When}\ x=-1\\[2ex]-1-a+b-1-1=(-1-1)(-1+1)Q(x)\\[2ex]-3-a+b=0\\[2ex]-a+b=3\ \ \ ------(2)\\[2ex](1)+(2)\Rightarrow 2b=2\Rightarrow b=1\\[2ex](1)-(2)\Rightarrow 2a=-4\Rightarrow a=-2\\[2ex]{{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)\\[2ex]\therefore Q(x)=\displaystyle \frac{{{{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1}}{{{{x}^{2}}-1}}\\[2ex]\text{When }Q(x)\text{ is divided by }x+2\text{, }\\[2ex]\text{the remainder is }Q(-2).\\[2ex]\therefore Q(-2)=\displaystyle \frac{{{{{(-2)}}^{5}}-2{{{(-2)}}^{3}}+{{{(-2)}}^{2}}+(-2)-1}}{{{{{(-2)}}^{2}}-1}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ =-5\end{array}$

7 (a). The binary operation $\odot$ on $R$ be defined by $x\odot y=x+y+10xy$ Show that the binary operation is commutative. Find the values $b$ such that $ (1\odot b)\odot b=485$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}x\odot y=x+y+10xy\\[2ex]y\odot x=y+x+10yx\\[2ex]\ \ \ \ \ \ \ \ =x+y+10xy\\[2ex]\therefore x\odot y=y\odot x\\[2ex]\therefore \ \text{The binary operation is commutative}\text{.}\\[2ex]1\odot b=1+b+10b\\[2ex](1\odot b)\odot b=(1+b+10b)\odot b\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(1+b+10b)+b+10(1+b+10b)b\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+22b+110{{b}^{2}}\\[2ex](1\odot b)\odot b=485\\[2ex]1+22b+110{{b}^{2}}=485\\[2ex]110{{b}^{2}}+22b-484=0\\[2ex]5{{b}^{2}}+b-22=0\\[2ex](5b+11)(b-2)=0\\[2ex]b=-\displaystyle \frac{{11}}{5}\ \text{or}\ b=2\end{array}$

7 (b). If, in the expansion of $(1 + x)^m (1 - x)^n$, the coefficient of $x$ and $x^2$ are $-5$ and $7$ respectively, then find the value of $m$ and $n$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}{{(1+x)}^{m}}{{(1-x)}^{n}}=\left( {1+{}^{m}{{C}_{1}}x+{}^{m}{{C}_{2}}{{x}^{2}}+...} \right)\left( {1-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+...} \right)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+\left( {{}^{m}{{C}_{1}}-{}^{n}{{C}_{1}}} \right)x+\left( {{}^{m}{{C}_{2}}-{}^{m}{{C}_{1}}{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}} \right){{x}^{2}}+...\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+\left( {m-n} \right)x+\left( {\displaystyle \frac{{m(m-1)}}{2}-mn+\displaystyle \frac{{n(n-1)}}{2}} \right){{x}^{2}}+...\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+\left( {m-n} \right)x+\left( {\displaystyle \frac{{{{m}^{2}}-2mn+{{n}^{2}}-(m+n)}}{2}} \right){{x}^{2}}+...\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+\left( {m-n} \right)x+\left( {\displaystyle \frac{{{{{(m-n)}}^{2}}-(m+n)}}{2}} \right){{x}^{2}}+...\\[2ex]\text{By the problem,}\\[2ex]\ \ \ \ m-n=-5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(1)\\[2ex]\ \ \ \displaystyle \frac{{{{{(m-n)}}^{2}}-(m+n)}}{2}=7\\[2ex]\therefore \displaystyle \frac{{{{{(-5)}}^{2}}-(m+n)}}{2}=7\\[2ex]\ \ \ 25-(m+n)=14\\[2ex]\ \ \ m+n=11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(2)\\[2ex](1)+(2)\Rightarrow 2m=6\Rightarrow m=3\\[2ex](1)-(2)\Rightarrow -2n=-16\Rightarrow n=8\end{array}$

8 (a). Find the solution set in $R$ for the inequation $2x (x + 2)\ge (x + 1) (x + 3)$ and illustrate it on the number line.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}2x(x+2)\ge (x+1)(x+3)\\[2ex]2{{x}^{2}}+4x\ge {{x}^{2}}+4x+3\\[2ex]{{x}^{2}}-3\ge 0\\[2ex]\left( {x+\sqrt{3}} \right)\left( {x-\sqrt{3}} \right)\ge 0\\[2ex]\left( {x+\sqrt{3}\ge 0\ \text{and}\ x-\sqrt{3}\ge 0} \right)\ \text{or}\ \left( {x+\sqrt{3}\le 0\ \text{and}\ x-\sqrt{3}\le 0} \right)\ \\[2ex]\left( {x\ge -\sqrt{3}\ \text{and}\ x\ge \sqrt{3}} \right)\ \text{or}\ \left( {x\le -\sqrt{3}\ \text{and}\ x\le \sqrt{3}} \right)\ \\[2ex]\therefore x\ge \sqrt{3}\ \text{or}\ x\le -\sqrt{3}\\[2ex]\therefore \ \text{Solution Set = }\left\{ {x\ |\ x\le -\sqrt{3}\ \text{or}\ x\ge \sqrt{3}} \right\}\\[2ex]\text{Number Line}\end{array}$

8 (b). If the $ {{m}^{{\text{th}}}}$ term of an A.P. is $ \displaystyle \frac{1}{n}$ and $ {{n}^{{\text{th}}}}$ term is $ \displaystyle \frac{1}{m}$ where $m\ne n$, then show that $u_{mn} = 1$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Let the first and the common difference }\\[2ex]\text{of the give A}\text{.P}\text{. be }a\ \text{and }d\ \text{respectively}\text{.}\\[2ex]\text{By the problem,}\\[2ex]{{u}_{m}}=\displaystyle \frac{1}{n}\\[2ex]a+(m-1)d=\displaystyle \frac{1}{n}\\[2ex]na+mnd-nd=1\ \ \ \ \ \ -----(1)\\[2ex]{{u}_{n}}=\displaystyle \frac{1}{m}\\[2ex]a+(n-1)d=\displaystyle \frac{1}{m}\\[2ex]ma+mnd-md=1\ \ \ \ -----(2)\\[2ex](1)-(2)\Rightarrow a(n-m)-(n-m)d=0\\[2ex]\therefore (n-m)(a-d)=0\\[2ex]\text{Since}\ m\ne n,n-m\ne 0.\\[2ex]\therefore a-d=0\Rightarrow a=d\\[2ex]\text{By equation (2), }\\[2ex]am+mnd-md=1\ \Rightarrow mnd=1\Rightarrow mna=1\\[2ex]\therefore {{u}_{{mn}}}=a+(mn-1)d\\[2ex]\ \ \ \ \ \ \ \ =d+mnd-d\\[2ex]\ \ \ \ \ \ \ \ =1\end{array}$

9 (a). The sum of the first two terms of a geometric progression is $12$ and the sum of the first four terms is $120$. Calculate the two possible values of the fourth term in the progression.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Let the first and the common ratio }\\[2ex]\text{of the give G}\text{.P}\text{. be }a\ \text{and }r\ \text{respectively}\text{.}\\[2ex]\text{By the problem,}\\[2ex]{{u}_{1}}+{{u}_{2}}=12\\[2ex]a+ar=12\\[2ex]a\left( {1+r} \right)=12\ \ \ \ \ \ \ -----(1)\\[2ex]{{u}_{1}}+{{u}_{2}}+{{u}_{3}}+{{u}_{4}}=120\\[2ex]12+{{u}_{3}}+{{u}_{4}}=120\\[2ex]{{u}_{3}}+{{u}_{4}}=108\\[2ex]a{{r}^{2}}+a{{r}^{3}}=108\\[2ex]a{{r}^{2}}\left( {1+r} \right)=108-----(2)\\[2ex]\therefore \displaystyle \frac{{a{{r}^{2}}\left( {1+r} \right)}}{{a\left( {1+r} \right)}}=\displaystyle \frac{{108}}{{12}}\\[2ex]\ \ \ {{r}^{2}}=9\Rightarrow r=\pm 3\\[2ex]\text{When}\ r=-3,\ a\left( {1-3} \right)=12\ \Rightarrow -6\\[2ex]\therefore {{u}_{4}}=a{{r}^{3}}=-6{{(-3)}^{3}}=162\\[2ex]\text{When}\ r=3,\ a\left( {1+3} \right)=12\ \Rightarrow 3\\[2ex]\therefore {{u}_{4}}=a{{r}^{3}}=3{{(3)}^{3}}=81\end{array}$

9 (b). Given that $ A=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)$. If $A + A' = I$ where $I$ is a unit matrix of order $2$, find the value of $\theta$ for $ \displaystyle 0{}^\circ <\theta <90{}^\circ $.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}A=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\[2ex] {\sin \theta } & {\cos \theta } \end{array}} \right)\\[2ex]{A}'=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\[2ex] {-\sin \theta } & {\cos \theta } \end{array}} \right)\\[2ex]\text{By the problem,}\\[2ex]A+{A}'=I\\[2ex]\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\[2ex] {\sin \theta } & {\cos \theta } \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\[2ex] {-\sin \theta } & {\cos \theta } \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 & 0 \\[2ex] 0 & 1 \end{array}} \right)\\[2ex]\left( {\begin{array}{*{20}{c}} {2\cos \theta } & 0 \\[2ex] 0 & {2\cos \theta } \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 & 0 \\[2ex] 0 & 1 \end{array}} \right)\\[2ex]\therefore 2\cos \theta =1\\[2ex]\ \ \ \cos \theta =\displaystyle \frac{1}{2}\\[2ex]\ \ \ \theta =60{}^\circ \end{array}$

10 (a). The matrix $A$ is given by $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 3 \\ 4 & 5 \end{array}} \right)$.
(a) Prove that $A^2 = 7A + 2I$ where $I$ is the unit matrix of order $2$.
(b) Hence, show that $ \displaystyle {{A}^{{-1}}}=\frac{1}{2}~\left( {A-7I} \right)$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}A=\left( {\begin{array}{*{20}{c}} 2 & 3 \\[2ex] 4 & 5 \end{array}} \right)\\[2ex]{{A}^{2}}=\left( {\begin{array}{*{20}{c}} 2 & 3 \\[2ex] 4 & 5 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2 & 3 \\[2ex] 4 & 5 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {2\times 2+3\times \;4} & {2\times \;3+3\times \;5} \\[2ex] {4\times \;2+5\times \;4} & {4\times \;3+5\times \;5} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {16} & {21} \\[2ex] {28} & {37} \end{array}} \right)\\[2ex]7A+2I=7\left( {\begin{array}{*{20}{c}} 2 & 3 \\[2ex] 4 & 5 \end{array}} \right)+2\left( {\begin{array}{*{20}{c}} 1 & 0 \\[2ex] 0 & 1 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {14} & {21} \\[2ex] {28} & {35} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} 2 & 0 \\[2ex] 0 & 2 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {16} & {21} \\[2ex] {28} & {37} \end{array}} \right)\\[2ex]\therefore {{A}^{2}}=7A+2I\\[2ex]\therefore A\cdot A\cdot {{A}^{{-1}}}=7A\cdot {{A}^{{-1}}}+2I\cdot {{A}^{{-1}}}\\[2ex]\therefore A\cdot I=7I+2{{A}^{{-1}}}\\[2ex]\therefore A=7I+2{{A}^{{-1}}}\\[2ex]\therefore 2{{A}^{{-1}}}=A-7I\\[2ex]\therefore {{A}^{{-1}}}=\displaystyle \frac{1}{2}\left( {A-7I} \right)\end{array}$

10 (b). Draw a tree diagram to list all possible outcomes when four fair coins are tossed simultaneously. Hence determine the probability of getting:
(a) all heads,
(b) two heads and two tails,
(c) more tails than heads,
(d) at least one tail,
(e) exactly one head.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\therefore \ \ \text{Number of possible outcomes}\ =16\\[2ex](\text{i})\ \text{The set of favourable outcomes for getting all heads}\ \text{=}\left\{ {(H,H,H,H)} \right\}\\[2ex]\ \ \ \ \text{Number of favourable outcomes = 1}\\[2ex]\ \ \ \ P\text{ (getting all heads) =}\displaystyle \frac{1}{{16}}\\[2ex](\text{ii})\ \text{The set of favourable outcomes for getting two heads and two tails}\\[2ex]\ \ \ \ \ \text{=}\left\{ {(H,H,T,T),\text{ }(H,T,H,T),\text{ }(H,T,T,H),\text{ }(T,H,H,T),\text{ }(T,H,T,H),\text{ }(T,T,H,H)} \right\}\\[2ex]\ \ \ \ \text{Number of favourable outcomes = 6}\\[2ex]\ \ \ \ P\text{ (getting two heads and two tails) =}\displaystyle \frac{6}{{16}}=\displaystyle \frac{3}{8}\\[2ex](\text{iii})\ \text{The set of favourable outcomes for getting more tails than heads}\\[2ex]\ \ \ \ \ \text{=}\left\{ {(H,T,T,T),\text{ }(T,H,T,T),\text{ }(T,T,H,T),\text{ }(T,T,T,H)} \right\}\\[2ex]\ \ \ \ \text{Number of favourable outcomes = 4}\\[2ex]\ \ \ \ P\text{ (getting more tails than heads) =}\displaystyle \frac{4}{{16}}=\displaystyle \frac{1}{4}\\[2ex](\text{iv})\ P\text{ (getting at least one tail)}=1-P\text{ (no tail)}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-P\text{ (all head)}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-\displaystyle \frac{1}{{16}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{15}}{{16}}\\[2ex](\text{v})\ \text{The set of favourable outcomes for getting exactly one head}\\[2ex]\ \ \ \ \ \text{=}\left\{ {(H,T,T,T),\text{ }(T,H,T,T),\text{ }(T,T,H,T),\text{ }(T,T,T,H)} \right\}\\[2ex]\ \ \ \ \text{Number of favourable outcomes = 4}\\[2ex]\ \ \ \ P\text{ (getting exactly one head) =}\displaystyle \frac{4}{{16}}=\displaystyle \frac{1}{4}\end{array}$

SECTION (C)
(Answer Any THREE questions.)

11 (a). $PQR$ is a triangle inscribed in a circle. The tangent at $P$ meet $RQ$ produced at $T$,and $PC$ bisecting $\angle RPQ$ meets side $RQ$ at $C$. Prove $\triangle TPC$ is isosceles.

(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}\angle TPC=\beta +\gamma \\[2ex]\angle R=\gamma \ \ \ [\angle \ \text{between tangent and chord}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment }\!\!]\!\!\text{ }\\[2ex]\text{Since }PC\ \text{bisects}\ \angle RPQ,\ \\[2ex]\beta =\alpha \\[2ex]\therefore \angle TPC=\alpha +\angle R\\[2ex]\text{In}\ \triangle RPC,\ \angle PCT=\alpha +\angle R\\[2ex]\angle TPC=\angle PCT\\[2ex]\therefore \,\triangle TPC\ \text{is isosceles}\text{.}\end{array}$

11 (b). In $\triangle ABC$, $D$ is a point of $AC$ such that $AD = 2CD$. $E$ is on $BC$ such that $DE \parallel AB$. Compare the areas of $\triangle CDE$ and $\triangle ABC$. If $\alpha (ABED) = 40$, what is $\alpha(ΔABC)$?

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ AD=\text{ }2CD\ \text{ }\!\![\!\!\text{ given }\!\!]\!\!\text{ }\\[2ex]\ \ \ DE\parallel AB\\[2ex]\therefore \ \triangle CAB\sim\triangle CDE\\[2ex]\therefore \displaystyle \frac{{\alpha (\triangle CAB)}}{{\alpha (\triangle CDE)}}=\displaystyle \frac{{A{{C}^{2}}}}{{C{{D}^{2}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{{(AD+CD)}}^{2}}}}{{C{{D}^{2}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{{(2CD+CD)}}^{2}}}}{{C{{D}^{2}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{9C{{D}^{2}}}}{{C{{D}^{2}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9\\[2ex]\therefore \displaystyle \frac{{\alpha (\triangle CAB)}}{{\alpha (\triangle CAB)-\alpha (\triangle CDE)}}=\displaystyle \frac{9}{{9-1}}\\[2ex]\therefore \displaystyle \frac{{\alpha (\triangle CAB)}}{{\alpha (ABED)}}=\displaystyle \frac{9}{8}\\[2ex]\therefore \alpha (\triangle CAB)=\displaystyle \frac{9}{8}\alpha (ABED)\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{9}{8}\times 40\\[2ex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =45\ \text{sq-unit}\end{array}$

12 (a). If $L, M, N,$ are the middle points of the sides of the $\triangle ABC$, and $P$ is the foot of perpendicular from $A$ to $BC$. Prove that $L, N, P, M$ are concyclic.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Since }AP\bot BC\text{ and }M\text{ is the midpoint of }AC\text{, }\\[2ex]\text{a circle with centre }M\text{ and diameter }AC\text{ }\\[2ex]\text{will pass through }P\text{.}\\[2ex]\therefore MP\text{ = }MC\text{ }\!\![\!\!\text{ radii of }\odot M\text{ }\!\!]\!\!\text{ }\\[2ex]\therefore \gamma \ \text{=}\ \phi \text{.}\\[2ex]\text{Since }L\text{ and }N\text{ are the midpoints of }AB\text{ and }BC\text{, }\\[2ex]LN\parallel AC\ \text{and }LN=\displaystyle \frac{1}{2}AC.\\[2ex]\text{Similarly }LM\parallel BC\ \text{and }LM=\displaystyle \frac{1}{2}BC.\\[2ex]LMCN\ \text{is a parallelogram}\text{.}\\[2ex]\therefore \gamma \ \text{=}\theta \Rightarrow \phi \ \text{=}\theta \\[2ex]\text{Since}\ \phi +\angle MPN=\text{ }180{}^\circ ,\\[2ex]\ \theta +\angle MPN=\text{ }180{}^\circ ,\\[2ex]L,\ N,\ P,\ M\ \text{are concyclic}\text{.}\end{array}$

12 (b). Solve the equation $\displaystyle \sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ for $ \displaystyle 0{}^\circ \le \theta \le 360{}^\circ $.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\sqrt{3}\cos \theta +\sin \theta =\sqrt{2},\ 0{}^\circ \le \theta \le 360{}^\circ \\[2ex]\text{Let}\ R\cos \alpha =\sqrt{3}\ \text{and}\ R\sin \alpha =1\ \\[2ex]\text{where }R>0\ \text{and}\ \alpha <90{}^\circ .\\[2ex]\therefore {{R}^{2}}{{\cos }^{2}}\alpha +{{R}^{2}}{{\sin }^{2}}\alpha =3+1\\[2ex]\therefore \ {{R}^{2}}\left( {{{{\cos }}^{2}}\alpha +{{{\sin }}^{2}}\alpha } \right)=4\\[2ex]\therefore \ {{R}^{2}}=4\Rightarrow R=2\text{ }\\[2ex]\ \ \ \displaystyle \frac{{R\sin \alpha }}{{R\cos \alpha }}=\displaystyle \frac{1}{{\sqrt{3}}}\\[2ex]\therefore \ \tan \alpha =\ \displaystyle \frac{1}{{\sqrt{3}}}\Rightarrow \alpha =30{}^\circ \\[2ex]\text{Now}\ \ \ \sqrt{3}\cos \theta +\sin \theta \\[2ex]\ \ \ \ \ \ \ \ =R\cos \theta \cos \alpha +R\sin \theta \sin \alpha \\[2ex]\ \ \ \ \ \ \ \ =R\left( {\cos \theta \cos \alpha +\sin \theta \sin \alpha } \right)\\[2ex]\ \ \ \ \ \ \ \ =R\cos \left( {\theta -\alpha } \right)\\[2ex]\ \ \ \ \ \ \ \ =2\cos \left( {\theta -30{}^\circ } \right)\\[2ex]\therefore 2\cos \left( {\theta -30{}^\circ } \right)=\sqrt{2}\\[2ex]\therefore \cos \left( {\theta -30{}^\circ } \right)=\displaystyle \frac{{\sqrt{2}}}{2}\\[2ex]\therefore \theta -30{}^\circ =45{}^\circ \ \text{or}\ \theta -30{}^\circ =315{}^\circ \\[2ex]\therefore \theta =75{}^\circ \ \text{or}\ \theta =345{}^\circ \end{array}$

13 (a). In $\triangle ABC, AB = x, BC = x + 2$, $AC = x - 2$ where $x > 4$, prove that $ \displaystyle \cos A=\frac{{x-8}}{{2(x-2)}}$. Find the integral values of $x$ for which $A$ is obtuse.

(5 marks)

Show/Hide Solution

$ \displaystyle \vartriangle ABC,AB=x,BC=x+2,AC=x-2,\ x>4$

$ \displaystyle \cos A=\displaystyle \frac{{A{{B}^{2}}+A{{C}^{2}}-B{{C}^{2}}}}{{2\cdot AB\cdot AC}}$

$ \displaystyle \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}+{{{(x-2)}}^{2}}-{{{(x+2)}}^{2}}}}{{2\cdot x\cdot (x-2)}}$

$ \displaystyle \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}+{{x}^{2}}-4x+4-{{x}^{2}}-4x-4}}{{2\cdot x\cdot (x-2)}}$

$ \displaystyle \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{2}}-8x}}{{2\cdot x\cdot (x-2)}}$

$ \displaystyle \ \ \ \ \ \ \ \ =\displaystyle \frac{{x(x-8)}}{{2x(x-2)}}$

$ \displaystyle \ \ \ \ \ \ \ \ =\displaystyle \frac{{x-8}}{{2(x-2)}}$

$ \displaystyle \text{Since} A\ \text{is obtuse}.$

$ \displaystyle \cos A<0$

$ \displaystyle \displaystyle \frac{{x-8}}{{2(x-2)}}<0$

$ \displaystyle \text{Since}\ x>4,\ x-2>2.$

$ \displaystyle \therefore x-8<0\Rightarrow x<8$

$ \displaystyle \therefore 4
$ \displaystyle \therefore \ \text{The integral value of }x\text{ are }5,\ 6\ \text{and }7.$

13 (b). The sum of the perimeters of a circle and square is $k$, where $k$ is some constant. Using calculus, prove that the sum of their areas is least, when the side of the square is double the radius of the circle.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\begin{array}{*{20}{l}} {\text{ Let the side-length of a square be }x} \\[2ex] {\text{ and the radius of the circle be }r} \\[2ex] {\text{ Sum of perimeters }=k(\text{ given })} \\[2ex] {4x+2\pi r=k} \\[2ex] {\therefore r=\displaystyle \frac{{k-4x}}{{2\pi }}} \\[2ex] {\text{ Let the sum of the areas be }A.} \\[2ex] {\therefore A={{x}^{2}}+\pi {{r}^{2}}} \end{array}\\[2ex]\begin{array}{*{20}{l}} {\therefore A={{x}^{2}}+\pi {{{\left( {\displaystyle \frac{{k-4x}}{{2\pi }}} \right)}}^{2}}} \\[2ex] {\therefore A={{x}^{2}}+\displaystyle \frac{{{{{(k-4x)}}^{2}}}}{{4\pi }}} \\[2ex] {\displaystyle \frac{{dA}}{{dx}}=2x+\displaystyle \frac{{2(-4)(k-4x)}}{{4\pi }}} \\[2ex] {\quad =2\left( {x+\displaystyle \frac{{4x-k}}{\pi }} \right)} \\[2ex] {\quad =\displaystyle \frac{2}{\pi }[(\pi +4)x-k]} \end{array}\\[2ex]\begin{array}{*{20}{l}} {\displaystyle \frac{{dA}}{{dx}}=0\text{ when }\displaystyle \frac{2}{\pi }[(\pi +4)x-k]=0} \\[2ex] {\therefore (\pi +4)x-k=0\Rightarrow x=\displaystyle \frac{k}{{\pi +4}}} \\[2ex] {\displaystyle \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\displaystyle \frac{{2(\pi +4)}}{\pi }>0} \\[2ex] {\therefore A\text{ is minimum when }x=\displaystyle \frac{k}{{\pi +4}}} \end{array}\\[2ex]\therefore r=\displaystyle \frac{1}{{2\pi }}\left[ {k-\displaystyle \frac{{4k}}{{\pi +4}}} \right]\\[2ex]\ \ \ \ \ =\displaystyle \frac{1}{{2\pi }}\left[ {\displaystyle \frac{{\pi k+4k-4k}}{{\pi +4}}} \right]\\[2ex]\ \ \ \ \ =\displaystyle \frac{1}{2}\left( {\displaystyle \frac{k}{{\pi +4}}} \right)\\[2ex]\ \ \ \ \ =\displaystyle \frac{x}{2}\\[2ex]\therefore x=2r\\[2ex]\text{Hence the sum of their areas is least,}\\[2ex]\text{when the side of the square is double }\\[2ex]\text{the}\ \text{radius of the circle}\text{.}\end{array}$

14 (a). The vector $ \overrightarrow{{OA}}$ has magnitude $39$ units and has the same direction as $ \displaystyle 5\hat{i}+12\hat{j}$. The vector $ \overrightarrow{{OB}}$ has magnitude $25$ units and has the same direction as $ \displaystyle -3\hat{i}+4\hat{j}$. Express $ \overrightarrow{{OA}}$ and $ \overrightarrow{{OB}}$ in terms of $ \hat{i}$ and $\hat{j}$ and find the magnitude of $ \overrightarrow{{AB}}.$

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{*{20}{l}} {\text{ Let }\vec{p}=5\hat{\imath }+12\hat{\jmath }\text{ and }\vec{q}=-3\hat{\imath }+4\hat{\jmath }} \\[2ex] \begin{array}{l}\therefore \ |\vec{p}|=\sqrt{{{{5}^{2}}+{{{12}}^{2}}}}=\sqrt{{169}}=13\text{ and }\\[2ex]\ \ |\vec{q}|=\sqrt{{{{{(-3)}}^{2}}+{{4}^{2}}}}=\sqrt{{25}}=5\end{array} \\[2ex] \begin{array}{l}\therefore \hat{p}=\displaystyle \frac{{\vec{p}}}{{|\vec{p}|}}=\displaystyle \frac{1}{{13}}(5\hat{\imath }+12\hat{\jmath })\text{ and }\\[2ex]\ \ \ \hat{q}=\displaystyle \frac{{\vec{q}}}{{|\vec{q}|}}=\displaystyle \frac{1}{5}(-3\hat{\imath }+4\hat{\jmath })\end{array} \\[2ex] {\ \ \ |\overrightarrow{{OA}}|=39\text{ and }\overrightarrow{{OA}}\text{ has the same direction }\hat{p}.} \\[2ex] \begin{array}{l}\therefore \overrightarrow{{OA}}=39\hat{p}\\[2ex]\ \ \ \ \ \ \ \ =39\times \displaystyle \frac{1}{{13}}(5\hat{\imath }+12\hat{\jmath })=15\hat{\imath }+36\hat{\jmath }\\[2ex]\ \ \ \ \ \ \ \ =15\hat{\imath }+36\hat{\jmath }\\[2ex]\begin{array}{*{20}{l}} \begin{array}{l}\text{ Similarly, }\\[2ex]\ \ |\overrightarrow{{OB}}|\ =25\text{ and }\overrightarrow{{OB}}\text{ has the same direction }\hat{q}\text{ }\text{. }\end{array} \\[2ex] \begin{array}{l}\therefore \overrightarrow{{OB}}=25\hat{q}\\[2ex]\ \ \ \ \ \ \ \ =25\times \displaystyle \frac{1}{5}(-3\hat{\imath }+4\hat{\jmath })=-15\hat{\imath }+20\hat{\jmath }\\[2ex]\ \ \ \ \ \ \ \ =-15\hat{\imath }+20\hat{\jmath }\end{array} \\[2ex] \begin{array}{l}\therefore \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\[2ex]\ \ \ \ \ \ \ \ =(-15\hat{\imath }+20\hat{\jmath })-(15\hat{\imath }+36\hat{\jmath })\\[2ex]\ \ \ \ \ \ \ \ =-30\hat{\imath }-16\hat{\jmath }\end{array} \\[2ex] \begin{array}{l}\therefore \ |\overrightarrow{{AB}}|\ =\sqrt{{{{{(-30)}}^{2}}+{{{(-16)}}^{2}}}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ =\sqrt{{1156}}\\[2ex]\ \ \ \ \ \ \ \ \ \ \ =34\end{array} \end{array}\end{array} \end{array}$

14 (b). Find the coordinates of the stationary points of the curve $y = x\ln x - 2x$. Determine whether it is a maximum or a minimum point.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{*{20}{l}} {\text{Curve : }y=x\ln x-2x} \\[2ex] \begin{array}{l}\displaystyle \frac{{dy}}{{dx}}=x\left( {\displaystyle \frac{1}{x}} \right)+\ln x-2\\[2ex]\,\ \ \ \ =\ln x-1\end{array} \\[2ex] {\displaystyle \frac{{dy}}{{dx}}=0\text{ when }\ln x-1=0} \\[2ex] \begin{array}{l}\text{ln }x=1\\[2ex]x=e\left( {\ln x={{{\log }}_{e}}x} \right)\end{array} \\[2ex] \begin{array}{l}\text{When }x=e,\\[2ex]y=e\ln e-2e\\[2ex]\ \ \ =-e\end{array} \\[2ex] {\therefore \text{ The stationary point is }(e,-e)} \\[2ex] \begin{array}{l}\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\displaystyle \frac{1}{x}\\[2ex]{{\left. {\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=e}}}=\displaystyle \frac{1}{e}>0\end{array} \\[2ex] {\therefore (e,-e)\text{ is a minimum point}\text{. }} \end{array}$

နားလည်လွယ်ကူစေရန် illustration ထည့်ပေးခြင်း ဖြစ်သည်။ ဖြေဆိုသည့်အခါ ပုံထည့်ဆွဲပေးရန်မလိုပါ။

2020

MATRICULATION EXAMINATION

Sample Question Set (3)

MATHEMATICS Time allowed: 3hours
WRITE YOUR ANSWERS IN THE ANSWER BOOKLET.
SECTION (A)
(Answer ALL questions.)

1 (a). Let $ \displaystyle f:R\backslash \{\pm 2\}\to R$ be a function defined by $ \displaystyle f(x)=\frac{{3x}}{{{{x}^{2}}-4}}$.Find the positive value of $z$ such that $f(z) = 1$.

(3 marks)

(b). If the polynomial $x^3 - 3x^2 + ax - b$ is divided by $(x - 2 )$ and $(x + 2)$, the remainders are $21$ and $1$ respectively. Find the values of $a$ and $b$.

(3 marks)

2(a). Find the middle term in the expansion of $(x^2 - 2y)^{10}$.

(3 marks)

(b). In a sequence if $u_1=1$ and $u_{n+1}=u_n+3(n+1)$, find $u_5$ .

(3 marks)

(b). A bag contains tickets, numbered $11, 12, 13, ...., 30$. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket is

(i) a multiple of $7$

(ii) greater than $15$ and a multiple of $5$.

(3 marks)

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

(3 marks)

5(a). Solve the equation $2 \sin x \cos x -\cos x + 2\sin x - 1 = 0$ for $ \displaystyle 0{}^\circ \le x\le \text{ }360{}^\circ $.

(3 marks)

(b). Differentiate $ \displaystyle y=\frac{1}{{\sqrt[3]{x}}}$ from the first principles.

(3 marks)

SECTION (B)
(Answer Any FOUR questions.)

(5 marks)

(b). Given that $x^5 + ax^3 + bx^2 - 3 = (x^2 - 1) Q(x) - x - 2$, where $Q(x)$ is a polynomial. State the degree of $Q (x)$ and find the values of $a$ and $b$. Find also the remainder when $Q (x)$ is divided by $x + 2$.

(5 marks)

7 (a). The binary operation $\odot$ on $R$ be defined by $x\odot y=x+y+10xy$ Show that the binary operation is commutative. Find the values $b$ such that $ (1\odot b)\odot b=485$.

(5 marks)

(b). If, in the expansion of $(1 + x)^m (1 – x)^n$, the coefficient of $x$ and $x^2$ are $-5$ and $7$ respectively, then find the value of $m$ and $n$.

(5 marks)

8 (a). Find the solution set in $R$ for the inequation $2x (x + 2)\ge (x + 1) (x + 3)$ and illustrate it on the number line.

(5 marks)

(b). If the $ {{m}^{{\text{th}}}}$ term of an A.P. is $ \displaystyle \frac{1}{n}$ and $ {{n}^{{\text{th}}}}$ term is $ \displaystyle \frac{1}{m}$ where $m\ne n$, then show that $u_{mn} = 1$.

(5 marks)

9 (a). The sum of the first two terms of a geometric progression is $12$ and the sum of the first four terms is $120$. Calculate the two possible values of the fourth term in the progression.

(5 marks)

(b). Given that $ A=\left( {\begin{array}{*{20}{c}} {\cos \theta } & {-\sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array}} \right)$. If $A + A' = I$ where $I$ is a unit matrix of order $2$, find the value of $\theta$ for $ \displaystyle 0{}^\circ <\theta <90{}^\circ $.

(5 marks)

10 (a). The matrix $A$ is given by $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 3 \\ 4 & 5 \end{array}} \right)$.
(i) Prove that $A^2 = 7A + 2I$ where $I$ is the unit matrix of order $2$.
(ii) Hence, show that $ \displaystyle {{A}^{{-1}}}=\frac{1}{2}~\left( {A-7I} \right)$.

(5 marks)

(b). Draw a tree diagram to list all possible outcomes when four fair coins are tossed simultaneously. Hence determine the probability of getting:
(i) all heads,
(ii) two heads and two tails,
(iii) more tails than heads,
(iv) at least one tail,
(v) exactly one head.

(5 marks)

SECTION (C)
(Answer Any THREE questions.)

11 (a). $PQR$ is a triangle inscribed in a circle. The tangent at $P$ meet $RQ$ produced at $T$,and $PC$ bisecting $\angle RPQ$ meets side $RQ$ at $C$. Prove $\triangle TPC$ is isosceles.

(5 marks)

(b). In $\triangle ABC$, $D$ is a point of $AC$ such that $AD = 2CD$. $E$ is on $BC$ such that $DE \parallel AB$. Compare the areas of $\triangle CDE$ and $\triangle ABC$. If $\alpha (ABED) = 40$, what is $\alpha(ΔABC)$?

(5 marks)

12 (a). If $L, M, N,$ are the middle points of the sides of the $\triangle ABC$, and $P$ is the foot of perpendicular from $A$ to $BC$. Prove that $L, N, P, M$ are concyclic.

(5 marks)

(b). Solve the equation $\displaystyle \sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ for $ \displaystyle 0{}^\circ \le \theta \le 360{}^\circ $.

(5 marks)

13 (a). In $\triangle ABC, AB = x, BC = x + 2$, $AC = x – 2$ where $x > 4$, prove that $ \displaystyle \cos A=\frac{{x-8}}{{2(x-2)}}$. Find the integral values of $x$ for which $A$ is obtuse.

(5 marks)

(b). The sum of the perimeters of a circle and square is $k$, where $k$ is some constant. Using calculus, prove that the sum of their areas is least, when the side of the square is double the radius of the circle.

(5 marks)

(b). Find the coordinates of the stationary points of the curve $y = x\ln x - 2x$. Determine whether it is a maximum or a minimum point.

(5 marks)

Target Mathematics ၏ အစဉ်အလာအတိုင်း တက္ကသိုလ်ဝင်တန်း စာမေးပွဲကို ဝင်ရောက်ဖြေဆိုကြမည့် ကျောင်းသား/သူတို့အတွက် လေ့ကျင့်ရန် မေးခွန်းတစ်စုံ တင်ပြလိုက်ပါသည်။ လေ့ကျင့်ဖြေဆိုကြစေလိုပါသည်။ အဖြေများကို နောက်ရက်တွင် ဆက်လက် ဖော်ပြပေးပါမည်။

1. Find the unit vector in the direction of $ \displaystyle \overrightarrow{{PQ}}$ where $ \displaystyle P$ and $ \displaystyle Q$ are points $ \displaystyle (2, 3)$ and $ \displaystyle (7, – 9)$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ P\operatorname{and}\ Q\ \text{are points}\ (2,3)\ \operatorname{and}\ (7,-9).\\\\\therefore \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right)\ \operatorname{and}\ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} 7 \\ {-9} \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\ \overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-9} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ {-12} \end{array}} \right)\\\\\therefore \ \ \left| {\ \overrightarrow{{PQ}}} \right|=\sqrt{{{{5}^{2}}+{{{\left( {-12} \right)}}^{2}}}}=13\\\\\therefore \ \ \text{The unit vector in }\\\ \ \ \ \text{the direction of}\ \ \ \ =\displaystyle \frac{{\overrightarrow{{PQ}}}}{{\left| {\ \overrightarrow{{PQ}}} \right|}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{13}}\left( {\begin{array}{*{20}{c}} 5 \\ {-12} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {\displaystyle \frac{5}{{13}}} \\ {-\displaystyle \frac{{12}}{{13}}} \end{array}} \right)\end{array}$

2. Given that $ \displaystyle \overrightarrow{{OP}}=\widehat{\text{i}}+2\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OQ}}=7\widehat{\text{i}}-4\widehat{\text{j}}$. Find the position vector of a point $ \displaystyle R$ which lies on the line $ \displaystyle PQ$ such that $ \displaystyle PR : RQ = 2 : 1$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\widehat{\text{i}}+2\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OQ}}=7\widehat{\text{i}}-4\widehat{\text{j}}\\\\\ \ \ \ \ PR:RQ=2:1\\\\\ \ \ \ \ \text{By section formula},\ \\\\\ \ \ \ \ \ \overrightarrow{{OR}}=\displaystyle \frac{{\left( {1\times \overrightarrow{{OP}}} \right)+\left( {2\times \overrightarrow{{OQ}}} \right)}}{{1+2}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{3}\ \left[ {\widehat{\text{i}}+2\widehat{\text{j}}+2\left( {7\widehat{\text{i}}-4\widehat{\text{j}}} \right)} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ =5\widehat{\text{i}}-2\widehat{\text{j}}\end{array}$

3. If $ \displaystyle \overrightarrow{{OA}}=-5\widehat{\text{i}}+6\widehat{\text{j}}$, $ \displaystyle \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OC}}=9\widehat{\text{i}}+4\widehat{\text{j}}$, show that $ \displaystyle A, B$ and $ \displaystyle C$ are collinear.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=-5\widehat{\text{i}}+6\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=9\widehat{\text{i}}+4\widehat{\text{j}}\\\ \ \ \ \ \\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {2\widehat{\text{i}}+5\widehat{\text{j}}} \right)-\left( {-5\widehat{\text{i}}+6\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =7\widehat{\text{i}}-\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {9\widehat{\text{i}}+4\widehat{\text{j}}} \right)-\left( {2\widehat{\text{i}}+5\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =7\widehat{\text{i}}-\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{BC}}\\\\\therefore \ \ \ A,B\ \operatorname{and}\ C\ \text{are collinear}.\end{array}$

4. Given that $ \displaystyle \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right)$, $ \displaystyle \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)$ and $ \displaystyle \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right)$. If $ \displaystyle P, Q$ and $ \displaystyle R$ are collinear, find the value of $ \displaystyle k$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{QR}}=\overrightarrow{{OR}}-\overrightarrow{{OQ}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\\\\\ \ \ \ \text{By the problem,}\ \\\\\ \ \ P,\ Q\ \operatorname{and}\ R\ \text{are collinear}.\\\\\therefore \ \ \text{Let}\ \overrightarrow{{PQ}}=h\overrightarrow{{QR}}\\\\\therefore \ \ \left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)=h\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\\\\\ \ \ \ \left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {5h} \\ {3h} \end{array}} \right)\\\\\therefore \ \ 3h=3\\\\\ \ \ \ h=1\\\\\ \ \ \ -2-k=5h\\\\\ \ \ k=-2-5h\\\\\ \ \ k=-7\\\ \ \ \end{array}$

5. Using a vector method, show that the points $ \displaystyle A (– 8, 10), B (– 1, 9)$ and $ \displaystyle C (6, 8)$ are collinear and hence find the ratio $ \displaystyle AB : BC$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=\left( {\begin{array}{*{20}{c}} {-8} \\ {10} \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OB}}=\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OC}}=\left( {\begin{array}{*{20}{c}} 6 \\ 8 \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-8} \\ {10} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-1} \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 6 \\ 8 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-1} \end{array}} \right)\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{BC}}\\\\\ \ \ A,\ B\ \operatorname{and}\ C\ \text{are collinear and }\\\\\ \ \ AB:BC=1:1\ \end{array}$

6. If $ \displaystyle \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)$, $ \displaystyle \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)$ and $ \displaystyle \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)$, show that $ \displaystyle \Delta PQR$ is a right triangle.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ 6 \end{array}} \right)\\\\\therefore \ PQ=\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{6}^{2}}}}=\sqrt{{40}}\\\\\ \ \ \ \overrightarrow{{QR}}=\overrightarrow{{OR}}-\overrightarrow{{OQ}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {14} \\ {-2} \end{array}} \right)\\\\\therefore \ QR=\sqrt{{{{{14}}^{2}}+{{{\left( {-2} \right)}}^{2}}}}=\sqrt{{200}}\\\\\ \overrightarrow{{PR}}=\overrightarrow{{OR}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {12} \\ 4 \end{array}} \right)\\\\\therefore \ \ PR=\sqrt{{{{{12}}^{2}}+{{4}^{2}}}}=\sqrt{{160}}\\\\\ \ \ \ P{{Q}^{2}}+P{{R}^{2}}=40+160=200\\\\\ \ \ \ Q{{R}^{2}}=200\\\\\therefore \ \ P{{Q}^{2}}+P{{R}^{2}}=\ Q{{R}^{2}}\\\\\therefore \ \ \Delta PQR\ \text{is a right triangle}\text{.}\end{array}$

7. If the position vectors of the points $ \displaystyle A, B$ and $ \displaystyle C$ are $ \displaystyle 9\widehat{\text{i}}+6\widehat{\text{j}}$, $ \displaystyle 4\widehat{\text{i}}+3\widehat{\text{j}}$ and $\displaystyle -5\widehat{\text{i}}+8\widehat{\text{j}}$ respectively, show that the $ \displaystyle \Delta ABC$ is an obtuse triangle.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=9\widehat{\text{i}}+6\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=4\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=-5\widehat{\text{i}}+8\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-5\widehat{\text{i}}-3\widehat{\text{j}}\\\\\therefore \ \ \ A{{B}^{2}}={{\left( {-5} \right)}^{2}}+{{\left( {-3} \right)}^{2}}=34\\\\\therefore \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =-9\widehat{\text{i}}+5\widehat{\text{j}}\\\\\therefore \ \ \ B{{C}^{2}}={{\left( {-9} \right)}^{2}}+{{5}^{2}}=106\\\\\therefore \ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-14\widehat{\text{i}}+2\widehat{\text{j}}\\\\\therefore \ \ \ A{{C}^{2}}={{\left( {-14} \right)}^{2}}+{{2}^{2}}=200\\\\\therefore \ \ \ A{{B}^{2}}+B{{C}^{2}}=140\\\\\therefore \ \ \ A{{C}^{2}}>A{{B}^{2}}+B{{C}^{2}}\\\\\therefore \ \ \ \Delta ABC\ \text{is an obtuse triangle}\text{.}\end{array}$

8. The position vectors of the points A, B, and C are $ \displaystyle \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)$, $ \displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)$ and $ \displaystyle \left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)$ respectively. Prove that $ \displaystyle \Delta ABC$ is an isosceles right triangle.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \overrightarrow{{OA}}=\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right),\\\\\ \ \ \overrightarrow{{OB}}=\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right),\\\\\ \ \ \overrightarrow{{OC}}=\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)\\\\\ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ {-6} \end{array}} \right)\\\\\therefore \ \ AB=\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{{\left( {-6} \right)}}^{2}}}}=\sqrt{{40}}\\\\\ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-4} \\ 8 \end{array}} \right)\\\\\therefore \ \ BC=\sqrt{{{{{\left( {-4} \right)}}^{2}}+{{8}^{2}}}}=\sqrt{{80}}\\\\\ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-6} \\ 2 \end{array}} \right)\\\\\therefore \ \ AC=\sqrt{{{{{\left( {-6} \right)}}^{2}}+{{2}^{2}}}}=\sqrt{{40}}\\\\\therefore \ \ AB=AC\\\\\ \ \ \ A{{B}^{2}}+A{{C}^{2}}=40+40=80=B{{C}^{2}}\\\\\therefore \ \ \ \Delta ABC\ \text{is an isosceles right triangle}\text{.}\end{array}$

9. $ \displaystyle OABC$ is a parallelogram such that $ \displaystyle \overrightarrow{{OA}}=5\widehat{\text{i}}+3\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OC}}=-2\widehat{\text{i}}+\widehat{\text{j}}$. Find the unit vector in the direction of $ \displaystyle \ \overrightarrow{{OB}}$ and $ \displaystyle \ \overrightarrow{{AC}}$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=5\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=-2\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ OABC\ \text{is a parallelogram}.\\\\\therefore \ \ \ \overrightarrow{{OB}}=\overrightarrow{{OA}}+\overrightarrow{{OC}}\ \ \ \left( {\because \text{parallelogram rule}\text{.}} \right)\\\\\ \ \ \ \ \overrightarrow{{OB}}=5\widehat{\text{i}}+3\widehat{\text{j}}-2\widehat{\text{i}}+\widehat{\text{j}}=3\widehat{\text{i}}+4\widehat{\text{j}}\\\\\therefore \ \ \ OB=\sqrt{{{{3}^{2}}+{{4}^{2}}}}=5\\\\\therefore \ \ \ \text{the unit vector in }\\\ \ \ \ \ \text{the direction of}\ \overrightarrow{{OB}}=\displaystyle \frac{{\overrightarrow{{OB}}}}{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{5}\left( {3\widehat{\text{i}}+4\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \text{Again}\ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\ \ \\\ \ \ \ \ \overrightarrow{{AC}}=\left( {-2\widehat{\text{i}}+\widehat{\text{j}}} \right)-\left( {5\widehat{\text{i}}+3\widehat{\text{j}}} \right)=-7\widehat{\text{i}}-2\widehat{\text{j}}\\\\\therefore \ \ \ AC=\sqrt{{{{{\left( {-7} \right)}}^{2}}+{{{\left( {-2} \right)}}^{2}}}}=\sqrt{{53}}\\\\\therefore \ \ \ \text{the unit vector in }\\\ \ \ \ \ \text{the direction of}\ \overrightarrow{{AC}}=\displaystyle \frac{{\overrightarrow{{AC}}}}{{AC}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{{53}}}}\left( {-7\widehat{\text{i}}-2\widehat{\text{j}}} \right)\end{array}$

10. Given that the position vectors of the points $ \displaystyle A, B$ and $ \displaystyle C$ relative to origin $ \displaystyle O$ are $ \displaystyle -\widehat{\text{i}}+\widehat{\text{j}}$, $ \displaystyle 5\widehat{\text{i}}+\widehat{\text{j}}$ and $ \displaystyle p\widehat{\text{i}}+q\widehat{\text{j}}$ respectively. If $ \displaystyle \Delta ABC$ is equilateral, find the possible values of $ \displaystyle p$ and $ \displaystyle q$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=-\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=5\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=p\widehat{\text{i}}+q\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =6\widehat{\text{i}}\\\\\therefore \ \ \ AB=6\\\\\ \ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {p-5} \right)\widehat{\text{i}}+\left( {q-1} \right)\widehat{\text{j}}\\\\\ \ \ \ \ BC=\sqrt{{{{{\left( {p-5} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}\\\\\ \ \ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\left( {p+1} \right)\widehat{\text{i}}+\left( {q-1} \right)\widehat{\text{j}}\\\\\ \ \ \ \ AC=\sqrt{{{{{\left( {p+1} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}\\\\\ \ \ \ \ \text{Since }\Delta ABC\ \text{is equilateral,}\\\\\ \ \ \ \ AB=BC=AC=6.\\\\\therefore \ \ \sqrt{{{{{\left( {p-5} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}=6\\\\\ \ \ \ {{\left( {p-5} \right)}^{2}}+{{\left( {q-1} \right)}^{2}}=36\\\\\ \ \ \ {{p}^{2}}+{{q}^{2}}-10p-2q=10\ ---(1)\\\\\ \ \ \ \text{Similarly,}\\\\\ \ \ \ \sqrt{{{{{\left( {p+1} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}=6\\\\\ \ \ \ {{\left( {p+1} \right)}^{2}}+{{\left( {q-1} \right)}^{2}}=36\\\\\ \ \ \ {{p}^{2}}+{{q}^{2}}+2p-2q=34\ ---(2)\\\\\ \ \ \ \text{By (2)}-\text{(1),}\\\text{ }\\\ \ \ \ 12p=24\\\\\therefore \ \ p=2\\\\\ \ \ \ \text{Substituting}\ p=2\text{ in (1),}\\\text{ }\\\therefore \ \ 4+{{q}^{2}}-20-2q=10\\\ \\\ \ \ \ {{q}^{2}}-2q=26\\\\\ \ \ \ {{q}^{2}}-2q+1=27\\\\\ \ \ \ {{(q-1)}^{2}}=27\\\\\ \ \ \ q-1=\pm \sqrt{{27}}\\\\\ \ \ \ q=1\pm 3\sqrt{3}\end{array}$

About Target Mathematics

Readers' Opinions

Popular Posts

Blog Archive

Follow on Twitter

Followers