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 $f(z) = 1$.
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(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.
4(b). If $ \displaystyle 3~\overrightarrow{{OA}}-2\overrightarrow{{OB}}-\overrightarrow{{OC}}~=\vec{0}$, show that the points $A, B$ and $C$ are collinear.
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.
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 $x + 2$.
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$.
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 $n$.
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 $.
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)$.
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.
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 $\alpha(ΔABC)$?
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 $.
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.
$ \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}}.$
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)
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)
(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)
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)
(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.)
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)
(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)
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)
(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.
MATRICULATION EXAMINATION 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$ and $ \displaystyle g$ are defined by $ \displaystyle f(x)=x+1$, and $ \displaystyle g(x)=2x^2-x+3$. Find the values of $ \displaystyle x$ which satisfy the equation $ \displaystyle (f\circ g)(x)=4x+1$.
1.(b) The expression $ \displaystyle 2x^2+5x-3$ leaves a remainder of $ \displaystyle 2p^2-3p$ when divided by $ \displaystyle 2x-p$. Find the values of $ \displaystyle p$.
2.(a) Find and simplify the coefficient of $ \displaystyle x^7$ in the expansion of $ \displaystyle {{\left( {{{x}^{2}}+\frac{2}{x}} \right)}^{8}},x\ne 0$.
3.(a) The matrices $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 5 \end{array}} \right)$ and $ \displaystyle B=\left( {\begin{array}{*{20}{c}} x & y \\ 0 & z \end{array}} \right)$ are such that $ \displaystyle AB=A+B$. Find the values of $ \displaystyle x, y$ and $ \displaystyle z$.
3.(b) A die is thrown. If the probability of getting a number not less than x is $ \displaystyle \frac{2}{3}$, find $ \displaystyle x$.
(3 marks)
Show/Hide Solution
Set of possible outcomes = $\ \displaystyle \left\{ {1,2,3,4,5,6} \right\}$
Number of possible outcomes = $\ \displaystyle 6$
$ \displaystyle P(\text{a number not less than}\,x)=\frac{2}{3}=\frac{4}{6}$
Number of favourable outcomes = $\ \displaystyle 4$
Since $ \displaystyle P(\text{a number not less than}\,3)=\frac{4}{6}$,
$ \displaystyle \therefore \ \ \ x=3$.
4.(a) $ \displaystyle AT$ and $ \displaystyle BT$ are tangents to the circle $ \displaystyle ABC$ at $ \displaystyle B$. Prove that $ \displaystyle \angle BTX=2\angle ACB$.
4.(b) The coordinates of $ \displaystyle A, B$ and $ \displaystyle C$ are $ \displaystyle (1,0), (4,2)$ and $ \displaystyle (5,4)$ respectively. Use vector method to determine the coordinates of $ \displaystyle D$ if $ \displaystyle ABCD$ is a parallelogram.
6.(a) The functions $ \displaystyle f$ and $ \displaystyle g$ are defined by $ \displaystyle f(x)=2x-1$ and $ \displaystyle g(x)=4x+3$. Find $ \displaystyle \left( {g\circ f} \right)(x)$ and $ \displaystyle {{g}^{{-1}}}(x)$ in simplified form. Show also that $ \displaystyle {{\left( {g\circ f} \right)}^{{-1}}}(x)=\left( {{{f}^{{-1}}}\circ {{g}^{{-1}}}} \right)(x)$.
6.(b) The expression $ \displaystyle ax^3 - x^2 + bx - 1$ leaves the remainders of $ \displaystyle - 33$ and $ \displaystyle 77$ when divided by $ \displaystyle x + 2$ and $ \displaystyle x - 3$ respectively. Find the values of $ \displaystyle a$ and $ \displaystyle b$, and the remainder when divided by $ \displaystyle x - 2$.
$ \displaystyle \therefore \ \ \ f(x)$ leaves the remainders of $ \displaystyle 23$ when divided by $ \displaystyle x - 2$
7.(a) A binary operation $ \displaystyle \odot$ on $ \displaystyle R$ is defined by $ \displaystyle x \odot y = (3y - x)^2 - 8y^2$. Show that the binary operation is commutative. Find the possible values of $ \displaystyle k$ such that $ \displaystyle 2 \odot k = -31$.
7.(b) If the coefficients of $ \displaystyle x^r$ and $ \displaystyle x^{r+2}$ in the expansion of $ \displaystyle (1 + x)^{2n}$ are equal, show that $ \displaystyle r = n -1$.
8.(a) Find the solution set in $ \displaystyle R$ of the inequation $ \displaystyle x^2-3x+2\le 0$ by algebraic method and illustrate it on the number line.
8.(b) Find the sum of the first $ \displaystyle 12$ terms of the $ \displaystyle A.P$. $ \displaystyle 44, 40, 36,$ ... . Find also the sum of the terms between the $ \displaystyle {{12}^{{\text{th}}}}$ term and the $ \displaystyle {{26}^{{\text{th}}}}$ term of that $ \displaystyle A.P.$
9.(a) The sum of the first $ \displaystyle n$ terms of a certain sequence is given by $ \displaystyle {{S}_{{_{n}}}}={{2}^{n}}-1$. Find the first three terms of the sequence and express the $ \displaystyle {{n}^{{\text{th}}}}$ term in terms of $ \displaystyle n.$
9.(b) Using the definition of inverse matrix, find the inverse of the matrix $ \displaystyle \left( {\begin{array}{*{20}{c}} 3 & 1 \\ 2 & 1 \end{array}} \right)$.
(5 marks)
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Let $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 3 & 1 \\ 2 & 1 \end{array}} \right)$ and $\displaystyle {{A}^{{-1}}}=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$
By the definition of the inverse of the matrix,
$ \displaystyle A{{A}^{{-1}}}=I,\ $ where $ \displaystyle I$ is a unit matrix of order $ \displaystyle 2$.
10.(a) Find the inverse of the matrix $ \displaystyle \left( {\begin{array}{*{20}{c}} 5 & 6 \\ 7 & 8 \end{array}} \right)$. Use it to determine the coordinate of the point of intersection of the lines $ \displaystyle 5x + 6y=7$ and $ \displaystyle 8y + 7x = 10$.
$ \displaystyle \therefore \ \ \ $ Number of favourable outcomes = $ \displaystyle 9$
$ \displaystyle \begin{array}{l}\therefore \ \ \ \ P(\text{score on }{{1}^{{\text{st}}}}\text{ die is prime and score on }{{2}^{{\text{nd}}}}\ \text{is even)}\\\text{ }\\\ \ \ =\displaystyle \frac{9}{{36}}\\\\\ \ \ =\displaystyle \frac{1}{4}\end{array}$
SECTION (C) (Answer any THREE questions.)
11.(a) $ \displaystyle PT$ is a tangent and $ \displaystyle PQR$ is a secant to a circle. A circle with $ \displaystyle T$ as centre and radius $ \displaystyle TQ$ meets $ \displaystyle QR$ again at $ \displaystyle S$. Prove that $ \displaystyle \angle RTS=\angle RPT$
11.(b) In the diagram, $ \displaystyle P$ is the point on $ \displaystyle AC$ such that $ \displaystyle AP=3PC$, $ \displaystyle R$ is the point on $ \displaystyle BP$ such that $ \displaystyle BR=2RP$ and $ \displaystyle QR\parallel AC$. Given that $ \displaystyle \alpha \left( {\Delta BPA} \right)=36\ \text{c}{{\text{m}}^{\text{2}}}$, calculate $ \displaystyle \alpha \left( {\Delta BPC} \right)$ and $ \displaystyle \alpha \left( {\Delta BRQ} \right)$.
14.(a) In the quadrilateral $ \displaystyle ABCD$, $ \displaystyle M$ and $ \displaystyle N$ are the midpoints of $ \displaystyle AC$ and $ \displaystyle BD$ respectively, prove that $ \displaystyle \overrightarrow{{AB}}+\ \overrightarrow{{CB}}+\ \overrightarrow{{AD}}+\ \overrightarrow{{CD}}=4\ \overrightarrow{{MN}}$.
