Trigonometry (Practice Problems)

1.        Given that $ \displaystyle \sin 16{}^\circ =p,$ determine the following in terms of $ \displaystyle p$.

(a) $ \displaystyle \sin 196{}^\circ $.

(b) $ \displaystyle \cos 16{}^\circ $.

(c) $ \displaystyle \tan 32{}^\circ $.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \sin 16{}^\circ =p\\\\\text{(a)}\ \ \ \sin 196{}^\circ =\sin (180{}^\circ +16{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin 16{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-p\\\\\\\text{(b)}\ \ \ \cos 16{}^\circ \ \ =\sqrt{{1-{{{\sin }}^{2}}16{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{{1-{{p}^{2}}}}\\\\\\\text{(c)}\ \ \ \tan 16{}^\circ \ \ =\displaystyle \frac{{\sin 16{}^\circ }}{{\cos 16{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{p}{{\sqrt{{1-{{p}^{2}}}}}}\\\\\ \ \ \ \ \ \ \tan 32{}^\circ \ \ =\displaystyle \frac{{2\tan 16{}^\circ }}{{1-{{{\tan }}^{2}}16{}^\circ }}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{2p}}{{\sqrt{{1-{{p}^{2}}}}}}}}{{1-\displaystyle \frac{{{{p}^{2}}}}{{1-{{p}^{2}}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2p}}{{\sqrt{{1-{{p}^{2}}}}}}\times \displaystyle \frac{{1-{{p}^{2}}}}{{1-2{{p}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{2p\sqrt{{1-{{p}^{2}}}}}}{{1-2{{p}^{2}}}}\end{array}$

2.        Simplify $ \displaystyle \frac{{\sqrt{{1-{{{\cos }}^{2}}2\theta }}}}{{\cos (-\theta )\cos (90{}^\circ +\theta )}}$ completely, given that $ \displaystyle 0{}^\circ <\theta <90{}^\circ $.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ 1-{{\cos }^{2}}2\theta ={{\sin }^{2}}2\theta \\\\\ \ \ \ \ \ \cos (-\theta )=\cos \theta \\\\\ \ \ \ \ \ \cos (90{}^\circ +\theta )=-\sin \theta \\\\\therefore \ \ \ \ \ \displaystyle \frac{{\sqrt{{1-{{{\cos }}^{2}}2\theta }}}}{{\cos (-\theta )\cos (90{}^\circ +\theta )}}\\\\\ \ \ \ \ \ \ \ =\displaystyle \frac{{\sqrt{{{{{\sin }}^{2}}2\theta }}}}{{\cos \theta \left( {-\sin \theta } \right)}}\\\\\ \ \ \ \ \ \ \ =-\displaystyle \frac{{\sin 2\theta }}{{\cos \theta \sin \theta }}\\\\\ \ \ \ \ \ \ \ =-\displaystyle \frac{{2\sin \theta \cos \theta }}{{\cos \theta \sin \theta }}\\\\\ \ \ \ \ \ \ \ =-2\end{array}$

3.        In the diagram $\displaystyle \vartriangle TPS,\ PR\bot TS$. If $\displaystyle TR=2,RS = 3$ and $\displaystyle PR=1$, find the value of

(a) $ \displaystyle \sin T$ and $ \displaystyle \cos T$

(b) $ \displaystyle \sin S$ and $ \displaystyle \cos S$

Hence without using tables, calculate the value of $ \displaystyle \cos (S+T)$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \vartriangle TPS,\ PR\bot TS\\\\\ \ \ \ \ TR=2,RS=3,\ PR=1\\\\\therefore \ \ \ PT=\sqrt{{{{2}^{2}}+{{1}^{2}}}}=\sqrt{5}\\\\\ \ \ \ \ PS=\sqrt{{{{3}^{2}}+{{1}^{2}}}}=\sqrt{{10}}\\\\\therefore \ \ \ \sin T=\displaystyle \frac{1}{{\sqrt{5}}},\ \cos T=\displaystyle \frac{2}{{\sqrt{5}}}\\\\\ \ \ \ \ \sin S=\displaystyle \frac{1}{{\sqrt{{10}}}},\ \cos S=\displaystyle \frac{3}{{\sqrt{{10}}}}\\\\\therefore \ \ \ \cos \left( {S+T} \right)=\cos S\cos T-\sin S\sin T\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{3}{{\sqrt{{10}}}}\times \displaystyle \frac{2}{{\sqrt{5}}}-\displaystyle \frac{1}{{\sqrt{{10}}}}\times \displaystyle \frac{1}{{\sqrt{5}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{5}{{\sqrt{{50}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{5}{{5\sqrt{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{2}}}\end{array}$

4.        If $ \displaystyle \sin x-\cos x=\frac{3}{4}$, calculate the value of $ \displaystyle \sin 2x$ without using tables.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \sin x-\cos x=\displaystyle \frac{3}{4}\\\\\therefore \ \ {{\left( {\sin x-\cos x} \right)}^{2}}=\displaystyle \frac{9}{{16}}\\\\\ \ \ \ {{\sin }^{2}}x-2\sin x\cos x+{{\cos }^{2}}x=\displaystyle \frac{9}{{16}}\\\\\ \ \ \ {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x=\displaystyle \frac{9}{{16}}\\\\\therefore \ \ 1-\sin 2x=\displaystyle \frac{9}{{16}}\\\\\therefore \ \ \sin 2x=\displaystyle \frac{7}{{16}}\end{array}$

5.        Simplify $ \displaystyle \frac{{4\sin x\cos x}}{{2{{{\sin }}^{2}}x-1}}$ to a single trigonometric ratio. Hence, find the value of $ \displaystyle \frac{{4\sin 15{}^\circ \cos 15{}^\circ }}{{2{{{\sin }}^{2}}15{}^\circ -1}}$ in surd form.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{4\sin x\cos x}}{{2{{{\sin }}^{2}}x-1}}\\\\=\ \ \displaystyle \frac{{2\left( {2\sin x\cos x} \right)}}{{-\left( {1-2{{{\sin }}^{2}}x} \right)}}\\\\=\ \ \displaystyle \frac{{2\sin 2x}}{{-\cos 2x}}\\\\=\ \ -2\tan 2x\\\\\therefore \ \ \displaystyle \frac{{4\sin 15{}^\circ \cos 15{}^\circ }}{{2{{{\sin }}^{2}}15{}^\circ -1}}\\\\=\ -2\tan 30{}^\circ \\\\=\ -\displaystyle \frac{2}{{\sqrt{3}}}\end{array}$

6.        Points $ \displaystyle P(-\sqrt{7},3)$ and $ \displaystyle S(a,b)$ are on the Cartesian plane, as shown in figure. If $ \displaystyle \angle POR=\angle POS=\theta $, $ \displaystyle \angle POR=\angle POS=\theta $ and $ \displaystyle OS=6$,

(a) $ \displaystyle \tan \theta$

(b) $ \displaystyle \sin (-\theta )$

(c) $ \displaystyle a$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ P=(-\sqrt{7},3)\\\\\therefore \ \ \ \ OP=\sqrt{{7+9}}=4\\\\\therefore \ \ \ \ \sin \theta =\displaystyle \frac{3}{4}\\\\\text{(a)}\ \ \tan \theta =-\displaystyle \frac{3}{{\sqrt{7}}}\\\\\text{(b)}\ \ \sin \left( {-\theta } \right)=-\sin \theta =-\displaystyle \frac{3}{4}\\\\\text{(c)}\ \ \cos 2\theta =1-2{{\sin }^{2}}\theta =1-\displaystyle \frac{9}{8}=-\displaystyle \frac{1}{8}\\\\\ \ \ \ \ \text{Since}\ \cos 2\theta =\displaystyle \frac{a}{6},\ \\\\\ \ \ \ \ \displaystyle \frac{a}{6}=-\displaystyle \frac{1}{8}\\\\\therefore \ \ \ a\ =-\displaystyle \frac{3}{4}\end{array}$

7.        If $ \displaystyle \sin A = p$ and $ \displaystyle \cos A = q,$

(a) Show that $ \displaystyle \tan 2A=\frac{{2pq}}{{{{q}^{2}}-{{p}^{2}}}}$

(b) Express $ \displaystyle p^4-q^4$ as a single trigonometric ratio.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \sin A=p,\ \ \cos A=q\\\\\therefore \ \ \ \ \tan A=\displaystyle \frac{p}{q}\ \\\\\text{(a)}\ \ \tan 2A=\ \ \displaystyle \frac{{2\tan A}}{{1-{{{\tan }}^{2}}A}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{2\left( {\displaystyle \frac{p}{q}} \right)}}{{1-{{{\left( {\displaystyle \frac{p}{q}} \right)}}^{2}}}}\ \ \\\ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{\displaystyle \frac{{2p}}{q}}}{{\displaystyle \frac{{{{q}^{2}}-{{p}^{2}}}}{{{{q}^{2}}}}}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{2p}}{q}\times \displaystyle \frac{{{{q}^{2}}}}{{{{q}^{2}}-{{p}^{2}}}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \ \displaystyle \frac{{2pq}}{{{{q}^{2}}-{{p}^{2}}}}\\\\\text{(b)}\ \ {{p}^{4}}-{{q}^{4}}=\left( {{{p}^{2}}+{{q}^{2}}} \right)\left( {{{p}^{2}}-{{q}^{2}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {{{{\sin }}^{2}}A+{{{\cos }}^{2}}A} \right)\left( {{{{\sin }}^{2}}A-{{{\cos }}^{2}}A} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1\left( {{{{\cos }}^{2}}A-{{{\sin }}^{2}}A} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 2A\end{array}$

8.       In the diagram below, $ \displaystyle T(x,p)$ is a point in the third quadrant and it is given that $ \displaystyle \sin \alpha =\frac{p}{{\sqrt{{1+{{p}^{2}}}}}}$.

(a) Show that $ \displaystyle x =-1$.

(b) Find $ \displaystyle \cos 2\alpha $ in terms of $ \displaystyle p$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{(a)}\ \ \sin \alpha =\displaystyle \frac{p}{{\sqrt{{1+{{p}^{2}}}}}}\ \ \ \left[ {\because \text{given}} \right]\\\\\ \ \ \ \ \sin \alpha =\displaystyle \frac{p}{h}\ \ \ \left[ {\text{diagram}} \right]\\\\\therefore \,\ \ \displaystyle \frac{p}{h}=\displaystyle \frac{p}{{\sqrt{{1+{{p}^{2}}}}}}\\\\\therefore \ \ h=\sqrt{{1+{{p}^{2}}}}\\\\\ \ \ \ \text{Since}\ {{x}^{2}}+{{p}^{2}}={{h}^{2}},\\\\\ \ \ \ {{x}^{2}}+{{p}^{2}}=1+{{p}^{2}}\\\\\therefore \ \ {{x}^{2}}=1\\\\\therefore \ \ x=\pm 1\\\\\ \ \ \ \text{Since}\ \alpha \,\ \text{lies in }{{\text{3}}^{{\text{rd}}}}\text{ quadrant,}\\\\\ \ \ x=-1.\\\\\\\text{(b)}\ \ \cos 2\alpha =1-2{{\sin }^{2}}\alpha \\\\\therefore \,\ \ \ \cos 2\alpha =1-\displaystyle \frac{{2{{p}^{2}}}}{{1+{{p}^{2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{1-{{p}^{2}}}}{{1+{{p}^{2}}}}\end{array}$

9.        If $ \displaystyle \sin 76{}^\circ = x$ and $ \displaystyle \cos 76{}^\circ = y$, show that $ \displaystyle x^2- y^2 = \sin 62{}^\circ$.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \sin 76{}^\circ =x,\ \cos 76{}^\circ =y\\\\\ \ \ {{x}^{2}}-{{y}^{2}}={{\sin }^{2}}76{}^\circ -{{\cos }^{2}}76{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\left( {{{{\cos }}^{2}}76{}^\circ -{{{\sin }}^{2}}76{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos \left( {2\times 76{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 152{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos \left( {90{}^\circ +62{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\left( {-\sin 62{}^\circ } \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin 62{}^\circ \end{array}$

10.      $ \displaystyle T(8, k)$ is a point in the first quadrant as shown in figure. If $ \displaystyle \tan β = \frac{1}{4},$ determine, without using tables,

(a) the value of $ \displaystyle k$.

(b) the value of $ \displaystyle \sin β$ in surd form.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\,\ \ \tan \beta =\displaystyle \frac{1}{4}\ \ \left[ {\text{given}} \right]\\\\\,\ \ \tan \beta =\displaystyle \frac{k}{8}\ \ \left[ {\text{diagram}} \right]\\\\\therefore \displaystyle \frac{k}{8}=\displaystyle \frac{1}{4}\\\\\therefore \ k=2\\\\\ \ \ \text{Since}\ O{{T}^{2}}={{8}^{2}}+{{k}^{2}},\\\\\ \ \ O{{T}^{2}}={{8}^{2}}+{{2}^{2}}=68\\\\\therefore \ OT=\sqrt{{68}}=2\sqrt{{17}}\\\\\therefore \ \sin \beta =\displaystyle \frac{k}{{OT}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{2}{{2\sqrt{{17}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{{17}}}}\end{array}$

Integration by Substitution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{f(g(x)){g}'(x)}}\ dx\\\\\,\ \ \ \text{Let}\ u=g(x).\\\\\therefore \ \ \displaystyle \frac{{du}}{{dx}}={g}'(x)\\\\\therefore \ \ du={g}'(x)dx\end{array}$ $\displaystyle \begin{array}{l}\begin{array}{|r|l|l|l|c|c|c|c|c|c|} \hline { \displaystyle \therefore \int{{f(g(x)){g}'(x)}}\ dx=\int{{f(u)}}\ du} \\ \hline \end{array}\end{array}$

1.       Evaluate

$\displaystyle \text{(a)}\ \ \ \ \int{{(2{{x}^{3}}-3x+1)(2x-1)}}\ dx$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{(2{{x}^{3}}-3x+1)(2x-1)}}\ dx\\\\\,\ \ \ \text{Let}\ u=2{{x}^{3}}-3x+1.\\\\\therefore \ \ du=(6x-3)dx\\\\\therefore \ \ du=3(2x-1)dx\\\\\therefore \ \ (2x-1)dx=\displaystyle \frac{1}{3}du\\\\\therefore \ \ \displaystyle \int{{(2{{x}^{3}}-3x+1)(2x-1)}}\ dx\\\\\ =\ \displaystyle \int{{u\cdot }}\displaystyle \frac{1}{3}du\\\\\ =\ \displaystyle \frac{1}{3}\displaystyle \int{{u\ }}du\\\\\ =\ \displaystyle \frac{1}{6}{{u}^{2}}+C\\\\\ =\ \displaystyle \frac{1}{6}{{\left( {2{{x}^{3}}-3x+1} \right)}^{2}}+C\end{array}$

$ \displaystyle \text{(b)}\ \ \ \ \int{{\frac{{{{{\left( {2-\sqrt{x}} \right)}}^{5}}}}{{\sqrt{x}}}}}\ dx$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {2-\sqrt{x}} \right)}}^{5}}}}{{\sqrt{x}}}}}\ dx\\\\\,\ \ \ \text{Let}\ u=2-\sqrt{x}.\\\\\therefore \ \ du=-\displaystyle \frac{1}{{2\sqrt{x}}}dx\\\\\therefore \ \ \displaystyle \frac{1}{{\sqrt{x}}}dx=-2du\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {2-\sqrt{x}} \right)}}^{5}}}}{{\sqrt{x}}}}}\ dx\\\\\ =\ -2\displaystyle \int{{{{u}^{5}}\ }}du\\\\\ =\ -2\left( {\displaystyle \frac{1}{6}{{u}^{6}}} \right)+C\\\\\ =\ -\displaystyle \frac{1}{3}{{u}^{6}}+C\\\\\ =\ -\displaystyle \frac{1}{3}{{\left( {2-\sqrt{x}} \right)}^{6}}+C\end{array}$

$\displaystyle (\text{c})\ \ \ \ \int{{\frac{{{{x}^{2}}}}{{5{{x}^{3}}-2}}}}\ dx$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}}}{{5{{x}^{3}}-2}}}}\ dx\\\\\,\ \ \ \text{Let}\ u=5{{x}^{3}}-2.\\\\\therefore \ \ du=15{{x}^{2}}\ dx\\\\\therefore \ \ {{x}^{2}}\ dx=\displaystyle \frac{1}{{15}}du\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}}}{{5{{x}^{3}}-2}}}}\ dx\\\\\ =\ \displaystyle \frac{1}{{15}}\displaystyle \int{{\displaystyle \frac{1}{u}\ }}du\\\\\ =\ \displaystyle \frac{1}{{15}}\ln |u|+C\\\\\ =\ \displaystyle \frac{1}{{15}}\ln |5{{x}^{3}}-2|+C\end{array}$

$ \displaystyle (\text{d)}\ \ \ \ \int{{\frac{{{{e}^{{3x}}}}}{{1+{{e}^{x}}}}}}\ dx$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{e}^{{3x}}}}}{{1+{{e}^{x}}}}}}\ dx\\\\=\ \ \displaystyle \int{{\displaystyle \frac{{{{e}^{{2x}}}\cdot {{e}^{x}}}}{{1+{{e}^{x}}}}}}\ dx\\\\=\ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {{{e}^{x}}} \right)}}^{2}}\cdot {{e}^{x}}}}{{1+{{e}^{x}}}}}}\ dx\\\\\,\ \ \ \text{Let}\ u=1+{{e}^{x}},\ \text{then}\ {{e}^{x}}=u-1\\\\\therefore \ \ du={{e}^{x}}\ dx\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{e}^{{3x}}}}}{{1+{{e}^{x}}}}}}\ dx\\\\\ =\ \displaystyle \int{{\displaystyle \frac{{{{{\left( {u-1} \right)}}^{2}}}}{u}}}\ du\\\\\ =\ \displaystyle \int{{\displaystyle \frac{{{{u}^{2}}-2u+1}}{u}}}\ du\\\\\ =\ \displaystyle \int{{\left( {u-2+\displaystyle \frac{1}{u}} \right)}}\ du\\\\=\ \displaystyle \int{u}\ du-2\displaystyle \int{{du}}\ +\displaystyle \int{{\displaystyle \frac{1}{u}}}\ du\\\\\ =\ \displaystyle \frac{1}{2}{{u}^{2}}-2u+\ln |u|+C\\\\\ =\ \displaystyle \frac{1}{2}{{\left( {1+{{e}^{x}}} \right)}^{2}}-2\left( {1+{{e}^{x}}} \right)+\ln |1+{{e}^{x}}|+C\\\\\ =\ \displaystyle \frac{1}{2}\left( {1+2{{e}^{x}}+{{e}^{{2x}}}} \right)-2\left( {1+{{e}^{x}}} \right)+\ln |1+{{e}^{x}}|+C\\\\\ =\ \displaystyle \frac{1}{2}+{{e}^{x}}+\displaystyle \frac{1}{2}{{e}^{{2x}}}-2-2{{e}^{x}}+\ln |1+{{e}^{x}}|+C\\\\\ =\ \displaystyle \frac{1}{2}{{e}^{{2x}}}-{{e}^{x}}+\ln |1+{{e}^{x}}|-\displaystyle \frac{3}{2}+C\\\\\ =\ \displaystyle \frac{1}{2}{{e}^{{2x}}}-{{e}^{x}}+\ln |1+{{e}^{x}}|+\ k\ \ \ \left( {k=C-\displaystyle \frac{3}{2}} \right)\end{array}$

$ \displaystyle (\text{e})\ \ \ \ \int{{\frac{{{{{(\ln x)}}^{8}}+1}}{x}}}\ dx$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{{(\ln x)}}^{8}}+1}}{x}}}\ dx\\\\\,\ \ \ \text{Let}\ u=\ln x,\ \\\\\therefore \ \ du=\displaystyle \frac{1}{x}\ dx\\\\\therefore \ \ \displaystyle \int{{\displaystyle \frac{{{{{(\ln x)}}^{8}}+1}}{x}}}\ dx\\\\\therefore \ \ \displaystyle \int{{\left[ {{{{(\ln x)}}^{8}}+1} \right]\displaystyle \frac{1}{x}}}\ dx\\\\\ =\ \displaystyle \int{{\left( {{{u}^{8}}+1} \right)}}\ du\\\\=\ \displaystyle \int{{{{u}^{8}}}}\ du+2\displaystyle \int{{du}}\ \\\\\ =\ \displaystyle \frac{1}{9}{{u}^{9}}+2u+C\\\\\ =\ \displaystyle \frac{1}{9}{{\left( {\ln x} \right)}^{9}}+2\left( {\ln x} \right)+C\end{array}$

$ \displaystyle \text{(f)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}-8x+3}}{{x+2}}}}\ dx$

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}-8x+3}}{{x+2}}}}\ dx\\\\\ \ \ =\ \ \displaystyle \int{{\displaystyle \frac{{{{{\left( {x+2} \right)}}^{2}}-12(x+2)+23}}{{x+2}}}}\ dx\\\\\ \ \ \ \ \ \ \text{Let}\ u=x+2\\\\\therefore \ \ \ \ \ du=dx\\\\\therefore \ \ \ \ \ \ \displaystyle \int{{\displaystyle \frac{{{{x}^{2}}-8x+3}}{{x+2}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\displaystyle \frac{{{{u}^{2}}-12u+23}}{u}}}\ du\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {u-12+\displaystyle \frac{{23}}{u}} \right)}}\ du\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {u-12+\displaystyle \frac{{23}}{u}} \right)}}\ du\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{u}^{2}}-12u+23\ln |u|+C\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{\left( {x+2} \right)}^{2}}-12\left( {x+2} \right)+23\ln |x+2|+C\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{x}^{2}}-10x+23\ln |x+2|-22+C\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{2}{{x}^{2}}-10x+23\ln |x+2|+k\ \left[ {k=C-22} \right]\end{array}$

😍 ဆရာမေလး သင္တာနဲ႔တြဲၾကည့္ ပိုမွတ္မိတာေပါ့ 😍

2019 Matriculation Examination : Math Paper Solution

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$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)=x+1,\ \ \\\\\ \ \ \ \ g(x)=2{{x}^{2}}-x+3\\\\\ \ \ \ \ \left( {f\circ g} \right)(x)=4x+1\\\\\therefore \ \ \ f\left( {2{{x}^{2}}-x+3} \right)=4x+1\\\\\therefore \ \ \ 2{{x}^{2}}-x+3+1=4x+1\\\\\therefore \ \ \ 2{{x}^{2}}-5x+3=0\\\\\therefore \ \ \ (x-1)(2x-3)=0\\\\\therefore \ \ \ x=1\ (\text{or})\ x=\displaystyle \frac{3}{2}\end{array}$

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$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let}\ f(x)=2{{x}^{2}}+5x-3.\ \\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ f\left( {\displaystyle \frac{p}{2}} \right)=2{{p}^{2}}-3p\\\\\therefore \ \ \ 2{{\left( {\displaystyle \frac{p}{2}} \right)}^{2}}+5\left( {\displaystyle \frac{p}{2}} \right)-3=2{{p}^{2}}-3p\\\\\therefore \ \ \ \displaystyle \frac{{{{p}^{2}}}}{2}+\displaystyle \frac{{5p}}{2}-3=2{{p}^{2}}-3p\\\\\therefore \ \ \ {{p}^{2}}+5p-6=4{{p}^{2}}-6p\\\\\therefore \ \ \ 3{{p}^{2}}-11p+6=0\\\\\therefore \ \ \ (3p-2)(p-3)=0\\\\\therefore \ \ \ p=\displaystyle \frac{2}{3}\ (\text{or})\ p=3\end{array}$

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 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{In the expansion of}{{\left( {{{x}^{2}}+\displaystyle \frac{2}{x}} \right)}^{8}},x\ne 0\\\\\ \ \ \ \ {{(r+1)}^{{\text{th}}}}\ \text{term}={}^{8}{{C}_{r}}{{\left( {{{x}^{2}}} \right)}^{{8-r}}}{{\left( {\displaystyle \frac{2}{x}} \right)}^{r}}\\\\\therefore \ \ \ {{(r+1)}^{{\text{th}}}}\ \text{term}={}^{8}{{C}_{r}}{{2}^{r}}{{x}^{{16-3r}}}\\\\\therefore \ \ \ \text{For}\ {{x}^{7}},\ 16-3r=7\\\\\therefore \ \ \ r=3\\\\\therefore \ \ \ \text{Coefficient of}\ {{x}^{7}}={}^{8}{{C}_{3}}{{2}^{3}}\\\\\therefore \ \ \ \text{Coefficient of}\ {{x}^{7}}=\displaystyle \frac{{8\times 7\times 6}}{{1\times 2\times 3}}\times 8=448\end{array}$

2.(b)      Find the sum of all even numbers between $ \displaystyle 69$ and $ \displaystyle149.$

(3 marks)

Show/Hide Solution

All even numbers between $ \displaystyle 69$ and $ \displaystyle149.$ are $ \displaystyle 70, 72, 74,$ ... $ \displaystyle 148$. It is an $ \displaystyle A.P$ with $ \displaystyle a=70$, $ \displaystyle d=2$ and $ \displaystyle {{u}_{{_{n}}}}=148$. $ \displaystyle \begin{array}{l}\therefore \ \ \ \ a+\left( {n-1} \right)d=148\\\\\therefore \ \ \ \ 70+\left( {n-1} \right)2=148\\\\\therefore \ \ \ \ n=40\\\\\therefore \ \ \ \ \text{Required Sum}\ ={{S}_{{40}}}\\\\\ \ \ \ \ \ {{S}_{n}}=\displaystyle \frac{n}{2}\left( {a+l} \right)\\\\\therefore \ \ \ \ {{S}_{{40}}}=\displaystyle \frac{{40}}{2}\left( {70+148} \right)\\\\\therefore \ \ \ \ {{S}_{{40}}}=4360\end{array}$

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 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ A=\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 5 \end{array}} \right),\ B=\left( {\begin{array}{*{20}{c}} x & y \\ 0 & z \end{array}} \right)\\\\\ \ \ \ AB=A+B\,\ \ \left[ {\text{given}} \right]\\\\\therefore \,\ \ \left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 5 \end{array}} \right)\ \left( {\begin{array}{*{20}{c}} x & y \\ 0 & z \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 2 & 0 \\ 0 & 5 \end{array}} \right)+\ \left( {\begin{array}{*{20}{c}} x & y \\ 0 & z \end{array}} \right)\\\\\therefore \,\ \ \left( {\begin{array}{*{20}{c}} {2x+0} & {2y+0} \\ 0 & {0+5z} \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} {2+x} & y \\ 0 & {5+z} \end{array}} \right)\\\\\therefore \ \ \ \left( {\begin{array}{*{20}{c}} {2x} & {2y} \\ 0 & {5z} \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} {2+x} & y \\ 0 & {5+z} \end{array}} \right)\\\\\therefore \ \ \ 2x=2+x\Rightarrow x=2\\\\\ \ \ \ \ 2y=y\Rightarrow y=0\\\\\ \ \ \ \ 5z=5+z\Rightarrow z=\displaystyle \frac{5}{4}\end{array}$

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$.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \gamma =\alpha =\beta \ \ \ \ \ \ (\angle \ \text{between tangents and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{=}\angle \ \text{in alternate segments)}\\\\\ \ \ \ \text{But}\ \theta =\alpha +\beta \ \ \ (\text{exterior }\angle \ \text{of a }\vartriangle \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{=}\ \text{sum of opposite interior }\angle \text{s)}\\\\\therefore \ \ \theta =\gamma +\gamma =2\gamma \\\\\therefore \ \ \angle BTX=2\angle ACB\end{array}$

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.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ A=(1,0),\ B=(4,2),\ C=(5,4)\\\\\ \ \ \ \ \text{Let}\ D=(a,b).\\\\\ \ \ \ \ ABCD\ \text{is a parallelogram}\text{.}\\\\\therefore \ \ \ \overrightarrow{{AD}}=\overrightarrow{{BC}}\\\\\therefore \ \ \ \overrightarrow{{OD}}-\overrightarrow{{OA}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \left( {\begin{array}{*{20}{c}} a \\ b \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right)\\\\\therefore \ \ \ \ \left( {\begin{array}{*{20}{c}} {a-1} \\ b \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right)\\\\\therefore \ \ \ a-1=2\Rightarrow a=2\\\\\ \ \ \ \ b=2\end{array}$

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

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ 2\cos x\sin x=\sin x,\ 0{}^\circ \le x\le 360{}^\circ \\\\\ \ \ \ \ 2\cos x\sin x-\sin x=0\\\\\ \ \ \ \ \sin x\left( {2\cos x-1} \right)=0\\\\\ \ \ \ \ \sin x=0\ \left( {\text{or}} \right)\ 2\cos x-1=0\\\\\ \ \ \ \ \sin x=0\ \left( {\text{or}} \right)\ \cos x=\displaystyle \frac{1}{2}\\\\(\text{i})\ \ \text{For}\ \sin x=0,\\\\\ \ \ \ \ x=0{}^\circ \ \ \left( {\text{or}} \right)\ \ x=180{}^\circ \ \ \left( {\text{or}} \right)\ \ x=360{}^\circ \\\\(\text{i})\ \ \text{For}\ \cos x=\displaystyle \frac{1}{2},\\\\\ \ \ \ \ x=60{}^\circ \ \ \left( {\text{or}} \right)\ \ x=300{}^\circ \end{array}$

5.(b)      Differentiate $ \displaystyle x^3+2x$ with respect to $ \displaystyle x$ from the first principle.

(3 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Let}\ y={{x}^{3}}+2x.\\\\\therefore \ \ \ \ y+\delta y={{\left( {x+\delta x} \right)}^{3}}+2\left( {x+\delta x} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{3}}+3{{x}^{2}}\left( {\delta x} \right)+3x{{\left( {\delta x} \right)}^{2}}+{{\left( {\delta x} \right)}^{3}}+2x+2\left( {\delta x} \right)\\\\\therefore \ \ \ \delta y=\ \left( {y+\delta y} \right)-y\\\\\ \ \ \ \ \ \ \ \ \ \ \ =3{{x}^{2}}\left( {\delta x} \right)+3x{{\left( {\delta x} \right)}^{2}}+{{\left( {\delta x} \right)}^{3}}+2\left( {\delta x} \right)\\\\\therefore \ \ \ \displaystyle \frac{{\delta y}}{{\delta x}}=3{{x}^{2}}+3x\left( {\delta x} \right)+{{\left( {\delta x} \right)}^{2}}+2\end{array}$ $ \displaystyle \therefore \ \ \ \displaystyle \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{\delta y}}{{\delta x}}$ $ \displaystyle \therefore \ \ \ \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\left[ {3{{x}^{2}}+3x\left( {\delta x} \right)+{{{\left( {\delta x} \right)}}^{2}}+2} \right]$ $ \displaystyle \therefore \ \ \ \frac{{dy}}{{dx}}=3{{x}^{2}}+2$

SECTION (B)
(Answer any FOUR questions.)

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)$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ f(x)=2x-1,\ \ g(x)=4x+3\\\\\ \ \ \ \ \ \ \ \left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\\\therefore \ \ \ \ \ \ \left( {g\circ f} \right)(x)=g\left( {2x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left( {2x-1} \right)+3\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8x-1\\\\\ \ \ \ \ \ \ \text{Let}\ {{g}^{{-1}}}(x)=y,\ \text{then}\\\\\ \ \ \ \ \ \ \ g(y)=x\\\\\ \ \ \ \ \ \ \ 4y+3=x\\\\\therefore \ \ \ \ \ \ y=\displaystyle \frac{{x-3}}{4}\\\\\therefore \ \ \ \ \ \ {{g}^{{-1}}}(x)=\displaystyle \frac{{x-3}}{4}\\\\\ \ \ \ \ \ \ \text{Let}\ \ {{\left( {g\circ f} \right)}^{{-1}}}(x)=z,\ \text{then}\\\\\ \ \ \ \ \ \ \left( {g\circ f} \right)(z)=x\\\\\ \ \ \ \ \ \ \ 8z-1=x\\\\\therefore \ \ \ \ \ \ z=\displaystyle \frac{{x+1}}{8}\\\\\therefore \ \ \ \ \ \ {{\left( {g\circ f} \right)}^{{-1}}}(x)=\displaystyle \frac{{x+1}}{8}\ \ \ ---(1)\\\\\ \ \ \ \ \ \ \text{Let}\ {{f}^{{-1}}}(x)=w,\ \text{then}\\\\\ \ \ \ \ \ \ \ f(w)=x\\\\\ \ \ \ \ \ \ \ 2w-1=x\\\\\therefore \ \ \ \ \ \ w=\displaystyle \frac{{x+1}}{2}\\\\\therefore \ \ \ \ \ \ {{f}^{{-1}}}(x)=\displaystyle \frac{{x+1}}{2}\\\\\therefore \ \ \ \ \ \left( {{{f}^{{-1}}}\circ {{g}^{{-1}}}} \right)(x)={{f}^{{-1}}}\left( {{{g}^{{-1}}}(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {\displaystyle \frac{{x-3}}{4}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{x-3}}{4}+1}}{2}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x+1}}{8}\,---(2)\\\\\therefore \ \ \ \ \ \ \text{By}\ (1)\ \operatorname{and}\ (2),\ \\\\\ \ \ \ \ \ \ {{\left( {g\circ f} \right)}^{{-1}}}(x)=\left( {{{f}^{{-1}}}\circ {{g}^{{-1}}}} \right)(x)\end{array}$

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$.

(5 marks)

Show/Hide Solution

$ \displaystyle \text{Let}\ f(x)=a{{x}^{3}}-{{x}^{2}}+bx-1.$ $ \displaystyle f(x)$ leaves the remainders of $ \displaystyle - 33$ when divided by $ \displaystyle x + 2$ $ \displaystyle \begin{array}{l}\therefore \ \ \ f(-2)=-33\\\\\therefore \ \ \ a{{(-2)}^{3}}-{{(-2)}^{2}}+b(-2)-1=-33\\\\\therefore \ \ \ -8a-4-2b-1=-33\ \\\\\therefore \ \ \ 4a+b=14\ \ \ \ ---(1)\end{array}$ Similarly, $ \displaystyle f(x)$ leaves the remainders of $ \displaystyle 77$ when divided by $ \displaystyle x -3$ $ \displaystyle \begin{array}{l}\therefore \ \ \ f(3)=77\\\\\therefore \ \ \ a{{(3)}^{3}}-{{(3)}^{2}}+b(3)-1=77\\\\\therefore \ \ \ 27a-9+3b-1=77\ \\\\\therefore \ \ \ 9a+b=29\ \ \ \ ---(2)\ \\\\\ \ \ \ \text{By}\ (2)-(1),\\\\\ \ \ \ 5a=\ 15\\\\\therefore \ \ a=3\\\\\therefore \ \ 4\left( 3 \right)+b=14\ \\\\\therefore \ \ b=2\ \ \\\\\therefore \ \ f(x)=3{{x}^{3}}-{{x}^{2}}+2x-1.\ \\\\\therefore \ \ f(2)=3{{(2)}^{3}}-{{(2)}^{2}}+2(2)-1=23\ \end{array}$ $ \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$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ x\odot y={{(3y-x)}^{2}}-8{{y}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9{{y}^{2}}-6xy+{{x}^{2}}-8{{y}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-6xy+{{y}^{2}}\\\\\therefore \ \ \ y\odot x={{(3x-y)}^{2}}-8{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9{{x}^{2}}-6xy+{{y}^{2}}-8{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-6xy+{{y}^{2}}\\\\\therefore \ \ \ x\odot y=y\odot x\\\ \end{array}$ Therefore, the binary operation is commutative. $ \displaystyle \begin{array}{l}\ \ \ \ \ x\odot y={{x}^{2}}-6xy+{{y}^{2}}\\\\\ \ \ \ \ 2\odot k=-31\\\\\therefore \ \ \ {{2}^{2}}-6(2)(k)+{{k}^{2}}=-31\\\\\therefore \ \ \ {{k}^{2}}-12k+35=0\\\\\therefore \ \ \ (k-7)(k-5)=0\\\\\therefore \ \ \ k=7\ (\text{or})\ k=5\end{array}$

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$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ {{(r+1)}^{{\text{th}}}}\text{ term in the expansion of}\ {{(1+x)}^{{2n}}}\\\\\ \ \ \ =\ \ {}^{{2n}}{{C}_{r}}{{x}^{r}}\\\\\therefore \ \ \ \ \ \text{Coefficient of }{{x}^{r}}={}^{{2n}}{{C}_{r}}\\\\\ \ \ \ \ \ \ \text{Coefficient of }{{x}^{{r+2}}}={}^{{2n}}{{C}_{{r+2}}}\\\\\ \ \ \ \ \ \ \text{By the problem, }\\\\\ \ \ \ \ \ \ {}^{{2n}}{{C}_{r}}={}^{{2n}}{{C}_{{r+2}}}\\\\\therefore \ \ \ \ \ r=r+2\ \text{which is impossible}\text{.}\\\\\ \ \ \ \ \ \text{But we have }{}^{{2n}}{{C}_{r}}={}^{{2n}}{{C}_{{2n-r}}}.\\\\\therefore \ \ \ \ {}^{{2n}}{{C}_{{2n-r}}}={}^{{2n}}{{C}_{{r+2}}}\\\\\therefore \ \ \ \ 2n-r=r+2\\\\\therefore \ \ \ \ 2r=2n-2\\\\\therefore \ \ \ \ r=n-1\end{array}$

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.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ {{x}^{2}}-3x+2\le 0\\\\\therefore \ \ \ \ \ \ \left( {x-1} \right)(x-2)\le 0\\\\\therefore \ \ \ \ \ \ \text{There are two possibilities that}\\\\(\text{i})\ \ \ \ x-1\le 0\ \operatorname{and}\ x-2\ge 0\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{or}\\\\(\text{ii})\ \ \ \ x-1\ge 0\ \operatorname{and}\ x-2\le 0\ \\\\\\(\text{i})\ \ \ \ x-1\le 0\ \operatorname{and}\ x-2\ge 0\\\\\therefore \ \ \ \ \ \ x\le 1\ \operatorname{and}\ x\ge 2\end{array}$ $ \displaystyle \therefore\ \ \ \ $ There is no point on the number line which satifies both conditions. $ \displaystyle \begin{array}{l}(\text{ii})\ \ \ x-1\ge 0\ \operatorname{and}\ x-2\le 0\ \\\\\therefore \ \ \ \ \ x\ge 1\ \operatorname{and}\ x\le 2\ \end{array}$ $ \displaystyle \therefore \ \ \ \ \text{Solution set =}\left\{ {x|1\le x\le 2} \right\}$ 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.$

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ 44,40,36,...\ \text{is an}\ A.P.\\\\\therefore \ \ a=44,\\\\\ \ \ \ d=40-44=-4\\\\\ \ \ \ {{S}_{n}}=\displaystyle \frac{n}{2}\left\{ {2a+(n-1)d} \right\}\\\\\ \ \ \ {{S}_{{12}}}=\displaystyle \frac{{12}}{2}\left\{ {2\left( {44} \right)+(12-1)\left( {-4} \right)} \right\}\\\\\ \ \ \ {{S}_{{12}}}=264\\\\\ \ \ \ \text{Let the required sum be}\ S.\\\\\therefore \ \ S={{u}_{{13}}}+{{u}_{{14}}}+{{u}_{{15}}}+...+{{u}_{{25}}}\\\\\therefore \ \ S=\displaystyle \frac{{13}}{2}\left( {{{u}_{{13}}}+{{u}_{{25}}}} \right)\\\\\ \ \ \ \ \ \ \ \ =\displaystyle \frac{{13}}{2}\left( {a+12d+a+24d} \right)\\\\\ \ \ \ \ \ \ \ \ =13\left( {a+18d} \right)\\\\\ \ \ \ \ \ \ \ \ =13\left[ {44+18\left( {-4} \right)} \right]\\\\\ \ \ \ \ \ \ \ \ =-364\end{array}$

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.$

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ {{S}_{{_{n}}}}={{2}^{n}}-1\\\\\therefore \ \ {{S}_{1}}={{2}^{1}}-1=1\\\\\ \ \ \ {{S}_{2}}={{2}^{2}}-1=3\\\\\ \ \ \ {{S}_{3}}={{2}^{3}}-1=7\\\\\therefore \ \ {{u}_{1}}={{S}_{1}}=1\\\\\ \ \ \ {{u}_{2}}={{S}_{2}}-{{S}_{1}}=2\\\\\ \ \ \ {{u}_{3}}={{S}_{3}}-{{S}_{2}}=4\\\\\therefore \ \ \ {{u}_{n}}={{S}_{n}}-{{S}_{{n-1}}}\\\\\ \ \ \ \ \ \ \ \ =\left( {{{2}^{n}}-1} \right)-\left( {{{2}^{{n-1}}}-1} \right)\\\\\ \ \ \ \ \ \ \ \ ={{2}^{n}}-{{2}^{{n-1}}}\\\\\ \ \ \ \ \ \ \ \ ={{2}^{n}}\left( {1-\displaystyle \frac{1}{2}} \right)\\\\\ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{2}^{n}}}}{2}\\\\\ \ \ \ \ \ \ \ \ ={{2}^{{n-1}}}\end{array}$

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)

Show/Hide Solution

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$. $ \displaystyle \begin{array}{l}\therefore \ \ \left( {\begin{array}{*{20}{c}} 3 & 1 \\ 2 & 1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \left( {\begin{array}{*{20}{c}} {3a+c} & {3b+d} \\ {2a+c} & {2b+d} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\therefore \ \ 3a+c=1----(1)\\\\\,\ \ \ 2a+c=0----(2)\\\\\ \ \ (1)-(2)\Rightarrow a=1\\\\\therefore \ \ 2\times 1+c=0\Rightarrow c=-2\\\\\text{Similarly},\\\\\ \ \ \ 3b+d=0----(3)\\\\\,\ \ \ 2b+d=1----(4)\\\\\ \ \ (3)-(4)\Rightarrow b=-1\\\\\therefore \ \ 3\left( {-1} \right)+d=0\Rightarrow d=3\\\\\therefore \ \ {{A}^{{-1}}}=\left( {\begin{array}{*{20}{c}} 1 & {-1} \\ {-2} & 3 \end{array}} \right)\end{array}$

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$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Let}\ A=\ \left( {\begin{array}{*{20}{c}} 5 & 6 \\ 7 & 8 \end{array}} \right).\\\\\therefore \ \ \ \ \det A=40-42=-2\ne 0\\\\\therefore \ \ \ \ {{A}^{{-1}}}\ \text{exists}\text{.}\ \\\\\therefore \ \ \ \ {{A}^{{-1}}}=\displaystyle \frac{1}{{\det A}}\left( {\begin{array}{*{20}{c}} 8 & {-6} \\ {-7} & 5 \end{array}} \right)\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{1}{2}\left( {\begin{array}{*{20}{c}} 8 & {-6} \\ {-7} & 5 \end{array}} \right)\\\\\ \ \ \ \ \left. \begin{array}{l}\ 5x+6y=7\\\ 8y+7x=10\end{array} \right\}\ \ \ \ \left( {\text{given}} \right)\\\\\ \ \ \ \ \left. \begin{array}{l}\ 5x+6y=7\\\ 7x+8y=10\end{array} \right\}---(1)\\\\\ \ \ \ \ \text{Transforming into matrix form,}\\\\\ \ \ \ \ \left( {\begin{array}{*{20}{c}} 5 & 6 \\ 7 & 8 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ =\left( {\begin{array}{*{20}{c}} 7 \\ {10} \end{array}} \right)---(2)\\\\\ \ \ \ \ \text{Let}\ X=\ \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ \operatorname{and}\ B=\left( {\begin{array}{*{20}{c}} 7 \\ {10} \end{array}} \right).\\\\\ \therefore \ \ \text{The matrix equation becomes,}\\\\\ \ \ \ \ AX=B\\\\\therefore \ \ \ {{A}^{{-1}}}AX={{A}^{{-1}}}B\\\\\ \ \ \ \ IX={{A}^{{-1}}}B\\\\\therefore \ \ \ X={{A}^{{-1}}}B\\\\\therefore \ \ \ \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=-\displaystyle \frac{1}{2}\left( {\begin{array}{*{20}{c}} 8 & {-6} \\ {-7} & 5 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 7 \\ {10} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{1}{2}\left( {\begin{array}{*{20}{c}} {56-60} \\ {-49+50} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{1}{2}\left( {\begin{array}{*{20}{c}} {-4} \\ 1 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 2 \\ {-\displaystyle \frac{1}{2}} \end{array}} \right)\\\\\therefore \ \ x=2,\ \ y=-\displaystyle \frac{1}{2}\end{array}$ Therefore, the point of intersection of the two lines is $ \displaystyle \left( {2,-\frac{1}{2}} \right)$.

10.(b)    Construct the table of outcomes for rolling two dice.

Find the probability of an outcome in which the score on the first die is less than that on the second die.

Find also the probability that the score on first die is prime and the score on the second is even.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ {{\text{2}}^{{\text{nd}}}}\ \text{die}\\ \ {{\text{1}}^{{\text{st}}}}\ \text{die}\ \ \ \begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & {(1,1)} & {(1,2)} & {(1,3)} & {(1,4)} & {(1,5)} & {(1,6)} \\ \hline 2 & {(2,1)} & {(2,2)} & {(2,3)} & {(2,4)} & {(2,5)} & {(2,6)} \\ \hline 3 & {(3,1)} & {(3,2)} & {(3,3)} & {(3,4)} & {(3,5)} & {(3,6)} \\ \hline 4 & {(4,1)} & {(4,2)} & {(4,3)} & {(4,4)} & {(4,5)} & {(4,6)} \\ \hline 5 & {(5,1)} & {(5,2)} & {(5,3)} & {(5,4)} & {(5,5)} & {(5,6)} \\ \hline 6 & {(6,1)} & {(6,2)} & {(6,3)} & {(6,4)} & {(6,5)} & {(6,6)} \\ \hline\end{array}\end{array}$ $ \displaystyle \therefore \ \ \ $ Number of possible outcomes = $ \displaystyle 36$        Set of favourable outcomes for the score in which the score on the first die is less than that on the second die = $ \displaystyle \begin{array}{l}\{(1,2),(1,3),(1,4),(1,5),(1,6),\\\ (2,3),(2,4),(2,5),(2,6),(3,4),\\\ (3,5),(3,6),(4,5),(4,6),(5,6)\}\end{array}$ $ \displaystyle \therefore \ \ \ $ Number of favourable outcomes = $ \displaystyle 15$ $\displaystyle \therefore \ P(\text{score on }{{1}^{{\text{st}}}}\text{ die } < \text{score on }{{2}^{{\text{nd}}}}\text{ die) }=\displaystyle \frac{{15}}{{36}}=\displaystyle \frac{5}{{12}}$        Set of favourable outcomes for the score in which the score on first die is prime and the score on the second is even =$ \displaystyle \begin{array}{l}\{(2,2),(2,4),(2,6),(3,2),(3,4),\\\ (3,6),(5,2),(5,4),(5,6)\}\end{array}$ $ \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$

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \text{In}\ \Delta PTQ,\\\\\ \ \ \alpha =\gamma ~+\angle RPT\\\\\ \ \ \beta =\angle R~+\angle RTS\\\\\ \ \ \text{In}\ \odot T,TQ=TS\ \left( {\text{radii}} \right)\\\\\therefore \ \alpha =\ \beta \\\\\therefore \ \gamma ~+\angle RPT=\ \angle R~+\angle RTS\\\\\ \ \ \text{Since}\ \gamma =\angle R,\ \ (\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segments)}\\\\\ \ \angle RTS=\angle RPT\end{array}$

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)$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ AP=3PC\\\\\ \ \ BR=2RP\\\\\ \ \ QR\parallel AC\\\\\ \ \ \alpha (\Delta BPA)=36\ \text{c}{{\text{m}}^{2}}\\\\\ \ \ \text{Since }\Delta BPA\ \operatorname{and}\ \Delta BPC\ \\\ \ \ \text{have the same altitude,}\\\\\,\ \ \displaystyle \frac{{\alpha \left( {\Delta BPC} \right)}}{{\alpha \left( {\Delta BPA} \right)}}=\displaystyle \frac{{PC}}{{AP}}\\\\\therefore \displaystyle \frac{{\alpha \left( {\Delta BPC} \right)}}{{36}}=\displaystyle \frac{{PC}}{{3PC}}\\\\\therefore \alpha \left( {\Delta BPC} \right)=12\ \text{c}{{\text{m}}^{2}}\\\\\ \ \ \text{Since }QR\parallel AC,\\\\\ \ \ \Delta BRQ\sim \Delta BPA\\\\\therefore \displaystyle \frac{{\alpha \left( {\Delta BRQ} \right)}}{{\alpha \left( {\Delta BPA} \right)}}=\displaystyle \frac{{B{{R}^{2}}}}{{B{{P}^{2}}}}\\\\\therefore \displaystyle \frac{{\alpha \left( {\Delta BRQ} \right)}}{{\alpha \left( {\Delta BPA} \right)}}=\displaystyle \frac{{B{{R}^{2}}}}{{{{{(BR+RP)}}^{2}}}}\\\\\therefore \displaystyle \frac{{\alpha \left( {\Delta BRQ} \right)}}{{\alpha \left( {\Delta BPA} \right)}}=\displaystyle \frac{{{{{\left( {2RP} \right)}}^{2}}}}{{{{{(2RP+RP)}}^{2}}}}\\\ \ \ \left[ {\because BR=2RP} \right]\\\\\therefore \displaystyle \frac{{\alpha \left( {\Delta BRQ} \right)}}{{36}}=\displaystyle \frac{4}{9}\\\\\therefore \alpha \left( {\Delta BRQ} \right)=16\ \text{c}{{\text{m}}^{2}}\ \ \text{ }\end{array}$

12.(a)    Prove that the quadrilateral formed by producing the bisectors of the interior angles of any quadrilateral is cyclic.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\text{Given}\ :\ \text{Quadrilateral }ABCD,\\\ \ \ \ \ \ \ \ \ \ \ AP,BP,CR\ \operatorname{and}\ DR\\\ \ \ \ \ \ \ \ \ \ \ \text{are interior angle bisectors}\\\ \ \ \ \ \ \ \ \ \ \ \text{of respective vertices}\text{.}\\\\\text{To Prove : }PQRS\ \text{is cyclic}\text{.}\\\\\text{Proof}\ :\ \ \ \ AP,BP,CR\ \operatorname{and}\ DR\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{are interior angle bisectors}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{of respective vertices}\text{.}\\\\\ \ \ \ \ \ \ \ \ \therefore \ \ \ {{\alpha }_{1}}={{\alpha }_{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\beta }_{1}}={{\beta }_{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\gamma }_{1}}={{\gamma }_{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\delta }_{1}}={{\delta }_{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ABCD\ \text{is a}\ \text{quadrilateral}\text{.}\\\\\ \ \ \ \ \ \ \ \ \therefore \ \ \ {{\alpha }_{1}}+{{\alpha }_{2}}+\ {{\beta }_{1}}+{{\beta }_{2}}+{{\gamma }_{1}}+{{\gamma }_{2}}+\ {{\delta }_{1}}+{{\delta }_{2}}=360{}^\circ \\\\\ \ \ \ \ \ \ \ \ \therefore \ \ \ 2{{\alpha }_{1}}+2{{\beta }_{1}}+2{{\gamma }_{1}}+2{{\delta }_{1}}=360{}^\circ \\\\\ \ \ \ \ \ \ \ \ \therefore \ \ \ {{\alpha }_{1}}+{{\beta }_{1}}+{{\gamma }_{1}}+{{\delta }_{1}}=180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{In}\ \Delta ABP,\angle P+{{\alpha }_{1}}+{{\beta }_{1}}=180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{In}\ \Delta CRD,\angle R+{{\gamma }_{1}}+{{\delta }_{1}}=180{}^\circ \\\\\ \ \ \ \ \ \ \ \therefore \ \ \ \angle P+\angle R+{{\alpha }_{1}}+{{\beta }_{1}}+{{\gamma }_{1}}+{{\delta }_{1}}=360{}^\circ \\\\\ \ \ \ \ \ \ \ \therefore \ \ \ \angle P+\angle R+180{}^\circ =360{}^\circ \\\\\ \ \ \ \ \ \ \ \therefore \ \ \ \angle P+\angle R=180{}^\circ \\\\\ \ \ \ \ \ \ \ \therefore \ \ \ PQRS\ \text{is cyclic}\text{.}\ \ \ \ \ \ \ \end{array}$

12.(b)    If $ \displaystyle \alpha +\beta +\gamma =180{}^\circ $, prove that $ \displaystyle \tan \frac{\alpha }{2}\tan \frac{\beta }{2}+\tan \frac{\beta }{2}\tan \frac{\gamma }{2}+\tan \frac{\alpha }{2}\tan \frac{\gamma }{2}=1$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \alpha +\beta +\gamma =180{}^\circ \\\\\therefore \ \ \ \displaystyle \frac{{\alpha +\beta +\gamma }}{2}=90{}^\circ \\\\\therefore \ \ \ \displaystyle \frac{\alpha }{2}+\displaystyle \frac{\beta }{2}+\displaystyle \frac{\gamma }{2}=90{}^\circ \\\\\therefore \ \ \ \displaystyle \frac{\alpha }{2}+\displaystyle \frac{\beta }{2}=90{}^\circ -\displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \tan \left( {\ \displaystyle \frac{\alpha }{2}+\displaystyle \frac{\beta }{2}} \right)=\tan \left( {90{}^\circ -\displaystyle \frac{\gamma }{2}} \right)\\\\\therefore \ \ \displaystyle \frac{{\tan \left( {\displaystyle \frac{\alpha }{2}} \right)+\tan \left( {\displaystyle \frac{\beta }{2}} \right)}}{{1-\tan \left( {\displaystyle \frac{\alpha }{2}} \right)\tan \left( {\displaystyle \frac{\beta }{2}} \right)}}=\cot \displaystyle \frac{\gamma }{2}\\\\\therefore \ \ \displaystyle \frac{{\tan \left( {\displaystyle \frac{\alpha }{2}} \right)+\tan \left( {\displaystyle \frac{\beta }{2}} \right)}}{{1-\tan \left( {\displaystyle \frac{\alpha }{2}} \right)\tan \left( {\displaystyle \frac{\beta }{2}} \right)}}=\displaystyle \frac{1}{{\tan \displaystyle \frac{\gamma }{2}}}\\\\\therefore \ \ \tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\gamma }{2}=1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}\\\\\therefore \ \ \ \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\gamma }{2}=1\end{array}$ ========================================= $ \displaystyle \alpha=0{}^\circ,\beta=0{}^\circ, \gamma=180{}^\circ$ $ \displaystyle \alpha=0{}^\circ,\beta=180{}^\circ, \gamma=0{}^\circ$ $ \displaystyle \alpha=180{}^\circ,\beta=0{}^\circ, \gamma=0{}^\circ$ ျဖစ္ရပ္အတြက္ အဓိပၸာယ္ မသတ္မွတ္ႏိုင္ပါ။ Credit : Dr. Aung Kyaw                  ဆရာ ေသာ္ဇင္ထြန္း

13.(a)    In $ \displaystyle \Delta ABC$, if $ \displaystyle \angle B=\angle A+15{}^\circ $, $ \displaystyle \angle C=\angle B+15{}^\circ $ and $ \displaystyle BC=6$, find $ \displaystyle AC$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \text{In}\ \Delta ABC,\ BC=6\\\\\ \ \ \angle B=\angle A+15{}^\circ \\\\\ \ \ \angle C=\angle B+15{}^\circ \\\\\therefore \ \ \angle C=\angle A+15{}^\circ +15{}^\circ \\\\\ \ \ \ \ \ \ \ \ =\angle A+30{}^\circ \\\\\ \ \ \text{Since}\,\angle A+\angle B+\angle C=180{}^\circ ,\\\\\ \ \ \angle A+\angle A+15{}^\circ +\angle A+30{}^\circ =180{}^\circ \\\\\therefore \ 3\angle A=135{}^\circ \\\\\therefore \ \angle A=45{}^\circ \\\\\therefore \ \angle B=60{}^\circ \ \operatorname{and}\ \angle C=75{}^\circ \\\\\ \ \ \text{By the law of sines,}\\\\\ \ \ \displaystyle \frac{{AC}}{{\sin B}}=\displaystyle \frac{{BC}}{{\sin A}}\\\\\therefore \ \ AC=\displaystyle \frac{{BC\sin B}}{{\sin A}}\\\\\therefore \ \ AC=\displaystyle \frac{{BC\sin 60{}^\circ }}{{\sin 45{}^\circ }}\\\\\therefore \ \ AC=6\times \displaystyle \frac{{\sqrt{3}}}{2}\times \sqrt{2}=3\sqrt{6}\end{array}$

13.(b)    If $ \displaystyle y = \ln (\sin^3 2x)$, then prove that If $ \displaystyle 3(\frac{{d}^{2}y}{d{x}^{2}}) + (\frac{dy}{dx})^2+36=0$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ y=\ln ({{\sin }^{3}}2x)\ \ \\\ \\\ \ \ \ \ =\ln {{(\sin 2x)}^{3}}\\\\\ \ \ \ \ =3\ln (\sin 2x)\end{array}$ $ \displaystyle \ \ \ \frac{{dy}}{{dx}}=\frac{3}{{\sin 2x}}\cdot \frac{d}{{dx}}(\sin 2x)$ $ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{3}{{\sin 2x}}\cdot \cos 2x\cdot \frac{d}{{dx}}(2x)$ $ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{6\cos 2x}}{{\sin 2x}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ =6\cot 2x$ $ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(-{{\operatorname{cosec}}^{2}}2x)\frac{d}{{dx}}(2x)$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =-12{{\operatorname{cosec}}^{2}}2x$ $ \displaystyle \therefore 3(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}})+{{(\frac{{dy}}{{dx}})}^{2}}+36$ $ \displaystyle \begin{array}{l}=3(-12{{\operatorname{cosec}}^{2}}2x)+{{(6\cot 2x)}^{2}}+36\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\cot }^{2}}2x+1)\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\operatorname{cosec}}^{2}}2x)\ \ \ \ \left[ {1+{{{\cot }}^{2}}2x={{{\operatorname{cosec}}}^{2}}2x} \right]\\\\=0\end{array}$

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}}$.

(5 marks)

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Let }O\text{ be the origin}\text{.}\\\\\therefore \ \ \ \ \overrightarrow{{AB}}+\ \overrightarrow{{CB}}+\ \overrightarrow{{AD}}+\ \overrightarrow{{CD}}\\\\\ \ \ =\ \overrightarrow{{OB}}-\overrightarrow{{OA}}+\ \overrightarrow{{OB}}-\overrightarrow{{OC}}+\ \overrightarrow{{OD}}-\overrightarrow{{OA}}+\ \overrightarrow{{OD}}-\overrightarrow{{OC}}\\\\\ \ \ =2\left( {\overrightarrow{{OB}}+\overrightarrow{{OD}}-\overrightarrow{{OA}}-\overrightarrow{{OC}}} \right)\ \\\\\ \ \ \ \ \ \text{Since }M\text{ and }N\text{ are the midpoints of }AC\text{ and }BD\text{.}\\\\\ \ \ \ \ \ \overrightarrow{{OM}}=\displaystyle \frac{1}{2}\ \left( {\overrightarrow{{OA}}+\ \overrightarrow{{OC}}} \right)\\\\\ \ \ \ \ \ \overrightarrow{{ON}}=\displaystyle \frac{1}{2}\ \left( {\overrightarrow{{OB}}+\ \overrightarrow{{OD}}} \right)\\\\\therefore \ \ \ \ \ \overrightarrow{{MN}}=\ \overrightarrow{{ON}}-\overrightarrow{{OM}}\\\\\,\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{2}\left( {\overrightarrow{{OB}}+\overrightarrow{{OD}}-\overrightarrow{{OA}}-\overrightarrow{{OC}}} \right)\\\text{ }\\\therefore \ \ \ \ \ 4\overrightarrow{{MN}}=2\left( {\overrightarrow{{OB}}+\overrightarrow{{OD}}-\overrightarrow{{OA}}-\overrightarrow{{OC}}} \right)\\\\\therefore \ \ \ \ \overrightarrow{{AB}}+\ \overrightarrow{{CB}}+\ \overrightarrow{{AD}}+\ \overrightarrow{{CD}}=4\overrightarrow{{MN}}\end{array}$

14.(b)    Find the normals to the curve $ \displaystyle xy+2x-y=0$ that are parallel to the line $ \displaystyle 2x + y = 0$.

(5 marks)

Show/Hide Solution

ပံုဆြဲရန္မလိုပါ $\displaystyle \begin{array}{l}\ \ \ \ \ \ \text{Curve}\ \text{: }xy+2x-y=0\\\\\ \ \ \ \ \ \text{Differentiate with respect to }x.\\\\\ \ \ \ \ \ x\displaystyle \frac{{dy}}{{dx}}+y+2-\displaystyle \frac{{dy}}{{dx}}=0\\\\\therefore \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=\displaystyle \frac{{2+y}}{{1-x}}\\\\\therefore \ \ \ \ \text{gradient of tangent to the curve}=\displaystyle \frac{{2+y}}{{1-x}}\\\\\therefore \ \ \ \ \text{gradient of normal to the curve}=-\displaystyle \frac{1}{{\displaystyle \frac{{dy}}{{dx}}}}=\displaystyle \frac{{x-1}}{{y+2}}\\\ \ \ \ \ \ \ \\\ \ \ \ \ \ \text{Line}\ \text{: }2x+y=0\Rightarrow y=-2x\\\\\therefore \ \ \ \ \text{gradient of line = }-2\ \\\\\ \ \ \ \ \ \text{Since normals }\parallel \ \text{given line}\\\\\ \ \ \ \ \ \text{gradient of normal}=\text{gradient of line}\text{.}\\\\\therefore \ \ \ \ \displaystyle \frac{{x-1}}{{y+2}}=-2\\\\\therefore \ \ \ \ x-1=-2y-4\\\\\therefore \ \ \ \ x=-2y-3\\\\\ \ \ \ \ \ \text{Substituting }x=-2y-3\\\ \ \ \ \ \text{in curve equation,}\\\\\ \ \ \ \ \ (-2y-3)y+2(-2y-3)-y=0\\\ \\\therefore \ \ \ \ {{y}^{2}}+4y+3=0\\\\\therefore \ \ \ \ (y+3)(y+1)=0\\\\\therefore \ \ \ \ y=-3\ (\text{or})\ y=-1\\\\\therefore \ \ \ \ \text{When }y=-3,x=3\\\\\ \ \ \ \ \text{When }y=-1,x=-1\\\\\ \ \ \ \ \text{The equation of normal line at}\\\ \ \ \ \ \ \left( {-1,-1} \right)\ \text{is}\\\\\ \ \ \ \ y+1=-2\left( {x+1} \right)\\\\\therefore \ \ \ \ 2x+y+3=0\\\\\text{ }\ \ \ \ \ \text{The equation of normal line at}\\\ \ \ \ \ \ \left( {3,-3} \right)\ \text{is}\\\\\ \ \ \ \ y+3=-2\left( {x-3} \right)\\\\\therefore \ \ \ \ 2x+y-3=0\end{array}$