1.        A manufacturer wants to design an open box having a square base and a surface area of $ \displaystyle 108$ square inches, as shown in figure . What dimensions will produce a box with maximum volume?
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$ \displaystyle \begin{array}{l}\text{ Total}\ \text{surface}\ \text{area =108 i}{{\text{n}}^{\text{2}}}\\\\\text{ (area of base)+(area of four sides) = 108}\\\\\ \ \ \ {{x}^{2}}+4xh=108\end{array}$

$ \displaystyle \ \ \ \ h=\frac{{108-{{x}^{2}}}}{{4x}}$

$ \displaystyle \begin{array}{l}\text{ Let the volume of the box be }V.\\\\\therefore \ \ V={{x}^{2}}h\end{array}$

$ \displaystyle \therefore \ \ V={{x}^{2}}\left( {\frac{{108-{{x}^{2}}}}{{4x}}} \right)=27x-\frac{1}{4}{{x}^{3}}$

$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=27-\frac{3}{4}{{x}^{2}}$

$ \displaystyle \ \ \ \ \frac{{dV}}{{dx}}=0\ \text{when}$

$ \displaystyle \ \ \ \ 27-\frac{3}{4}{{x}^{2}}=0$

$ \displaystyle \therefore \ \ {{x}^{2}}=36\Rightarrow x=6\,\ \ \left[ {\because x>0} \right]$

$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=-\frac{3}{2}x$

$ \displaystyle \therefore \ \ {{\left. {\frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=-\frac{3}{2}(6)=-9<0$

$ \displaystyle \therefore \ \ V\ \text{is maximum when }x=6\ \text{cm}\text{.}$

$ \displaystyle \therefore \ \ h=\frac{{108-{{6}^{2}}}}{{24}}=3\ \text{cm}$

$ \displaystyle \begin{array}{l}\therefore \ \ \text{The box has maximum volume when }\\\ \ \ \ x=6\ \text{cm and }h=3\ \text{cm}.\end{array}$


2.         Which points on the curve $ \displaystyle y = 4 − x^2$ are closest to the point $ \displaystyle (0, 2)?$
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$ \displaystyle \begin{array}{l}\text{ Let (}x,y)\ \text{be the required point on the curve}\text{.}\\\\\text{ Curve: }y=4-{{x}^{2}}\\\\\ \ \ \ \text{Let the distance between (0,2) and (}x,y)\ \text{be}\ s.\\\\\therefore \ \ \ s=\sqrt{{{{{(x-0)}}^{2}}+{{{(y-2)}}^{2}}}}\ \left[ {\text{using distance formula}} \right]\\\\\therefore \ \ \ s=\sqrt{{{{x}^{2}}+{{{(4-{{x}^{2}}-2)}}^{2}}}}\\\\\therefore \ \ \ s=\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\end{array}$

$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{4{{x}^{3}}-6x}}{{2\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$

$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$

$ \displaystyle \ \ \ \ \ \frac{{ds}}{{dx}}=0\ \text{when}$

$ \displaystyle \ \ \ \ \ \frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}=0$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Since }\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\ne 0,\\\\\ \ \ \ \ x(2{{x}^{2}}-3)=0\end{array}$

$ \displaystyle \therefore \ \ \ x=0\ (\text{or) }x=-\sqrt{{\frac{3}{2}}}\ (\text{or)}\ x=\sqrt{{\frac{3}{2}}}\ $

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}=\frac{{2{{x}^{6}}-9{{x}^{4}}+24{{x}^{2}}-12}}{{{{{\left( {{{x}^{4}}-3{{x}^{2}}+4} \right)}}^{{\frac{3}{2}}}}}}$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=0}}}=-\frac{3}{2}<0$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=-\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$

$ \displaystyle \therefore \ \ s\text{ is maximum volume when }$

$ \displaystyle \ \ \ \ x=-\sqrt{{\frac{3}{2}}}\text{ and }x=-\sqrt{{\frac{3}{2}}}.$

$ \displaystyle \therefore \ \ \text{When }x=-\sqrt{{\frac{3}{2}}},y=\frac{5}{2}.$

$ \displaystyle \ \ \ \ \text{When }x=\sqrt{{\frac{3}{2}}},y=\frac{5}{2}$

$ \displaystyle \therefore \ \ \text{The closest ponits on the curve are}$

$ \displaystyle \ \ \ \ \left( {-\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right)\text{and }\left( {\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right).$


3.        A rectangular page is to contain $ \displaystyle 24$ square inches of print. The margins at the top and bottom of the page are to be $ \displaystyle 1 \frac{1}{2}$ inches, and the margins on the left and right are to be $ \displaystyle 1$ inch (see Figure). What should the dimensions of the page be so that the least amount of paper is used?

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$ \displaystyle \begin{array}{l} \ \ \ \text{Let the length and width of the printed }\\\ \ \ \text{area be }x\ \text{and }y\text{ respectively}\text{.}\\\\\text{ By the problem, }xy=24\end{array}$

$ \displaystyle \therefore \ \ y=\frac{{24}}{x}$

$ \displaystyle \ \ \ \ \text{Let the area of the paper needed be }A.$

$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(y+2)} \end{array}$

$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(\frac{{24}}{x}+2)} \end{array}=30+2x+\frac{{72}}{x}$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2-\frac{{72}}{{{{x}^{2}}}}$

$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when }2-\frac{{72}}{{{{x}^{2}}}}=0.$

$ \displaystyle \begin{array}{l}\therefore \ \ \ {{x}^{2}}=36\\\\\ \ \ \ \ x=6\ \ \ (\because x>0)\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{144}}{{{{x}^{3}}}}$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=\frac{{144}}{{{{6}^{3}}}}>0$

$ \displaystyle \therefore \ \ A\text{ is maximum volume when }x=6$

$ \displaystyle \therefore \ \ \text{ }y=\frac{{24}}{6}=4\ \text{cm}.$

$ \displaystyle \therefore \ \ \text{The dimenssions of the paper}=4\ \text{cm}\times 6\text{cm}.$


4.         Two posts, one $ \displaystyle 12$ feet high and the other $ \displaystyle 28$ feet high, stand $ \displaystyle 30$ feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?


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$ \displaystyle \begin{array}{l} \ \ \ \text{Let the length of wire required to attach 12-foot post be }y\ \ \ \\\ \ \ \ \text{and that required to attach 28-foot post be }z\text{.}\\\\\ \ \ \text{Let the distance of the stack from 12-foot post be }x.\\\\\therefore \ \ {{x}^{2}}+{{12}^{2}}={{y}^{2}}\Rightarrow y=\sqrt{{144+{{x}^{2}}}}\\\\\ \ \ \ {{(30-x)}^{2}}+{{28}^{2}}={{z}^{2}}\Rightarrow z=\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\ \ \ \ \text{Let the total length of wire required be }W.\end{array}$

$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {W=x+y} \end{array}=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.$

$ \displaystyle \therefore \ \ \ \frac{{dW}}{{dx}}=\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$

$ \displaystyle \ \ \ \ \ \frac{{dW}}{{dx}}=0\ \text{when }\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}=0.$

$ \displaystyle \therefore \ \ \ \frac{x}{{\sqrt{{144+{{x}^{2}}}}}}=\frac{{30-x}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ x\sqrt{{{{x}^{2}}-60x+1684}}=(30-x)\sqrt{{{{x}^{2}}+144}}\\\\\ \ \ \ \ {{x}^{2}}({{x}^{2}}-60x+1684)={{(30-x)}^{2}}(x2+144)\\\\\ \ \ \ \ {{x}^{4}}-60{{x}^{3}}+1684{{x}^{2}}={{x}^{4}}-60{{x}^{3}}+1044{{x}^{2}}-8640x+129,600\\\\\ \ \ \ \ 640{{x}^{2}}+8640x-129,600=0\\\\\ \ \ \ \ 2{{x}^{2}}+27x-405=0\\\\\ \ \ \ \ (x-9)(2x+45)=0\\\\\ \ \ \ \ \text{Since }x>0,\ x=9\\\\\ \ \ \ \ W=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\therefore \ \ \ \text{When }x=0,\\\\\ \ \ \ \ W=\sqrt{{144}}+\sqrt{{1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =12+41.04=53.04\\\\\ \ \ \ \ \text{When }x=9,\\\\\ \ \ \ \ W=\sqrt{{144+81}}+\sqrt{{81-540+1684}}\\\\\ \ \ \ \ \ \ \ \ =15+35=50\\\\\ \ \ \ \ \text{When }x=30,\\\\\ \ \ \ \ W=\sqrt{{144+900}}+\sqrt{{900-1800+1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =32.31+28=60.31\\\\\therefore \ \ \ W\text{ is maximum when }x=9.\\\\\therefore \ \ \ \text{The}\ \text{wires}\ \text{should}\ \text{be}\ \text{staked}\ \text{at}\ \text{9}\ \text{feet}\ \text{from}\ \text{the}\ \text{12-foot}\ \text{pole}.\end{array}$


5.       The sum of the perimeters of a circle and square is $ \displaystyle k,$ where $ \displaystyle k$ is some constant. Prove that the sum of their areas is least, when the side of the square is double the radius of the circle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the radius of the circle be }r\ \text{and the length }\\\ \ \ \ \ \text{of each side of the square be }x.\\\\\ \ \ \ \ \text{Sum of perimeters}=k\ \ \left[ {\text{given}} \right]\\\\\therefore \ \ \ \ 4x+\text{ }2\pi r=k\end{array}$

$ \displaystyle \therefore \ \ \ r=~\frac{{k-4x}}{{2\pi }}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the total are be }A.\\\\\therefore \ \ \ A={{x}^{2}}+\pi {{r}^{2}}\end{array}$

$ \displaystyle \therefore \ \ \ A={{x}^{2}}+\pi {{\left( {\frac{{k-4x}}{{2\pi }}} \right)}^{2}}$

$ \displaystyle \therefore \ \ \ A={{x}^{2}}+\frac{{{{{(k-4x)}}^{2}}}}{{4\pi }}$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2x+\frac{{2(-4)(k-4x)}}{{4\pi }}$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2\left( {x+\frac{{4x-k}}{\pi }} \right)$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=\frac{2}{\pi }\left[ {(\pi +4)x-k} \right]$

$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when}\ \frac{2}{\pi }\left[ {(\pi +4)x-k} \right]=0$

$ \displaystyle \therefore \ \ \ x=\frac{k}{{\pi +4}}$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{2(\pi +4)}}{\pi }>0$

$ \displaystyle \therefore \ \ \ A\ \text{is minimum when }x=\frac{k}{{\pi +4}}.$

$ \displaystyle \therefore \ \ \ r=\frac{1}{{2\pi }}[k-\frac{{4k}}{{\pi +4}}]$

$ \displaystyle \ \ \ \ \ \ \ \ =\frac{{\pi k+4k-4k}}{{2\pi (\pi +4)}}$

$ \displaystyle \ \ \ \ \ \ \ \ =~\frac{1}{2}\left( {\frac{k}{{\pi +4}}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ =~\frac{1}{2}x$

$ \displaystyle \begin{array}{l}\therefore \ \ x=2r\\\\\ \ \ \text{Hence the sum of their areas is least},\\\ \ \ \text{when the side of the square is double }\\\ \ \ \text{the radius of the circle}.\end{array}$


6.        Find two positive numbers that the sum of the first number cubed and the second number is $ \displaystyle 500$ and the product is a maximum.

7.        An open box of maximum volume is to be made from a square piece of material, $ \displaystyle 24$ inches on a side, by cutting equal squares from the corners and turning up the sides (see figure). How much material should be cut?


8.        A Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window (see figure). Find the dimensions of a Norman window of maximum area when the total perimeter is $ \displaystyle 16$ feet.

9.        Find the area of the largest isosceles triangle that can be inscribed in a circle of radius $ \displaystyle 6$ (see figure).
(a) Solve by writing the area as a function of $ \displaystyle h$.

(b) Solve by writing the area as a function of $ \displaystyle α$.

(c) Identify the type of triangle of maximum area.


10.     Twenty feet of wire is to be used to form two figures: equilateral triangle and square, how much wire should be used for each figure so that the total enclosed area is maximum?

11.     A right triangle is formed in the first quadrant by the $ \displaystyle x$- and $ \displaystyle y$-axes and a line through the point $ \displaystyle (1, 2)$ (see figure).

(a) Write the length $ \displaystyle L$ of the hypotenuse as a function of $ \displaystyle x.$

(b) Find the vertices of the triangle such that its area is a minimum.


တြက္ၾကည့္ပါ။ အေျဖမ်ားကို ေနာက္ရက္တြင္ တင္ေပး ပါမည္။

$ \displaystyle \odot O$ ဟာ unit circle (radius = $ \displaystyle 1$ unit) တစ္ခုျဖစ္ပါတယ္။ $ \displaystyle x$ radian ပမာဏရွိတဲ့ $\displaystyle \angle AOB$ ကို ဗဟိုမွာ တည္ေဆာက္ပါမယ္။ ဒါဆိုရင္ $ \displaystyle A$ ကေန $ \displaystyle OB$ ေပၚကို ေထာင့္မတ္က်တဲ့မ်ဥ္း $ \displaystyle AC$ အလ်ားက $ \displaystyle \sin x$ ျဖစ္သြားမယ္။ ဘာ့ေၾကာင့္လဲ ... $\displaystyle \frac{{AC}}{{OA}}=\sin x$ ျဖစ္လို႔ပါ။

အမွတ္ $ \displaystyle B$ မွာ $ \displaystyle \odot O$ ကို tangent ျဖစ္တဲ့ မ်ဥ္းတစ္ေၾကာင္းဆြဲပါမယ္။ ၎ tangent မ်ဥ္းက $ \displaystyle OA$ ဆက္ဆြဲမ်ဥ္းကို $ \displaystyle D$ ျဖတ္ပါမယ္ $ \displaystyle OB$ က radius ျဖစ္တာေၾကာင့္ $ \displaystyle DA\bot OB$ ျဖစ္ပါတယ္။

ေထာင့္မွန္ ႀတိဂံ $ \displaystyle \vartriangle OBD$ မွာ $ \displaystyle \frac{{OD}}{{OB}}=\tan x$ ျဖစ္တာေၾကာင့္ $ \displaystyle OD = \tan x$ ျဖစ္ပါတယ္။

တစ္ခါ ေထာင့္မွန္ ႀတိဂံ $ \displaystyle \vartriangle AOC$ မွာ $ \displaystyle \frac{{OC}}{{OA}}=\cos x$ ျဖစ္တာေၾကာင့္ $ \displaystyle OC = \cos x$ ျဖစ္ပါတယ္။

ဒါဆိုရင္ အခု $ \displaystyle \vartriangle AOB$ ရဲ့ ဧရိယာကို ရွာၾကည့္မယ္။


$ \displaystyle \ \ \ \ \text{Area}\ \text{of } \vartriangle AOB=\frac{1}{2}\times OB\times AC$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\times 1\times \sin x$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\sin x$

Sector $ \displaystyle AOB$ ရဲ့ ဧရိယာကို ရွာၾကည့္ဦးမယ္။


$ \displaystyle \ \ \ \ \text{Area}\ \text{of sector AOB}=\frac{1}{2}\times O{{B}^{2}}\times x$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\times 1\times x$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}x$

အခုတစ္ခါ $ \displaystyle \vartriangle OBD$ ရဲ့ ဧရိယာကို ရွာၾကည့္ဦးမယ္။


$ \displaystyle \ \ \ \ \text{Area}\ \text{of }\vartriangle OBD=\frac{1}{2}\times OB\times BD$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\times 1\times \tan x$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\tan x$

ပံုမွာ ျမင္ရတဲ့ အတိုင္း...

$ \displaystyle \ \ \ \ \text{Area}\ \text{of }\vartriangle OBD>\text{Area}\ \text{of sector AOB}>\text{Area}\ \text{of }\vartriangle AOB$

ျဖစ္တယ္ဆိုတာ သိသာေစပါတယ္။ ဒါေၾကာင့္ ...

$ \displaystyle \ \ \ \ \frac{1}{2}\tan x>\frac{1}{2}x>\ \frac{1}{2}\tan x$ ျဖစ္မွာေပါ့။ ဒါဆိုရင္ ...

$ \displaystyle \ \ \ \ \tan x>x>\sin x$

$ \displaystyle \ \ \ \ \frac{{\sin x}}{{\cos x}}>x>\sin x$ ျဖစ္မွာေပါ့။

ဒီေနရာမွာ $ \displaystyle \sin x, \cos x, \tan x$ တို႔ဟာ အနားေတြရဲ့ အလ်ားေတြ ျဖစ္လို႔ အေပါင္းကိန္းေတြ ျဖစ္တယ္လို႔ သိရပါမယ္။ မညီမွ်ျခင္း တစ္ခုလံုးကို $ \displaystyle \sin x$ နဲ႔ စားလိုက္ရင္ ...

$ \displaystyle \ \ \ \ \frac{1}{{\cos x}}>\frac{x}{{\sin x}}>1$ ျဖစ္ပါတယ္။

အေပါင္းကိန္းေတြရဲ့ မညီမွ်ျခင္း တစ္ခုမွာ တန္ဖိုးေတြကို ေျပာင္းျပန္လွန္ရင္ လကၡဏာက ဆန္႔က်င္ဘက္ကို ေျပာင္းရပါတယ္။ ဒါ့ေၾကာင့္ ...

$ \displaystyle \ \ \ \ \cos x<\frac{{\sin x}}{x}<1$ ဆိုတာ ဖိုး $ \displaystyle 0$ မဟုတ္တဲ့ မည့္သည့္ေထာင့္တန္ဖိုး $ \displaystyle x$ အတြက္ မဆို မွန္ကန္တဲ့ မညီမွ်ျခင္း တစ္ခုျဖစ္ပါတယ္။

$ \displaystyle x=0$ ျဖစ္သြားရင္ေတာ့ $ \displaystyle \sin x= \sin 0=0, \cos x= \cos 0=1 $ ျဖစ္ေပမယ့္ $ \displaystyle \frac{{\sin x}}{x}= \frac{0}{0}$ ဆိုေတာ့ မေရမရာ ျဖစ္သြားပါေတာ့တယ္။

ဒါကို graphically ေျဖရွင္းႏိုင္ပါတယ္။ အေပၚက geogebra applet မွာ $ \displaystyle x$ တန္ဖိုးကို ေလွ်ာ့ၾကည္ပါ့။ $ \displaystyle x$ က $ \displaystyle 0$ အနားကို ေရာက္လာေလေလ $ \displaystyle \sin x, x$ နဲ႔ $ \displaystyle \tan x$ တို႔ဟာ တစ္ထပ္ထဲ နီးပါ ျဖစ္လာပါေတာ့တယ္။ တနည္းဆိုေသာ္ $ \displaystyle \sin x\approx x\approx \tan x$ ျဖစ္လာတာေပါ့။ ဒါေၾကာင့္ ။ $ \displaystyle x$ က $ \displaystyle 0$ အနားကို ေရာက္လာတဲ့အခါ $ \displaystyle \frac{{\sin x}}{x}\approx 1$ ျဖစ္သြားပါတယ္။ ဒါကို Limit Notation နဲ႔ $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\sin x}}{x}=1$ လို႔ ေရးႏိုင္ပါတယ္။ အခ်ဳပ္ဆိုရရင္...

 (1) $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\sin x} \right)=0$
 (2) $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\cos x} \right)=1$
 (3) $ \displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\left( {\frac{{\sin x}}{x}} \right)=1$



$ \displaystyle f(x)=\frac{{{{x}^{3}}-1}}{{x-1}}$ ဆိုတဲ့ function တစ္ခုကို စဥ္စားၾကည့္ရေအာင္ $ \displaystyle f(x)$ ဟာ ထည့္လိုက္တဲ့ $ \displaystyle x$ တန္ဖိုးတိုင္းအတြက္ output (image) တစ္ခု ထုတ္ေပးမွာေပါ့။ 

အေပၚက slider ကို ဆြဲၾကည့္ပါ။ 

ဒါေပမယ့္ $ \displaystyle x=1$ ျဖစ္သြားရင္ေတာ့ $ \displaystyle f(x)$ ဟာ ဘာလုပ္ရမွန္းမသိေတာ့ပါဘူး။ ဘာလို႔လဲ ဆိုေတာ့ $ \displaystyle f(1)=\frac{{1-1}}{{1-1}}=\frac{0}{0}$ ဆိုတာကို $ \displaystyle f(x)$ က နားမလည္ေတာ့ဘူးေလ။ 

ဘယ္ကိန္းကို မဆို ာ့ $ \displaystyle 0$ နဲ႔ စားျခင္းက အဓိပါယ္မရွိလို႔ပါပဲ။ ဒါဆိုရင္ $ \displaystyle x$ က $ \displaystyle 1$ ျဖစ္လို႔ မရဘူးေပါ့။ $ \displaystyle x=1$ မျဖစ္ရဘူးဆိုရင္ ဘယ္ေလာက္အထိ ျဖစ္ခြင့္ ရွိပါသလဲ။ 

$ \displaystyle x=1$ မျဖစ္ဖို႔ပဲ လိုတာ။ ။ $ \displaystyle 1$ မဟုတ္တဲ့ ဘယ္ကိန္းစစ္မဆို ျဖစ္လို႔ ရပါတယ္။ 

ကၽြန္ေတ့ာ္ ေက်ာင္းသားေတြကို သင္လိုက္သလို ေျပာရရင္ ။ $ \displaystyle 1$ ကို ထိလို႔ မရဘူး ထိရင္ ေရွာ့ပဲ။ မထိနဲ႔လို႔ ေျပာတာ.. မကပ္နဲ႔ လို႔ မေျပာတဲ့ အတြက္ ။ $ \displaystyle 1$ ရဲ့ အနားကို ကပ္ႏိုင္ သေလာက္ထိ ကပ္ၾကည့္မယ္..။ 

တကယ္ေတာ့ ကပ္ႏိုင္သေလာက္ဆိုတာ ဘယ္ေလာက္လဲ တိတိက်က် တန္ဖိုးေတာ့ မရွိဘူးေပါ့။ ေအာက္က ဇယားကိုၾကည့္ပါ။


$ \displaystyle x$ က $ \displaystyle 1$ အနားကို နီးကပ္လာေလေလ ... $ \displaystyle f(x)$ က $ \displaystyle 3$ အနားကို ေရာက္လာေလေလေပါ့။ 

$ \displaystyle x=1$ ဆိုၿပီး တိုက္ရိုက္ထည့္ လိုက္ရင္ေတာ့ ေရွာ့ပဲ...။ စိတ္ရွည္ရွည္နဲ႔ ကပ္ၾကည့္ လိုက္ရင္ေတာ့ ရလဒ္တစ္ခု ထြက္လာတာေပါ့။ 

ဒါကို Calculus မွာ $ \displaystyle f(x)$ ရဲ့ limit လို႔ သတ္မွတ္ၿပီး သေကၤတ အားျဖင့္ $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,f(x)=3$ (The limit of f(x) as x approaches 1 is 3) လို႔ သတ္မွတ္ပါတယ္။ 

လက္ေတြ႔ပုစာၦေျဖရွင္းရာမွာေတာ့ ပုစာၦတိုင္းကို အခုလို ခ်ဥ္းကပ္ေျဖရွင္းဖို႔ မျဖစ္ႏိုင္ေတာ့တဲ့အတြက္ တိုက္ရိုက္အစားသြင္းၾကည့္တဲ့ အခါ $ \displaystyle f(1)=\frac{0}{0}$ (indeterminate form လို႔ ေခၚပါတယ္) ျဖစ္ရင္ discontinuity ကေန continuous condition (အစားသြင္းလို႔ ရမယ့္ အေျခအေနကို) ေျပာင္းလဲ ေျဖရွင္းၿပီးမွ တြက္ၾကရတာေပါ့။ 

အခုလို တြက္ပါမယ္။

$ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}=\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(x-1)({{x}^{2}}+x+1)}}{{x-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,({{x}^{2}}+x+1)\ \ \ \ \ \ \left[ {\text{continuous}\ \text{condition}} \right]$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3\end{array}$

Definition of a limit

If $ \displaystyle f(x)$ becomes arbitrarily close to a single number $ \displaystyle L$ as $ \displaystyle x$ approaches $ \displaystyle c$ from either side, then the limit of $ \displaystyle f(x),$ as $ \displaystyle x$ approaches $ \displaystyle c$, is $ \displaystyle L$. This limit is written as

$ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=L$

Limit ပုစာၦမ်ား တြက္ရမွာ သိထားရမယ့္ properties ေတြကို ေတာ့ 👉 ဒီေနရာမွာ 👈 ဖတ္ၾကည့္ပါ။

နေ့စဉ်လူနေမှု ဘဝတွင် Calculus ကို ဘယ်နေရာတွေမှာ သုံးသလဲ... သိလိုကြပါသည်။ Calculus မပါပဲ ရှုပ်ထွေးသော လုပ်ငန်းစဉ်များကို ဖြေရှင်းရန် မလွယ်ကူပါ...။ Calculus ဖြင့် ရှုပ်ထွေးခက်ခဲသော ပြသနာများကို တိကျစွာ အဖြေထုတ်ပေးနိုင်းပါသည်။ အဆင့်မြင့်နည်းပညာနှင့် Engineering subject များတွင် calculus သည် အဓိက အခန်းကဏ္ဍမှာ ပါဝင်ပါသည်။ 

ထို့ကြောင့် calculus သည် အဆင့်မြင့် နည်းပညာများကို လေ့လာရန် မဖြစ်မနေလိုအပ်သော သင်္ချာ၏ အစိပ်အပိုင်း ဖြစ်ပါသည်။ 

အောက်ပါ ဥပမားများကို ကြည့်လျင် Calculus ဘာကြောင့် လိုအပ်ရသလဲ ဆိုသည်ကို အတိုင်းအတာ တစ်ခုအထိနားလည် သဘောပေါက်လိမ့်မည်ဟု ယုံကြည်မိပါသည်။

 Without Calculus  With Calculus 
 Value of $ \displaystyle f(x)$ when $ \displaystyle x=c$
 
 Limit of Value of $ \displaystyle f(x)$ as $ \displaystyle x$ approaches $ \displaystyle c$

 Slope of a line

Slope of a curve  

 Secant line to a curve

 Tangent line to a curve

 Average rate of change between $ \displaystyle t=a$ and  $ \displaystyle t=b$


Instataneous rate of change between $ \displaystyle t=c$


 Curvature of a circle
 Curvature of a curve
 

 
 Height of a curve when  $ \displaystyle x=c$

Maximum height of a curve on an interval

Tangent plane to a sphere
Tangent plane to a surface

 Direction of a motion a long a line


Direction of a motion a long a curve

 Area of a rectangle

Area under a curve


Work done by a constant force
 
 Work done by a variable force
 
 Centre of a rectangle


Centroid of a region
 Length of a line segment

 Length of an acrc


 Surface area of a cylinder

 Surface area of a solid revolution

 Mass of a solid of a constant density

  Mass of a solid of a variable density

 Volume of a rectangular solid

 Volume of a region under a surface

 Sum of a finite number of a trems

 $ \displaystyle S={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}$

 Sum of an infinite number of a trems

$ \displaystyle S={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...$


1.        Prove that in any triangle $ \displaystyle ABC,$

(i) $ \displaystyle \sin (A+B) = \sin C.$

(ii) $ \displaystyle \cos(A+B) + \cos C = 0.$

(iii) $ \displaystyle \cos \frac{A+B}{2} = \sin \frac{C}{2}.$

(iv) $ \displaystyle \tan \frac{A+B}{2} = \cot \frac{C}{2}.$

Show/Hide Solution
(i) $ \displaystyle \text{Since}\ A+B+C=180{}^\circ ,$

$ \displaystyle \begin{array}{l}\therefore A+B=180{}^\circ -C\\\\\therefore \sin (A+B)=\sin (180{}^\circ -C)\\\\\therefore \sin (A+B)=\sin C\end{array}$

(ii) $ \displaystyle \text{Similarly, }\cos (A+B)=\cos (180{}^\circ -C)$

$ \displaystyle \begin{array}{l}\therefore \cos (A+B)=-\cos C\\\\\therefore \cos (A+B)+\cos C=0\end{array}$

(iii) $ \displaystyle \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\sin \frac{C}{2}$

(iv) $ \displaystyle \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\cot \frac{C}{2}$

2.        In any quadrilateral $ \displaystyle ABCD,$ prove that

(i) $ \displaystyle \sin (A+B) + \sin (C+D)=0.$

(ii) $ \displaystyle \cos (A+B) = \cos (C+D).$

Show/Hide Solution
(i) $ \displaystyle \text{In}\ \text{any}\ \text{quadrilateral}\ ABCD,$

$ \displaystyle \begin{array}{l}\ \ \ \ A+B+C+D=360{}^\circ \\\\\therefore \ \ A+B=360{}^\circ -(C+D)\\\\\therefore \ \ \sin (A+B)=\sin \left[ {360{}^\circ -(C+D)} \right]\\\\\therefore \ \ \sin (A+B)=-\sin (C+D)\\\\\therefore \ \ \sin (A+B)+\sin (C+D)=0\end{array}$

(ii) $ \displaystyle \ \text{Since }A+B+C+D=360{}^\circ ,$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ A+B=360{}^\circ -(C+D)\\\\\therefore \ \ \cos (A+B)=\cos \left[ {360{}^\circ -(C+D)} \right]\\\\\therefore \ \ \sin (A+B)=\cos (C+D)\end{array}$

3.        If $ \displaystyle A, B, C, D$ be the angles of a cyclic quadrilateral, taken in order, prove that

(i) $ \displaystyle \cos A + \cos B + \cos C + \cos D=0.$

(ii) $ \displaystyle \cos (180{}^\circ +A)+\cos (180{}^\circ +B)+\cos (180{}^\circ +C)-\sin (90{}^\circ +D)=0.$

Show/Hide Solution
(i) $ \displaystyle \text{Since }ABCD\ \text{is a cyclic quadrilateral},$

$ \displaystyle \begin{array}{l}\ \ \ \ A+C=180{}^\circ \Rightarrow A=180{}^\circ -C\\\\\ \ \ \ B+D=180{}^\circ \Rightarrow B=180{}^\circ -D\\\\\therefore \ \ \cos A=\cos (180{}^\circ -C)\Rightarrow \cos A=-\cos C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --(1)\\\\\ \ \ \ \text{Similarly,}\cos B=\cos (180{}^\circ -D)\Rightarrow \cos B=-\cos D--(2)\\\\\ \ \ \ (1)+(2)\Rightarrow \cos A+\cos B=-\cos C-\cos D\end{array}$

(ii) $ \displaystyle \ \cos (180{}^\circ +A)+\cos (180{}^\circ +B)+\cos (180{}^\circ +C)-\sin (90{}^\circ +D)$

$ \displaystyle \begin{array}{l}\ \ \ \ =-\cos A-\cos B-\cos C-\cos D\\\\ \ \ \ \ =-(\cos A+\cos B+\cos C+\cos D)\\\\ \ \ \ \ =0\ \ \ \ \ \left[ {\because \text{by}\ (\text{i})} \right]\end{array}$

4.        If $ \displaystyle \tan 35{}^\circ =x,$ prove that $ \displaystyle \frac{{\tan 145{}^\circ -\tan 125{}^\circ }}{{1+\tan 145{}^\circ \tan 125{}^\circ }}=\frac{{1-{{x}^{2}}}}{{2x}}.$

Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \ \tan 35{}^\circ =x\\\\\ \ \ \ \tan 145{}^\circ =\tan (180{}^\circ -35{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\tan 35{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-x\\\\\ \ \ \ \tan 125{}^\circ =\tan (90{}^\circ +35{}^\circ )\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cot 35{}^\circ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{{\tan 35{}^\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{x}$

$ \displaystyle \ \ \ \ \ \ \frac{{\tan 145{}^\circ -\tan 125{}^\circ }}{{1+\tan 145{}^\circ \tan 125{}^\circ }}$

$ \displaystyle \ \ \ \ =\ \ \frac{{-x-\left(\displaystyle {-\frac{1}{x}} \right)}}{{1-x\left(\displaystyle {-\frac{1}{x}} \right)}}$

$ \displaystyle \ \ \ \ =\ \ \frac{{\displaystyle \frac{{1-{{x}^{2}}}}{x}}}{2}$

$ \displaystyle \ \ \ \ =\ \ \frac{{1-{{x}^{2}}}}{{2x}}$

5.        Prove that

(i) $ \displaystyle \sin (A+B)\sin (A-B)={{\sin }^{2}}A-{{\sin }^{2}}B.$

(ii) $ \displaystyle \sin (A+B)\sin (A-B)={{\cos }^{2}}B-{{\cos }^{2}}A.$

(iii) $ \displaystyle \cos (A+B)\cos (A-B)={{\cos }^{2}}A-{{\sin }^{2}}B.$

(iv) $ \displaystyle \cos (A+B)\cos (A-B)={{\cos }^{2}}B-{{\sin }^{2}}A.$

Show/Hide Solution
(i)$ \displaystyle \ \ \ \sin (A+B)\sin (A-B)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ =(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\ \ \ \ \ \ ={{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B\\\\ \ \ \ \ \ ={{\sin }^{2}}A(1-{{\sin }^{2}}B)-(1-{{\sin }^{2}}A){{\sin }^{2}}B\\\\ \ \ \ \ \ ={{\sin }^{2}}A-{{\sin }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B\\\\ \ \ \ \ \ ={{\sin }^{2}}A-{{\sin }^{2}}B\end{array}$

(ii)$ \displaystyle \ \ \ \sin (A+B)\sin (A-B)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ =(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)\\\\ \ \ \ \ \ ={{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B\\\\ \ \ \ \ \ =(1-{{\cos }^{2}}A){{\cos }^{2}}B-{{\cos }^{2}}A(1-{{\cos }^{2}}B)\\\\\ \ \ \ \ ={{\cos }^{2}}B-{{\cos }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A+{{\cos }^{2}}A{{\cos }^{2}}B\\\\ \ \ \ \ \ ={{\cos }^{2}}A-{{\cos }^{2}}B\end{array}$

(iii)$ \displaystyle \ \ \ \cos (A+B)\cos (A-B)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ =(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)\\\\\ \ \ \ \ \ ={{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B\\\\\ \ \ \ \ \ ={{\cos }^{2}}A(1-{{\sin }^{2}}B)-(1-{{\cos }^{2}}A){{\sin }^{2}}B\\\\\ \ \ \ \ \ ={{\cos }^{2}}A-{{\cos }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B\\\\\ \ \ \ \ \ ={{\cos }^{2}}A-{{\sin }^{2}}B\end{array}$

(iv) $ \displaystyle \ \ \ \cos (A+B)\cos (A-B)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ =(\cos A\cos B-\sin A\sin B)(\cos A\cos B+\sin A\sin B)\\\\\ \ \ \ \ \ \ ={{\cos }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B\\\\\ \ \ \ \ \ \ =(1-{{\sin }^{2}}A){{\cos }^{2}}B-{{\sin }^{2}}A(1-{{\cos }^{2}}B)\\\\\ \ \ \ \ \ \ ={{\cos }^{2}}B-{{\sin }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B\\\\\ \ \ \ \ \ \ ={{\cos }^{2}}B-{{\sin }^{2}}A\end{array}$

6.        Prove that $ \displaystyle \frac{{\sin (A-B)}}{{\cos A\cos B}}+\frac{{\sin (B-C)}}{{\cos B\cos C}}+\frac{{\sin (C-A)}}{{\cos C\cos A}}=0.$

Show/Hide Solution
$ \displaystyle \ \ \ \ \frac{{\sin (A-B)}}{{\cos A\cos B}}=\frac{{\sin A\cos B-\cos A\sin B}}{{\cos A\cos B}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin A\cos B}}{{\cos A\cos B}}-\frac{{\cos A\sin B}}{{\cos A\cos B}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan A-\tan B$

$ \displaystyle \ \ \ \ \frac{{\sin (B-C)}}{{\cos B\cos C}}=\frac{{\sin B\cos C-\cos B\sin C}}{{\cos B\cos C}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin B\cos C}}{{\cos B\cos C}}-\frac{{\cos B\sin C}}{{\cos B\cos C}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan B-\tan C$

$ \displaystyle \ \ \ \ \frac{{\sin (C-A)}}{{\cos C\cos A}}=\frac{{\sin C\cos A-\cos C\sin A}}{{\cos C\cos A}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin C\cos A}}{{\cos C\cos A}}-\frac{{\cos C\sin A}}{{\cos C\cos A}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\tan C-\tan A$

$ \displaystyle \therefore \ \ \ \frac{{\sin (A-B)}}{{\cos A\cos B}}+\frac{{\sin (B-C)}}{{\cos B\cos C}}+\frac{{\sin (C-A)}}{{\cos C\cos A}}$

$ \displaystyle \ \ =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A$ $ \displaystyle \ \ =0$

7.        Prove that

(i) $ \displaystyle \frac{{\cos 17{}^\circ +\sin 17{}^\circ }}{{\cos 17{}^\circ -\sin 17{}^\circ }}=\tan 62{}^\circ .$

(ii) $ \displaystyle \tan 50{}^\circ =\tan 40{}^\circ +2\tan 10{}^\circ .$

(iii) $ \displaystyle \tan 70{}^\circ =2\tan 50{}^\circ +\tan 20{}^\circ .$

(iv) $ \displaystyle \tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A.$

Show/Hide Solution
(i) $ \displaystyle \ \ \ \ \frac{{\cos 17{}^\circ +\sin 17{}^\circ }}{{\cos 17{}^\circ -\sin 17{}^\circ }}$

$ \displaystyle \ \ \ \ \ \ =\frac{{\displaystyle \frac{{\cos 17{}^\circ }}{{\cos 17{}^\circ }}+\displaystyle \frac{{\sin 17{}^\circ }}{{\cos 17{}^\circ }}}}{{\displaystyle \frac{{\cos 17{}^\circ }}{{\cos 17{}^\circ }}-\displaystyle \frac{{\sin 17{}^\circ }}{{\cos 17{}^\circ }}}}$

$ \displaystyle \ \ \ \ \ \ =\frac{{1+\tan 17{}^\circ }}{{1+\tan 17{}^\circ }}$

$ \displaystyle \ \ \ \ \ \ =\frac{{\tan 45{}^\circ +\tan 17{}^\circ }}{{1+\tan 45{}^\circ \tan 17{}^\circ }}$

$ \displaystyle \ \ \ \ \ \ =\tan (45{}^\circ +17{}^\circ )$

$ \displaystyle \ \ \ \ \ \ =\tan 62{}^\circ $


(ii) $ \displaystyle \ \ \ \ \ \ \tan 50{}^\circ =\tan (40{}^\circ +10{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\frac{{\tan 40{}^\circ +\tan 10{}^\circ }}{{1-\tan 40{}^\circ \tan 10{}^\circ }}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ -\tan 50{}^\circ \tan 40{}^\circ \tan 10{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\tan 50{}^\circ \tan 40{}^\circ \tan 10{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\tan (90{}^\circ -40{}^\circ )\tan 40{}^\circ \tan 10{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\cot 40{}^\circ \tan 40{}^\circ \tan 10{}^\circ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +\tan 10{}^\circ +\frac{1}{{\tan 40{}^\circ }}(\tan 40{}^\circ \tan 10{}^\circ )$

$ \displaystyle \therefore \ \ \ \ \ \ \ \ \tan 50{}^\circ =\tan 40{}^\circ +2\tan 10{}^\circ $


(iii) $ \displaystyle \ \ \ \ \tan 70{}^\circ =\tan (50{}^\circ +20{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\frac{{\tan 50{}^\circ +\tan 20{}^\circ }}{{1-\tan 50{}^\circ \tan 20{}^\circ }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ -\tan 70{}^\circ \tan 50{}^\circ \tan 20{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ +\tan 70{}^\circ \tan 50{}^\circ \tan 20{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ +\tan (90{}^\circ -20{}^\circ )\tan 50{}^\circ \tan 20{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 50{}^\circ +\cot 20{}^\circ \tan 50{}^\circ \tan 20{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan 70{}^\circ =\tan 50{}^\circ +\tan 20{}^\circ +\frac{1}{{\tan 20{}^\circ }}(\tan 50{}^\circ \tan 20{}^\circ )$

$ \displaystyle \therefore \ \ \ \ \ \ \tan 70{}^\circ =2\tan 50{}^\circ +\tan 20{}^\circ $


(iv) $ \displaystyle \ \ \tan 3A=\tan (2A+A)$

$ \displaystyle \ \ \ \ \ \ \ \ \tan 3A=\frac{{\tan 2A+\tan A}}{{1-\tan 2A\tan A}}$

$ \displaystyle \ \ \ \ \ \ \ \ \tan 3A-\tan 3A\tan 2A\tan A=\tan 2A+\tan A$

$ \displaystyle \therefore \ \ \ \ \tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A$

8.        (i) If $ \displaystyle A+B=45{}^\circ, $ prove that $ \displaystyle (1+\tan A)(1+\tan B)=2.$

(ii) If $ \displaystyle A+B=45{}^\circ, $ prove that $ \displaystyle (\cot A - 1)(\cot B - 1)=2.$

(iii) If $ \displaystyle A-B=45{}^\circ, $ prove that $ \displaystyle (1+\tan A)(1+\tan B)=2\tan A.$

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(i) $ \displaystyle \ A+B=45{}^\circ $

$ \displaystyle \therefore \ \ \ \tan (A+B)=\tan 45{}^\circ =1$

$ \displaystyle \therefore \ \ \ \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=1$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \tan A+\tan B=1-\tan A\tan B\\\\\therefore \ \ \ \tan A+\tan B+\tan A\tan B=1\\\\\therefore \ \ \ 1+\tan A+\tan B+\tan A\tan B=2\\\\\therefore \ \ \ \left( {1+\tan A} \right)+\tan B\left( {1+\tan A} \right)=2\\\\\therefore \ \ \ \left( {1+\tan A} \right)(1+\tan B)=2\end{array}$


(ii) $ \displaystyle \ A+B=45{}^\circ $

$ \displaystyle \therefore \ \ \ \tan (A+B)=\tan 45{}^\circ =1$

$ \displaystyle \therefore \ \ \ \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=1$

$ \displaystyle \therefore \ \ \ \tan A+\tan B=1-\tan A\tan B$

$ \displaystyle \therefore \ \ \ \frac{1}{{\cot A}}+\frac{1}{{\cot B}}=1-\frac{1}{{\cot A\cot B}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Multiplying}\ \text{both}\ \text{sides}\ \text{with }\cot A\cot B,\\\\\ \ \ \ \ \cot B+\cot A=\cot A\cot B-1\\\\\ \ \ \ \ \cot A\cot B-\cot B-\cot A=1\\\\\ \ \ \ \ \cot A\cot B-\cot B-\cot A+1=2\\\\\therefore \ \ \ \cot B\left( {\cot A-1} \right)-\left( {\cot A-1} \right)=2\end{array}$


(iii)$ \displaystyle \ A-B=45{}^\circ $

$ \displaystyle \therefore \ \ \ \tan (A-B)=\tan 45{}^\circ =1$

$ \displaystyle \therefore \ \ \ \frac{{\tan A-\tan B}}{{1+\tan A\tan B}}=1$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \tan A-\tan B=1+\tan A\tan B\\\\\therefore \ \ \ 1+\tan A\tan B+\tan B=\tan A\\\\\ \ \ \ \ \text{Adding }\tan A\ \text{to both}\ \text{sides,}\ \\\\\ \ \ \ \ 1+\tan A+\tan A\tan B+\tan B=2\tan A\\\\\ \ \ \ \ \left( {1+\tan A} \right)+\tan B\left( {1+\tan A} \right)=2\tan A\end{array}$

9.        Prove that $ \displaystyle \frac{{\tan (A+B)}}{{\cot (A-B)}}=\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}.$

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$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan (A+B)=\frac{{\sin (A+B)}}{{\cos (A+B)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan (A+B)=\frac{{\sin A\cos B+\cos A\sin B}}{{\cos A\cos B-\sin A\sin B}}--(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \text{Similarly,}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \tan (A-B)=\frac{{\sin A\cos B-\cos A\sin B}}{{\cos A\cos B+\sin A\sin B}}--(2)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \text{By (1)}\times \text{(2), we get}\\\\\ \ \ \ \ \ \ \ \ \ \tan (A+B)\tan (A-B)\end{array}$

$ \displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A{{{\cos }}^{2}}B-{{{\cos }}^{2}}A{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A{{{\cos }}^{2}}B-{{{\sin }}^{2}}A{{{\sin }}^{2}}B}}$

$ \displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A(1-{{{\sin }}^{2}}B)-(1-{{{\sin }}^{2}}A){{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A(1-{{{\sin }}^{2}}B)-(1-{{{\cos }}^{2}}A){{{\sin }}^{2}}B}}$

$ \displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}A{{{\sin }}^{2}}B-{{{\sin }}^{2}}B+{{{\sin }}^{2}}A{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\cos }}^{2}}A{{{\sin }}^{2}}B-{{{\sin }}^{2}}B+{{{\cos }}^{2}}A{{{\sin }}^{2}}B}}$

$ \displaystyle \ \ \ =\ \ \frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{\tan (A+B)}}{{\cot (A-B)}}=\tan (A+B)\tan (A-B)$

$ \displaystyle \therefore \ \ \ \ \ \ \frac{{\tan (A+B)}}{{\cot (A-B)}}=\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}$

10.        If $ \displaystyle 3\tan A\tan B=1,$ prove that $ \displaystyle 2\cos (A+B)=\cos (A-B).$

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$ \displaystyle \ \ \ \ \ \ \ \ 3\tan A\tan B=1$

$ \displaystyle \ \ \ \ \ \ \ \tan A\tan B=\frac{1}{3}$

$ \displaystyle \ \ \ \ \ \ \ \cot A\cot B=3$

$ \displaystyle \ \ \ \ \ \ \ \frac{{\cos A\cos B}}{{\sin A\sin B}}=3$

$ \displaystyle \ \ \ \ \ \ \ \text{By Componendo - Dividendo,}$

$ \displaystyle \ \ \ \ \ \ \ \frac{{\cos A\cos B+\sin A\sin B}}{{\cos A\cos B-\sin A\sin B}}=\frac{{3+1}}{{3-1}}$

$ \displaystyle \ \ \ \ \ \ \ \frac{{\cos (A-B)}}{{\cos (A+B)}}=2$

$ \displaystyle \therefore \ \ \ \ 2\cos (A+B)=\cos (A-B)$

11.        Prove that $ \displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\frac{{\sin 2\theta }}{{\cos \theta }}=\operatorname{cosec}\theta $

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Solution

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta }}{{\sin \theta \cos \theta }}+\displaystyle \frac{{\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos \left( {2\theta -\theta } \right)}}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{1}{{\sin \theta }}\\\\=\ \ \ \operatorname{cosec}\theta \end{array}$


1.     A family has $ \displaystyle 4$ children. Draw a tree diagram to list all possible outcomes. If each outcomes is equally likely to occur, find the probability that the last two children are girls. Find also the probability that exactly two children are boys.

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           Let Boy be denoted by $ \displaystyle B$ and Girl be by $ \displaystyle G.$



$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =16$

      Set of favourable outcomes for the last two children are girls

$ \displaystyle \ \ \ \ \ \ \ \ = \{(B,B,G,G),(B,G,G,G),(G,B,G,G),(G,G,G,G)\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =4$

$ \displaystyle \therefore\ \ P$( the last two children are girls) $ \displaystyle = \frac{4}{16}= \frac{1}{4}$


      Set of favourable outcomes for exactly two children are boys

$ \displaystyle \ \ \ \ \ \ \ \ = \{(B,B,G,G),(B,G,B,G),(B,G,G,B),(G,B,B,G),(G,B,G,B),(G,G,B,B)\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =6$

$ \displaystyle \therefore\ \ P$( the last two children are girls) $ \displaystyle = \frac{6}{16}= \frac{3}{8}$


2.      How many $ \displaystyle 2$-digit numerals can you form from $ \displaystyle 2,3,5,6$ without repeating any digit? Find the probability of a numeral which is divisible by $ \displaystyle 2$. If one of these numerals is chosen at random, find the probability that it is divisible by $ \displaystyle 13.$

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$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =12$

      Set of favourable outcomes for a numeral which is divisible by $ \displaystyle 2$

$ \displaystyle \ \ \ \ \ \ \ \ = \{26,32,36,52,56,62\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =6$

$ \displaystyle \therefore\ \ P$( a numeral which is divisible by $ \displaystyle 2$) $ \displaystyle = \frac{6}{12}= \frac{1}{2}$


      Set of favourable outcomes for a numeral which is divisible by $ \displaystyle 13$

$ \displaystyle \ \ \ \ \ \ \ \ = \{26,52,65\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =3$

$ \displaystyle \therefore\ \ P$( a numeral which is divisible by $ \displaystyle 13$) $ \displaystyle = \frac{3}{12}= \frac{1}{4}$



3.      How many $ \displaystyle 3$-digit numerals can you form from $ \displaystyle 1, 0 , 5$ and $ \displaystyle 6$ without repeating any digit? Find the probability of numeral which is divisible by $ \displaystyle 5.$

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$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle =18$

      Set of favourable outcomes for a numeral which is divisible by $ \displaystyle 5$

$ \displaystyle \ \ \ \ \ \ \ \ = \{105,150,160,165,510,560,605,610,615,650\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =10$

$ \displaystyle \therefore\ \ P$( a numeral which is divisible by $ \displaystyle 5$) $ \displaystyle = \frac{10}{18}= \frac{5}{9}$



4.      A fair coin is tossed three times. Draw a tree diagram to determine the set of all possible outcomes. Hence find the probability of

(a) getting three heads.

(b) getting two heads and one tail in any order.

(c) getting at least one head.

(d) getting no head.

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$ \displaystyle \therefore\ \ $ Set of possible outcomes $ \displaystyle = \{HHH,HHT,HTH, HTT,THH,THT,TTH,TTT\} $

$ \displaystyle \therefore\ \ $Number of possible outcomes $ \displaystyle = 8$

(a)       Set of favourable outcomes for getting three heads $ \displaystyle = \{HHH \}$

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 1$

$ \displaystyle \therefore\ \ P$(getting three heads) $ \displaystyle = \frac{1}{8}$


(b)       Set of favourable outcomes for getting two heads and one tail in any order
$ \displaystyle \ \ \ \ \ \ = \{HHT,HTH,THH \}$

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 3$

$ \displaystyle \therefore\ \ P$(getting two heads and one tail in any order) $ \displaystyle = \frac{3}{8}$


(c)       Set of favourable outcomes for getting at least one head
$ \displaystyle \ \ \ \ \ \ = \{HHH,HHT,HTH, HTT,THH,THT,TTH \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 7$

$ \displaystyle \therefore\ \ P$(getting at least one head) $ \displaystyle = \frac{7}{8}$


(d)    Set of favourable outcomes for getting no head
$ \displaystyle \ \ \ \ \ \ = \{TTT \}$

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 1$

$ \displaystyle \therefore\ \ P$(getting no head) $ \displaystyle = \frac{1}{8}$


5.      Draw a table of all posssible outcomes for throwing two fair dice. Hence, calculate the probability that

(a) the product of the scores on the two dice is divisible by $ \displaystyle 6$ or $ \displaystyle 9.$

(b) both dice showing the same number.

(c) the sum of the score is a multiple of $ \displaystyle 3.$

(d) the total score on the two dice is prime.

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$ \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$

(a)     Set of favourable outcomes for the product of the scores on the two dice is divisible by $ \displaystyle 6$ or $ \displaystyle 9$

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,6),(2,3), (2,6), (3,2), (3,3), (3, 4), (3, 6),(4,3), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4, 6), (5,6), (6,1), (6,2), (6, 3), (6, 4), (6, 5), (6, 6)\} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle =16$

$ \displaystyle \therefore\ \ P$(the product of the scores on the two dice is divisible by $ \displaystyle 6$ or $ \displaystyle 9$) $ \displaystyle = \frac{16}{36}= \frac{4}{9}$


(b)     Set of favourable outcomes for both dice showing the same number

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,1),(2,2), (3,3), (4,4), (5,5), (6,6) \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 6$

$ \displaystyle \therefore\ \ P$(both dice showing the same number) $ \displaystyle = \frac{6}{36}= \frac{1}{6}$


(c)     Set of favourable outcomes for the sum of the score is a multiple of $ \displaystyle 3$

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,2),(1,5), (2,1), (2,4), (3,3),(3,6),\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4,2), (4,5), (5,1), (5,4), (6,3), (6,6) \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 12$

$ \displaystyle \therefore\ \ P$(both dice showing the same number) $ \displaystyle = \frac{12}{36}= \frac{1}{3}$


(d)     Set of favourable outcomes for the total score on the two dice is prime

$ \displaystyle \ \ \ \ \ \ \ \ = \{(1,1),(1,2), (1,4),(1,6),(2,1), (2,3), (2,5), \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5) \} $

$ \displaystyle \therefore\ \ $Number of favourable outcomes $ \displaystyle = 15$

$ \displaystyle \therefore\ \ P$(both dice showing the same number) $ \displaystyle = \frac{15}{36}= \frac{5}{12}$