Answer for 2019 Sample Question : Section (C)

ဒီေနရာမွာ တင္ေပးခဲ့ေသာ Section (C) ေမးခြန္းရဲ့ အေျဖျဖစ္ပါတယ္။ Section (A) ရဲ့ အေျဖကိုေတာ့ ဒီေနရာမွာ တင္ေပးခဲ့ၿပီး  Section (B) အေျဖကို ဒီေနရာမွာ တင္ေပးထားပါတယ္။

Section (C)

Solution

11.    (a) Two unequal circles are tangent externally at $ \displaystyle O$. $ \displaystyle AB$ the chord of the first circle is tangent to the second circle at $ \displaystyle C$, and $ \displaystyle AO$ meets this circle at $ \displaystyle E$. Prove that $ \displaystyle ∠BOC=∠COE$.

(5 marks)

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    Draw a common tangent at $ \displaystyle O$ to meet $ \displaystyle BC$ at $ \displaystyle D.$

$\displaystyle \begin{array}{l}\ \ \ \text{In}\odot P,\\\\\ \ \ \alpha =\angle A\ \ \ \text{( }\angle \text{ between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \text{in alternate segment)}\\\\\ \ \ OD=CD\ \ \text{(tangents from same exterior point)}\\\\\therefore \beta =\gamma \\\\\therefore \alpha +\beta =\angle A+\gamma \\\\\ \ \ \text{But }\alpha +\beta =\angle BOC\\\\\ \ \ \text{In}\ \vartriangle AOC,\\\text{ }\\\ \ \ \angle A+\gamma =\angle COE\\\\\therefore \angle BOC=\angle COE\end{array}$

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

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ AD=2CD\ \ \ (\text{given)}\\\\\ \ \ \text{Since}\ DE\parallel AB,\\\\\ \ \ \vartriangle CAB\sim \vartriangle CDE\end{array}$

$ \displaystyle \therefore \frac{{\alpha (\Delta CAB)}}{{\alpha (\Delta CDE)}}=\frac{{A{{C}^{2}}}}{{C{{D}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{(AD+CD)}}^{2}}}}{{C{{D}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{(2CD+CD)}}^{2}}}}{{C{{D}^{2}}}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9\\\\\ \ \ \text{Let }\alpha (\Delta CDE)=x\ \text{and}\ \alpha (\Delta CAB)=9x.\\\\\therefore \alpha (ABED)=9x-x=8x\\\\\therefore 8x=40\Rightarrow x=5\\\\\therefore \alpha (\Delta CAB)=9\times 5=45\ \text{sq-units}\text{.}\end{array}$

12.    (a) The position vectors, relative to an origin $ \displaystyle O$, of three nonlinear points $ \displaystyle A, B$ and $ \displaystyle C$ are $ -2\hat{\text{i}}+3\hat{\text{j}}$, $ 3\hat{\text{i}}+2\hat{\text{j}}$ and $ -\hat{\text{i}}-5\hat{\text{j}}$. Show that $ \displaystyle ΔABC$ is isosceles.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \overrightarrow{{OA}}=-2\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \overrightarrow{{OB}}=3\widehat{\text{i}}+2\widehat{\text{j}}\\\\\ \ \ \ \overrightarrow{{OC}}=-\widehat{\text{i}}-5\widehat{\text{j}}\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =3\widehat{\text{i}}+2\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =5\widehat{\text{i}}-\widehat{\text{j}}\\\\\therefore \ \ AB=\sqrt{{{{5}^{2}}+{{{(-1)}}^{2}}}}=\sqrt{{26}}\\\\\therefore \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =-\widehat{\text{i}}-5\widehat{\text{j}}-3\widehat{\text{i}}-2\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =-4\widehat{\text{i}}-7\widehat{\text{j}}\\\\\therefore \ \ BC=\sqrt{{{{{(-4)}}^{2}}+{{{(-7)}}^{2}}}}=\sqrt{{65}}\\\\\therefore \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-\widehat{\text{i}}-5\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =\widehat{\text{i}}-8\widehat{\text{j}}\\\\\therefore \ \ AC=\sqrt{{{{1}^{2}}+{{{(-8)}}^{2}}}}=\sqrt{{65}}\\\\\therefore \ \ AC=BC\\\\\therefore \ \ \ \Delta ABC\ \text{is an isosceles triangle}\text{.}\end{array}$

12.    (b) The tangent at the point $ \displaystyle C$ on the circle meets the diameter $ \displaystyle AB$ produced at $ \displaystyle T$, If $ \displaystyle ∠BCT=27°$, calculate $ \displaystyle ∠CTA$. If $ \displaystyle CT=t$ and $ \displaystyle BT=x$, prove that the radius of the circle is $ \large {\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}}$.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \angle BAC=\angle BCT=\text{ }27{}^\circ \ \ \ (\angle \text{ between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{=}\ \angle \ \text{in alternate segment)}\\\\\ \ \ \ \angle ACB=90{}^\circ \ \ \ \ \ \ \ (\angle \text{ in semicircle)}\\\\\ \ \ \ \text{In}\ \vartriangle \text{ACT,}\\\\\ \ \ \ \angle BAC+\angle ACT+\angle CTA=180{}^\circ \\\\\therefore \ \ 27{}^\circ +27{}^\circ +90{}^\circ +\angle CTA=180{}^\circ \\\\\therefore \ \ \angle CTA=36{}^\circ \\\\\ \ \ \ \text{Let}\ AB=2r,\text{where }r\ \text{is the radius of the circle}\text{.}\\\\\ \ \ \ \text{Since}\ AT\cdot BT=C{{T}^{2}},\\\\\ \ \ \ (2r+x)\cdot x={{t}^{2}}\\\\\therefore \ \ 2xr+{{x}^{2}}={{t}^{2}}\end{array}$

$ \displaystyle \therefore \ \ r=\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}$

13.    (a) Find the angles of a triangle, given that $ \displaystyle \angle A$ is obtuse and $ \sec (B+C)=\operatorname{cosec}(B-C)=2$.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{In}\ \vartriangle ABC,\angle A\ \text{is obtuse and}\\\\\ \ \ \ \sec (B+C)=\operatorname{cosec}(B-C)=2\end{array}$

$ \displaystyle \therefore \ \ \cos (B+C)=\sin (B-C)=\frac{1}{2}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ \angle A\ \text{is obtuse,}\\\\\ \ \ \angle B+\angle C=60{}^\circ \\\\\ \ \ \angle B-\angle C=30{}^\circ \\\\\therefore \ \ 2\angle B=90{}^\circ \Rightarrow \angle B=45{}^\circ \\\\\therefore \ \ 45{}^\circ +\angle C=60{}^\circ =\angle C=15{}^\circ \\\\\ \ \ \ \text{Since}\ \angle A+\angle B+\angle C=180{}^\circ ,\\\\\ \ \ \ \angle A+45{}^\circ +15{}^\circ =180{}^\circ \Rightarrow \angle A=120{}^\circ \end{array}$

13.    (b) Differentiate $ \displaystyle \frac{1}{\sqrt{x}}$ with respect to $ \displaystyle x$ from the first principles.

(5 marks)

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$ \displaystyle \ \ \ \ \text{Let}\ f(x)=\frac{1}{{\sqrt{x}}}.$

$ \displaystyle \therefore \ \ f(x+\delta x)=\frac{1}{{\sqrt{{x+\delta x}}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{1}{{\sqrt{{x+\delta x}}}}-\frac{1}{{\sqrt{x}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{\sqrt{x}-\sqrt{{x+\delta x}}}}{{\sqrt{x}\sqrt{{x+\delta x}}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{\sqrt{x}-\sqrt{{x+\delta x}}}}{{\sqrt{x}\sqrt{{x+\delta x}}}}\times \frac{{\sqrt{x}+\sqrt{{x+\delta x}}}}{{\sqrt{x}+\sqrt{{x+\delta x}}}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{x-x-\delta x}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{-\delta x}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore \ \ \frac{{f(x+\delta x)-f(x)}}{{\delta x}}=\frac{{-1}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}$

$ \displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$ \displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\sqrt{x}\sqrt{{x+0}}\left[ {\sqrt{x}+\sqrt{{x+0}}} \right]}}=-\frac{1}{{2x\sqrt{x}}}$

14.    (a) A cruise ship travels at a bearing of $ \displaystyle 45°$ at $ \displaystyle 15$ mph for $ \displaystyle 3$ hours, and changes course to a bearing of $ \displaystyle 120°$. It then travels $ \displaystyle 10$ mph for $ \displaystyle 2$ hours. Find the distance of the ship from its original position and also its bearing from the original position.

(5 marks)

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$ \displaystyle \begin{array}{l}AB=3\ \text{hours}\ \times 15\ \text{mph}=45\ \text{mi}\\\\BC=2\ \text{hours}\ \times 10\ \text{mph}=20\ \text{mi}\\\\{{\beta }_{2}}=45{}^\circ ,{{\beta }_{1}}=180{}^\circ -120{}^\circ =60{}^\circ \\\\\beta =45{}^\circ +60{}^\circ =105{}^\circ \\\\\text{By the law of cosines,}\\\\A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2(AB)(BC)\cos \beta \\\\\ \ \ \ \ \ \ \,\ =\text{ }{{45}^{2}}+{{20}^{2}}-2\left( {45} \right)\left( {20} \right)\cos 105{}^\circ \\\\\ \ \ \ \ \ \ \ \ =\text{ }2025+400+1800\cos 75{}^\circ ~\\\\\ \ \ \ \ \ \ \ \ =\text{ }2425+465.9\\\\\ \ \ \ \ \ \ \ \ =\text{ }2890.9\\\\AC~~~~=\text{ }53.77\text{ mi}\end{array}$

No	Log
1800 cos 75°	3.2553 $ \displaystyle \overline{1}.4130$
465.9	2.6683

$ \displaystyle \ \ \ \ \text{By the law of sines,}$

$ \displaystyle \ \ \ \ \frac{{\sin \alpha }}{{BC}}=\frac{{\sin \beta }}{{AC}}$

$ \displaystyle \ \ \ \ \sin \alpha ~~=~\frac{{BC}}{{AC}}\times \sin \beta $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{20}}{{53.77}}\times \sin 105{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{20}}{{53.77}}\times \sin 75{}^\circ \ (\sin 105{}^\circ =\sin 75{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\sin 21{}^\circ 3'$

$ \displaystyle \therefore \ \ \alpha =21{}^\circ 3'$

$ \displaystyle \therefore \ \ \theta =45{}^\circ +21{}^\circ 3'=66{}^\circ 3'$

No	Log
20 sin 75°	1.3010 $ \displaystyle \overline{1}.9849$
53.77	1.02859 1.7305
sin21°3′	$ \displaystyle \overline{1}.5554$

Hence, the ship is $ \displaystyle 53.77$ mi from the original position.

It is in the direction $ \displaystyle N 66° 3' E.$

  14.    (b) Find the two positive numbers $ \displaystyle x$ and $ \displaystyle y$ such that their sum is $ \displaystyle 60$ and $ \displaystyle xy^3$ is maximum.

(5 marks)

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ x+y=60\\\\\therefore \ \ \ \ y=60-x\\\\\ \ \ \ \ \ \text{Let}\ z=x{{y}^{3}}\\\\\therefore \ \ \ \ z=x{{\left( {60-x} \right)}^{3}}\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{dz}}{{dx}}={{\left( {60-x} \right)}^{3}}-3x{{\left( {60-x} \right)}^{2}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ ={{\left( {60-x} \right)}^{2}}\left( {60-x-3x} \right)\\\\\ \ \ \ \ \ \ \ \ \ ={{\left( {60-x} \right)}^{2}}\left( {60-4x} \right)\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{dz}}{{dx}}=0,\text{when}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ {{\left( {60-x} \right)}^{2}}\left( {60-4x} \right)=0\\\\\ \ \ \ \ x=60\ \ \text{or}\,\ x=15\\\\\ \ \ \ \ \text{When}\ x=60,\ y=0\ \text{which is impossible as }y>0.\\\\\ \ \ \ \ \text{When}\ x=15,\ y=45\ \text{which is possible}\text{.}\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-x} \right)}^{2}}-2\left( {60-x} \right)\left( {60-4x} \right)$

$ \displaystyle \ \ \ \ \ \text{When}\ x=15,$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-15} \right)}^{2}}-2\left( {60-15} \right)\left( {60-60} \right)$

$ \displaystyle \therefore \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-15} \right)}^{2}}<0$

$ \displaystyle \therefore \ \ \ \text{z is maximum}\ \text{when}\ x=15\ \text{and}\ y=45.$

Circles Theorem : Geogebra Applet

Theorem မွန္ကန္ခ်က္မ်ားကို လက္ေတြ႕ ၾကည့္ရန္ Slider (သို႔) Moveable Point မ်ာကို ေရႊ႕ၾကည့္ႏိုင္ပါသည္။

Theorem (1)

စက္၀ိုင္းတစ္ခု၏ အ၀န္းပိုင္း တစ္ခုမွ ဗဟိုတြင္ ခံေဆာင္ထားေသာ ေထာင့္သည္ ၎အ၀န္းပိုင္းကို အ၀န္းတြင္ ခံေဆာင္ထားေသာ ေထာင့္၏ ႏွစ္ဆရွိသည္။

 
 

Corollary (1.1)

အ၀န္းပိုင္း တစ္ခုမွ အျခားအ၀န္းပိုင္း တစ္ခုတြင္ ခံေဆာင္ထားေသာ အ၀န္းခံေထာင့္မ်ား တူညီၾကသည္။


Corollary (1.2)

စက္၀ိုင္းျခမ္းအတြင္း အ၀န္းခံေထာင့္သည္ ေထင့္မွန္ တစ္ခုျဖစ္သည္။



Corollary (1.3)

စက္၀ိုင္းတြင္းက် စတုဂံတစ္ခု၏ အတြင္းမ်က္ႏွာခ်င္းဆိုင္ ေထာင့္တစ္စံုသည္ ေထာင့္ေျဖာင့္ ျဖည့္ဘက္မ်ား ျဖစ္ၾကသည္။




Corollary (1.4)

စက္၀ိုင္းတြင္းက် စတုဂံတစ္ခု၏ အနားတစ္ဘက္ကို ဆက္ဆြဲ၍ ျဖစ္လာေသာ အျပင္ေထာင့္သည္ အတြင္း မ်က္ႏွာခ်င္းဆိုင္ေထာင့္ ႏွင့္ ညီသည္။

Area between Curves (IGCSE O LEVEL) - Problems and Solutions

If the fumction $ \displaystyle f(x)$ and $ \displaystyle g(x)$ intersect at $ \displaystyle x_1=a$ and $ \displaystyle x_2=b$, then the bounded area between $ \displaystyle f(x)$ and $ \displaystyle g(x)$ is

$ \displaystyle \int_{a}^{b}{{\left[ {f(x)-g(x)} \right]}}\ dx$

Where $ \displaystyle f(x)$ is the curve of upper boundry and $ \displaystyle g(x)$ is the curve of lower boundry.

ဆရာႀကီး ေဒါက္တာ ေရႊေက်ာ္ ၏ Drill For Exam Blog မွ Area under curve ပုစာၦမ်ား ျဖစ္ပါသည္။

Problem (1)

The diagram shows part of the curve $ \displaystyle y=9x^2−x^3,$ which meets the $ \displaystyle x$-axis at the origin $ \displaystyle O$ and at the point $ \displaystyle A$. The line $ \displaystyle y−2x+18=0$ passes through $ \displaystyle A$ and meets the $ \displaystyle y$-axis at the point $ \displaystyle B.$

(i) Show that, for $ \displaystyle x≥0, 9x^2−x^3≤108.$

(ii) Find the area of the shaded region bounded by the curve, the line $ \displaystyle AB$ and the $ \displaystyle y$-axis.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Line : }y-2x+18\Rightarrow y=2x-18\\\\\ \ \ \ \text{Curve : }y=9{{x}^{2}}-{{x}^{3}}\end{array}$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=18x-3{{x}^{2}}=3x(6-x)$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=0\ \text{when}\ 3x(6-x)=0$

$ \displaystyle {\therefore \ \ x=0\ \text{or }x=6}$

$ \displaystyle \therefore \ \ \text{When }x=0, y=0$

$ \displaystyle {\therefore \ \ \text{When }x=6,\ }$

$ \displaystyle {\ \ \ \ y=9({{6}^{2}})-{{{(6)}}^{3}}=108}$

$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-6x$

$ \displaystyle \therefore \ \ \text{When }x=6,\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-36<0$

$ \displaystyle \begin{array}{l}\therefore \ \ y=108\ \text{is a maximum value}\text{.}\\\\\therefore \ \ \text{For}\ x\ge 0,\ y\le 0.\\\\\ \ \ \ \text{According to the diagram, }\\\ \ \ \ \text{the line and curve intersect when }y=0.\\\\\therefore \ \ 2x-18=0\Rightarrow x=9.\\\\\ \ \ \ \text{Let the curve of upper boundry be }{{y}_{1}}\text{ and }\\\ \ \ \ \text{that of the lower boundry be }{{y}_{2}}\text{, and }\\\ \ \ \ \text{let the area of the shaded region be }A.\\\\\therefore \ \ {{y}_{1}}=9{{x}^{2}}-{{x}^{3}},\ {{y}_{2}}=2x-18\ \operatorname{and}\end{array}$

$ \displaystyle \ \ \ A=\int_{0}^{9}{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}}\ dx$

$ \displaystyle \ \ \ \ \ \ \ =\int_{0}^{9}{{\left( {9{{x}^{2}}-{{x}^{3}}-2x+18} \right)}}\ dx$

$ \displaystyle \ \ \ \ \ \ \ =\left. {\frac{9}{3}{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-\frac{{2{{x}^{2}}}}{2}+18x} \right]_{0}^{9}$

$ \displaystyle \ \ \ \ \ \ \ =\left. {3{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-{{x}^{2}}+18x} \right]_{0}^{9}$

$ \displaystyle \ \ \ \ \ \ \ =3{{(9)}^{3}}-\frac{{{{{(9)}}^{4}}}}{4}-{{(9)}^{2}}+18(9)$

$ \displaystyle \ \ \ \ \ \ \ =627.75$

Problem (2)

The diagram shows part of the curve $ \displaystyle y = 2\sin 3x.$ The normal to the curve $ \displaystyle y = 2\sin 3x$ at the point where $ \displaystyle x =\frac{\pi}{9}$ meets the $ \displaystyle y$-axis at the point $ \displaystyle P.$

(i) Find the coordinates of $ \displaystyle P.$

(ii) Find the area of the shaded region bounded by the curve, the normal and the $ \displaystyle y$-axis.

Show/Hide Solution

$ \displaystyle \ \ \ \text{Curve : }y=2\sin 3x,$

$ \displaystyle \ \ \ \text{When }x=\frac{\pi }{9},$

$ \displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$

$ \displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$

$ \displaystyle \ \ \ \ \ =\sqrt{3}$

$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=6\cos 3x$

$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$

$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$

$ \displaystyle \ \ \ y=\sqrt{3}-\frac{1}{3}\left( {x-\frac{\pi }{9}} \right)$

$ \displaystyle \ \ \ y=\sqrt{3}+\frac{\pi }{{27}}-\frac{x}{3}$

$ \displaystyle \ \ \ \text{When the normal cuts the }y\text{-axis,}\ x=0.$

$ \displaystyle \therefore \ y=\sqrt{3}+\frac{\pi }{{27}}$

$ \displaystyle \therefore \ P=(0,\sqrt{3}+\frac{\pi }{{27}})=(0,1.85)$

$ \displaystyle \ \ \ \text{Let the area of the shaded region be }A.$

$ \displaystyle \therefore A=\left. {\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)x-\frac{{{{x}^{2}}}}{6}+\frac{2}{3}\cos 3x} \right]_{0}^{{\frac{\pi }{9}}}$

$ \displaystyle \therefore A=\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)\frac{\pi }{9}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{2}{3}\cos 3\left( {\frac{\pi }{9}} \right)-\frac{2}{3}\cos (0)$

$ \displaystyle \therefore A=\frac{{\sqrt{3}\pi }}{9}+\frac{{{{\pi }^{2}}}}{{243}}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{1}{3}-\frac{2}{3}=2.912$

 Problem (3)

The diagram shows part of the curve $ \displaystyle y=\sin \frac{1}{2}x$. The tangent to the curve at the point $ \displaystyle P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right)$ cuts the $ \displaystyle x$-axis at the point $ \displaystyle Q$. (i) Find the coordinates of $ \displaystyle Q$. (ii) Find the area of the shaded region bounded by the curve, the tangent and the $ \displaystyle x$-axis.

Show/Hide Solution

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

$ \displaystyle \ \ \ \ y=0\Rightarrow \sin \frac{1}{2}x=0$

$ \displaystyle \therefore \ \ \frac{1}{2}x=0\ \text{or}\ \frac{1}{2}x=\pi \ \ (\text{for first half cycle)}$

$ \displaystyle \therefore \ \ x=0\ \text{or}\ x=2\pi $

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{1}{2}x$

$ \displaystyle \ \ \ \ \text{At }P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right),\frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{{3\pi }}{4}=-\frac{{\sqrt{2}}}{4}$

$ \displaystyle \therefore \ \ \text{Equation of tangent at }P\ \text{is}$

$ \displaystyle \ \ \ \ y=\frac{{\sqrt{2}}}{2}-\frac{{\sqrt{2}}}{4}\left( {x-\frac{{3\pi }}{2}} \right)$

$ \displaystyle \therefore \ \ y=\frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x$

$ \displaystyle \ \ \ \ \text{When the tangent cuts the }x\text{-axis, }y=0.$

$ \displaystyle \therefore \ \ \frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x=0$

$ \displaystyle \therefore \ \ x=2+\frac{{3\pi }}{2}$

$ \displaystyle \therefore \ \ Q=\left( {2+\frac{{3\pi }}{2},0} \right)=(6.71,0)$

$ \displaystyle \ \ \ \ \text{Area of }\Delta PQR=\frac{1}{2}\cdot PQ\cdot PR$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}(2)\left( {\frac{{\sqrt{2}}}{2}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \ \text{Area of yellow region}=\int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\left( {\sin \frac{1}{2}x} \right)dx}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left. {-2\cos \frac{1}{2}x} \right]_{{\frac{{3\pi }}{2}}}^{{2\pi }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {\cos \pi -\cos \frac{{3\pi }}{4}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {-1+\frac{{\sqrt{2}}}{2}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2-\sqrt{2}$

$ \displaystyle \ \ \ \ \text{Let the required area be }A.$

$ \displaystyle \therefore \ \ A=\text{Area of }\Delta PQR-\text{Area of yellow region}$

$ \displaystyle \therefore \ \ A=\frac{{\sqrt{2}}}{2}-2+\sqrt{2}=\frac{{3\sqrt{2}}}{2}-2$

ဆက္လက္ ေဖၚျပပါမည္

Circles : Problems and Solutions

Problem (1)

Given : $ \displaystyle SPT$ is the tangent to the circle at $ \displaystyle P$.

             $ \displaystyle PQ$ and $ \displaystyle RQ$ are the chords of the circle.

             $ \displaystyle PM\bot RQ$ and $ \displaystyle RN\bot SPT$.

Prove : $ \displaystyle MN\parallel PQ$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \angle QPT=\angle QRP\ (\angle \ \text{between tangent and chord }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \ \text{in alternate segment})\\\\\ \ \ \ \ \text{Since}\ PM\bot RQ\ \text{and}\ RN\bot SPT,\ \\\\\ \ \ \ \ \angle PMR=\angle PNR=90{}^\circ \\\\\therefore \ \ \ \angle PMR+\angle PNR=180{}^\circ \\\\\therefore \ \ \ PMRN\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ \angle PNM=\angle QRP\ \ (\angle \ \text{in same arc)}\\\\\therefore \ \ \ \angle PNM=\angle QPT\\\\\ \ \ \ \ \text{Since}\ \angle PNM\ \text{and}\ \angle QPT\ \text{are alternating angles,}\\\\\ \ \ \ \ MN\parallel PQ.\ \end{array}$

Problem (2)

In the fgure, $ \displaystyle AP$ is a tangent to the circle at $ \displaystyle A$ and $ \displaystyle AP$ is parallel to $ \displaystyle BQ$. Prove that

(i) $ \displaystyle ∆ABC$ is similar to $ \displaystyle ∆AQB,$

(ii) $ \displaystyle AB^2 = AQ × AC.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \alpha =\beta \ (\angle \ \text{between tangent and chord }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \ \text{in alternate segment})\\\\\ \ \ \ \ \text{Since}\ AP\parallel BQ,\ \ \alpha =\theta .\\\\\ \ \ \ \ \text{In }\Delta ABC\ \text{and}\ \Delta AQB,\\\\\ \ \ \ \ \alpha =\theta \ \text{(proved)}\\\\\ \ \ \ \ \delta =\delta \ \ \text{(common }\angle \text{)}\\\\\therefore \ \ \ \Delta ABC\sim \Delta AQB\ \ (\text{AA corollary)}\end{array}$

$ \displaystyle \therefore \ \ \ \frac{{AB}}{{AQ}}=\frac{{AC}}{{AB}}$

$ \displaystyle \therefore \ \ \ A{{B}^{2}}=AQ\times AC$

Problem (3)

In the fgure, $ \displaystyle O$ is the centre of the circle, $ \displaystyle PQ$ is a diameter and $ \displaystyle AB$ is a chord which is parallel to $ \displaystyle PQ.$ $ \displaystyle AQ$ and $ \displaystyle OB$ intersect at $ \displaystyle X$. Prove that $ \displaystyle ∠BXQ = 3∠PQA$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ AB\parallel PQ,\\\\\ \ \ \ \beta =\phi .\ \ \ \text{(}\because \text{alternating }\angle \text{)}\\\\\ \ \ \ \alpha =\theta .\ \ \ \text{(}\because \text{alternating }\angle \text{)}\end{array}$

$ \displaystyle \ \ \ \ \alpha =\frac{1}{2}\phi \ \ \ \text{(}\because \text{inscribed }\angle =\frac{1}{2}\text{central }\angle \text{)}$

$ \displaystyle \begin{array}{l}\therefore \ \ \phi =\beta =2\alpha =2\theta \\\\\ \ \ \ \text{In }\Delta ABX,\ \\\\\ \ \ \ \gamma =\beta +\alpha \\\\\ \ \ \ \ \ \ =2\alpha +\alpha \\\\\ \ \ \ \ \ \ =3\alpha \\\\\ \ \ \ \ \ \ =3\theta \ \ \ \ (\because \alpha =\theta )\\\\\therefore \ \ \angle BXQ=3\angle PQA\end{array}$

Problem (4)

In the fgure, $ \displaystyle BD$ and $ \displaystyle CE$ are tangents to the circle, of which $ \displaystyle AB$ is a diameter and $ \displaystyle ACD$ is a straight line. Prove that

(i) $ \displaystyle \angle ABC=\angle ECD$

(ii) $ \displaystyle BE = ED.$

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$ \displaystyle \begin{array}{l}\ \ \ \text{It is obvious that}\\\\\ \ \ \beta =\gamma \ \ (\because \angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment})\\\\\ \ \ \text{But, }\gamma =\theta \ \ (\because \text{vertically opposite }\angle \text{s)}\\\\\therefore \ \beta =\theta \\\\\therefore \angle ABC=\angle ECD\\\\\ \ \ \ \text{Since }AB\ \text{is a diameter, }\\\\\ \ \ \ \angle ACB=90{}^\circ \ \ (\because \angle \ \text{in semicircle)}\\\\\therefore \ \ \angle BCD=90{}^\circ \\\\\ \ \ \ \text{Furthermore, }AB\ \text{is a diameter and }BD\ \text{is a tangent,}\\\\\ \ \ \ AB\bot BD.\\\\\therefore \ \ \Delta ABD\sim \Delta ACB\sim \Delta BCD\\\\\therefore \ \ \beta =\delta \\\\\therefore \ \ \theta =\delta \\\\\therefore \ \ \Delta BCD\ \text{is an isosceles triangle with base }CD.\\\\\therefore \ \ EC=ED\\\\\ \ \ \ \text{Since}\ BE=EC,\ \ (\because \text{tangents from same exterior point)}\\\\\ \ \ \ BE=ED\end{array}$

Problem (5)

 In the figure, $ \displaystyle ABCD$ and $ \displaystyle APRD$ are two circles intersecting at $ \displaystyle A$ and $ \displaystyle D$. $ \displaystyle ARC$ and $ \displaystyle BPD$ are straight lines. $ \displaystyle DR$ produced meets $ \displaystyle BC$ at $ \displaystyle S$. Prove that $ \displaystyle PR$ is parallel to $ \displaystyle BC.$

Show/Hide Solution

In circle $ \displaystyle APRD, \alpha = \phi \ \ \ (\angle \text{s}\ \text{in}\ \text{same}\ \text{arc})$

Similarly, In circle $ \displaystyle ABCD, \alpha = \beta \ \ \ (\angle \text{s}\ \text{in}\ \text{same}\ \text{arc})$

$ \displaystyle \therefore\ \phi= \beta$

Since $ \displaystyle \phi$ and $ \displaystyle \beta$ are corresponding angles,

$ \displaystyle PR \parallel BC$

Problem (6)

In the figure, $ \displaystyle HAE$ is the tangent to the circle at $ \displaystyle H, BH = BE$ and $ \displaystyle KH$ is the angle bisector of $ \displaystyle ∠BHE$ and it cuts the circle at $ \displaystyle D.$ Given that $ \displaystyle BD$ produced meets $ \displaystyle HE$ at $ \displaystyle A,$ prove that

(i) $ \displaystyle HD = BD,$

(ii) $ \displaystyle A, D, K$ and $ \displaystyle E$ are concyclic.

Show/Hide Solution

$ \displaystyle \begin{array}{l}\ \ \ \beta =\phi ,\ \ \ (\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment})\end{array}$

$ \displaystyle \ \ \ \text{Since}\ KH\ \text{is}\ \text{the}\ \text{angle}\ \text{bisector}\ \text{of}\ ∠BHE,$

$ \displaystyle \ \ \ \theta = \phi$

$ \displaystyle \therefore \theta = \beta$

$ \displaystyle \therefore HD = BD$

$ \displaystyle \ \ \ \text{Since}\ BH = BE,$

$ \displaystyle \ \ \ \varepsilon=\theta + \phi$

$ \displaystyle \therefore \varepsilon=\theta + \beta$

$ \displaystyle \ \ \ \text{In}\ \triangle BDH,$

$ \displaystyle \ \ \ \alpha=\theta + \beta$

$ \displaystyle \therefore \varepsilon=\alpha$

$ \displaystyle \text{Since}\ \alpha +\delta =180{}^\circ ,$

$ \displaystyle \ \ \ \varepsilon +\delta =180{}^\circ ,$

$ \displaystyle \therefore \ A,D,K,E\ \text{are concyclic}\text{.}$

Calculus - Limits

Limit ပုစ္ဆာများ တွက်ရာတွင် သိရှိနားလည်ထားရမည့် ဂုဏ်သတ္တိများ ကို လေ့လာသိရှိပြီးမှ ပုစ္ဆာများကို တွက်လျှင် ပို၍ နားလည်လွယ်ကူစေပါသည်။

1.      $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}$

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$ \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 }}}\,\left[ {{{x}^{2}}+x+1} \right]$

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3$

2.      $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{\sqrt[3]{x}-1}}$

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$ \displaystyle \ \ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{\sqrt[3]{x}-1}}\ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{{\left( {\sqrt[3]{x}} \right)}}^{3}}-{{1}^{3}}}}{{\sqrt[3]{x}-1}}$

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,{{\left( {\sqrt[3]{x}} \right)}^{2}}+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {\sqrt[3]{x}} \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( 1 \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3$

3.      $ \displaystyle \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}-9}}{{{{x}^{2}}+\sqrt{3}x-6}}\ $

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$ \displaystyle \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}-9}}{{{{x}^{2}}+\sqrt{3}x-6}}\ \ =\underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{({{x}^{2}}-3)({{x}^{2}}+3)}}{{(x-\sqrt{3})(x+2\sqrt{3})}}$

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

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

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{{(\sqrt{3}+\sqrt{3})(3+3)}}{{(\sqrt{3}+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{{12\sqrt{3}}}{{3\sqrt{3}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 4$

4.      $ \displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{{{x}^{2}}-x-6}}\ $

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$ \displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{{{x}^{2}}-x-6}}\ =\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{(x+2)({{x}^{2}}+5)}}{{(x+2)(x-3)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+5}}{{x-3}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{4+5}}{{-2-3}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{9}{5}$

5.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{3{{x}^{3}}-x-6}}\ $

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$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{3{{x}^{3}}-x-6}}\ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}\left( {1+\frac{2}{x}+\frac{5}{{{{x}^{2}}}}+\frac{{10}}{{{{x}^{3}}}}} \right)}}{{{{x}^{3}}\left( {3-\frac{1}{{{{x}^{2}}}}-\frac{6}{{{{x}^{3}}}}} \right)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{1+\frac{2}{x}+\frac{5}{{{{x}^{2}}}}+\frac{{10}}{{{{x}^{3}}}}}}{{3-\frac{1}{{{{x}^{2}}}}-\frac{6}{{{{x}^{3}}}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{1+0+0+0}}{{3-0-0}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{3}$

6.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{{-3}}}+5{{x}^{{-2}}}+8}}{{5{{x}^{{-3}}}-{{x}^{{-1}}}-2}}\ $

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$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{{-3}}}+5{{x}^{{-2}}}+8}}{{5{{x}^{{-3}}}-{{x}^{{-1}}}-2}}\ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+\frac{5}{{{{x}^{2}}}}+8}}{{\frac{5}{{{{x}^{3}}}}-\frac{1}{x}-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+\frac{5}{{{{x}^{2}}}}+8}}{{\frac{5}{{{{x}^{3}}}}-\frac{1}{x}-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{0+0+8}}{{0-0-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-4$

7.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{4}}+3{{x}^{3}}-7{{x}^{2}}-12x-4}}{{{{x}^{3}}-8}}\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{4}}+3{{x}^{3}}-7{{x}^{2}}-12x-4}}{{{{x}^{3}}-8}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}\left( {2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}} \right)}}{{{{x}^{3}}\left( {1-\frac{8}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{x\left( {2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}} \right)}}{{\left( {1-\frac{8}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,x\cdot \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}}}{{1-\frac{8}{{{{x}^{3}}}}}}\ $

$ \displaystyle =\infty $

8.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+3x+2}}{{{{x}^{3}}+32{{x}^{2}}-5x-20}}\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+3x+2}}{{{{x}^{3}}+32{{x}^{2}}-5x-20}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}\left( {1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}} \right)}}{{{{x}^{3}}\left( {1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\left( {1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}} \right)}}{{x\left( {1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{1}{x}\cdot \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}}}{{1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}}}\ $

$ \displaystyle =0$

9.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\left[ {\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\times \frac{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}} \right]\ $

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

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{\sqrt{{{{x}^{2}}\left( {1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}} \right)}}+\sqrt{{{{x}^{2}}\left( {1+\frac{1}{{{{x}^{2}}}}} \right)}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{x\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+x\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{x\left( {\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}} \right)}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{2}{{\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}}}$

$ \displaystyle =\frac{2}{{\sqrt{{1+0+0}}+\sqrt{{1+0}}}}$

$ \displaystyle =\frac{2}{{\sqrt{1}+\sqrt{1}}}$

$ \displaystyle =1$

10.      $ \displaystyle \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+3+...+n}}{{{{n}^{2}}}}$

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Since $ \displaystyle 1 + 2 + 3 + … + n$ is an arithmetic series with the first term $ \displaystyle 1$ and the common difference $ \displaystyle 1$ having $ \displaystyle n$ terms.

$ \displaystyle \therefore 1+2+3+\ldots +n=\frac{n}{2}(1+n)$

$ \displaystyle \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+3+...+n}}{{{{n}^{2}}}}$

$ \displaystyle =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{n+{{n}^{2}}}}{2}}}{{{{n}^{2}}}}$$ \displaystyle =\frac{1}{2}\cdot \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{n+{{n}^{2}}}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\cdot \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{n+{{n}^{2}}}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\underset{{n\to \infty }}{\mathop{{\cdot \lim }}}\,\frac{{{{n}^{2}}\left( {\frac{1}{n}+1} \right)}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\underset{{n\to \infty }}{\mathop{{\cdot \lim }}}\,\frac{{\frac{1}{n}+1}}{1}$

$ \displaystyle =\frac{1}{2}\left( 1 \right)$

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

11.      $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+4+...+{{2}^{{n-1}}}}}{{{{2}^{n}}+1}}$

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Since $ \displaystyle 1 + 2 + 4 + … + 2^{n – 1}$ is a geometric series with the first term $ \displaystyle 1$ and the common ratio $ \displaystyle 2$ having $ \displaystyle n$ terms.

$ \displaystyle \therefore 1+2+4+\ldots +{{2}^{{n-1}}}=\frac{{1({{2}^{n}}-1)}}{{2-1}}={{2}^{n}}-1$

$ \displaystyle \ \ \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+4+...+{{2}^{{n-1}}}}}{{{{2}^{n}}+1}}\ \ \ \ $

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{2}^{n}}-1}}{{{{2}^{n}}+1}}$

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

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1-\frac{1}{{{{2}^{n}}}}}}{{1+\frac{1}{{{{2}^{n}}}}}}$

$ \displaystyle \ \ \ =\frac{{1-0}}{{1+0}}$

$ \displaystyle \ \ \ =1$

12.      $ \displaystyle \ \underset{{t\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{t}+{{t}^{2}}}}{{2t-{{t}^{2}}}}$

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Let $ \displaystyle x=\sqrt{t}$ then $ \displaystyle t = x^2$

As $ \displaystyle t\to \infty, t\to \infty $

$ \displaystyle \therefore \ \ \ \ \ \ \underset{{t\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{t}+{{t}^{2}}}}{{2t-{{t}^{2}}}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{x+{{x}^{4}}}}{{2x-{{x}^{4}}}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}(\frac{1}{{{{x}^{3}}}}+1)}}{{{{x}^{4}}(\frac{2}{{{{x}^{3}}}}-1)}}$

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

$ \displaystyle \ \ \ =\frac{{0+1}}{{0-1}}$

$ \displaystyle \ \ \ =-1$

13.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-{{4}^{{-x}}}}}{{{{4}^{x}}+{{4}^{{-x}}}}}$

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$ \displaystyle \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-{{4}^{{-x}}}}}{{{{4}^{x}}+{{4}^{{-x}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-\displaystyle \frac{1}{{{{4}^{x}}}}}}{{{{4}^{x}}+\displaystyle \frac{1}{{{{4}^{x}}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\left( {{{4}^{x}}-\displaystyle \frac{1}{{{{4}^{x}}}}} \right)\times \displaystyle \frac{1}{{{{4}^{x}}}}}}{{\left( {{{4}^{x}}+\displaystyle \frac{1}{{{{4}^{x}}}}} \right)\times \displaystyle \frac{1}{{{{4}^{x}}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{1-\displaystyle \frac{1}{{{{4}^{{2x}}}}}}}{{1+\displaystyle \frac{1}{{{{4}^{{2x}}}}}}}$

$ \displaystyle \begin{array}{l}=\ \ \displaystyle \frac{{1-0}}{{1+0}}\\\\=1\end{array}$

Calculus : Approximation

1.       Given that $ \displaystyle y = 3x^3 - 7x^2 + 8$, find the value of $ \displaystyle \frac{dy}{dx}$ at the point $ \displaystyle (2, 4)$. Hence, find the approximate increase in $ \displaystyle x$ which will cause $ \displaystyle y$ to increase from $ \displaystyle 4$ to $ \displaystyle 4.04$.

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$ \displaystyle \begin{array}{l}\ \ \ y=3{{x}^{3}}-7{{x}^{2}}+8\\\\\ \ \ \text{Let}\ ({{x}_{0}},{{y}_{0}})=(2,4)\ \text{and}\ {{y}_{1}}=4.04\\\\\therefore \ \delta y={{y}_{1}}-{{y}_{0}}\\\\\ \ \ \ \ \ \ \ =4.04-4\\\\\ \ \ \ \ \ \ \ =0.04\end{array}$

$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=9{{x}^{2}}-14x$

$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(2,4)}}}=9{{(2)}^{2}}-14(2)=8$

$ \displaystyle \ \ \ \text{By linear approximation,}$

$ \displaystyle \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}\cdot \delta x$

$ \displaystyle \begin{array}{l}\therefore \ 0.04\simeq 8\cdot \delta x\\\\\therefore \delta x\simeq 0.005\\\\\therefore \ x\ \text{is increased by 0}\text{.005}\text{.}\end{array}$

2.       Given that $ \displaystyle y = 2x^2 + 3x$, find the approximate percentage change in $ \displaystyle y$ when $ \displaystyle x$ decreases from $ \displaystyle 2$ to $ \displaystyle 1.97$.

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$ \displaystyle \begin{array}{l}\ \ \ \ y=2{{x}^{2}}+3x\\\\\ \ \ \ \text{Let}\ {{x}_{0}}=2\ \text{and}\ {{x}_{1}}=1.97\\\\\therefore \ \ \delta x=1.97-2=-0.03\\\\\ \ \ \ \text{When}\ {{x}_{0}}=2,\ {{y}_{0}}=14\end{array}$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=4x+3$

$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{2}}=4(2)+3=11$

$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =11(-0.03)\\\\\ \ \ \ \ \ \ \ \ =-0.33\\\\\ \ \ \ \text{Approximate percentage change in }y\end{array}$

$ \displaystyle \ \ \ \ =\frac{{\delta y}}{{{{y}_{0}}}}\times 100 $ %

$ \displaystyle \ \ \ \ =\frac{{-0.33}}{{14}}\times 100$ %

$ \displaystyle \ \ \ \ =-2.36$ %

$ \displaystyle \therefore \ \ \text{ }y\ \text{is decreased by}\ 2.36$ %.

3.       Use approximation to approximate the following values $ \displaystyle \sqrt{80}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ y=\sqrt{x}\ \text{and}\ {{x}_{0}}=81\\\\\therefore \ \ {{y}_{0}}=9\ \ \\\\\ \ \ \ \text{Let}\ {{x}_{1}}=\ 80.\\\\\therefore \ \ {{y}_{1}}=\sqrt{{80}}\\\\\therefore \ \ \delta x=80-81=-1\end{array}$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{{2\sqrt{x}}}$

$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{81}}}=\frac{1}{{2\sqrt{{81}}}}=\frac{1}{{18}}$

$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{{18}}(-1)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =-0.056\\\\\therefore \ \ {{y}_{1}}={{y}_{0}}+\delta y=9-0.056=8.944\\\\\therefore \ \ \sqrt{{80}}\simeq 8.944\end{array}$

4.       The side of a square is $ \displaystyle 5\ \text{cm}$ . How much will the area of the square increase when the side expands by $ \displaystyle 0.01\ \text{cm}$?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of the side of a square be }x\\\\\ \ \ \ \operatorname{and}\ \text{the area of the square be }A.\\\\\therefore \ \ A={{x}^{2}}\\\\\ \ \ \ \text{Let }{{x}_{0}}=5\ \text{then}\ {{A}_{0}}=\ 25\\\\\ \ \ \ \text{When the sides expand 0}\text{.01 cm, }\\\\\ \ \ \ \delta x=0.01\ \text{cm}\text{.}\end{array}$

$ \displaystyle \ \ \ \ \frac{{dA}}{{dx}}=2x$

$ \displaystyle \ \ \ \ {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dA}}{{dx}}} \right|}_{5}}=2(5)=10$

$ \displaystyle \ \ \ \ \delta A\simeq {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =10(0.01)\\\\\ \ \ \ \ \ \ \ \ =0.1\ \text{c}{{\text{m}}^{2}}\\\\\therefore \ \ \text{The area of the square is increased by }0.1\ \text{c}{{\text{m}}^{2}}.\end{array}$

5.       If the sides of a square are increased by $ \displaystyle 20$ %, by what percentage does the area of the square increase?

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$ \displaystyle \ \ \ \ \text{Solution (1) - Using Approximation}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\ \ \ \ \text{and}\ \text{the area be }A.\text{ }\end{array}$

$ \displaystyle \therefore \ \ A={{x}^{2}}.$

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

$ \displaystyle \ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}$

$ \displaystyle \ \ \ \ \frac{{\delta x}}{x}=\text{20 } \text{% }=0.2\Rightarrow \delta x=0.2x$

$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dx}}\times \delta x$

$ \displaystyle \begin{array}{l}\ \ \ \ \delta A\simeq 2x\times 0.2x\\\\\therefore \ \ \delta A\simeq 0.4{{x}^{2}}\\\\\ \ \ \ \delta A\simeq 0.4A\end{array}$

$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.4}=\text{40}\ \text{%}$

$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 40}\ \text{%.}$

$ \displaystyle \ \ \ \ \text{Solution (2) - Without Calculus}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\\\ \ \ \ \text{and}\ \text{the area be }A.\\\\\therefore \ \ {{A}_{0}}={{x}^{2}}\\\\\ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}\\\\\ \ \ \ {{x}_{1}}=x+0.2x=1.2x\\\\\therefore \ \ {{A}_{1}}={{(1.2x)}^{2}}=1.44{{x}^{2}}\\\\\ \ \ \ \delta A\simeq {{A}_{1}}-{{A}_{0}}=0.44{{x}^{2}}\\\\\therefore \ \ \ \delta A=0.44A\end{array}$

$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}=\text{0}\text{.44}=\text{44}\ \text{% }$

$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 44}\ \text{%.}$

6.       One side of a rectangle is three times the other. If the perimeter increases by $ \displaystyle 2$ % what is the percentage increase in area?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{As one side of the rectangle is three times the other},\\\ \ \ \ \text{let the length and the breadth be 3}x\ \text{and}\ x.\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ \text{the perimetre and area of the rectangle be }\\\ \ \ \ p\ \text{and }A\ \text{respectively}\text{.}\end{array}$

$ \displaystyle \therefore \ \ p=2(3x+x)=8x\ \text{and}\ A=3{{x}^{2}}.$

$ \displaystyle \therefore \ \ x=\frac{p}{8}\Rightarrow A=3{{\left( {\frac{p}{8}} \right)}^{2}}=\frac{{3{{p}^{2}}}}{{64}}$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dp}}=\frac{{3p}}{{32}}$

$ \displaystyle \ \ \ \ \text{When the perimeter is incresed by 2}$ %,

$ \displaystyle \ \ \ \ \frac{{\delta p}}{p}=\text{2 %}=0.02\Rightarrow \delta p=0.02p$

$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dp}}\times \delta p$

$ \displaystyle \ \ \ \ \delta A\simeq \frac{{3p}}{{32}}\times 0.02p$

$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{32\times 2}}\times 0.02\times 2$

$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{64}}\times 0.04$

$ \displaystyle \ \ \ \ \delta A\simeq 0.04A$

$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.04}=\text{4}$ %

$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 4}$ %.

7.      The kinetic energy $ \displaystyle K$ of a body of mass $ \displaystyle m$ moving with speed $ \displaystyle v$ is given by $ \displaystyle K =\frac{1}{2}mv^2$ . If a body’s speed is increased by $ \displaystyle 1.5$ % what is the approximate percentage change in the kinetic energy?

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$ \displaystyle \ \ \ \ \text{Kinetic Energy}=\ K,\text{ mass}=m,\text{ speed}=v$

$ \displaystyle \ \ \ \ K=\frac{1}{2}m{{v}^{2}}$

$ \displaystyle \ \ \ \ \text{If a body's speed is incresed by 1.5}$ %,

$ \displaystyle \ \ \ \ \frac{{\delta v}}{v}=\text{1}\text{.5}\ \text{%}=0.015\Rightarrow \delta v=0.015v$

$ \displaystyle \ \ \ \ \frac{{dK}}{{dv}}=mv$

$ \displaystyle \ \ \ \ \delta K\simeq \frac{{dK}}{{dv}}\times \delta v$

$ \displaystyle \begin{array}{l}\ \ \ \ \delta K\simeq mv\times \delta v\\\\\therefore \ \ \delta K\simeq mv\times 0.015v\end{array}$

$ \displaystyle \therefore \ \ \delta K\simeq \frac{1}{2}m{{v}^{2}}\times 0.03$

$ \displaystyle \therefore \ \ \delta K\simeq K\times 0.03$

$ \displaystyle \therefore \ \ \frac{{\delta K}}{K}\simeq 0.03=3$ %

$ \displaystyle \therefore \ \ \text{The kinetic energy is increased by}$ 3%.

8.       The two equal sides of an isosceles triangle with fixed base $ \displaystyle b$ are decreasing at the rate of $ \displaystyle 3\ \text{cm}$ per second. How fast is the area decreasing when the two equal sides are equal to the base ?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of two equal sides be }x\ \text{and }\\\ \ \ \ \text{the the length of}\ \text{altitude of the triangle on base}\ b\text{ be }h.\text{ }\end{array}$.

$ \displaystyle \therefore \ \ \text{By Pythagoras' } \text{Theorem,}$

$ \displaystyle \ \ \ \ h=\sqrt{{{{x}^{2}}-{{{\left( {\frac{b}{2}} \right)}}^{2}}}}=\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}$

$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dx}}{{dt}}=3\ \text{cm/s.}$

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

$ \displaystyle \therefore \ \ A=\frac{1}{2}bh=\frac{1}{2}b\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dx}}=\frac{1}{2}b\times \frac{{2x}}{{2\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}=\frac{{bx}}{{2\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}$

$ \displaystyle \ \ \ \ \text{When }x=b,\frac{{dA}}{{dx}}=\frac{{{{b}^{2}}}}{{2\sqrt{{{{b}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}=\frac{{{{b}^{2}}}}{{2b\sqrt{3}}}=\frac{b}{{\sqrt{3}}}$

$ \displaystyle \ \ \ \ \text{By Chain Rule,}\frac{{dA}}{{dt}}=\frac{b}{{\sqrt{3}}}\times 3=\sqrt{3}b\ \text{c}{{\text{m}}^{2}}\text{/s.}$

$ \displaystyle \therefore \ \ \text{The area of the triangle is decreasing at a rate of }\sqrt{3}b\ \text{c}{{\text{m}}^{2}} \text{/s.}$

9.       The volume of the cube is increasing at a rate of $ \displaystyle 9\ \text{cm}^3/ \text{s}$. How fast is the surface area increasing when the edge is $ \displaystyle 10\ \text{cm}$ long ?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of each sides be }x,\ \\\ \ \ \ \text{the volume be }V\text{and the surface area be }A.\end{array}$

$ \displaystyle \therefore \ \ V={{x}^{3}}\ \text{and}\ A=6{{x}^{2}}$

$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=3{{x}^{2}}\ \text{and}\ \frac{{dA}}{{dx}}=12x.$

$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dV}}{{dt}}=9\ \text{c}{{\text{m}}^{3}}\text{/s}\text{.}$

$ \displaystyle \ \ \ \ \text{Since }\frac{{dV}}{{dt}}=\frac{{dV}}{{dx}}\times \frac{{dx}}{{dt}},\ \ \ \ \ \left[ {\text{Chain Rule}} \right]$

$ \displaystyle \ \ \ \ 3{{x}^{2}}\times \frac{{dx}}{{dt}}=9\Rightarrow \frac{{dx}}{{dt}}=\frac{3}{{{{x}^{2}}}}$

$ \displaystyle \ \ \ \ \text{Similarly }\frac{{dA}}{{dt}}=\frac{{dA}}{{dx}}\times \frac{{dx}}{{dt}}$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dt}}=12x\left( {\frac{3}{{{{x}^{2}}}}} \right)=\frac{{36}}{x}$

$ \displaystyle \ \ \ \ \text{When }x=10\ \text{cm},\ \frac{{dA}}{{dt}}=\frac{{36}}{{10}}=3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$

$ \displaystyle \therefore \ \ \text{The surface area of the cube is increasing at a rate of }3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$

No.8 ႏွင့္ No.9 သည္ Composite Function မ်ား ျဖစ္ေနေသာေၾကာင့္ Chain Rule ကို သံုးရပါသည္။

Practice :The Remainder Theorem and The Factor Theorem

1.       It is given that $ \displaystyle f(x)=x^3+ax^2+bx-48$. When $ \displaystyle f(x)$ is divided by $ \displaystyle x - 3$ the remainder is $ \displaystyle 6$. Given that $ \displaystyle f'(1)=0$ , find the value of $ \displaystyle a$ and of $ \displaystyle b$.

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       $ \displaystyle f(x)=x^3+ax^2+bx-48$

       When $ \displaystyle f(x)$ is divided by $ \displaystyle x-3$, the remainder is $ \displaystyle 6$.

$ \displaystyle \begin{array}{l}\therefore \ \ f(3)=6\\\\\ \ \ \ {{(3)}^{3}}+a{{(3)}^{2}}+b(3)-48=6\\\\\ \ \ \ 3a+b=9\,\ \ -----(1)\\\\\ \ \ \ {f}'(x)=3{{x}^{2}}+2ax+b\\\\\ \ \ \ {f}'(1)=0\ \ \ \ [\text{given}]\\\\\ \ \ \ 3{{(1)}^{2}}+2a(1)+b=0\\\\\ \ \ \ 2a+b=-3\,\ \ ----(2)\end{array}$

      By equation (1) - equation (2),

$ \displaystyle \begin{array}{l}\ \ \ \ \ a=12\\\\\therefore \ \ \  24+b=-3\, \Rightarrow b=-27\end{array}$

2.       If $ \displaystyle f(x) = ax^2 + bx + c$ leaves remainders $ \displaystyle 1 , 25 , 1$ on division by $ \displaystyle x - 1 , x + 1 , x - 2$ respectively, show that $ \displaystyle f(x)$ is a perfect square.

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$ \displaystyle \begin{array}{l}\ \ \ f(x)=a{{x}^{2}}+bx+c\\\\\ \ \ f(1)=1\\\\\ \ \ a{{(1)}^{2}}+b(1)+c=1\\\\\ \ \ a+b+c=1\ \ \ \ ----(1)\\\\\ \ \ f(-1)=25\\\\\ \ \ a{{(-1)}^{2}}+b(-1)+c=1\\\\\ \ \ a-b+c=25\ ----(2)\\\\\ \ \ f(2)=1\\\\\ \ \ a{{(2)}^{2}}+b(2)+c=1\\\\\ \ \ 4a+2b+c=1\ \ ---(3)\\\\\ \ \ \text{By equation (1)}-\text{equation (2),}\\\\\ \ \ 2b=-24\Rightarrow b=-12\\\\\ \ \ \text{Substituting }b=-12\ \text{in equations (1)}\operatorname{and}\text{(3),}\\\\\ \ \ a-12+c=1\ \Rightarrow a+c=13\ \ \ \ \ --(4)\\\\\ \ \ 4a-24+c=1\ \Rightarrow 4a+c=25--(5)\\\\\ \ \ \text{By equation (5)}-\text{equation (4),}\\\\\ \ \ 3a=12\Rightarrow a=4\\\\\ \ \ \text{Substituting }a=4,b=-12\ \text{in equations (1), }\\\\\ \ \ 4-12+c=1\ \Rightarrow c=9\\\\\therefore f(x)=4{{x}^{2}}-12x+9={{(2x-3)}^{2}}\\\\\therefore f(x)\ \text{is a perfect square}\text{.}\end{array}$

3.       It is given that $ \displaystyle x-2$ is a factor of $ \displaystyle f(x)=x^3+kx^2-8x-8$ where $ \displaystyle k$ is an integer.

(i) Find the value of the integer $ \displaystyle k$.

(ii) Using your of $ \displaystyle k$, find the non-integer roots of the equation $ \displaystyle f(x)=0$ in the form $ \displaystyle a\pm \sqrt{b}$, where $ \displaystyle a$ and $ \displaystyle b$ are integers.

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{3}}+k{{x}^{2}}-8x-8\\\\\ \ \ \ \text{Since}\ x-2\ \text{is a factor of }f(x),\\\\\ \ \ \ f(2)=0\\\\\therefore \ \ {{(2)}^{3}}+k{{(2)}^{2}}-8(2)-8=0\\\\\therefore \ \ 8+4k-16-8=0\\\\\therefore \ \ k=4\\\\\therefore \ \ f(x)={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ \text{Let}\ f(x)=(x-2)({{x}^{2}}+ax+b)\\\\\therefore \ \ (x-2)({{x}^{2}}+ax+b)={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ {{x}^{3}}+(a-2){{x}^{2}}+(b-2a)x-2b={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ a-2=4\Rightarrow a=6\\\\\ \ \ \ 2b=8\Rightarrow b=4\\\\\therefore \ \ f(x)=(x-2)({{x}^{2}}+6x+4)\\\\\ \ \ \ \text{When }f(x)=0,\\\\\ \ \ \ (x-2)({{x}^{2}}+6x+4)=0\\\\\ \ \ \ (x-2)({{x}^{2}}+6x+9-5)=0\\\\\ \ \ \ (x-2)\left[ {{{{(x+3)}}^{2}}-5} \right]=0\\\\\ \ \ \ x=2\ \text{or }{{(x+3)}^{2}}=5\\\\\therefore \ \ x=2\ \text{or }x=-3\pm \sqrt{5}\\\\\therefore \ \ \text{The non-integer roots of }f(x)=0\ \text{are }-3\pm \sqrt{5}.\end{array}$

4.       A function f is such that $ \displaystyle f(x)=4x^3+4x^2+ax+b$. It is given that $ \displaystyle 2x−1$ is a factor of both $ \displaystyle f(x)$ and $ \displaystyle f'(x)$.

(i) Find the value of $ \displaystyle a$ and $ \displaystyle b$.

(ii) Hence find the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x+3$.

(iii) Express $ \displaystyle f(x)$ in the form $ \displaystyle f(x)=(2x−1)(px^2+qx+r)$, where $ \displaystyle p, q$ and $ \displaystyle r$ are integers to be found.

(iv) find the values of $ \displaystyle x$ for which $ \displaystyle f(x) = 0$.

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)=4{{x}^{3}}+4{{x}^{2}}+ax+b\\\\\therefore \ \ {f}'(x)=12{{x}^{2}}+8x+a\\\\\ \ \ \ 2x-1\ \text{is a factor of both}\ f(x)\ \text{and}\ {f}'(x).\end{array}$

$ \displaystyle \therefore \ \ f\left( {\frac{1}{2}} \right)=0$

$ \displaystyle \therefore \ \ 4{{\left( {\frac{1}{2}} \right)}^{3}}+4{{\left( {\frac{1}{2}} \right)}^{2}}+a\left( {\frac{1}{2}} \right)+b=0$

$ \displaystyle \ \ \ \ \frac{1}{2}+1+\frac{a}{2}+b=0$

$ \displaystyle \therefore \ \ \ a+2b=-3$

$ \displaystyle \ \ \ \ \ \text{And}\ {f}'\left( {\frac{1}{2}} \right)=0$

$ \displaystyle \therefore \ \ \ 12{{\left( {\frac{1}{2}} \right)}^{2}}+8\left( {\frac{1}{2}} \right)+a=0$

$ \displaystyle \begin{array}{l}\therefore \ \ \ a=-7\\\\\therefore \ \ \ -7+2b=-3\ \Rightarrow b=2\\\\\therefore \ \ \ f(x)=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\ \ \ \ \ \text{When}\ f(x)\ \text{is divided by}x+3,\\\\\ \ \ \ \ \text{the remainder is}\ f(-3).\\\\\therefore \ \ \ f(-3)=4{{(-3)}^{3}}+4{{(-3)}^{2}}-7(-3)+2=-49\\\\\ \ \ \ \ \text{When}\ f(x)=(2x-1)(p{{x}^{2}}+qx+r),\\\\\ \ \ \ \ \ (2x-1)(p{{x}^{2}}+qx+r)=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\ \ \ \ \ \ 2p{{x}^{3}}+(2q-p){{x}^{2}}+(2r-q)x-r=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\therefore \ \ \ \ r=-2\\\\\ \ \ \ \ \ 2r-q=-7\Rightarrow -4-q=-7\Rightarrow q=3\\\\\ \ \ \ \ \ 2q-p=4\Rightarrow 6-p=4\Rightarrow p=2\\\\\therefore \ \ \ \ f(x)=(2x-1)(2{{x}^{2}}+3x-2)\\\\\ \ \ \ \ \text{When}\ f(x)=0,(2x-1)(2{{x}^{2}}+3x-2)=0\\\\\therefore \ \ \ \ (2x-1)(2x-1)(x+2)=0\end{array}$

$ \displaystyle \therefore \ \ \ \ x=\frac{1}{2}\ \text{or}\ x=-2$

5.       The remainder when the expression $ \displaystyle x^3+9x^2+bx+c$ is divided by $ \displaystyle x-2$ is twice the remainder when the expression is divided by $ \displaystyle x-1$. Show that $ \displaystyle c = 24$. Given that $ \displaystyle x+8$ is a factor of $ \displaystyle x^3+9x^2+bx+24$, show that the equation $ \displaystyle x^3+9x^2+bx+c=0$ has only one real root.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ f(x)={{x}^{3}}+9{{x}^{2}}+bx+c\\\\\ \ \ \ \text{By the problem},\\\\\ \ \ \ f(2)=2f(1)\\\\\ \ \ \ {{2}^{3}}+9\cdot {{2}^{2}}+2b+c=2\left( {{{1}^{3}}+9\cdot {{1}^{2}}+b+c} \right)\\\\\ \ \ \ 8+36+2b+c=2+18+2b+2c\\\\\therefore \ \ c=24\\\\\ \ \ \ \text{Since}\ x+8\ \text{is a factor of}\ {{x}^{3}}+9{{x}^{2}}+bx+24,\\\\\ \ \ \ f(-8)=0\\\\\ \ \ \ {{(-8)}^{3}}+9{{(-8)}^{2}}+b(-8)+24=0\\\\\ \ \ \ {{(-8)}^{3}}+9{{(-8)}^{2}}+b(-8)+24=0\\\\\therefore \ \ b=11\\\\\therefore \ f(x)={{x}^{3}}+9{{x}^{2}}+11x+24\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{2}}+\ \ x+\ \ 3\\x+8\ \ \overline{\left){\begin{array}{l}{{x}^{3}}+9{{x}^{2}}+11x+24\\\underline{{{{x}^{3}}+8{{x}^{2}}}}\\\ \ \ \ \ \ \ \ {{x}^{2}}+11x\\\ \ \ \ \ \ \ \ \underline{{{{x}^{2}}+\ \ 8x}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x+24\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{3x+24}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$ \displaystyle \therefore \ \ f(x)=(x+8)({{x}^{2}}+x+3)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =(x+8)\left[ {{{x}^{2}}+2\left( {\frac{1}{2}} \right)x+\frac{1}{4}+\frac{{11}}{4}} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =(x+8)\left[ {{{{\left( {x+\frac{1}{2}} \right)}}^{2}}+\frac{{11}}{4}} \right]$

$ \displaystyle \ \ \ \ \text{When}\ f(x)=0,(x+8)\left[ {{{{\left( {x+\frac{1}{2}} \right)}}^{2}}+\frac{{11}}{4}} \right]=0$

$ \displaystyle \therefore \ \ x=-8\ \text{or}\ {{\left( {x+\frac{1}{2}} \right)}^{2}}=-\frac{{11}}{4}$

$ \displaystyle \ \ \ \ \text{Since}\ {{\left( {x+\frac{1}{2}} \right)}^{2}}\ge 0,{{\left( {x+\frac{1}{2}} \right)}^{2}}=-\frac{{11}}{4}\ \text{is impossible}\text{.}$

$ \displaystyle \therefore \ \ \text{There is no other real solution for }{{x}^{3}}+9{{x}^{2}}+11x+24=0.$

6.       Given that the remainder when $ \displaystyle f(x) = x^3 - x^2 + ax$ is divided by $ \displaystyle x + a$, where $ \displaystyle a > 0$, is twice the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x - 2a$, find the value of $ \displaystyle a$. Find also the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x – 2$.

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{3}}-{{x}^{2}}+ax\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ f(-a)=2f(2a)\\\\\ \ \ \ -{{a}^{3}}-{{a}^{2}}-{{a}^{2}}=2\left( {8{{a}^{3}}-4{{a}^{2}}+2{{a}^{2}}} \right)\\\\\ \ \ \ 17{{a}^{3}}-2{{a}^{2}}=0\\\\\ \ \ \ {{a}^{2}}(17a-2)=0\end{array}$

$ \displaystyle \ \ \ \ \text{Since}\ a>0,17a-2=0\Rightarrow a=\frac{2}{{17}}$

$ \displaystyle \therefore \ \ f(x)={{x}^{3}}-{{x}^{2}}+\frac{2}{{17}}x$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{When}\ f(x)\ \text{is divided by}\ x-2,\\\ \ \ \ \text{the remainder is}\ f(2).\end{array}$

$ \displaystyle \therefore \ \ f(2)={{(2)}^{3}}-{{(2)}^{2}}+\frac{2}{{17}}(2)=\frac{{72}}{{17}}$

7.       Given that $ \displaystyle f(x) = x^{2n} - ( p + 1)x^2 + p$, where $ \displaystyle n$ and $ \displaystyle p$ are positive intergers. Show that $ \displaystyle x - 1$ is a factor of $ \displaystyle f(x)$ for all values of $ \displaystyle p$. When $ \displaystyle p = 4$, find the value of $ \displaystyle n$ for which $ \displaystyle x - 2$ is a factor of $ \displaystyle f(x)$ and, for this case, hence factorise $ \displaystyle f(x)$ completely.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)={{x}^{{2n}}}-(p+1){{x}^{2}}+p\ \text{where}\ n,p\in {{J}^{+}}\\\\\ \ \ \ \ f(1)={{(1)}^{{2n}}}-(p+1){{(1)}^{2}}+p\\\\\ \ \ \ \ \text{Since}\ n,p\in {{J}^{+}},{{(1)}^{{2n}}}=1\ \text{and}-(p+1)=-p-1.\\\\\therefore \ \ \ f(1)=1-p-1+p=0\\\\\therefore \ \ \ x-1\ \text{is a factor of }f(x)\ \text{for all values of}\ p.\\\\\ \ \ \ \ \text{When}\ p=4,f(x)={{x}^{{2n}}}-5{{x}^{2}}+4.\\\\\ \ \ \ \ x-2\ \text{is a factor of }f(x).\\\\\therefore \ \ \ f(2)=0\\\\\ \ \ \ \ {{2}^{{2n}}}-5\cdot {{2}^{2}}+4=0\\\\\therefore \ \ \ {{2}^{{2n}}}=16\Rightarrow n=2\end{array}$

8.       The polynomial $ \displaystyle f(x)$ is given by $ \displaystyle f(x) = ax^3 + 11x^2 + cx - 60$ , where $ \displaystyle a$ and $ \displaystyle c$ are constants. If the roots of $ \displaystyle f(x) = 0$ are $ \displaystyle 2 , - 3$ and $ \displaystyle k$, find the values of $ \displaystyle a , c$ and $ \displaystyle k$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)=a{{x}^{3}}+11{{x}^{2}}+cx-60\\\\\ \ \ \ \ \text{Since the roots of}\ f(x)=0\ \text{are}\ 2,-3\ \text{and}\ k,\\\\\ \ \ \ \ a{{x}^{3}}+11{{x}^{2}}+cx-60=(x-2)(x+3)(x-k)\\\\\therefore \ \ \ a{{x}^{3}}+11{{x}^{2}}+cx-60={{x}^{3}}+(1-k){{x}^{2}}-(6+k)x+6k\\\\\ \ \ \ \ \text{Equating the corresponding terms,}\\\\\ \ \ \ \ a=1,\\\\\ \ \ \ \ 6k=-60\Rightarrow k=-10\\\\\ \ \ \ \ c=-(6+k)=-(6-10)=4\end{array}$

9.       Given that $ \displaystyle 4x^4 - 9a^2x^2 + 2(a^2 - 7) x - 18$ is exactly divisible by $ \displaystyle 2x - 3a$, show that $ \displaystyle a^3 - 7a - 6 = 0$ and hence find the possible values of $ \displaystyle a$.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)=4{{x}^{4}}-9{{a}^{2}}{{x}^{2}}+2({{a}^{2}}-7)x-18\\\\\ \ \ \ \ \text{Since }f(x)\ \text{is exactly divisible by}\ 2x-3a,\end{array}$

$ \displaystyle \ \ \ \ \ f\left( {\frac{{3a}}{2}} \right)=0.$

$ \displaystyle \therefore \ \ \ 4{{\left( {\frac{{3a}}{2}} \right)}^{4}}-9{{a}^{2}}{{\left( {\frac{{3a}}{2}} \right)}^{2}}+2({{a}^{2}}-7)\left( {\frac{{3a}}{2}} \right)-18=0$

$ \displaystyle \ \ \ \ \ \ 4\left( {\frac{{81{{a}^{4}}}}{{16}}} \right)-9{{a}^{2}}\left( {\frac{{9{{a}^{2}}}}{4}} \right)+3{{a}^{3}}-21a-18=0$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ {{a}^{3}}-7a-6=0\\\\\ \ \ \ \ \text{Let }g(a)={{a}^{3}}-7a-6\\\\\ \ \ \ \ g(-1)=-1+7-6=0\\\\\therefore \ \ \ \ a+1\ \text{is a factor of }g(a).\\\\\therefore \ \ \ \ \text{Let }g(a)=(a+1)({{a}^{2}}+pa+q),\\\ \ \ \ \ \ \text{where }p\ \text{and }q\ \text{are any real constants}\text{.}\\\\\therefore \ \ \ g(a)={{a}^{3}}+(p+1){{a}^{2}}+(p+q)a+q\\\\\therefore \ \ \ {{a}^{3}}-7a-6={{a}^{3}}+(p+1){{a}^{2}}+(p+q)a+q\\\\\ \ \ \ \ \text{Equating the corresponding terms,}\\\\\ \ \ \ \ p+1=0\Rightarrow p=-1\ \ q=-6\\\\\therefore \ \ \ {{a}^{3}}-7a-6=(a+1)({{a}^{2}}-a-6)\\\\\therefore \ \ \ {{a}^{3}}-7a-6=(a+1)(a+2)(a-3)\\\\\ \ \ \ \text{When }{{a}^{3}}-7a-6=0,\\\\\ \ \ \ (a+1)(a+2)(a-3)=0\\\\\therefore \ \ a=-1\ \text{or }a=-2\ \text{or }a=3.\end{array}$

10.      Solve the equation $ \displaystyle 4x^3 + 3x^2 -16x = 12$. Hence find the value of $ \displaystyle x$ such that $ \displaystyle 4e^{3x} + 3e^{2x} - 16e^x = 12$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x=12\\\\\therefore \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x-12=0\\\\\ \ \ \ \ \text{Let}\ f(x)=4{{x}^{3}}+3{{x}^{2}}-16x-12\\\\\ \ \ \ \ f(2)=4{{(2)}^{3}}+3{{(2)}^{2}}-16(2)-12\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =32+12-32-12\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =0\\\\\therefore \ \ \ (x-2)\ \text{is a factor of }f(x).\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{x}^{2}}+11x+\ \ 6\\\ \ \ \ \ x-2\overline{\left){\begin{array}{l}4{{x}^{3}}+3{{x}^{2}}-16x-12\\\underline{{4{{x}^{3}}-8{{x}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ 11{{x}^{2}}-16x\\\ \ \ \ \ \ \ \ \ \ \underline{{11{{x}^{2}}-22x}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x-12\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{6x-12}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ f(x)=(x-2)\left( {4{{x}^{2}}+11x+\ \ 6} \right)\\\\\therefore \ \ f(x)=(x-2)(x+2)(4x+3)\\\\\ \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x-12=0\\\\\therefore \ \ (x-2)(x+2)(4x+3)=0\end{array}$

$ \displaystyle \therefore \ \ x=2\ \text{or}\ x=-2\ \text{or}\ x=-\frac{3}{4}$

$ \displaystyle \begin{array}{l}\ \ \ \ 4{{e}^{{3x}}}+3{{e}^{{2x}}}-16{{e}^{x}}=12\\\\\ \ \ \ 4{{({{e}^{x}})}^{3}}+3{{({{e}^{x}})}^{2}}-16{{e}^{x}}-12=0\end{array}$

$ \displaystyle \therefore \ \ {{e}^{x}}=2\ \text{or}\ {{e}^{x}}=-2\ \text{or}\ {{e}^{x}}=-\frac{3}{4}$

$ \displaystyle \ \ \ \ \text{Since}\ {{e}^{x}}>0\ \text{for all }x\in R,$

$ \displaystyle \ \ \ \ {{e}^{x}}=-2\ \text{or}\ {{e}^{x}}=-\frac{3}{4}\ \text{is impossible}$

$ \displaystyle \therefore \ \ {{e}^{x}}=2\Rightarrow x=\ln 2\ $

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