Current Affairs 11th Class

Notes - Mathematics Olympiads - Complex Number

Complex Number   Complex Numbers: "Complex number is the combination of real and imaginary number".   Definition: A number of the form$x+iy,$, where $x,y\in R$ and $i=\sqrt{-1}$ is called a complex number and (i) is called iota. A complex number is usually denoted by z and the set of complex number is denoted by C.   $\Rightarrow C=\{x+iy:x\in R,\,Y\in R,\,i=\sqrt{-1}\}$   For example: $5+3i,$ $-1+i,$ $0+4i,$ $4+0i$ etc. are complex numbers.   Note: Integral powers of iota (i)   since $i=\sqrt{-1}$ hence we have ${{i}^{2}}=-1,$ and ${{i}^{4}}=1.$   Conjugate of a complex number: If a complex number $z=a+i\,b,$ $(a,b)\in R,$ then its conjugate is defined as $\overline{z}=a-ib$                              Hence, we have   $\operatorname{Re}(z)=\frac{z+\overline{z}}{2}$ and $\operatorname{Im}(z)=\frac{z-\overline{z}}{2i}$   $\Rightarrow$ Geometrically, the conjugate of z is the reflection or point image of z in the real axis.   e.g.                     (i)         $z=3-4i$                         $z=3-(-4)=3+4i$                         (ii)        $z=2+5i$                         $\overline{z}=2-5i$                         (iii)       $\overline{z}=5i$                         $\overline{z}=-5i$
• Operation (i.e. Addition and Multiplication) of Complex Number
e.g.       ${{z}_{1}}=a+ib=(a,b)$ ${{z}_{2}}=c+id=(c,d)$ Then ${{z}_{1}}+{{z}_{2}}=(a+c,b+d)=(a+c)+i(b+d)$ ${{z}_{1}}.{{z}_{2}}=(a,b)(c,d)=(ac-bd,bc+ad)$ ${{z}_{1}}-{{z}_{2}}=(a-c,b-d)\Rightarrow {{z}_{1}}-{{z}_{2}}=(a-c,b-a)$
• Some Properties of Conjugate Number
(a)        $(\overline{z})=z$                                                          (b)        $z+\overline{z}$ if & only if z is purely real   (c)        $z+-\overline{z}$ iff z  is purely imaginary            (d)        $z+\overline{z}=2\operatorname{Re}(z)=2p$ Real part of z   (e)        $z-\overline{z}=2i\,\,lm(z)=2i$ Imaginary part of z        (f)         $\overline{{{z}_{1}}+{{z}_{2}}}={{\overline{z}}_{1}}\pm {{\overline{z}}_{2}}$   (g)        $\overline{{{z}_{1}}.{{z}_{2}}}={{\overline{z}}_{1}}.{{\overline{z}}_{2}}$                                         (h)        $\left( \frac{\overline{{{z}_{1}}}}{{{z}_{2}}} \right)=\overline{\frac{{{z}_{1}}}{{{z}_{2}}}},{{z}_{2}}\ne 0.$   (i)         $\text{f}i\,\,z=\text{f}({{z}_{1}})$ then $\overline{z}=\text{f}(\overline{{{z}_{1}}})$                              (j)         $(\overline{{{z}^{4}}})={{(\overline{z})}^{4}}$   (k)${{z}_{1}}{{\overline{z}}_{2}}+{{\overline{z}}_{1}}{{z}_{2}}=2\operatorname{Re}({{z}_{1}}.{{\overline{z}}_{2}})=2.Re({{z}_{1}}.{{\overline{z}}_{2}})$
• Square Roots of a Complex Number
Let z = x + iy. Let the square root of a complex number $z=x+iy$ is $u+iv$ i.e. $\sqrt{x+iy}=u+iv$                         ......... (1)   squaring both sides we have $x+iy={{(u+iv)}^{2}}={{u}^{2}}+{{(iv)}^{2}}+2.u.(iv)$ $={{u}^{2}}-{{v}^{2}}+2iuv$   Equating real & imaginary part, we have $x={{u}^{2}}-{{v}^{2}}$                                        ........ (2) $y=2uv$                                                          ........ (3)   Now, ${{u}^{2}}+{{v}^{2}}=\sqrt{{{({{u}^{2}}+{{v}^{2}})}^{2}}+4{{u}^{2}}{{v}^{2}}}$   $=\sqrt{{{x}^{2}}+{{y}^{2}}}$                                           ........ (4)   Solving (2) & (4), we have   $2{{u}^{2}}=\sqrt{{{x}^{2}}+{{y}^{2}}+x}$   ${{u}^{2}}=\sqrt{\frac{{{x}^{2}}+{{y}^{2}}}{2}}+\frac{x}{2}$                          $\therefore \,\,\,u=\pm \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}-x}{2}}$   Similarly, $v=\pm \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}-x}{2}}$   From (3), we can determine the sign of xy as, if xy > 0 then x & y will be the same sign.   $\sqrt{x+iy}=\pm \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+x}{2}},+i\sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}-x}{2}}$
• If $\mathbf{xy<0}$ then square root of $\mathbf{z}\sqrt{\mathbf{x+iy}}$
$=\pm \left( \sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+x}{2}}-i\sqrt{\frac{\sqrt{{{x}^{2}}+{{y}^{2}}-x}}{2}} \right)$   In short form it can be written. i.e.        $xy>0$   Then square root of $z=x+iy$ i.e. $\sqrt{x+iy}=\pm \left( \sqrt{\frac{\left| z \right|+x}{2}}+i\sqrt{\frac{\left| z \right|-x}{2}} \right)$ e.g. Find the square root of 7+24i   Giveen $z=\sqrt{x+iy}=\sqrt{7+24i}$   $\left| \,z\, \right|=\sqrt{{{7}^{2}}+{{(24)}^{2}}}=\sqrt{625}=25$   Square root of $z=7+24i$   $=\pm \sqrt{\frac{25+7}{2}}=i\sqrt{\frac{25-7}{2}}=\pm \left( \sqrt{\frac{32}{2}}+\sqrt{\frac{18}{2}} \right)$   $=\pm (4+i3)=(4+3)$ or$-(4+3i)$
• Polar form of a complex number
Let us consider 0 as origin & OX as x-axis & OY as Y-axis.     Let $z=x+iy$ is a complex no. It is represented as $P(x,y).$ more...

Notes - Mathematics Olympiads - Continuity Differentiability

Continuity and Differentiability of a Function   Introduction   The word 'continuous' means without any break or gap. If the graph a function has no break or gap or jump, then it is said to be continuous. A function which is not continuous is called a discontinuous function. While studying graphs of functions, we see that graphs of functions sinx, x, cosx, etc. are continuous on R but greatest integer functions [x] has break at every integral point, so it is not continuous. Similarly tanx, cotx, seex, 1/x etc. are also discontinuous function on R.   Continuous Function                            Discontinuous Function                                       Continuity of a Function at a Point   A function f(x) is said to be continuous at a point x = a of its domain if and only if it satisfies the following three condition:   (i)         f(a) exist. ('a' line in the domain of f) (ii)        $\underset{x+a}{\mathop{\lim }}\,$ f(x) exist e.e; $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,$ f(x)= $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,$ f(x) or R.H.L= L.H.L (iii)       $\underset{x+a}{\mathop{\lim }}\,$ f(x) = f(a) (limit equals the value of function)   Continuity from Left and Right A function f(x) is said to be continuous at a point x = a of its domain if and only if it satisfies the following three condition:   (i)         f(a) exist. ('a' line in the domain of f) (ii)        $\underset{x+a}{\mathop{\lim }}\,$ f(x) exist e.e; $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,$ f(x)= $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,$ f(x) or R.H.L= L.H.L (iii)       $\underset{x+a}{\mathop{\lim }}\,$ f(x) = f(a) (limit equals the value of function)   Cauchy's Definition of Continuity A function f is said to be continuous at a point a of its domain D if for every $\varepsilon >o$ there exists $\delta >o$ (dependent of$\varepsilon$) such that $\left| x-a \right|<\delta$ $\Rightarrow \left| \text{f}(x)-\text{f(a)} \right|<\varepsilon$   Comparing this definition with the definition of limit we find that f(x) is continuous at x = a if $\underset{x\to a}{\mathop{\lim }}\,$f(x) exists and is equal to f(a) i.e; $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,$ f(a) $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,$ f(x).   Continuity from Left and Right             Function f(x) is said to be   (i)         Left continuous at x= a if $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\,\,\text{f(x)=f(a)}\text{.}$ (ii)        Right continuous at x = a if $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\,\,\text{f(x)=f(a)}\text{.}$ Thus a function f(x) is continuous at a point x = a if it is left continuous as well as right continuous at x = a.     Differentiation   The function, f(x) is differentiable at point P, iff there exists a unique tangent at point P. In other words, f(x) is differentiable at point P iff the curve does not have pas a corner point i.e; the function is not differentiable at those more...

Notes - Mathematics Olympiads - Trigonometry

Trigonometry   Key Points to Remember Trigonometry: It is derived/contained from two greek words trigon and metron means that the measurement of three sides of the triangle.     A triangle has three vertex, three angles and three sides in above figure. vertex be A, Band C Angle be $\angle ABC,$ $\angle BCA,$ and $\angle CAB,$or $\angle BAC$ According to sides, type of triangle be (a)        Equilateral Triangle: All sides are equal (b)        Isosceles Triangle: Two sides are equal (c)        Scalene Triangle: All sides are different. According to Angle, There are three types of triangles: (a)        Acute Angle Triangle              (b)        Obtuse Angle Triangle              (c)        Right angle Triangle
• Point to Remember
Sum of all angles in a triangle is $180{}^\circ$
• Trigonometric Ratio
In right angle triangle, basically, there are three triagonometrical ratio: sin, cosine and tangent. i.e. total trigonometrical ratio are six: sin, cos, tan, cosec, sec, cot Let us consider a right angle triangle ABC in which     Then $\angle A=90{}^\circ -\theta$ $\sin \theta =\frac{P}{h}=\frac{AB}{AC},\,$$\cos ec\,\theta =\frac{h}{P}$ $\cos \,\theta =\frac{b}{h},$ $sec\theta =\frac{h}{b}$ $\tan \theta =\frac{P}{b},$ $\cot \theta =\frac{b}{P}$ Trigonometrical ratio shows the relation between angle and sides of the triangle.
• Product of trigonometrical ratio
 $\sin \theta \times \operatorname{cosec}\theta =1$ $\tan \theta \times \cot \theta =1$ $\cos \theta \times \sec \theta =1$

• Some Important Formula
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$            …….. (1) $LHS={{\left( \frac{P}{h} \right)}^{2}}+{{\left( \frac{b}{h} \right)}^{2}}=\frac{{{P}^{2}}}{{{h}^{2}}}+\frac{{{b}^{2}}}{{{h}^{2}}}$
 ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$ ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$
$=\frac{{{P}^{2}}+{{b}^{2}}}{{{h}^{2}}}=\frac{{{h}^{2}}}{{{h}^{2}}}=1$ [By pythagoras theorem]     Dividing equation (1) on both sides by ${{\cos }^{2}}q,$we have $\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }+\frac{{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }=\frac{1}{{{\sin }^{2}}\theta }$ $\Rightarrow {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta$ $\Rightarrow {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ Again, dividing (1) by ${{\sin }^{2}}\theta$, we have             $\frac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }+\frac{{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }=\frac{1}{{{\sin }^{2}}\theta }$             $\Rightarrow 1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta$             $\Rightarrow \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$              ${{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta -1$   Angle: Inclination between two sides (arms), is said to be an angle, e.g.   Angle may be positive or negative When angle be measured in anticlockwise direction then it is positive otherwise negative.
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Notes - Mathematics Olympiads - Limits and Derivatives

Limits and Derivatives   Key Points to Remember   A number, ($\ell$ is said to be the limit of the function $\text{y=f}(x)$ at$x=a$, then $\exists$ a positive number, $\in \,\,>\,\,0$ corresponding to the small positive number $\delta \,>\,\,0$ such that $\left| \text{f(x)-}\left. \ell \right| \right.\,<\,\varepsilon ,$ provided $\left| x-\left. a \right| \right.\,<\,\delta$ Evidently, $\underset{x\to \,a}{\mathop{\lim }}\,\,\,f(x)=\ell$ We have to learn how to evaluate the limit of the function $y=f(x)$ Only three type of function can be evaluate the limit of the function.
• Algebraic function
• Trigonometric function
• Logarithmic and exponential function
• Now, basically, there are three method to evaluate the limit of the function.
• By Formula Method
i.e.        $\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}$
• Factorisation Method: In this matter numerator and denominator are factorised. The common factors are cancelled and the rest is the result.
• Rationalisation Method: Rationalisation is followed when we have fractional overs $\left( like\frac{1}{2},\frac{1}{3}etc. \right)$ on expressions in numerator or denominator in both. After rationalisation the terms are factorised which on cancellation gives the result.

• Q. $\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-27}}{\mathbf{x-3}}\mathbf{=}\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-(3}{{\mathbf{)}}^{\mathbf{3}}}}{\mathbf{x-3}}\mathbf{=3(3}{{\mathbf{)}}^{\mathbf{3-1}}}$
•   $={{3.3}^{2}}=27$
• $\underset{\mathbf{x1}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{15}}}\mathbf{-1}}{{{\mathbf{x}}^{\mathbf{10}}}\mathbf{-1}}$
•   Dividing numerator and denomiator by (x -1), when   $=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{15}}-{{1}^{15}}}{\frac{x-1}{\frac{{{x}^{^{10}}}-{{(1)}^{10}}}{x-1}}}=\frac{15.{{(1)}^{15-1}}}{10.{{(1)}^{10-1}}}=\frac{15}{10}=\frac{3}{2}$
• $\underset{\mathbf{x0}}{\mathop{\mathbf{lim}}}\,\frac{\sqrt{\mathbf{1+x-1}}}{\mathbf{x}}\mathbf{=}\underset{\mathbf{x0}}{\mathop{\mathbf{lim}}}\,\frac{\mathbf{(}\sqrt{\mathbf{1+x-1)}}}{\mathbf{x}}\mathbf{\times }\frac{\sqrt{\mathbf{1+x}}\mathbf{+1}}{\sqrt{\mathbf{1+x}}\mathbf{+1}}$
•               $=\underset{x\to 0}{\mathop{\lim }}\,\frac{1+x-1}{x(\sqrt{1+x+1)}}[\therefore \,\,{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)]=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{(\sqrt{1+x+1)}};$ applying limit, $x\to 0$,we have $=\frac{1}{1+1}=\frac{1}{2}$
• $\underset{\mathbf{x2}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+4}}{\mathbf{x-2}}\mathbf{=}\underset{\mathbf{x2}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}{{\mathbf{2}}^{\mathbf{2}}}}{\mathbf{x-2}}$
•                           $\underset{x\to 2}{\mathop{\lim }}\,\frac{(x+2)(x-2)}{x-2}=2+2=4$   Q,        $\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-81}}{\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-5x-3}}\mathbf{=}\underset{\mathbf{x3}}{\mathop{\mathbf{lim}}}\,\frac{{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-(3}{{\mathbf{)}}^{\mathbf{4}}}}{\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-6x+x-3}}$   $\underset{x\to 3}{\mathop{\lim }}\,\frac{{{({{x}^{2}})}^{2}}-{{({{3}^{2}})}^{2}}}{2x(x-3)+1(x-3)}=\underset{x\to 3}{\mathop{\lim }}\,\frac{({{x}^{2}}-9)-({{x}^{2}}+9)}{(x-3)(2x+1)}=\underset{x\to 3}{\mathop{\lim }}\,\frac{(x+3)(x-3)({{x}^{2}}+9)}{(x-3)(2x+1)}$   Applying limit as $x\to 3$, we have   $=\frac{(3+3)({{3}^{2}}+9)}{(2\times 3+1)}=\frac{6\times 18}{7}=\frac{108}{7}$
• Properties of Limits
If $\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)=}\ell$and $\underset{x\to a}{\mathop{\lim }}\,\,\,g\text{(x)=m}$ then the following results are true:   (a)        $\underset{x\to a}{\mathop{\lim }}\,\,\left[ f(x)\pm \,g\text{(x)} \right]\text{=}\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}\pm \underset{x\to a}{\mathop{\lim }}\,\,\,g(x)=\ell +m$   (b)        $\underset{x\to a}{\mathop{\lim }}\,\,\left\{ k.\text{f(x)} \right\}\text{=k}\text{.}\,\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}\text{.}\,=k.\ell .$   (c)        $\underset{x\to a}{\mathop{\lim }}\,\,\left\{ k.\text{f(x)} \right\}\text{=k}\text{.}\,\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}\text{.}\,\underset{x\to a}{\mathop{\lim }}\,\,\,g(x)=\ell +m.$               (d)        $\underset{x\to a}{\mathop{\lim }}\,\,\left\{ \frac{\text{f(x)}}{g(x)} \right\}\text{=}\frac{\underset{x\to a}{\mathop{\lim }}\,\,\,\text{f(x)}}{\underset{x\to a}{\mathop{\lim }}\,\,\,g\text{(x)}}=\frac{\ell }{m}[i\text{f}\,\,\text{m}\ne \text{0 }\!\!]\!\!\text{ }$               (e)        $\underset{x\to a}{\mathop{\lim }}\,\,\text{f(x)=+}\infty$or$\text{-}\,\infty$, then $\underset{x\to a}{\mathop{\lim }}\,\,\frac{1}{\text{f(x)}}\text{=0}$               (f)         $\underset{x\to a}{\mathop{\lim }}\,\,\,\log \left\{ g(x) \right\}\,=\log \left[ \underset{x\to a}{\mathop{\lim \,g(x)}}\, \right]=\log \,m,\,m>0\frac{1}{\text{f(x)}}\text{=0}$               (g)        $\underset{x\to a}{\mathop{\lim }}\,\,\,{{\left[ f(x) \right]}^{g(x)}}\,={{\{\underset{x\to a}{\mathop{\lim }}\,\,\text{f(x) }\!\!\}\!\!\text{ }}^{\underset{x\to a}{\mathop{\lim \,\,\,g(x)}}\,}}={{\ell }^{m}}$
• Remember these results

• ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$
• ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$
• ${{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}+ab+{{b}^{2}})$
• ${{a}^{4}}-{{b}^{4}}={{({{a}^{2}})}^{2}}-{{({{b}^{2}})}^{2}})=({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})-(a+b)(a-b)({{a}^{2}}+{{b}^{2}})$
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• $\sum{n=1+2+3+4....n=\frac{n(n+1)}{2}}$
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• $\sum{{{n}^{2}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}....{{n}^{2}}=\frac{n(n+1)(2n+1)}{6}}$
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• ${{\sum{{{n}^{3}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}....{{n}^{3}}=\left\{ \frac{n(n+1)}{2} \right\}}}^{2}}$
•
• $\log (1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.....\infty$
•
• $\log (1-x)=-x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.....to\,\infty$
•   10. ${{e}^{x}}=1+x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}+.....to\infty$   11. ${{e}^{-x}}=1-x+\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}+\frac{{{x}^{4}}}{4}-.....to\infty$
• $\frac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+.....to\,\infty ,$ when x<1
•   13.  ${{a}^{x}}=1+x\log a+\frac{{{(x\log a)}^{2}}}{2}+.....to\infty$
• $\sin x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}......to\infty$
•     15. $\cos x=1-\frac{{{x}^{2}}}{2}+\frac{{{x}^{4}}}{4}-\frac{{{x}^{6}}}{6}+......to\infty$   16. $\tan x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+.....to\infty$   17. ${{\tan }^{-1}}x=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{{{x}^{7}}}{7}+.....0,\frac{\pi }{x}\le x\le \frac{\pi }{4}$
• Evaluation of limit of Trigonometrically Functions
Some basic formula,
• $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \,x}{x}=1$
• $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan \,x}{x}=1$
• $\underset{x\to 0}{\mathop{\lim }}\,\cos \,x=1$
• $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x}{x}=1$
• $\underset{x\to 0}{\mathop{\lim }}\,\frac{ta{{m}^{-1}}x}{x}=1$
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• Some useful results of evaluation of limit of logarthmic and expontential function

• $\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+x)}{x}=1$

• $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{x}}-1}{x}={{\log }_{e}}a$
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Notes - Mathematics Olympiads - Relation and Function

Relation and Function   Let $A=\{1,\,2,\,3,4,\}$, $B=\{2,\,3\}$ $A\times B=\{1,\,2,3,\,4,\}\times \{2,3\}=\{(1,2),(2,2),(3,2),(4,2),(1,3),(2,3),(3,3),(4,3)\}$   Let we choose an arbitrary set:   $R=[(1,2),(2,2),(1,3),(4,3)]$ Then R is said to be the relation between a set A to B.   Definition   Relation R is the subset of the Cartesian Product$A\times B$. It is represented as $R=\{(x,y):x\in A\,$ and $y\in B\}$ {the 2nd element in the ordered pair (x, y) is the image of 1st element}   Sometimes, it is said that a relation on the set A means the all members / elements of the relation R be the elements / members of $A\text{ }\times \text{ }A$. e.g.      Let $A=\{1,\,2,\,3\}$ and a relation R is defined as $R=\{(x,y):x<y$ where $x,y\in A\}$   Sol.     $\because$$\mathbf{A=\{1,}\,\mathbf{2,}\,\mathbf{3\}}$ $A\times A=\{(1,1),(2,2),(3,3),(2,1),(3,1),(1,2),(3,2),(1,3),(2,3)\}$             $\because \,\,\,\,R=\,\,\,\because x<y$             $\because \,\,\,\,R=\{(x,y):x<y,and\,x,y\in A\}=\{(1,2),(2,3),(1,3)$   Note: Let a set A has m elements and set B has n elements. Then $n(A\times B)$ be $m\times n$elements so, total number of relation from A to B or between A and B be${{2}^{m\times n}}$.
• A relation can be represented algebraically either by Roster method or set builder method.
Types of Relation
• Void Relation: A relation R from A to B is a null set, then R is said to be void or empty relation.

• Universal Relation: A relation on a set A is said to be universal relation, if each element of A is related to or associated with every element of A.

• Identity Relation: A relation ${{I}_{x}}\{(x,x):x\in A\}$ on a set A is said to be identity relation on A.

• Reflexive relation: A relation Ron the set A is said to be the reflexive relation. If each and every element of set A is associated to itself. Hence, R is reflexive iff $(a,\,a)\in R\,\,\forall \,\,a\in A$.
e.g.       $\Rightarrow \,\,A=\{1,\,2,\,3,\,4\}$ $R=\{(1,\,3),(1,1),(2,3),(3,2),(2,2),(3,1),(3,3),(4,4)\}$is a reflexive relation on f.   Sol.     Yes, because each and every element of A is related to itself in R.
• Symmetric relation: A relation R on a set A is said to be symmetric relation iff. $(x,y)\in R\Rightarrow (y,x)\in R\,\,\forall \,\,x,y\in A$
i.e. $x\,R\,y\Rightarrow y\,R\,x\,\,\forall \,\,x,y\in A$ $\because$ xRy is read as x is R-related to y.
• Anti-symmetric relation: A relation which is not symmetric is said to be anti-symmetric relation.

• Transitive relation: Let A be any non-empty set. A relation R on set A is said to be transitive relation R iff $(x,y)\in R$ and $(y,z)\in R$ then $(x,z)\in R\,\,\forall \,\,x,y,z\in R.$
i.e.       $xRy$ and $yRz\Rightarrow xRz\,\,\forall \,\,x,\,y,\,z\in R$.
• Let $\mathbf{A=\{1,}\,\mathbf{2,}\,\mathbf{3,}\,\mathbf{4\}}$
• $A\times A=\{(1,,1),(2,1),(3,1),(4,1),(1,2),(2,2),(3,2),(4,2),(1,3),(2,3),(3,3),(4,3),(1,4),(2,4),(3,4),(4,4),$ ${{R}_{1}}=(1,1),(2,2),(3,2),(2,3),(3,3),(4,4)$ ${{R}_{2}}=(2,2),(1,3),(3,3),(3,1),(1,1)$ ${{R}_{3}}=(1,1),(2,2),(3,4),(3,3),(4,4)$   State about${{\mathbf{R}}_{\mathbf{1}}}$, ${{\mathbf{R}}_{\mathbf{2}}}$ and${{\mathbf{R}}_{\mathbf{3}}}$. Are they reflexive, symmetric, anti-symmetric or transitive relations?   Sol.     ${{R}_{1}}$ is symmetric as well as transitive relation for ${{R}_{2}}$. ${{R}_{2}}$is not reflexive because $(4,4)\in more... Notes - Mathematics Olympiads - Set Theory Set Theory A Set A set is the collection of things which is well-defined. Here well-defined means that group or collection of things which is defined distinguishable and distinct e.g. Let A is the collection of the group M = {cow, ox, book, pen, man}. Is this group collection is a set or not? Actually this is not a set. Because it is collection of things but it cannot be defined in a single definition. For example, A = {1, 2, 3, 4, 5,... n} Here, collection A is a set because A is group or collection of natural numbers. e.g. A = {a, e, i, o, u}= {x/x : vowel of English alphabet} Distinguish Between a Set and a Member e.g. A = {2, 5, 8, 7} \[2\in A$. It is read as 2 belongs to set A. Or 2 is the member/element of set A. $6\notin A$- It means 6 doesn't belong to set A. A set is represented by capital letter of English/Greek Alphabet. $\alpha ,\beta ,\gamma ,\delta$or A, B, C, U, S etc. and its all elements is closed with { } bracket. Hence what is the difference between 2 and {2} $\because$2 can be element of a set whereas {2} represents a set whose one elements is 2.   Note:    A set is the collection of distinguish and distinct things A set can be written/represented in the two form   (a)        Tabular form/Roaster form (b)        Set builder form e.g.
• $A=\{a,e,i,o,u\}\to$Tabular form/roaster form
• = {$x/x:$ a vowel of english letter} = It is said to be set-builder form Note:    $x:x$or $x/x$is read as x such that $x$
• A = {2, 4, 6, 8, 10} = {$x/x$, is an even numbers $\le$10}
•   Type of Sets   (i)         Null Set: A set which has no element, is said to be a null set and denoted by $\phi \,or\,\{\}$ e.g.       (a) a set of three-eyed men (b) Set of real solution of the equation ${{x}^{2}}+1=0,$ ${{x}^{2}}=-1\Rightarrow x=\sqrt{-1}=$imaginary   (ii)        Singleton Set: A set having single element is said to be singleton set. e.g.       A = {2}   (iii)       Finite Set: A set having definite and countable elements, is said to be finite set e.g.       $A=\{a,e,i,o,u\}$   (iv)       Infinite Set: A set having uncountable and indefinite elements is said to be infinite set. e.g.       a set of stars.   (v)        Equal set: Two or more than two sets are said to be equal sets if they have the same elements. e.g.       $A=\{1,\,2,\,5,\,7,\,8\}$, $B=\{2,\,5,8,1,7\}$. So, $A=B$
• Operations on Sets
• Union-operation
• Intersection operation
• Complement operation
• Difference operation
• e.g.       $A=\{1,\,2,\,3,\,5\}$, $B=\{2,\,5,7,\,8\}$                         $A\,\bigcup B=\{1,\,2,3,5,7,8\}$, $A\,\bigcap \,B=\{2,\,5\}$
• Union Set: Let A and B be two non-empty sets then A union B i.e. $A\,\bigcup B$is a set more...

Notes - Mathematics Olympiads - Sequence Series

Sequence and series
• Sequence: A sequence is a mapping of or function whose domain is the set of natural number and its range is the set of real number or complex number. But we will study the real number sequence only.
e.g.       (a) ${{x}_{1}},\,{{x}_{2}},{{x}_{3}}.....{{x}_{n}}.$   (b)        2, 4, 6, 8, 10..... (c)        1, 4, 7, 10, 13....... etc. These above example is a sequence of real no.   Note:    Sequence is also said to be progression. Generally there are three type of sequence (a)        Arithmatic progression or sequence (A.P.) (b)        Geometric progression (G.P.) (c)        Harmonic prograssion (H.P.)
• Arithmetic sequence: A sequence (Sn) is said to be an arithmetic sequence if difference between any term and its proceeding term give the constant quantity.
e.g. Sn: 3,8,13,18, 23..... we choose any term $13-5=5$ $23-18=5$ etc. The constant quantity is said to common difference (c.d.) in A.P.
• Remember Some Points about A.P.
(a)        ${{n}^{th}}$ term whose 1st term and common difference is given in A.P. is written as ${{t}_{n}}=a+(n-1).d$ where a = 1st term and d = comman difference n = no. of terms in the A.P.   (b)        Sum of the ${{n}^{th}}$ term in A.P. be   $Sn=\frac{n}{2}\{2a+(n-1).d\}=\frac{n}{2}\{a+a+(n-1).d\}$   $=\frac{n}{2}\{a+{{t}_{n}}\}=\frac{n}{2}\{1st\,\,term+\,last\,\,term\}$   (c)        Single arithmetic mean between two given quantity is $A.M=\frac{a+b}{2}$ (d)        For n arithmetic mean between two terms a and b. $\therefore$ So, the total no. of terms = n+ 2 Last term $=b=a+(n+2-1).d=a+(n+1).d$ Where d = comman difference in A.P.   (e)        For choosing five term in A.P. We take them as $a-2b,a-b,a+b$and $(a+2b)$ etc.
• Some very useful results: If the last term, ${{t}_{n}}$ is in the linear form. i.e. ${{t}_{n}}=an+b.$ Here, n is considered as variable then the series so formed is said to be in A.P. Similarly if last tern, ${{t}_{n}}=a{{n}^{2}}+bn+c$ (i.e. in quadratic form), then the series so formed is said to be in A.P.
If ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,{{a}_{4}}....{{a}_{n}}$is said to be in A.P. Then   (a)        ${{a}_{1}}\pm \text{k},\,{{a}_{2}}\pm \text{k},....$will be in A.P.. Where k be any const quantity (b)        Even $\text{k}.{{a}_{2}},\text{k}.{{a}_{2}},\text{k}.{{a}_{3}}.....$will be said to be in A.P. (c)        $\frac{{{a}_{1}}}{\text{k}},\frac{{{a}_{2}}}{\text{k}},\frac{{{a}_{3}}}{\text{k}}...$is said to be in A.P. (d)        If ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,{{a}_{4}},\,{{a}_{5}}....$be in A.P. and also ${{b}_{1}},\,{{b}_{2}},\,{{b}_{3}},\,{{b}_{4}},\,{{b}_{5}}....$be in A.P. then the resulting sequence whose elements are corresponding element addition or multiplication of given sequences is said to be in A.P.             $\Rightarrow (b-a)=(n+1).d$             $d=\frac{b-a}{(n+1)}$             Hence, 1st A.M. between a and b   $=a+d\,\,\,\,\,=a+\left( \frac{b-a}{n+1} \right)$ 2nd A.M. between a and $=a+2d\,\,\,\,\,=a+2.\left( \frac{b-a}{n+1} \right)$ 2nd A.M. between a and b   ${{n}^{th}}$ A.M. between a and $b=a+n\frac{(b-a)}{n+1}$   (e)        Sum of the n-natural number is written as             $\text{Sn=}\frac{\text{n}(n+1)}{2},$ where $n\in N$   (f)         Sum of the square of first n-natural number is written as             $\text{S}{{\text{n}}^{\text{2}}}\text{=}\frac{\text{n}(n+1)(2n+1)}{6}$   (g)        Sum of the cube of the 1st n-natural number be             $\text{S}{{\text{n}}^{3}}\text{=}{{\left( \frac{n(n+1)}{2} \right)}^{2}}$ more...

Notes - Mathematics Olympiads - Two Dimensional Geometry

Two Dimensional Geometry (Coordinate and Straight Line)   Key Points to Remember
• Coordinate Geometry: It is the branch of mathematics in which deal with relation between two variable in algebraic form. It is 1st coined by French Mathematician Rene Descarties.
Let P(x, y) be any point   x$\to$ abscissa y$\to$ ordinate
• Some Basic Formula:
• Distance Formula:
(a)        The distance between two points $A({{x}_{1}},\,{{y}_{1}})$ & $B({{x}_{2}},\,{{y}_{2}})$   $Ab=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$             (b)        Distance between the origin 0(0, 0) and the point P(x, y) is OP                         $op=\sqrt{{{x}^{2}}+{{y}^{2}}}$   e.g.$A=(5,3)\,\,\,B=(-2,5)$ $\therefore \,\,\,AB$             $=\sqrt{{{(-2-5)}^{2}}{{(5-3)}^{2}}}=\sqrt{49+4}=\sqrt{53}$
• Section Formula: The coordinate of the point P(x, y) dividing the line segment joining the two points $A({{x}_{1}},{{y}_{1}})$ and $B({{x}_{2}},y{{ }_{2}})$ internally in the ratio m:n are given by
$x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},$ $y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},$         When P divides AB in the ratio m:n then   $x=\frac{m{{x}_{2}}-n{{x}_{1}}}{m-n},$ $y=\frac{m{{y}_{2}}-n{{y}_{1}}}{m-n},$             When P divides AB in the ratio 1:1 i.e. P is the mid point of AB   $\therefore \,\,\,\text{P}\equiv \text{(x,y)=}\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{2}}+{{y}_{1}}}{2} \right)$
• Area of triangle: A, B & C be the vertices of the triangle ABC such that $A\equiv ({{x}_{1}},{{y}_{1}}),$ $B\equiv ({{x}_{2}},{{y}_{2}}),$ & $C\equiv ({{x}_{3}},{{y}_{3}}),$
Area of ABC                         $=\frac{1}{2}\{{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})\}$             [Using determinant form]
• Area of a quadrilateral: The ara of the quadraleteral, whose vertices are $A({{x}_{1}},{{y}_{1}}),$ $B({{x}_{2}},{{y}_{2}}),$ $C({{x}_{3}},{{y}_{3}}),$& $D({{x}_{4}},{{y}_{4}}),$ is
Note: The rule for writting the area of a quadrilateral is the same as that of a triangle. Similarly, we can find the area of a polygon of n sides with vertices ${{A}_{1}}({{x}_{1}},{{y}_{1}}),{{A}_{2}}({{x}_{2}},{{y}_{2}})...{{A}_{n}}({{x}_{n}},{{y}_{n}})$is     If ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\,\,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ and ${{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0$ are the equation of the triangle. Then the area of the triangle be     Where ${{c}_{1}},\,{{c}_{2}},\,{{c}_{3}}$ be the co-factor of ${{c}_{1}},\,{{c}_{2}},\,{{c}_{3}}$ in the determinants: $\therefore \,\,\,{{c}_{1}}={{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}},$ ${{c}_{2}}={{a}_{3}}.{{b}_{1}}-{{a}_{1}}{{b}_{3}}$ & ${{c}_{3}}={{a}_{1}}.{{b}_{2}}-{{a}_{2}}{{b}_{1}}.$
• Condition of collinearity of three points: The three point $A({{x}_{1}},{{y}_{1}}),$ $B({{x}_{2}},{{y}_{2}})$ and $C({{x}_{3}},{{y}_{3}})$ are collinear iff $\Delta ABC=0$
[In determinant form]   Solved Problem
• Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear.
• Sol.      Let $A\equiv (a,b+c)$ $B\equiv (b,c+a)$ $C\equiv (c,a+b),$ be three point. To show the collinear of the points A, B and C. Area of $\Delta ABC$ should be zero. Now,         [Two columns are identical]   $=(a+b+c)\times 0=0$   Hence more...

Notes - Mathematics Olympiads - Pair Sraight Line

Pair and Straight Line   Key Points to Remember   A hemogenous of equation of second degree of the form $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents the pair of straight lines which passes through the origin.   (a)        If the lines are real and distinct then ${{h}^{2}}>ab$   (b)        If the lines are real and coincidents if ${{h}^{2}}=ab.$   (c)        If the lines are imaginary then ${{h}^{2}}<ab.$ Let $y={{m}_{1}}x\to (1)$ and $y={{m}_{2}}x\to (2)$ are two lines which are passing through the origin. Then $(y-{{m}_{1}}x)(y-{{m}_{2}}x)\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}$   ${{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}={{y}^{2}}+\frac{2h}{b}xy+\frac{a}{b}{{x}^{2}}$   Equation the coefficient of same variable of the$\frac{2h}{b}=({{m}_{1}}+{{m}_{2}})$we have, $({{m}_{1}}+{{m}_{2}})=\frac{2h}{b}$ & ${{m}_{1}}.{{m}_{2}}=\frac{a}{b}$
• Angle between the pair of straight lines
Let q be the angle between the two given pair of straight line $a{{x}^{2}}+2hxy+b{{y}^{2}}=0,$ which are passing through origin is written as             $\tan \theta =\pm \frac{2\sqrt{{{h}^{2}}}-ab}{a+b}$ for acute angle,   $\tan \theta =\left| \frac{2\sqrt{{{h}^{2}}}-ab}{a+b} \right|$   In cosine form   $\cos \theta =\frac{a+b}{\sqrt{{{(a+b)}^{2}}+4{{h}^{2}}}}$   Note:    If two lines are coincident             i.e.       $\theta =0{}^\circ$ or $180{}^\circ$      then $\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=0$   Hence ${{h}^{2}}=ab\Rightarrow {{h}^{2}}=a$   (b)        If two lines are perpendicular   i.e.        $\theta =90{}^\circ ,$ i.e. $\tan 90{}^\circ =\infty$   then Loosly, it can be written   $\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=\infty =\frac{1}{0}$   $a+b=0$ i.e. $\therefore$Sum of the coefficient of ${{x}^{2}}$ and ${{y}^{2}}$ respectively is zero.   (c)        The equation of pair of straight lines passing through the origin and perpendicular to the given equation of pair of straight lines$/a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ is written as $b{{x}^{2}}-2hxy+a{{x}^{2}}=0$ The general equation of second degree in x and y be $a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$ …….(i) represents a pair of straight lines iff $abc+2fgx-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$     (a)        If $\Delta \ne 0$ & ${{h}^{2}}-ab\ge 0$ then this general equation of second degree in x & y represents the equation of hyperbola. (b)        If $\Delta \ne 0$ & ${{h}^{2}}-ab\le 0$ then this pair of lines represent the equation ellipse. (c)        If $\Delta \ne 0$ & ${{h}^{2}}-ab=0$  then this pair of lines represent the equation of parabola (d)        If $a=b=1$ & $h=0$ then this represents the equation of circle.
• Some points to remember
(a)        Angle between the lines:- If $\theta$ is the angle between the two lines:             $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$  …… (1)             then $\tan \theta =\pm \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|$             $\Rightarrow \,\,\,\theta ={{\tan }^{-1}}\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|$   Note:    It is the same form which is obtained by the pair of straight lines passing through the origin.   (b)        Point of intersection of lines: The point of intersection of line (1) is obtained by the partially differentiation of $\text{f}\equiv a{{x}^{\text{2}}}+b{{y}^{2}}+2hxy+2gx+2\text{f}y+c=0$ w.r.t x and y respective & making             $\frac{\partial \text{f}}{\partial x}=0$                            ….. (i)             &  $\frac{\partial \text{f}}{\partial y}=0$                        ….. (ii) Then solve these equations and hence, we will obtain the value ofx and y, which are written as             $(x,y)=\left( \frac{bg-h}{{{h}^{2}}-ab},\frac{a\text{f-gh}}{{{h}^{2}}-ab} \right)$ Here,             $\frac{\partial \text{f}}{\partial x}=2ax+2hy+2g=0$ & more...

Notes - Mathematics Olympiads - Circles

Circles   Key Points to Remember   Circle: A circle is the locus of the points which move in the plane such that the its distance from a fixed point always remain constant, is said to be the circle. The fixed point is said to be the centre of the circle and its distance is said to be the radius of the circle.     Let C (O, r) is a circle with centre 0 & radius r. A be any point it. $\therefore$      OA = radius of the circle
• Standard Equation of the Circle: The standard equation of the circle whose centre be (h, k) and radius, a be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{a}^{2}}$
When centre be considered as the origin & radius be a, then equation of the circle is written as ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ e.g. find the equation of the circle whose centre is (2, 3) and radius is 6 units   Sol.      Let P(x, y) be any point on the circle by distance formula ${{(x-2)}^{2}}+{{(y-3)}^{2}}={{(6)}^{2}}$ ${{x}^{2}}+{{y}^{2}}-4x-6y+13=36$ ${{x}^{2}}+{{y}^{2}}-4x-6y-23=0$ Which is the required equation of the circle.
• General Equation of the Circle
Since, the general equation of the second degree be$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$..... (1) Condition for the circle. (i)         a = b should be unity (ii)        product of xy term be zero.   Here equation (1) becomes the general equation of the circle. i.e. The general equation of the circle be written as   ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$   Its radius $=\sqrt{{{g}^{2}}+{{\text{f}}^{2}}-c}$ centre of the circle $\equiv (-g,-\text{f})$
• Nature of the circle
(i)         If ${{g}^{2}}+{{\text{f}}^{2}}-c>0,$ then equation of the circle represents the real circle with the centre $(-g,-\text{f}).$ (ii)        If ${{g}^{2}}+{{\text{f}}^{2}}-c=0$ i.e. radius of the circle is zero. Then the equation of the circle represent point whose co-ordinate be $(-g,-\text{f}).$ (iii)       If ${{g}^{2}}+{{\text{f}}^{2}}-c<0$ i.e. radius of the circle is imaginary but its centre, $(-g,-\text{f}).$  be real. This type of circle is not possible to draw in the plane.
• Different form of the circle
(i)         Circle with centre, (a, b) and which touches the x-axis. Since, the circle touch the x-axis then radius of the circle is equal to the y-ordinate of the centre of the circle. i.e. Radius of the circle = b Hence, equation of the circle is   ${{({{x}^{2}}-a)}^{2}}+{{(y-b)}^{2}}={{b}^{2}}$ ${{x}^{2}}+{{y}^{2}}-2ax-2by+{{a}^{2}}+{{b}^{2}}={{b}^{2}}$ $\Rightarrow {{x}^{2}}+{{y}^{2}}-2ax-2by+{{a}^{2}}=0$   (ii)        Circle with centre, (a, b) which touches the y-axis. Since, equation of the circle touches the y- axis. i.e. the radius of the circle is equao to the x- ordinate of the centre of the circle. i.e. Radius of the circle is             ${{({{x}^{2}}+a)}^{2}}+{{(y-b)}^{2}}={{a}^{2}}$             $\Rightarrow {{x}^{2}}+{{y}^{2}}-2ax-2by+{{a}^{2}}+{{b}^{2}}={{a}^{2}}$                ${{x}_{2}}+{{y}_{2}}-2ax-2by+{{b}_{2}}=0$   (iii)       Circle with radius a and which touches both the coordinate axis. Since, when centre more...

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