Quadratic equation: An equation of the form \[a{{x}^{2}}+bx+c=0\] where a, b, and c \[\in \] R and \[a\ne 0\]is called a quadratic equation.
Note: (i) An equation of degree 2 is called a quadratic equation.
(ii) The quadratic equation is of the form\[a{{x}^{2}}+bx+c=0\].
Solution or roots of a quadratic equation: If p(x) = 0 is a quadratic equation, then the zeros of the polynomial p(x) are called the solutions or roots of the quadratic equation P(x) = 0.
Note: (i) Since the degree of a quadratic equation is 2, if has 2 roots or solutions.
(ii) \[x=a\]is the root of \[p\left( x \right)=0,\]if \[p\left( a \right)=0.\]
(iii) Finding the roots of a quadratic equation is called solving the quadratic equation.
Methods of solving a quadratic equation: There are different methods of solving a quadratic equation.
(a) Factorization method by:
(i) Splitting the middle term (ii) Completing the square
(b) Using formula for roots \[(\alpha ,\beta )=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
(i) Splitting the middle term: Consider the quadratic equation \[a{{x}^{2}}+bx+c=0\].
Step 1: Find the product of the coefficient of \[{{x}^{2}}\] and the constant term i.e., ac.
Step 2: (a) If ac is positive, then choose two factors of ac, whose sum is equal to b (the coefficient of the middle term).
(b) If ac is negative, then choose two factors of ac, whose difference is equal to b (the coefficient of the middle term).
Step 3: Express the middle terms as the sum (or difference) of the two factors obtained in step 2. [Now the quadratic equation has 4 terms]
Step 4: Express the given quadratic equation as a product of two binomials, and solve them. The two values obtained in step 2 are the roots of the given quadratic equation.
Elementary question - 1: Solve \[{{x}^{2}}+7x+12=0\]
Answer; ac = 1.12 = 12: Factors are 1, 2, 3, 4, 6, 12
The combination which gives middle term b =7 is 3 + 4 = 7
\[\therefore \] Written as \[{{x}^{2}}+3x+4x+12=0\]
or \[x\left( x+3 \right)+4\left( x+3 \right)=0\text{ }or\text{ }\left( x+4 \right)\left( x+3 \right)=0\]
(ii) Completing the square: In some cases, where the given quadratic equation can be solved by factorization, a suitable term is added and subtracted. Then terms are regrouped in such a manner that a square is completed by three of the terms. The equation is then solved using factorization method.
Usually, the term added and subtracted is the square of half the coefficient of x.
Formula method: The roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] are given by\[\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a},\]provided \[{{b}^{2}}-4ac\ge 0\]. This formula for finding the roots of a quadratic equation is called the quadratic formula.
Note: The roots of the quadratic equation using the quadratic formula are \[x=\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[x=\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
Roots of Biquadratic Equation
Any biquadratic equation, \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0,\] will have four roots. If a, p, y and 5 more...
QUADRATIC INEQUATION
FUNDAMENTALS
Quadratic In equations
Consider the quadratic equation\[a{{x}^{2}}+bx+c=0.\,\,\,(a\ne 0)\]where a, b, and c are real numbers.
The quadratic in equations related to \[a{{x}^{2}}+bx+c=0\]are \[a{{x}^{2}}+bx+c<0\]and \[a{{x}^{2}}+bx+c>0\].
Assume that a > 0 (This assumption is always valid because if a<0, we can always multiply the in equation by (– 1) to get a > 0.)
For example, \[-2{{x}^{2}}+3x+2<0\]can be written as \[-2{{x}^{2}}-3x-2>0\]
Note: The change in the sign of the inequality, when it is multiplied by (– 1).
Then following cases arise:
Case – 1: If\[{{b}^{2}}-4ac>0\], then the equation \[a{{x}^{2}}+bx+c=0\]has real and unequal roots. Let \[\alpha \]and \[\beta (\alpha <\beta )\]be the roots. Then,
\[\therefore a{{x}^{2}}+bx+c=a(x-\alpha )(x-\beta )\]
If\[x<\alpha \], then \[\left( x-\alpha \right)<0\] and \[(x-\beta )<0\]
\[\therefore a{{x}^{2}}+bx+c>0\]
If a\[\alpha <x<\beta \], then \[\left( x-\alpha \right)>0\] and \[(x-\beta \text{)}<0\]
\[\therefore a{{x}^{2}}+bx+c>0\]
Case – 2: If\[{{b}^{2}}-4ac=0\], then \[a{{x}^{2}}+bx+c=0\]has real and equal roots.
Let \[\alpha \] be the equal roots.\[\Rightarrow a{{x}^{2}}+bx+c=a(x-\alpha ')(x-\alpha ')\]
If\[x<\alpha '\]. Then, \[(x-\alpha ')<0\]. \[\therefore a{{x}^{2}}+bx+c>0\].
However, if\[x>\alpha '\], then,\[\left( x-\alpha ' \right)>0\].
\[\therefore a{{x}^{2}}+bx+c>0\].
Case – 3: If\[{{b}^{2}}-4ac<0\], then \[a{{x}^{2}}+bx+c=0\]has no real roots.
In this case, \[a{{x}^{2}}+bx+c>0,\forall x\in R\].
The above three cases (case 1 to case 3) may be summarized as:
If\[\alpha <x<\beta \], then \[\left( x-\alpha \right)\left( x-\beta \right)<0\] and vice - versa.
If\[~x<\alpha \] and\[x>\beta \left( \alpha <\beta \right)\], then \[\left( x-\alpha \right)\left( x-\beta \right)>0\] and vice - versa.
Sample Question: Solve the in equation\[{{x}^{2}}-7x+12<0\].
Solution: Given in equation is \[{{x}^{2}}-7x+12<0\Rightarrow {{x}^{2}}-3x-4x+12<0\]
\[\Rightarrow \left( x-3 \right)\left( x-4 \right)<0;\Rightarrow \left( x-3 \right)<0\] and \[\left( x-4 \right)>0\]……………..(1)
Or. \[\left( x-3 \right)>0\] and \[\left( x-4 \right)<0\]…………...(2)
In set notation, ‘and’ means intersection whereas ‘or’ means UNION.
From (1), x < 3 and x > 4
\[\therefore \] There is not overlap and this is a NULL set \[\phi \]………. (1)
From \[\left( 2 \right),x>3\] and \[x<4\Rightarrow x\in (3,\infty )\cap x\in (-\infty ,4)\Rightarrow x\in (3,4)\]
Clearly, the intersection set is (3, 4)
Finally, we have union of NULL set \[\phi \] in equation (1) and (3, 4) in equation (2) \[\Rightarrow \]
Solution \[=\phi \cup (3,4)=(3,4)\]
Elementary Questions -1
Solve for \[x:{{x}^{2}}-3x+2\ge 0\]
Solution:
Given in equation is \[{{x}^{2}}-3x+2\ge 0\].
\[\Rightarrow (x-1)(x-2)\ge 0\Rightarrow x-1\ge 0\] and \[x-2\ge 0\] or \[x-1\le 0\]and \[x-2\le 0\]
\[\Rightarrow x\ge 1\]and \[x\ge 2\Rightarrow x\in [1,\infty ]\cap [2,\infty )\Rightarrow x\in [2,\infty )\]…………..(1)
Or,
\[x\le 1\]and \[x\le 2\Rightarrow x\in [-\infty ,1)\cap [-\infty ,2)\Rightarrow x\in (-\infty ,1]\].............(2)
(1) & (2) are combined as \[x\in (-\infty ,1]\cup [2,\infty )\]
Hence, the solution for the given in equation is \[x\in (-\infty ,1]\cup [2,\infty )\].
This can be seen on number line as follows:
From \[(1);x\in [2,\infty )\]
OR,
From \[(2);x\in (-\,\infty ,1]\]
OR, means UNION\[\Rightarrow x\in (-\infty ,1]\cup more...
LINEAR EQUATION IN TWO VARIABLES
FUNDAMENTALS
While solving the problems, in most cases, first we need to frame an equation. Therefore, we will learn how to frame and solve equations sometimes. Framing an equation is more crucial aspect after which solving the equation may be quite easy.
Algebraic Expression
Expression of the form, \[3x,(3x+6),(2x-6y),3{{x}^{2}}+3\sqrt[3]{y},\frac{7{{x}^{6}}}{3}\sqrt{y}\] are algebraic expressions. \[3x\] and 6 are the terms of \[\left( 3x+6 \right)\] and 2x and 6y are the terms of\[2x-6y\]. Algebraic expressions are made of numbers, symbols and the basic arithmetical operations. In the term 3x, 3 is the numerical coefficient of x and x is the variable coefficient of 3.
The following step are involved in solving an equation.
Step – 1: Always ensure that the unknown quantities are on the LHS and the known quantities or constants on the RHS.
Step – 2: Add all the terms containing the unknowns on the LHS and all the known on the RHS. So that each side of the equation contains only one term.
Step – 3: Divide both sides of the equation by the coefficient of the unknown.
Example: lf\[4x+15=35\], then find the value of x
Solution: Step – 1: Group the known quantities as the RHS of the equation, i.e.,\[~4x=35-15\]
Step – 2: Simplify the numbers on the RHS\[\Rightarrow 4x=20\].
Step – 3: Since 4 is the coefficient of x, divide both the sides of the equation by 4.
Linear equation in two variables
An equation of the form\[ax+by=c\], where \[a\ne 0,b\ne 0\]and a, b and c are real numbers is known as a linear equation in two variables x and y.
A pair of linear equations in two variables
Two linear equations in the same two variables (x, y) are called a pair of linear equations.
General form of a pair of linear equations in two variables
The general form of a pair of linear equations in two variables is \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\], where \[{{a}_{1}},{{a}_{2}},{{b}_{1}},{{b}_{2}},{{c}_{1}}\] and \[{{c}_{2}}\] are real numbers such that \[a_{1}^{2}+b_{1}^{2}\ne 0\] and \[a_{2}^{2}+b_{2}^{2}\ne 0\].
Methods of solving a pair of linear equations in two variables: A pair of linear equations in two variables can be solved by (i) Graphical method (ii) Algebraic method.
(i) Graphical method: The graph of a linear equation is a straight line. The graph of a pair of linear equations in two variables is represented by two lines.
(a) If the two lines coincide, then the pair of linear equations has infinitely many solutions (each point on the line being a solution), and is said to be dependent or consistent.
(b) If the lines are parallel, then the pair of linear equations has no solution (no common point) and is said to be inconsistent.
In other words, there are three types of solutions of a pair of linear equations in two variables
(a) Unique solution (b) Infinitely many solutions (c) No solutions
(ii) Algebraic more...
STATISTICS
INTRODUCTION
Data
The word ‘data’ means, information in the form of numerical figures or a set of given facts. For example, the percentage of marks scored by 10 students of a class in a test are: 36, 80, 65, 75, 94, 48, 12, 64, 88 and 98.
Row Data
Data obtained from direct observation is called raw data,
The marks obtained by 100 students in a. monthly test is an. example of raw data or ungrouped.
Intact, little can be inferred from this data. However, arranging the marks in ascending order in the above example is a step towards making raw data more meaningful.
Grouped Data
To present the data in a more meaningful way, we condense the data into convenient number of classes or groups, generally not exceeding 10 and not less than 5. This helps us in perceiving at a glance, certain salient features of data.
Tabulation or Presentation of Data
A systematically arrangement of the data in a tabular form is called ‘tabulation’ or ‘presentation’ of the data. This groping results in a table called ‘frequency table’ that indicates the number of scores within each group.
Individual Series: Any raw data that is collected, forms an individual series.
Example:
The weights of 10 students.
36, 35, 32, 40, 65, 48, 54, 71, 62 and 33
Discrete Series: A discrete series is formulated from raw data. Here, the frequency of the observations are taken into consideration.
Example: Given below is the data showing the number of computers in 15 families of a locality.
1, 1, 2, 3, 2, 1, 4, 3, 2, 2, 1, 1, 1, 1, 4
Arranging the data in ascending order:
1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4
To count, we can use tally marks. We record tally marks in bunches of five, the fifth one crossing the other four diagonally, i.e.
Thus, we may prepare the following frequency table.
Figures having the same shape (not necessarily the same size) are called similar figures. Same shapes ensure that the corresponding angles are equal and their corresponding sides are proportional.
Congruent figures:
Figures having the same shape and the same size are called congruent figures. Here, apart from angles, corresponding sides are also equal
Similar Triangles:
Two triangles are said to be similar, if their corresponding angles are equal and corresponding sides are proportional.
e.g., If in \[\Delta \,ABC\]and \[\Delta \,PQR\]
\[\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R\]
and \[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR},\] then, \[\Delta \text{ }ABC\sim \Delta \,PQR;\] where symbol \[\sim \] is read as ‘is similar to’.
When two triangles (\[\Delta \text{ }ABC\]and \[\Delta \text{ }DEF\]as below) are similar, then all above results about angles and ratio of sides hold good. However, in questions, when you are asked to prove similarly, you can either prove:
(i) \[\angle A=\angle D,\angle B=\angle E\] (called AA similarly)
Or
(ii) \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\] (called SSS similarity)
Or
(iii) \[\frac{AB}{PQ}=\frac{BC}{QR}\] and \[\angle B=\angle Q\] (called SAS similarity)
Any one of the above three, would be sufficient for proving similarity.
Conversely: If \[\Delta \text{ }ABC\]is similar to\[\Delta \text{ }PQR\], then
\[\angle A=\angle D;\angle B=\angle E;\angle C=\angle Q\] and \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\]
When two triangles are congruent (notation for congruent is \[\cong \])
Then, \[\angle A=\angle D;\text{ }\angle B=\angle E;\text{ }\angle C=\angle F\]
and, \[AB=DE,BC=EF\] and \[CA=FD\]
However, to prove congruency, we need to prove any one of the following only:
(i) \[\text{ }AB=DE,\text{ }BC=EF\And CA=FD\](SSS congruency)
or
(ii) \[AB=DE;\text{ }BC=EF\And \angle B=\angle E\] (SAS congruency)
or
(iii) \[\angle A=\angle D;\angle B=\angle E\] and \[AB=DE\](ASA congruency)
How to look for similarity and congruency of angles
While looking for similarity and congruency, you should not only see external appearance but also which of the corresponding angles are equal (or, which of corresponding sides are in the same ratio).
In the above figure, \[\angle B=\angle F;\text{ }\angle C=\angle E\]and \[\angle A\cong \angle D;\] thus \[\Delta \,ACB\cong \Delta \,DEF\](and not\[\Delta \,ABC\cong \Delta \,DEF\])
Mathematical statement of the theorem\[\frac{AD}{DB}=\frac{AE}{EC}\] (where\[DE\parallel ~BC\])
i.e., if in \[\Delta \text{ }ABC\]as shown above, \[DE\parallel BC\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}\]
Converse of Basic Proportionality Theorem:
Mathematical statement of the theorem if in \[\Delta \,ABC\](as shown above),
\[\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow DE\parallel BC)\]
A circle is the locus of points in a plane which are at a fixed distance from a fixed point.
The fixed point is called the centre of the circle and the fixed distance is the radius of the circle and is denoted as ‘r’.
In the figure, OR is a radius of the circle ‘r’.
PQ is a diameter of the circle. OP and OQ are also the radii of the circle.
PQ = diameter (d) = 2r
The perimeter of the circle is called the circumference of the circle (C).
The circumference of the circle is n times the diameter, i.e. \[C=\pi d=2\pi r\]
Interior and exterior points of a circle.
In the figure with centre O, P, Q and R are three points as shown above lying in the plane of the circle. The points O and P are in the interior of the circle. The point Q is located on the circumference of the circle, whereas point R is located outside the circle.
As \[OP<r,P\]is a point in the interior of the circle.
If \[OQ=r,Q\]is a point on the circumference of the circle and is said to belong to the circle.
As \[OR>r,R\]is a point in the exterior of the circle.
CHORD
The line segment joining any two points on the circumference of a circle is called chord of the circle. In the figure \[\overline{PQ}\] and \[\overline{RS}\]are the chords.
PQ passes through centre O, hence it is a diameter of the circle. Diameter is the longest chord of the circle. It divides the circle into two equal parts and each part is called semi-circle.
Now, we will study certain theorems and also commit them to our memory.
Theorems and properties on chords
Theorem 1
The perpendicular bisector of a chord of a circle passes through the centre of the circle.
Given: RS is a chord of a circle with centre O. N is the midpoint of chord PQ.
Draw a line \[\bot \]to RS passing through ‘N’.
O will lie on line NOM.
Theorem 2
One and only one circle exists through three non – collinear points.
Given: P, Q and R are three non – collinear points as shown in the adjoining figure:
In order to locate the centre, join PQ and QR.
Draw \[\bot \] bisectors of PQ and a QR and let them meet at ‘O’ as show below.
Now, OP = OQ = OR = r = radius and you can draw the circle.
Theorem 3
Two equal chords of a circle are equidistant from the centre more...
AREA OF CIRCLE
FUNDAMENTALS
Perimeter and Area of a Circle
The distance covered by travelling once around a circle is called its perimeter, and in case of a circle, it is usually called its circumference. The circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter \[\pi \] (read as ‘pi’), m other words,
\[\frac{circumference}{diameter}=\pi \]
Or \[circumference=\pi \times diameter\]
\[=\pi \times 2r\] (where r is the radius of the circle) \[=2\pi r\]
The great Indian mathematician Aryabhatta (A. D. 476 – 550) gave an approximate value of \[\pi \] He stated that \[\pi =\frac{62832}{20000}\], which is nearly equal to 3.1416. It is also interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan (1887 – 1920) of India, mathematicians have been able to calculate the value of n correct to million places of decimals. As you known from Chapter I of Class IX, n is an irrational number and its decimal expansion is non – terminating and non – recurring (non – repeating). However, for practical purposes, we generally take the value of \[\pi \] as \[\frac{22}{7}\] or 3.14, approximately.
Areas of Sector and Segment of a Circle
The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called sector of the circle and the portion (or part) of the circular regions enclosed between a chord and the corresponding arc is called a segment of the circle.
Thus, in figure shaded region OAPB is a sector of the circle with centre O. \[\angle AOB\]is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is (\[360{}^\circ -\angle AOB\]).
Now, look at figure in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment.
Remarks: When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.
Calculating area of sector of a circle
Let OAPB be a sector of a circle with centre O and radius r (see figure). Let the degree measure of \[\angle AOB\] be \[\theta \].
You know that area of a circle (in fact of a circular regions or disc) is \[\pi {{r}^{2}}\].
more...
CO-ORDINATE GEOMETRY
FUNDAMENTALS
Co – ordinate geometry is a branch of science which establishes relationship between the position r a point in a plane and pair of algebraic numbers, called its co – ordinates.
Cartesian Co – ordinates
Let us draw coordinate axes with ‘O’ as origin. In Cartesian co-ordinates, the position of a point P - determined by knowing its horizontal and vertical distance from origin.
Draw PM and PN perpendiculars on OX and OY respectively. OM is called the x co – ordinate or abscissa of the point P. ON is called the y co – ordinate or the ordinate of the point P.
The abscissa and ordinate of a point are known as co – ordinates of the point P.
If OM = x, ON = y, then the co-ordinates of the point P are (x, y)
OX and OY are called as x – axis and y – axis, respectively and together, they are known as co – ordinate axes.
Origin ‘O’ is intersection of the axes of co – ordinates. The co – ordinates of the point O are O (0, 0)
The distance of the point P from y – axis is called its abscissa. In the figure OM is the abscissa.
The distance of the point P from x – axis is called its ordinate. ON is the ordinate in the figure.
Quadrant
A quadrant is \[\frac{1}{4}\] part of a plane divided by the co – ordinate axes.
(a) XOY is called the first quadrant, (or Ist quadrant)
(b) YOX', the second. (II nd quadrant)
(c) X'OY', the third. (III rd quadrant)
(d) Y'OX, the fourth. (IV th quadrant)
Sign Convention in different quadrants
(a) In the first quadrant, both co-ordinates i.e., abscissa and ordinate of a point are positive.
(b) In the second quadrant, abscissa is negative and ordinate is positive.
(c) In the third quadrant, for a point, both abscissa and ordinate are negative.
(d) In the fourth quadrant, the abscissa is positive and the ordinate is negative.
TRIGONOMETRY
Systems of Measurement of an Angle
Circular System
In this system, the angle is measured in radians.
Radian: The angle subtended by an arc length APB equal to the radius of a circle at its centre is defined of one radian (see figure). It is written as \[{{1}^{c}}\]. (‘c’ denotes radian)
Relation between the Units
Look at the circle in the above figure and note that,
\[360{}^\circ =2{{\pi }^{c}}\Rightarrow 90{}^\circ =\frac{{{\pi }^{c}}}{2}\] and \[45{}^\circ =\frac{{{\pi }^{c}}}{4}\]; Or, simply, \[90{}^\circ =\frac{\pi }{2},45{}^\circ =\frac{\pi }{4}\]
For convenience, the above relation can be written as, \[\frac{D}{90}=\frac{R}{\frac{\pi }{2}}\], where, D denotes degrees, and R radians.
Remember
Note: If no unit of measurement is shown for any angle, it is considered as radian.
Trigonometric Ratios
Let AOB be a right triangle with \[\angle AOB\] as \[90{}^\circ \]. Let \[\angle OAB\]be \[\theta \]. Notice that \[0{}^\circ <9<90{}^\circ \]. That is, \[\theta \] is an acute angle (see adjacent figure).
Six possible ratios among the three sides of the triangle AOB, are possible. They are called trigonometric ratios.
Sine of the angle \[\theta \] or, simply \[\text{sin}\theta \]: \[\sin \theta =\frac{Perpendicular}{Hypotenuse}=\frac{p}{h}=\frac{OB}{AB}\]
Cosine of the angle \[\theta \] or, simply \[\cos \theta \]: \[\cos \theta =\frac{Base}{Hypotenuse}=\frac{b}{h}=\frac{OA}{AB}\].
Tangent of the angle \[\theta \] or, simply \[tan\theta \]: \[\tan \theta =\frac{perpendicular}{base}=\frac{p}{b}=\frac{OB}{OA}\]
Cotangent of the angle \[\theta \] or, simply \[\cot \,\theta \]: \[\cot \theta =\frac{base}{perpendicular}=\frac{b}{p}=\frac{OA}{OB}\]
Cosecant of the angle \[\theta \] or, simply \[cosec\theta \]: \[co\sec \,\,\theta =\frac{hypetenuse}{perpendicular}=\frac{h}{p}=\frac{AB}{OB}\]
Secant of the angle \[\theta \] or, simply \[sec\theta \]: \[\sec \,\,\theta =\frac{h}{b}=\frac{AB}{OB}\]
PYTHAGOREAN TRIPLETS
Pythagorean Triplets are basically sides of a right A which obey Pythagoras theorem.
Examples are:
3, 4, 5 (and all their multiples in the form of 3k, 4k, 5k etc. for eg. 6, 8, 19 & etc.)
8, 15, 17 (and all their multiples in the form of 8k, 15k, 17k etc.)
9, 49, 41 (and all their multiples in the form of 9k, 49k, 41k etc.)
1, 2. 4, 2.6 (and all their multiples in the form of 1k, 2.4k, 2.6k etc.)
5, 12, 13 (and all their multiples in the form of 5k, 12k, 13k etc.)
7, 24, 25 (and all their multiples in the form of 7k, 24k, 25k etc.)
Example: If \[\cos \theta =\frac{2}{5}\], then find the values of \[\tan \,\theta ,\cos ec\,\theta .\]
Solution:
Given, \[\cos .\theta =\frac{2}{5}\]
Let PQR be the right triangle such that \[\angle QPR=\theta \] more...
PROBABILITY
FUNDAMENTAL
Probability - A Theoretical Approach
We know, in advance, that the coin can only land in one of two possible ways either head up or tail up (we dismiss the possibility of its ‘landing’ on its edge, which may be possible, for example, if it falls on sand). We can reasonably assume that each outcome, head or tail, is as likely to occur as the other. We refer to this by saying that the outcomes head and tail, are equally likely.
Considering another example of equally likely outcomes, suppose we throw a die once. For us, a die will always mean a fair die. What are the possible outcomes? They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So, the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6.
Are the outcomes of every experiment equally likely? Let us see. Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball without looking into the bag. What are the outcomes? Are the outcomes – a red ball and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you would agree that you are more likely to get a red ball than a blue ball. So, the outcomes (a red ball or a blue ball) are not equally likely. However, the outcome of drawing a ball of any colour from the bag is equally likely. So, all experiments do not necessarily have equally likely outcomes.
However, in this chapter, we will assume that all the experiments have equally likely outcomes. In class IX, GMO book, we defined the experiments or empirical probability P (E) of an event E as
\[\text{P(E)=}\frac{\text{Number of trials in which the event happened}}{\text{Total}\,\,\text{number}\,\,\text{of}\,\,\text{trials}}\]
The empirical interpretation of probability can be applied to every event associated with an experiment which can be repeated a large number of times. The requirement of repeating an experiment has some limitations, as it may be very expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die throwing experiments. But how about repeating the experiment of launching a satellite in order to compute the empirical probability of its failure during launching, or the repetition of the phenomenon of an earthquake to compute the empirical probability of a multi storied building getting destroyed in an earthquake.
In experiments where we are prepared to make certain assumptions, the repetition of an experiment can be avoided, as the assumptions help in directly calculating the exact (theoretical) probability, The assumption of equally likely outcomes (which is valid in many experiments, as in the two examples above, of a coin and of a die) is one such assumption that leads us to the following definition of probability of an event.
The theoretical probability of an event E, (also called more...
OR,
From \[(2);x\in (-\,\infty ,1]\]
OR, means UNION\[\Rightarrow x\in (-\infty ,1]\cup more... 
e.g., If in \[\Delta \,ABC\]and \[\Delta \,PQR\]
\[\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R\]
and \[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR},\] then, \[\Delta \text{ }ABC\sim \Delta \,PQR;\] where symbol \[\sim \] is read as ‘is similar to’.
(i) \[\angle A=\angle D,\angle B=\angle E\] (called AA similarly)
Or
(ii) \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\] (called SSS similarity)
Or
(iii) \[\frac{AB}{PQ}=\frac{BC}{QR}\] and \[\angle B=\angle Q\] (called SAS similarity)
Any one of the above three, would be sufficient for proving similarity.
Conversely: If \[\Delta \text{ }ABC\]is similar to\[\Delta \text{ }PQR\], then
\[\angle A=\angle D;\angle B=\angle E;\angle C=\angle Q\] and \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\]
In the above figure, \[\angle B=\angle F;\text{ }\angle C=\angle E\]and \[\angle A\cong \angle D;\] thus \[\Delta \,ACB\cong \Delta \,DEF\](and not\[\Delta \,ABC\cong \Delta \,DEF\])
Mathematical statement of the theorem\[\frac{AD}{DB}=\frac{AE}{EC}\] (where\[DE\parallel ~BC\])
i.e., if in \[\Delta \text{ }ABC\]as shown above, \[DE\parallel BC\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}\]
Converse of Basic Proportionality Theorem:
Mathematical statement of the theorem if in \[\Delta \,ABC\](as shown above),
\[\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow DE\parallel BC)\]
Given: RS is a chord of a circle with centre O. N is the midpoint of chord PQ.
Given: P, Q and R are three non – collinear points as shown in the adjoining figure:
Thus, in figure shaded region OAPB is a sector of the circle with centre O. \[\angle AOB\]is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is (\[360{}^\circ -\angle AOB\]).
Now, look at figure in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment.
Remarks: When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.
Calculating area of sector of a circle
Let OAPB be a sector of a circle with centre O and radius r (see figure). Let the degree measure of \[\angle AOB\] be \[\theta \].
You know that area of a circle (in fact of a circular regions or disc) is \[\pi {{r}^{2}}\].
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Draw PM and PN perpendiculars on OX and OY respectively. OM is called the x co – ordinate or abscissa of the point P. ON is called the y co – ordinate or the ordinate of the point P.
The abscissa and ordinate of a point are known as co – ordinates of the point P.
If OM = x, ON = y, then the co-ordinates of the point P are (x, y)
OX and OY are called as x – axis and y – axis, respectively and together, they are known as co – ordinate axes.
A quadrant is \[\frac{1}{4}\] part of a plane divided by the co – ordinate axes.
(a) XOY is called the first quadrant, (or Ist quadrant)
(b) YOX', the second. (II nd quadrant)
(c) X'OY', the third. (III rd quadrant)
(d) Y'OX, the fourth. (IV th quadrant)
Sign Convention in different quadrants
(a) In the first quadrant, both co-ordinates i.e., abscissa and ordinate of a point are positive.
(b) In the second quadrant, abscissa is negative and ordinate is positive.
(c) In the third quadrant, for a point, both abscissa and ordinate are negative.
(d) In the fourth quadrant, the abscissa is positive and the ordinate is negative.
Relation between the Units
Look at the circle in the above figure and note that,
\[360{}^\circ =2{{\pi }^{c}}\Rightarrow 90{}^\circ =\frac{{{\pi }^{c}}}{2}\] and \[45{}^\circ =\frac{{{\pi }^{c}}}{4}\]; Or, simply, \[90{}^\circ =\frac{\pi }{2},45{}^\circ =\frac{\pi }{4}\]
For convenience, the above relation can be written as, \[\frac{D}{90}=\frac{R}{\frac{\pi }{2}}\], where, D denotes degrees, and R radians.
Remember
Let AOB be a right triangle with \[\angle AOB\] as \[90{}^\circ \]. Let \[\angle OAB\]be \[\theta \]. Notice that \[0{}^\circ <9<90{}^\circ \]. That is, \[\theta \] is an acute angle (see adjacent figure).
Six possible ratios among the three sides of the triangle AOB, are possible. They are called trigonometric ratios.
Given, \[\cos .\theta =\frac{2}{5}\]
Let PQR be the right triangle such that \[\angle QPR=\theta \]
