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4

3

2 ● ●

1 ●

1 2 3 4 5 6


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BINARY OPERATION

A=[2,8] A={x/xϵ[2,8]}

B=[-4,3]

INEQUALITY

x-1=4

x-14 xϵ(5,∞) OR 52 OR xϵ(2, ∞) OR 22

x2 -5x +6


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x2 -5x +6 >0

(x-2)(x-3)>0

(x-2) (x-3)

+ +

+ -

- +

- -

xϵ (-∞, 2) OR xϵ (3, ∞)

(-∞, 2) U (3, ∞) i.e., xϵ (-∞, 2) U (3, ∞)

(-∞,2) 2 (2,3) 3 (3,∞)

(x-1)(x2 -5x +6) 0


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(-∞,1) 1 (1,2) 2 (2,3) 3 (3,∞)

(x-1)(x2 -5x +6)


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(1,2) U (3,∞)

(x-1)(x2 -5x +6) >0

(x-1) (x-2)(x-3)>0

+++ (3,∞)

+-- (1,2)

(1,2) U (3,∞)

FUNCTION

In mathematics, a binary relation on a set A is a collection of ordered pairs of elements of A. In

other words, it is a subset of the Cartesian product A2 = A × A. More generally, a binary

relation between two sets A and B is a subset of A × B.

Definition: Let A and B be nonempty sets. A binary relation (or just relation) from A to B is a subset

R ⊆ A × B.

A relation is a set of inputs and outputs, often written as ordered pairs (input, output). We can also

represent a relation as a mapping diagram or a graph.

A={a,b} B={1,2}

AxB={(a,1),(a,2),(b,1),(b,2)}

R1= ø R2={(a,1)} R3={(a,2)} R4={(b,1)}

R5={(b,2)} R6={(a,1),(a,2)} R7={(a,1),(b,1)} R8={(a,1),(b,2)}

R9={(a,2),(b,1)} R10={(a,2),(b,2)} R11={(b,1),(b,2)} R12={(a,1),(a,2),(b,1)}

R13={(a,1),(a,2),(b,2)} R14={(a,1),(b,1),(b,2)} R15={(a,2),(b,1),(b,2)} R16={(a,1),(a,2),(b,1),(b,2)}

Dom(R6)={a} Range(R6)={1,2}

Dom(R7)={a,b} Range(R7)={1}

Dom(R8)={a,b} Range(R8)={1,2}


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Dom(R12)={a,b} Range(R12)={1,2}

A function is a relation in which each input x (domain) has only one output y(range).

Examine whether the following is a function or not

A={0,2} B={x, y, t} R={(0,x),(2,y)}

A={1,2,3} B={3,4,5} R={(1,3),(2,4),(3,4)}

A={1,2,3} B={1,2,3} R={(1,1),(2,3),(3,2)}

A={0,2} B={4,6,8,10} R={(0,4),(0,6),(2,8)}

The mapping diagram of the relation

To check if a relation is a function, given a mapping diagram of the relation, use the following criterion:

1. each input has a connecting line (i.e., no input is left unconnected) and

2. each input has only one line connected to it,

then the outputs are a function of the inputs.

The above diagram does not represent a function with domain {1, 2, 3}, codomain {A, B, C, D} and set of

ordered pairs {(1,D), (2,B),(2,C), (3,C)}. The image is {B,C,D}.


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The above diagram represents a function with domain {1, 2, 3}, codomain {A, B, C, D} and set of ordered pairs

{(1,D), (2,C), (3,C)}. The image is {C,D}.

However, this second diagram does not represent a function. One reason is that 2 is the first element in more

than one ordered pair. In particular, (2, B) and (2, C) are both elements of the set of ordered pairs. Another

reason, sufficient by itself, is that 3 is not the first element (input) for any ordered pair. A third reason, likewise,

is that 4 is not the first element of any ordered pair.

In order to avoid the use of the informally defined concepts of "rules" and "associates", the above

intuitive explanation of functions is completed with a


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This is a function. You can tell by tracing from each x to

each y. There is only one y for each x; there is only one

arrow coming from each x.

Ha! Bet I fooled some of you on this one! This is a

function! There is only one arrow coming from each x;

there is only one y for each x. It just so happens that it's

always the same y for each x, but it is only that one y. So

this is a function; it's just an extremely boring function!

This one is not a function: there are two arrows coming

from the number 1; the number 1 is associated with

twodifferent range elements. So this is a relation, but it isnot a function.

Okay, this one's a trick question. Each element of thedomain that has a pair in the range is nicely well-behaved.

But what about that 16? It is in the domain, but it has no

range element that corresponds to it! This won't work! Sothen this is not a function. Heck, it ain't even a relation!


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The Vertical Line Test for Graphs

To determine whether y is a function of x, given a graph of a relation, use the following criterion: if every

vertical line you can draw goes through only 1 point, y is a function of x. If you can draw a vertical line

that goes through 2 or more points, y is not a function of x. This is called the vertical line test.

A={1,2,5} B={1,2,3,4}

R ={(1,2),(2,1),(5,2)}

4

3

2 ● ●

1 ●

1 2 3 4 5 6

R is a function

R ={(1,2),(2,1) ,(2,2),(5,2)}

4

3

2 ● ● ●

1 ●

1 2 3 4 5 6

R is not a function


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y=2x+5

X 1 2 3

Y 7 9 11

f(x)=2x+5, f(1)=7, f(2)=9, f(3)=11

R={(x,y)/ x+y=11, x,yϵ ℝ} y=11-x, A=(-∞,∞),B=(-∞,∞),R={…,(-2,13),…,(1,10),…,(5,6),…} R={(x,y)/ x+y0

+ +

+ -


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- +

- -

x x

1

1.1

1.5

1.9

1.99

1.999

3

3.1

3.5

3.9

3.99

3.999

3

2.9

2.5

2.1

2.01

2.001

5

4.9

4.5

4.1

4.01

4.001


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Continuous Function

A function is continuous at x=c if

(i) f(c) is defined

(ii) exists(iii) f(c)=

Example# f(x) c f(c) Continuity1 2 undefined 4 Discontinuous

2 3 0 does not exist Discontinuous3 0 undefined does not exist Discontinuous4

0 5 does not exist Discontinuous

5 x+5 2 7 7 Continuous


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Equation of a Straight Line in different forms


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(i) Slope-Intercept form

The equation of a straight line having slope m and making an intercept c on y-axis is

y = m x +c

(ii) Point-slope form

The equation of a straight line passing through the fixed point (x1, y1) and having slope m is

y -y1 = m (x -x1)

(iii) Two-point form

The equation of the line passing through two fixed points A (x1, y1) and B (x2,y2) is

y-y1= 1212

x x

y y

(x-x1)

(iv) Intercept form

The equation of the line cutting off intercepts a and b on the axes is

1b

y

a x

(v) General form

Every straight line can be represented by an equation of the first degree in x and y, and conversely everyfirst degree equation in x, y represents a straight line.

The equation A x +B y +C = 0 (where at least one of A and B is non-zero) is called the general form.


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a=2, b=c=3, m=-3/2, P1(2,0) P2(1,3/2), P3(0,3) A=3, B=2, C=-6

for m=-3/2, c=3, the form y=mx+c gives

y=-3/2x+3

for m=-3/2, P1(2,0), the form y-y1=m(x-x1) gives

y-0=-3/2(x-2)

or y=-3/2x+3

for P1(2,0) and P2(1,3/2), the form y-y1= 1212

x x

y y

(x-x1) gives

y-0=21

02/3

(x-2)

or y=-3/2(x-2)

or y=-3/2x+3


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for m=-3/2, c=3, the form y=mx+c gives

y=-3/2x+3

The cost of 10 units is Rs. 7500 and the cost of 20 units is Rs. 13900. Find the cost line

Given: P1(10,7500) P2(20,13900)

∴ Cost Liney=640x+1100

Units

x

Cost

y

Change

in x

Change

in y

0

1

2

5

10

15

20

1100

1740

2380

4300

7500

10700

13900

1

1

3

5

5

5

640

640

1920

3200

3200

3200

Polynomial Function

y=a0+ a1x+ a2x2+ a3 x

3+… +an x

n

y= an xn +…+ a3 x

3+ a2x2+ a1x+ a0

y= a1x+a0

y= a2x2+ a1x+a0

y= a3 x3 +a2x

2+ a1x+a0


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Quadratic Equation (Parabola)

y= ax2+ b x+ c

y= x2- 5x+6

(i) y= x2- 8x+20

x2- 5x+6=0

x2

- 8x+20=h

X 0 1 2 3 4 5 6 7 8 9

Y 20 13 8 5 4 5 8 13 20 29

(ii) y=- x2+ 6x+2


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X 0 1 2 3 4 5 6

Y 2 7 10 11 10 7 2

(iii) y=4x2-20x+29

X -5 -3 -1 1 3 5 7 9 11


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Y 229 125 53 13 5 29 85 173 293

(i) a=1,b=-8,c=20 V(4,4)

(ii) a=-1, b=6, c=2 V(3,11)

(iii) a=4, b=-20, c=29 V(5/2, 4)


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Parabola: y=ax2+bx+c

Mirror Image: y= (-ax2-bx+c) - b2/2a

Allied Parabolae: y=a (x-h)2 + k,

where h=-b/2a and k=-(b2-4ac)/4a


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Consider a parabola f(x)=ax2+bx+c.

Substitution of f(x)=h in the equation of the parabola results in

ax2+bx+c = h

or ax2+bx+c-h = 0

Solving for x gives

The coordinates of vertex of the parabola is

V(

, ) and the equation of the tangent line to the vertex is

h(x) =

Now for a>0 and a point p(x, h) for h ≥ (i.e. the point on or above the line

h(x) = ) or for a0 and a point p(x, h) for h <

(i.e. the point below the line h(x) = ) or for a (i.e. the point above the line h(x) = ), the value of b2 is less than

4a(c-h) and therefore the values of x are non-real. Such non real values of x can be

written as:


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or x=m± n ,where m = and n

Definition: A non-real coor dinate can be defined as “a coordinate, either of whose

abscissa or ordinate, but not both, is a non-real”.

So the coordinates of points for such non-real values of x are non-real i.e., the

coordinates are (m ± n , h), which cannot be located in any plane.The equation of the mirror image of the parabola f(x)=ax2+bx+c about an axis of

reflection h(x) = (which is the tangent line to the vertex) can be computed

as follow:

Ordinate of a point on mirror image = 2 × ordinate of the vertex –(ax²+bx+c)

or y= 2 × - (ax² + bx+ c)

which is the equation of the mirror image.

Now substituting g(x)=h (the value of h for which the values of x of the parabola is

non-real) in the equation of the mirror image of the parabola results in

or 4ah=-b²+4ac-4a²x²-4abx-b²

or 4a²x²+4abx+4ah+2b²-4ac=0

or 2a²x²+2abx+2ah+b²-2ac=0

Solving for x gives

or x= m ± n

and therefore the coordinates of points on the mirror image are real for the value of h

for which the coordinates of the parabola are non-real and are of the form ( m±n, h). The

coordinates of points on mirror image can therefore be obtained from the coordinates of


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the parabola by removing from the abscissa of the coordinates of the parabola(or it can be obtained using the transformation, derived below):

Derivation of Linear Transformation T: ℂ² ℝ² (in terms of coordinates) for whichT(m ± n , h) = (m ± n, h)

Let be the matrix of linear transformation then

=

which gives

a (m + n ) + bh = m + _________________________________________ (1)c (m + n ) + dh = h _________________________________________ (2)and =

which gives

a (m - n ) + bh = m - _________________________________________ (3)c (m - n ) + dh = h _________________________________________ (4)Subtracting (3) from (1) gives

a (m + n ) + bh - a (m - n ) - bh = m + - m +or am + an – am + an = 2nor 2an = 2nor a = 1or a = = x =

or a = -

Adding (1) and (3) gives

a (m + n ) + bh + a (m - n ) + bh = m + + m -or am + an + bh + am – an + bh = 2mor 2am + 2bh = 2m

or am + bh = m

Substituting the value of a gives

m(-

) + bh = m

or bh = m + m or bh = m(1 + ) or b =

Subtracting (4) from (2) gives

c (m + n )- c (m - n )+ dh – dh = h – h


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or cm + cn - cm+ cn = 0or 2 cn = 0or c = 0

Adding (2) and (4) gives

c (m + n )+ c (m - n )+ dh + dh = h +hor 2cm + 2dh = 2h

or cm + dh =

Substituting the value of c gives

0 + dh = h

or d = 1

Since a = - , b = , c = 0 and d = 1, so the matrix of linear transformation can bewritten as

and therefore

T(x,y) =

or

T(x,y) = (-ix + y , y)

is the required transformation in terms of coordinates

So it can be concluded that the non-real coordinates of the parabola f(x)=ax2+bx+c i.e.,

(m± ni, h) follow the rule given below:

Rule: “ correspond to each non-real coordinate (m + n , h) or (m - n , h) for a

parabola f(x)=ax2+bx+c , there exists a real coordinate(m+n,h) or (m-n,h) on its

mirror image about an axis of reflection h(x) = ”

Following examples elaborate the above discussion:

Let us take as an example the parabola f(x)= x2 - 8x+ 20 having a=1 (a>0) and vertex

V(4,4). The equation of the tangent line to the vertex is therefore h(x)=4.


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Take as an another example the parabola f(x) = -x2 +8x-25 having a=-1(a


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about the axis of reflection h(x) =-9 , cuts the x-axis at x=4±3 i.e., at x=1 or x=7.The

coordinates of the x-intercepts of the mirror image are therefore (1, 0) and (7, 0) (see

columns 6 and 12 of table 2 and the related graph)

In the second example, a=-1 (a


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0

(0o)

π/6

(30o)

π/4

(45o)

π/3

(60o)

π/2

(90o)

2π/3

(120o)

3π/4

(135o)

5π/6

(150o)

π

(180o)

Sin 0 1/2 1/√2 √3/2 1 √3/2 1/√2 1/2 0

Cos 1 √3/2 1/√2 ½ 0 -1/2 -1/√2 -√3/2 -1

Tan 0 1/√3 1 √3 ∞ -√3 -1 -1/√3 0

Wk=r1/n

[Cos{(Ɵ+2kπ)/n}+i Sin{(Ɵ+2kπ)/n}]

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