Concept Maps Class X Real Numbers Polynomials Linear equations in two variables Quadratic equation Airthmetic Progression Similar Triangles Height & Distance Tengent Coordinate Geometry Trignometry Construction Area related to circle Surface area & volume Probability Statistics
Real Numbers Real Numbers
p,
1,1/2,–7/5...
2
Irrational Numbers
Rational Numbers
89
Integers Non-Integers
Terminating
.... ...............
Non-Terminating
....
-3
-2
-1
0
1
2
3
.... Application H.C.F
3 22 ! 5
L.C.M
10 3
Euclid's division lemma
Definition
(p ! q ! r ) ! H.C.F(p, q, r ) H.C.F.(p, q) ! H.C.F.(q, r ) ! H.C.F(p, r ) H.C.F. (p,q,r) =
(p ! q ! r ) ! L.C.M.(p, q, r ) L.C.M.(p, q) ! L.C.M.(q, r ) ! L.C.M.(p, r )
Polynomials Polynomials 2
F(x) = ax + bx + c F(x) = ax3 + bx2 + cx + d Remainder theorem
Value of polynomial
Degree of polynomial
Factor theorem
Replace 'x' with 'k' ® P(k)
Highest power of x in polynomial
If P(k) = 0 Dividend = divisor ´ quotient + remainder
Then 'k' is a root or zero of the polynomial.
Geometrical interpretation of roots of zeroes of polynomial.
" coefficien t of x n"1
b a
coefficien t of xn cons tan t term coefficien t of x n
Y
Y
1 root
2 roots
Y
1 root
0
X
Y
No roots
0
X
Y
3 roots
0 X'
0
Y'
X
X'
0
Y'
X
X'
Y'
X'
Y'
X'
X
Y'
d a
Linear equations in two variables Linear equation in two variables a1x + b1y + c1 = 0 a2x + b2y + c2 = 0
..(1) ..(2)
Equation of a straight line
Equations reducible to a pair of linear equations
) a2 + b 2 * 0 ## ax + by + c = 0 ® ( a * 0,b * 0 # a, b, c are real nos. '#
Methods to solve
& # # % # # $
2 3 5 4 + , 13, " , "2 x y x y
Let
y = 2x – 4
Y
2p + 3q = 13 5p – 4q = –2
1 1 , p, , q x y
Solution (x,y) ® point lying on straight line. Algebric method
Graphical method a1 b1 * a2 b2
Intersecting (intersect at 1 point)
2 X'
0
X
-4
one solution (consistent)
a1 b1 c1 , , - Infinite a 2 b2 c 2
Coincident (Coincide)
solution
Y' a1 b1 c1 , * - No solution a2 b2 c 2
Parallel (No intersection)
Substitution method " c1 " b1y x, {Substitute in ( 2)} a1 " a2c1 " a2b1y + a1b 2 y , " c 2a1
x,
b1c 2 " b2c1 a1b2 " a2b1
y,
a2c1 " a1c 2 a1b2 " a2b1
Elimination method Multiply b2 in (1) & b1 in (2) a1b2x + b1b2y + c1 b2 = 0 ..(3) b1 a2x + b2b1y + c2b1 = 0 ..(4) (3) -(4) .... (a1b2-b1a2)x + (c1b2-c2b1) = 0 x,
b1c 2 " b2c1 a1b2 " a2b1
y,
a2c1 " a1c 2 a1b2 " a2b1
(inconsistent)
Cross multiplication method y x 1 b1 a1 c1
b1
b2
b2
a2
c2
x y 1 , , b1c 2 " b2c1 c1a2 " a1c 2 a1b2 " a2b1 x,
b1c 2 " b2c1 a1b2 " a2b1
y,
a2c1 " a1c 2 a1b2 " a2b1
Quadratic equation Methods to solve
Quadratic Formula
Completing the square
Factorization
Convert to (x+a)2-b2 = 0
e.g. x2 – 3x – 4 = 0 2 Þ x + (–4 + 1)x + (–4´1) = 0 Þ (x – 4) (x + 1) = 0 Product of roots = –4 Sum of roots = 3
e.g. x2 + 4x = 0 Þ x2 + 4x + 4 – 4 = 0 Þ (x + 2)2 – 22 = 0
. ,
" b + (b2 " 4ac ) 2a
/,
" b " (b2 " 4ac ) 2a
Y
Roots Standard form
X'
0
X
(b2– 4ac ³ 0 ® real roots)
Y' Y
2
ax + bx + c = 0 ; 2
ax +bx + c = 0 ; a ¹ 0 p(x) = 0 "where p(x) is polynomial of degree 2" e.g. (x + 2)3 = x3 – 4 Þ x2 + 4x + 2 = 0 (after simplification)
If a and b are the roots of the above equation then : aa2 + ba + c = 0 ab2 + bb + c = 0
X'
0
X
(b2– 4ac = 0 ® equal roots)
Y'
"b
Y
\a+ b = c a ab = a
X'
APPLICATIONS 1. Speed =
OF QUADRATIC EQUATIONS
Dis tan ce Time
2. Area of figures 3. Flow rate ´ time = volume of water 4. Number or ages
0
: Y'
X
(b2– 4ac < 0 ® Complex roots)
Airthmetic Progression Note : Taken 3 terms in A.P. : (a – d), a, (a + d)
Terms Properties
Numbers in the series are called Terms.
Common difference
Taken 5 terms in A.P. : (a – 2d), (a – d), a, (a + d), (a + 2d)
Common difference 'd' can be zero, positive, negative. a2 – a1 = a3 – a2 = a4 – a3 = a5 – a4 = an-an-1 = d. 5 – 2 = 8 – 5 = 11 – 8 = 14 – 11 = 17 – 14 = 3.
a1 ® a1 = 2 ® 1st Term a2 ® a1 + d = 5 ® 2nd Term nth Term
Arithmetic Progressions
Tn = a + (n–1) d.
a3 ® a1 + 2d = 5 ® 3rd Term . . an ® a1 + (n–1)d = 74 ® 25th Term
Sum of n Terms
st
Sum of 1 n terms of an A.P. ® Sn Sn= a1 + a2 + a3 + .... + an- 1 + an. n(n + 1) Sn = 2
Sum of n natural nos.
RESULTS
2
(n – m + 1)th term
mth term from end
th
Find A.P. whose n term is given ? a1, a2, a3,..... an-1, an. form an A.P.
n
e.g. Sum of first 7! 8 , 28 7 natural nos. Þ
Find Tn when Sn is given
e.g. 2, 5, 8, 11, ..... up to 25 terms.
e.g. Tn = 3n + 5 put n = 1,2,3...... T1 = 8, T2 = 11, T3 = 14......
Tn = Sn – Sn-1 e.g. Sn = n2 + 2n \ Tn = 2n + 1
Condition of an A.P. If a, b, c are 3 terms of an A.P. then : a + c = 2b.
n
Sn= 2 [2a + (n-1)d] = 2 [a1 + an]. e.g. 25 S25 = 2 [2(2) + (25-1)3] = 950
Similar Triangles Similar Triangles
Means 'Same shape' e.g. All circles are similar. All squares are similar.
Criteria
If Ds are similar
Q
P
Then :
A
Area theorem
R
Pythagoras theorem
C
B
Q AA criterion : If ÐA = ÐP and ÐB = ÐQ then DABC ~ DPQR.
P
B
R C
B
A
If DABC ~ DPQR : 2
2
2
Area( 6 ABC) 5 AB 2 5 AC 2 5 BC 2 ,3 0 ,3 0 ,3 0 Area( 6 PQR ) 4 PQ 1 4 PR 1 4 QR 1
SSS criterion : AB
AC
BC
If PQ , PR , QR then DABC ~ DPQR
D
AB AC BC , , PQ PR QR
C
DABC is right angled D (ÐB = 90°) DABC ~ DBDC ~ DADB : BD2 = AD ´ BC. BC2 = CD ´ AC. AB2 = CA ´ AD. AB2 + BC2 = AC2 ® P.Th.
AB
BC
ÐB = ÐQ then DABC ~ DPQR
and
Proved by Basic Proportionality theorem (THALES THEOREM) A
SAS criterion :
C
ÐA = ÐP ÐB = ÐQ ÐC = ÐR
E
D
If P Q , Q R and
R
P
B
A
A
Q
If DE | | BC then, C
B
AD AE , DB EC
NOTE : DABC ~ DPQR doesn't mean DABC ~ DQPR.
Converse of B.P.T. : If
AD AE , DB EC
DE | | BC.
then
Height & Distance
Heights & Distance C
A
Elevation Horizontal A
q
B
Height of tower BC = AB ´ tan q (given AB & q)
Horizontal
Height of tower AB = tan q ´ BC (given a & BC)
Applications
Depression a C B a = Angle of Depression
q = Angle of elevation @ Navigation @ Land surveys @ Buildings @ Optics @ Statics @ Crystallography
Definition : Angle formed by the line of sight with the horizontal when the point is above the horizontal.
Definition : Angle formed by the line of sight with the horizontal when the point is below the horizontal.
e.g.
e.g. Height of tower AB =
b a
CD ! tan . ! tan / (tan / " tan . )
PD =
(given CD, a, b) a
b
AB ! tan / (tan . " tan / )
(given AB, a, b) b
e.g. Height of tower BD =
(tan / " tan . ) ! AB tan .
(given AB, a, b) a b
Height of tower
a
Tengent
QT= 1/2 QR
NO TANGENT
ONE TANGENT
P
Q
P
O
P
R
of tangents from an external point to a circle are equal . * Length
In DOQP & ORP = OQ = OR (radii), OP = OP \ DOQP 7 DORP Þ PQ = PR AND
TANGENT
In DPQR PQ=PR, hence ÐPQR = ÐPRQ = 1/2 (180 – q) Since ÐOQR = 90º, ÐOQR = ÐOQP – ÐPQR ÐOQR = 90º – (90 – 1/2q) =1/2q Þ2 ÐOQR = ÐQPR.
&ÐQPR + ÐQOR = 180º ÐQPO = ÐRPO ÐQOP = ÐROP
P Tangent
B
Coordinate Geometry
Distance formula
Area of Triangle
Section formula
Q (x2,y2)
y
y P (x1,y1)
A (x1,y1)
R (x1,O)
S (x2,O)
x
RS = (x2–x1) = PT SQ = y2 ; ST = y1 ; QT = (y2 – y1) Applying Pythagoras in DPTQ PQ2 = PT2 + QT2 = (x2 – x1)2 + (y2 – y1)2
C
S
R
O
( x 2 " x1)2 + ( y 2 " y1)2
(x2,y2) B
Q
T
x
O
Q
P
R
A(x1y1) B(x2,y2) C(x3,y3)
(m : n Internally)
{ m mx ++ mm x 1 2
1
2 1
2
, m1y 2 + m2 y1 m1 + m2
{
Area = 1/2 ´ base ´ Altitude
{
m1x 2 " m2 x1 , m1y 2 " m2 y1 m1 " m2 m1 " m2
Area = 1/2 [x1(y2 –y3) + x2(y3 –y1) +x3(y1-y2)]
{
If Area = 0,
(Distance formula)
Application of distance formula Verifying collinearity A(x1,y1), B(x2,y2), C(x3,y3) ® Find AB,BC,CA using distance formula ® If AB + BC = CA or BC + CA = AB or AB + CA = BC ® Then 3 pointare collinear
x
Coordinates of P(x–xy) =
(m : n externally) PQ =
A (x,y) C(x3,y3)
(x,y)
O
y
B (x2,y2)
T (x2,y1)
Verifying collinearity A(x1,y1), B(x2,y2), C(x3,y3), D(x4,y4) ® Find AB,BC,CD,DA using distance formula ® Find diagonals CA & BD ® Check with different properties of quadrilaterals
then points are collinear Verifying triangle formations A(x1,y1), B(x2,y2), C(x3,y3) ® Find AB,BC,CA using distance formula Check if AB + BC > CA BC + CA > AB AB + CA > BC ® If yes, it is triangle, check for right D using pythagoras therom ® If no, it is not a D
Complimentry Angles
Graphs
T-ratios sinA =
AB
cosecA = BC = P
B
AB
cosA = AC
H
AC
cosA = AC = H
tanA =
H
AC
BC P = AC H
secA = AB = B
BC P = AB B
cotA =
secA =
cos(90-A) =
AB B = BC P
sinA =
BC AC
BC AC
tanA =
AC
AB BC
cot(90-A) =
y
AC cosec(90-A)= AB
AB tan(90-A)= BC
AB sin(90-A) = AC
secA = AB BC AB
sec(90-A) =
BC AB
AC BC
x'
" 38 2
"8
8 2
"8 0 2
8
38 2
AC BC
cosecA =
x
TAN
y'
y C ÐA of DABC = ÐA of DAMP P BC MP sin A = = AC AP A Note ; M B The values of the trignometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Interrelationship between T-ratios
sinA = T R I G O N O M E TRY
1 cos ecA
cosA =
1 sec A
tanA =
1 cot A
C
90° B
A
2
Divide both sides by Identities
AC2 AB
BC2 AC2
2
BC2 1+
+
+ 1,
BC2 AB2
0.5 0
8 2
0.5
8
38 2
28
2
,
2
2
, 1 =sin A + cos A = 1
AC
2
BC2 AC2 AB2
2
2
=cot A+1 = cosec A
AC2
BC
2
2
= 1+tan2A = sec2A
AB
2
0°
30°
45°
60°
90°
sin
0 0
1 4
2 4
3 4
4 4
cos
4 4
3 4
2 4
1 4
0 4
tan
0 4
1 3
2 2
3 1
9
cosec
9
4 1
4 2
4 3
4 4
sec
4 4
4 3
4 2
4 1
9
cot
9
3 1
2 2
1 3
0 4
Simplified trigonometric values
x
-1
SIN
y'
AB + BC = AC
AB2
1
y 1 0.5 0
8 2
0.5
8
38 2
28
x
-1 y'
COS
Construction
Triangle Scaling
Division of a line segment in m:n(3:2)ratio
Constructing Tangents
SCALING A
B2
B3 B
4
A' A C'
B1
B5 x
B
A
B
C B1
Q
B5C'||B3C C'A'||CA
B
B
A
1. Scaling up 5/3
P
O2
B4
X
B5 x
2. Scaling down 3/4
B1
B2
B3 B
C
B
4
x
Alternate-Method
3. 2 circles intersect at Q & R
A' C B2
X
AC 3 AA 3 = = A 3 A 5 CB 2
2. O2 as centre & O1O2 as radius , draw a circle
A
B1
Join A3C such that A3C is parallel to A5B
DAA3C ~ DAA5B hence 1. Join PO1 and bisect it (at O2)
C
B
R
B3
5 BC' BB 5 = = 3 BC BB 3
B
2 C
C B2
AB AC BC = = A' B A ' C' BC'
B3C'||B4C C'A'||CA
3
O1
DABC ~ A'BC'
A
A
B3
B4
DABC ~ A'BC' A ' B A ' C' BC' 3 = = = AC BC 4 AB
B
A
4. Join PQ & PR (Tangents) , Y
B2
x
Y
A
B Join A3B2
A1 A2
A3
B2
B1 A
3
B
A1 A2
X DAA3C ~ DBB2C AA 3 AC 3 = = BB2 BC 2
B1
2
A3
X
Area related to circle
Sector of a circle
Major sector
Area of combination of plane figures
Q
e.g. q = 60° r = 3m
e.g.
2 units
A
B
2
Area = pR r
q
r Generally sector implies minor sector
B
A
C
Sol.
P Length of sector AB =
r
r + r + r + r = length of square Þ 4r = ; r = 1/2 Area of shaded reg. = Area of square – Area of 4 circles 2 = (2´2) – 4 ´ 8 ( 1 )
Area of sector : AOBP = ´ pR2 360
: ´ 2pR 360
e.g.
e.g.
60 ´ p ´ 32 360 38 Þ m 2
: ´ 2p ´ 3 360 Þ pm
D
4
AOBP =
AB =
=4–p
Note: For rotating wheel 2pRn = speed ´ Time = Distance covered n = number of rotations
e.g.
2 units
A
B
I
Area of segment (APB) = Area of sector (AOB) – Area of DAOB
IV
II III
e.g. A
C 2 units
Sol.
B
I IV
II III
C
D
Sol. Area I + III = A(ABCD) – Area of semicircle with AD & BC as diameter =4–p
D
Area of shaded reg. = Area of square ABCD –Area of (1+2+3+4) = 4 – 2(4 – p) = 2p – 4
Surface area & volume
Cuboid
Frustum Cylinder
Cube Cone
Hemisphere Sphere
l1-l r2
h
r a
b
h
h
1. T.S.A ® 2[lb +bh + hl] 2. C.S.A ® 2[bh +hl] 3. Volume ® l ´ b ´ h 4. Diagonal ® 2 + b2 + h2
a 2 1. T.S.A ® 6a 2 2. C.S.A ® 4a 3 3. Volume ® a 4. Diagonal ® 3a
NOTE : 1. C.S.A.® Curved surface area 2. T.S.A.® Total surface area 3. l ® Length 4. b ® Breadth 5. h ® Height 6. l ® Slant height 7. r ® Radius 8. a ® Side of cube
h
l
r1
r
l
a
h1
r
r r
r
2
1. T.S.A ® pr(l + r) 2. C.S.A ® prl 3. Volume ® 1 8 r 2h 3
1. T.S.A ® 2pr(r+h) 2. C.S.A ® 2prh 2 3. Volume ® pr h
Volume = Volume of cylinder – Volume of hemisphere
1. C.S.A = T.S.A. = 4pr2 2. Volume = 4/3pr3
1. C.S.A = 2pr 3 2. T.S.A. = 3pr 3. Volume = 2/3pr3
1. C.S.A = p1r1l1 – pr2(l1–l) 2 2 2. T.S.A. = pl(r1+r2) + p(r1 + r2 ) 2 2 3. Volume = 1/3ph(r1 + r2 + r1r2) 4. Slant height = h2 + (r1 " r2 )2
Total surface area = C.S.A. of 2 hemispheres + C.S.A of cylinder
Probability PROBABILITY Elementary event (E) : Only one outcome sum of the probabilities of all elementary events is 1.
Applications Gambling, insurance & statistics control theory.
0 £ P(E) £ 1
Complimentary Events(E) Not E = E, P(E) = 1 – P (E) ln case 1 :
No. of outcomes favourable to E P(E) = No. of all possible outcomes
EVENT(E) 1 P(1) = ; P(1) = 1 – 1 = 5 6 6 6 P(1) = not getting 1 ; = getting 2,3,4, 5 and 6
In case 1 P(E) > 4 =P(5) + P(6) =
2 1 = 6 3
P(E) = 4 = P(1) + P(2) + P(3) + P(4) =
4 2 = 6 3
In case 2 In case 3 Sum = 8 for (2,6) (3,5) (4,4) (5,3) (6,2) 5 P(sum 8) = 36
Probability of picking up black P(B) = 2 9
Certain event P(E) = 1 Impossible event P(E) = 0 In case 2 : P(sum 13) is impossible. hence, P (sum 13) = 0
1
2
36 posibilities 3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
3,1
5,2
5,3
5,4
5,5
5,6
Examples Case -1) One dice is rolled: P(1) = Probability of getting 1. Similarly P(2),P(3), P(4), P(5) & P(6) P(1) =
1 1 , 1,2,3,4,5,6 6
P(E) > 4 ® Probability of getting 5 & 6.
Case -2) Two dice are rolled: P(sum = 8) = Probability of getting two numbers whose sum is 8.
Case -3) Box with balls : A box contains 2 black, 4 green and 4 red balls.
Red
Red
Red
Black
Black
Red
Green
Green
Green
Statistics
Grouped Class of Intervals NO. of Students
Ungrouped 10-25
25-40
40-55
55-70
70-85
85-100
2
3
7
6
6
6
MODE
MEAN
The value of the observation having the max. frequency
; fx x= ;f
i i i
MODE
MEDIAN Intervals
No. of students
C.f.
n-C.f.
10-25
2
2
28
25-40
3
5
25
40-55
7
12
18
55-70
6
18
12
70-85
6
24
6
85-100
6
30
0
Class size (h) = 15 Max. frequency f1 = 7 Modal class = 40-55 Lower limit of modal class = 40 f0 = 3 (Previous class f value) f2 = 6 (next class f value)
n / 2 " c.f . ! h = 62.5 Median = l + f
x1, x2, ........ xn ® observations f1, f2, ........ fn ® frequencies th
5n 2 Average of 3 + 10 42 1
5 f1 " f0 2 Mode = l + 33 2f " f " f 00! h 4 1 0 21
Step deviation
x =
; fx ;f
i i
Assumed mean method
x = a + hu = 62
5 x = a + h 33 4
i
Class Intervel
fi
xi
10-25
2
25-40
3
40-55
; f d 20 = 62 ; f 01
i i i
di = x i – a
17.5
35
-30
32.5
97.5
-15
7
47.5
332.5
55-70
6
62.5
70-85
6
85-100
6 30
x = a + d 5 x = a + 3 3 4
; f x 20 =62 ; f 01
fi x i
xi " a h
i i i
fi ui
fi di
-2
-4
-60
-1
-3
-45
0
0
0
0
375
15
1
6
90
77.5
465
30
2
12
180
92.5
555
45
3
18
1860
ui =
n is even
th
3 Methods
Direct Method
th
5n 2 & 3 0 observation. 42 1 5 n + 12 3 0 observation. 4 2 1
Mode = 40 + 4/5 ´15 = 52 MEAN
n = 30 ; n/2 = 15 55-70 is median class lower limit of the median class (l) = 55 c.f. = cum frequency of median preceeding class
MEDIAN
;
f iu i = 29
270
;
fi di
= 435
n is odd
