1 Basic Maths
2
1 Binomial theorem (a – b) = a2 + b2 – 2ab
2
a – b2 = (a + b) (a – b)
2 2
(1 + x) = 1 + 2 × 1x + x 2 2
(r – a) (r + a) ≈ (r2 – a2)2
if x <<< 1 then
2 Multiplication
(1 + x) = 1 + 2x
a c ac , Ex.: 2 5 10
MR* feel = =
b d bd 3 7 21
(Career + love)2 = Carrier + 2 love
Because carrier >>> love Division
n
n
x+Δx Δx Δx
= Xn 1 + = xn 1 + n a/b ad 2/3 (2)(4) 8
x x = , Ex.: = =
c/d bc 3/4 (3)(3) 9
Δ X <<<< X.
n
(1 – x) = 1 – nx Addition
(1 – x)
–n
= 1 + nx a c ad ± bc
± =
–n b d bd
(1 + x) = 1 – nx
1 5 (1)(6) + (2)(5) 16 4
Q.1. The expression of gravitational potential Ex.: + = =
2 6 (2)(6) 12 = 3
GMm
energy is U =– R + h where, R is the
radius of the planet and h is the height 3 Componendo and
above the surface. Approximate the Dividendo Theorem
expression by using binomial expansion
when h«R. In Ratio and proportion problems. It states
that for any four numbers a, b, c, and d. if;
Sol. GMm GMm
U =–
R + h = – R (1 + h/R) a c
=
For h«R b d
–1 h Then,
(1 + h/R) ≈ 1–
R a+b c+d
Therefore, =
a–b c–d
GMm
U ≈– 1– h 4 AP series
R R
2 Important formulae Next term = Previous term + Common
difference
2
(a + b) = a2 + b2 + 2ab a , a + d , a+2d , a + 3d , a + 4d.....
,Common difference Common ratio;
d = n
th
term– (n–1)th term nth term
r=
Ex: 2, 5, 8, 11, 14, 17, ..... (n-1)th term
th
d = 5 – 2 = 3 n term;
n
th
term; Tn = arn-1,
no. of Sum of infinites terms;
an = a + (n-1) d term a , valid when r < 1.
s∞ =
1-r
^ th st Common
n 1
diff. Q.2. Find the sum of given infinite series
term term
For last term, an = l
(i) 1, 1 , 1 , 1 , 1 , ....
2 4 8 16
Sum of n term; -1 , 1 - 1 , 1 , - 1 , ....
(ii) 1, ,
2 4 8 16 32
no. of terms.
1 1
Sn =
n
2a + (n-1) d Sol. (i) r = 1/4 = 1 , Sum = = =2
2 1/2 2 1-1 1/2
2
n st th
� Sn = (1 term + n term) 1 Sum= 1 1 2
2 (ii) r = - , =3 =
2 1--12 2 3
NOTE:-
n = no. of terms not last term. 6 Quadratic equation
st
Sum of 1 n-natural numbers
n(n+1) a, b, &
2
1 + 2 + 3 + 4 + .... + n = ax + bx + c = 0 c are constant
2 in which a can not
Sum of Squares of 1
st
n-natural numbers be zero
12 + 22 + 32 + 42 + ..... + n2 -b b2-4ac
X=
n (n+1) (2n+1) 2a.
=
6 -b
Sum of roots, x1 + x2 =
st a
Sum of Cubes of 1 n-natural numbers
n(n + 1)
2 c
Products of roots, x1·x2=
= a
2
Q.3. Find roots of equation x2 – 5x + 6
5 GP series = 0; find value of a, b & c by comparing
with ax2 + bx + c = 0
Next term = Previous term × Common ratio Sol. a = 1, b = –5 & c = 6
2 3 4
a , ar , ar , ar , ar (-5)2- 4×1×6
- (-5)
X=
Ex: 16, 8, 4, 2, 1, 1/2, 1/4, so on 2×1
2
Physics
, X1 = 5 + 1 log101 = 0
= 3 (Taking + sign)
2
log103 = 0.48 ≈ 0.5
X2 = 2 (Taking – sign)
loge(sin90°) = 0
2
Q.4. x – 4x = 0; then find roots of equation.
log105 + log1020 = 2
Sol.
log103
2
x = 4x log23 = 48
=
x = 4 wrong log102 30
Concept of Anti-log
x(x - 4) = 0
log ex = Y
x = 0 ; x = 4 correct
By taking Anti-log
(convert into concept of power)
Q.5. x2 – 4x + 3 = 0; then find roots. y
x=e
2
Sol. x – 3x – x + 3 = 0
MR* ka tadka
x(x-3) –1 (x – 3) = 0 log → Concept of Power
(x - 3) (x – 1) = 0 Power
23 = 8 log 28 = 3
x = 3, x = 1 Base ↑Result
Base wahi rahega (Power Result
7 Logarithms interchange hoga)
logy x = log x on the base y 8 Exponents
loge x = 2.303 log10 x A quantity multiplied by itself one or more
Properties of Logarithms times is indicated by an exponent. The
exponent, shown as a superscript, denotes
loga (xy) = loga x + loga y how many times the multiplication occurs,
for example:
x
log = log x – log y x =x
1
y
x · x = x2
1
logy x = x · x · x = x3
logx y
x · x · x · x = x4 ...... so on
1/n 1
logex = loge x
n The concept of exponents with a few
n examples are shown in table:
logex = n logex
Base
logba × loga b = 1 x2 x3 x4
x
logaa = 1 1 12 = 1 13 = 1 14 = 1
2 22 = 4 23 = 8 24 = 16
Some numerical value of Logarithms
3 32 = 9 33 = 27 34 = 81
loge1 = 0 4 42 = 16 43 = 64 44 = 256
5 52 = 25 53 = 125 54 = 625
log102 = 0.30
3
Basic Maths
2
1 Binomial theorem (a – b) = a2 + b2 – 2ab
2
a – b2 = (a + b) (a – b)
2 2
(1 + x) = 1 + 2 × 1x + x 2 2
(r – a) (r + a) ≈ (r2 – a2)2
if x <<< 1 then
2 Multiplication
(1 + x) = 1 + 2x
a c ac , Ex.: 2 5 10
MR* feel = =
b d bd 3 7 21
(Career + love)2 = Carrier + 2 love
Because carrier >>> love Division
n
n
x+Δx Δx Δx
= Xn 1 + = xn 1 + n a/b ad 2/3 (2)(4) 8
x x = , Ex.: = =
c/d bc 3/4 (3)(3) 9
Δ X <<<< X.
n
(1 – x) = 1 – nx Addition
(1 – x)
–n
= 1 + nx a c ad ± bc
± =
–n b d bd
(1 + x) = 1 – nx
1 5 (1)(6) + (2)(5) 16 4
Q.1. The expression of gravitational potential Ex.: + = =
2 6 (2)(6) 12 = 3
GMm
energy is U =– R + h where, R is the
radius of the planet and h is the height 3 Componendo and
above the surface. Approximate the Dividendo Theorem
expression by using binomial expansion
when h«R. In Ratio and proportion problems. It states
that for any four numbers a, b, c, and d. if;
Sol. GMm GMm
U =–
R + h = – R (1 + h/R) a c
=
For h«R b d
–1 h Then,
(1 + h/R) ≈ 1–
R a+b c+d
Therefore, =
a–b c–d
GMm
U ≈– 1– h 4 AP series
R R
2 Important formulae Next term = Previous term + Common
difference
2
(a + b) = a2 + b2 + 2ab a , a + d , a+2d , a + 3d , a + 4d.....
,Common difference Common ratio;
d = n
th
term– (n–1)th term nth term
r=
Ex: 2, 5, 8, 11, 14, 17, ..... (n-1)th term
th
d = 5 – 2 = 3 n term;
n
th
term; Tn = arn-1,
no. of Sum of infinites terms;
an = a + (n-1) d term a , valid when r < 1.
s∞ =
1-r
^ th st Common
n 1
diff. Q.2. Find the sum of given infinite series
term term
For last term, an = l
(i) 1, 1 , 1 , 1 , 1 , ....
2 4 8 16
Sum of n term; -1 , 1 - 1 , 1 , - 1 , ....
(ii) 1, ,
2 4 8 16 32
no. of terms.
1 1
Sn =
n
2a + (n-1) d Sol. (i) r = 1/4 = 1 , Sum = = =2
2 1/2 2 1-1 1/2
2
n st th
� Sn = (1 term + n term) 1 Sum= 1 1 2
2 (ii) r = - , =3 =
2 1--12 2 3
NOTE:-
n = no. of terms not last term. 6 Quadratic equation
st
Sum of 1 n-natural numbers
n(n+1) a, b, &
2
1 + 2 + 3 + 4 + .... + n = ax + bx + c = 0 c are constant
2 in which a can not
Sum of Squares of 1
st
n-natural numbers be zero
12 + 22 + 32 + 42 + ..... + n2 -b b2-4ac
X=
n (n+1) (2n+1) 2a.
=
6 -b
Sum of roots, x1 + x2 =
st a
Sum of Cubes of 1 n-natural numbers
n(n + 1)
2 c
Products of roots, x1·x2=
= a
2
Q.3. Find roots of equation x2 – 5x + 6
5 GP series = 0; find value of a, b & c by comparing
with ax2 + bx + c = 0
Next term = Previous term × Common ratio Sol. a = 1, b = –5 & c = 6
2 3 4
a , ar , ar , ar , ar (-5)2- 4×1×6
- (-5)
X=
Ex: 16, 8, 4, 2, 1, 1/2, 1/4, so on 2×1
2
Physics
, X1 = 5 + 1 log101 = 0
= 3 (Taking + sign)
2
log103 = 0.48 ≈ 0.5
X2 = 2 (Taking – sign)
loge(sin90°) = 0
2
Q.4. x – 4x = 0; then find roots of equation.
log105 + log1020 = 2
Sol.
log103
2
x = 4x log23 = 48
=
x = 4 wrong log102 30
Concept of Anti-log
x(x - 4) = 0
log ex = Y
x = 0 ; x = 4 correct
By taking Anti-log
(convert into concept of power)
Q.5. x2 – 4x + 3 = 0; then find roots. y
x=e
2
Sol. x – 3x – x + 3 = 0
MR* ka tadka
x(x-3) –1 (x – 3) = 0 log → Concept of Power
(x - 3) (x – 1) = 0 Power
23 = 8 log 28 = 3
x = 3, x = 1 Base ↑Result
Base wahi rahega (Power Result
7 Logarithms interchange hoga)
logy x = log x on the base y 8 Exponents
loge x = 2.303 log10 x A quantity multiplied by itself one or more
Properties of Logarithms times is indicated by an exponent. The
exponent, shown as a superscript, denotes
loga (xy) = loga x + loga y how many times the multiplication occurs,
for example:
x
log = log x – log y x =x
1
y
x · x = x2
1
logy x = x · x · x = x3
logx y
x · x · x · x = x4 ...... so on
1/n 1
logex = loge x
n The concept of exponents with a few
n examples are shown in table:
logex = n logex
Base
logba × loga b = 1 x2 x3 x4
x
logaa = 1 1 12 = 1 13 = 1 14 = 1
2 22 = 4 23 = 8 24 = 16
Some numerical value of Logarithms
3 32 = 9 33 = 27 34 = 81
loge1 = 0 4 42 = 16 43 = 64 44 = 256
5 52 = 25 53 = 125 54 = 625
log102 = 0.30
3
Basic Maths
