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FEATURES OF THIS BOOK: • All mathematical formula are in one frame. • Required for those who are preparing for entrance examination. • Helpful for pre-university, UG, PG students. • Easy to understand and apply . • It helps students to solve numerical in physics and maths

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MATHEMATICS
USED IN
PHYSICS
ESSENTIAL FOR
AIIMS/NEET/JEE
ASPIRANTS
FEATURES OF THIS BOOK:

 All mathematical formula are in one frame.
 Required for those who are preparing for entrance examination.
 Helpful for pre-university, UG, PG students.
 Easy to understand and apply .
 It helps students to solve numerical in physics and maths .

,MECHANICS
ALGEBRA
Common Identities
(i) (a + b)2 = a2 + b2 + 2ab = (a – b)2 + 4ab
(ii) ( a – b)2 = a2 + b2 – 2ab = (a + b)2 – 4ab
(iii) a2 – b2 = (a + b) (a – b)
(iv) (a + b)3 = a3 + b3 + 3ab(a + b)
= a3 + b3 + 3a2b + 3ab2
(v) (a – b)3 = a3 – b3 – 3ab (a – b)
= a3 – b3 – 3a2 b + 3ab2
(vi) a3 + b3 = (a + b) (a2 – ab + b2)
= (a + b)3 – 3ab(a + b)
(vii) a3 – b3 = (a – b) (a2 + ab + b2)
= (a – b)3 + 3ab (a – b)
(viii) (a + b)2 + (a – b)2 = 2(a2 + b2)
(ix) (a + b)2 – (a – b)2 = 4ab.
QUADRATIC EQUATION
An algebraic equation of second order (highest power of variable is 2) is called a quadratic equation
e.g.
ax 2  bx  c= 0, a0
It has solution for two values of x which are given by

b  b2  4ac
x =
2a
The quantity b – 4ac, is called discriminant of the equation.
2


BINOMIAL THEOREM
(i) The binomial theorem for any positive value of n

(x  a)n  xn  nC1axn1  nC2a2 xn2 ..........  nCr ar xnr  .........  an

n!
where a is constant and n =
Cr r !(n  r)!

Here n! = n(n  1)(n  2)............. 3  2 1
So 5! = 5  4  3  2 1  120
n(n 1) n(n 1)(n  2)
(ii) (1 x)n  1  nx  x2  x3  .....
2! 3!

For x  1, we can neglect the higher power of x.

So (1 x)n ; 1  nx

Similarly, (1  x)n ; 1  nx

(1 x)n ; 1  nx

(1  x)n ; 1  nx
Here n may have any value.

, Mathematics Used in Physics
Ex. 1 Evaluate (1.01)

Sol. (1.01)1/2 = (1  0.01)1/2
1
; 1   0.01
2
= 1.005

ARITHMETIC PROGRESSION (A.P.)
A sequence like a, a + d, a + 2d, ............is called arithmetic progression. Here d is the common
difference.
(i) The nth term of an A.P. is given by
an = a  (n 1)d
(ii) The sum of first n term of an A.P. is given by

Sn =
n
2
 I term + last term  (a1  an )
n
2
Here a1 = a and an = a + (n – 1)d
n [2a  (n 1)d ]
 Sn =
 2

GEOMETRIC PROGRESSION (G.P.)
The progression like, a, ar, ar2, .......... is called geometric progression, here r is called geometric ratio or
common ratio.
(i) The nth term of G.P. is given by

arn1
an =
(ii) The sum of the first n terms of G.P. is given by
a(rn 1)
Sn  for (r  0)
(r 1)

(1  rn )
and Sn  a for (r  0)
(1 r)
(iii) The sum of infinite term of G.P. for r < 1, is given by

Ist term
S =
1  Geometric ratio

a
or S =
1 r

1 1 1
Ex. 2 Find sum of the progression; 1, , 
, , ......... 





2 4 8

a
Sol. We have S =
1 r
1
Here, a = 1, r 
2
1
 S = 2
1  1 /2

, MECHANICS
EXPONENTIAL SERIES 
n
 1 1 1 1
The value of e ; e = lim  1   1    .........
n n 1! 2! 3!
1 1 1
= 1  1    ....... 2.718
2 6 24
x x2 x3
 ex = 1     ..........
1! 2! 3!

x
x x2 x3
and e = 1    ..........
1! 2! 3!
LOGARITHMIC SERIES
4
x2 x3  x
loge (1 x) = x   4  ..............
2 3
 1 1 1
loge (2) = loge (1 1)  1     .............. 
2 3 4

x2 x3 x4
log(1  x) =  x     ............. 
2 3 4
TRIGONOMETRIC SERIES
3 5
sin x = x  x  x ...........
3! 5!
2 4
cos x = 1  x  x ...........
2! 4!
LOGARITHMS
For a positive real number a and a rational number m, we have, am = b. The another way of expressing
the same fact in that of logarithms of b to the base a is m
i.e., loga b = m
There are two bases of logarithms that are used these days. One is base e and the other base 10. The
logarithms to base e are called natural logarithms. The logarithms to base 10 are called the common
logarithms.
Thus we can write
(i) 1000 on the base of 10 as 103, and in logarithms it is; log101000 = 3.
(ii) Similarly ex = y can be written as
logey = lny = x
Here loge  ln
loga1 = 0 ; log1010 = 1; log10 2 = 0.693; loge10 = 2.303

LAWS OF LOGARITHMS
Ist Law loga (mn) = loga m  loga n

 m 
IInd Law loga   = log a m  loga n
 n 

IIIrd Law loga (m)n = n loga m

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