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x ( f

1 is also periodic with same period T.

If f(x) is periodic with period T, f(x) is also periodic with same period T.

Period of x – [x] is 1. Period of algebraic functions x , x², x³ + 5 etc. does't exist.

If limf(x)

x→a does not exist, then we can not remove this discontinuity. So this become a non-removable discontinuity or essential discontinuity.

If f is continuous at x = c and g is discontinuous at x

= c, then

(a) f + g and f – g are discontinuous (b) f.g may be continuous

If f and g are discontinuous at x = c, then f + g, f – g and fg may still be continuous.

Point functions (domain and range consists one value only) is not a continuous function.

If a function is differentiable at a point, then it is continuous also at that point.

i.e., Differentiability ⇒ Continuity, but the converse need not be true.

If a function 'f' is not differentiable but is continuous at x = a, it geometrically implies a sharp corner or kink at x = a.

If f(x) and g(x) both are not differentiable at x = a then the product function f(x).g(x) can still be differentiable at x = a.

If f(x) is differentiable at x = a and g(x) is not differentiable at x = a then the sum function f(x) + g(x) is also not differentiable at x = a.

If f(x) and g(x) both are not differentiable at x = a, then the sum function may be a differentiable function.

Differentiation and Applications of Derivatives : dx

dy is dx

d (y) in which dx

d is simply a symbol of operation and not 'd' divided by dx.

If f´(x0) = ∞, the function is said to have an infinite derivative at the point x0. In this case the line tangent to the curve of y = f(x) at the point x0 is perpendicular to the x-axis.

Of all rectangles of a given perimeter, the square has the largest area.

All rectangles of a given area, the squares has the least perimeter.

A cone of maximum volume that can be inscribed in a sphere of a given radius r, is of height

3 r 4 .

A right circular cylinder of maximum volume that can be inscribed in a square of radius r, is of height

3 r 2 .

If at any point P(x1, y1) on the curve y = f(x), the tangent makes equal angle with the axes, then at the point P, ψ =

4 π or

4

3π. Hence, at P tan ψ = dx dy = ±1 Indefinite Integral :

If F1(x) and F2(x) are two antiderivatives of a function f(x) on an interval [a, b], then the difference between them is a constant.

The signum function has an antiderivative on any interval which does not contain the point x = 0, and does not possess an anti=derivative on any interval which contains the point.

The antiderivative of every odd function is an even function and vice-versa.

CALCULUS

Mathematics Fundamentals

MATHS

If In =

∫

^{xn.eaxdx, then In = xn}_a^{eax –} _a^nIn–1

If In =

∫

^{(logx)dx, then In} = x log x – x If In =

∫

_log¹_x^{dx, then}

In = log(logx) + logx + )

! 2 .(

2 ) x (log ² +

)

! 3 .(

3 ) x (log ³ + ...

If In =

∫

^{(logx)ndx; then In = x(logx)n – n.In–1}

Successive integration by parts can be performed when one of the functions is xⁿ (n is positive integer) which will be successively differentiated and the other is either of the following sin ax, cos ax, e^–ax, (x +a)^m which will be successively integrated.

Chain rule :

∫

^{u.vdx= uv1 – u´v2 + u"v3} – u"'v4 + ....

+ (–1)^{n – 1}u^n–1vn + (–1)ⁿ

∫

^un.vdx

where uⁿ stands for n^th differential coefficient of u and vn stands for n^th integral of v.

∫

^xeaxsin(bx + c)dx = ₂ ^ax₂ b a

xe

+ [a sin(bx + c) – b cos(bx + c)] – ₂ ^ax_{2 2}

) b a (

e

+ [(a² – b²)sin (bx + c) – 2ab cos (bx + c)] + k

∫

^xeaxcos(bx + c)dx = _{2 2}

ax

b a

e . x

+ [a cos(bx + c) – b sin(bx + c)] – ₂ ^ax_{2 2}

) b a (

e

+ [(a² – b²)cos (bx + c) – 2ab sin (bx + c)] + k

∫

^xeaxsin(bx + c)dx

= ^x_{2 2}

b ) a (log

a

+ [(loga)sin(bx + c) – b cos(bx + c)] + k

∫

^xeaxcos(bx + c)dx

= ^x_{2 2}

b ) a (log

a

+ [(loga)cos(bx + c) + b sin(bx + c)] + k

∫

_c^a_cos^cos_x^x₊⁺_d^b_sin^sin_x^xdx

= _{2 2}

d c

bd ac

+

+ x + _{2 2} d c

bc ad

+

− log |c cos x + d sinx| + k.

Reduction formulae for I(n,m) =

∫

_cos^sinmnx_x ^{dx is}

I(n,m) = 1 m

1

− .

x cos

x sin

1 m

1 n

−

−

– (m–1) ) 1 n ( −

.I(n–2, m – 2) Definite Integral and Area Under Curves :

The number f(c) = −

∫

b a f(x)dx )

a b (

1 is called the

mean value of the function f(x) on the interval [a, b].

If m and M are the smallest and greatest values of a function f(x) on an interval [a, b], then m(b – a) ≤

∫

a^bf(x)dx≤ M(b – a).

If f²(x) and g²(x) are integrable on [a, b], then

∫

a^bf(x)g(x)dx ≤

2 / b 1

a

2(x)dx

f 



 





∫

a^{b 1/2}

2(x)dx

g 



 





∫

Change of variables : If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously differentiable on the interval [t1, t2] and a = φ(t₁), b = φ(t₂), then

∫

a^bf(x)dx=

∫

t₁^t2f(^φ(t)^φ´(t)dt

Let a function f(x, α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d. Then for any α ∈[c, d], if I(α) =

∫

a^bf(x,^α)dx, then I´(α) =

∫

a^bf´(x,^α)dx, where I´(α) is the derivative of I(α) w.r.t. α and f´(x, α) is the derivative of f(x, α) w.r.t α, keeping x constant.

∫

a^bf´(x)dx= (b – a)

∫

0¹f[(b⁻a)t⁺a]dt

∫

a^bf(x)₊f(a₊b₋x) dx ) x (

f =

2 1(b – a)

The area of the region bounded by y² = 4ax, x² = 4by is 3

ab

16 sq. unit.

The area of the region bounded by y² = 4ax and y = mx is ²₃

m 3

a

8 sq. unit.

The area of the region bounded by y² = 4ax and its latus-rectum is

3 a 8 ²

sq. unit.

The area of the region bounded by one arch of sin(ax) or cos (ax) and x-axis is 2/sq. unit.

Area of the ellipse (x²/a²) + (y²/b²) = 1 is πab sq. unit.

Area of region bounded by the curve y = sin x, x-axis and the line x = 0 and x = 2π is 4 sq. unit.

Complex Numbers :

|z| ≥ |Re(z)| ≥ Re(z) and |z| ≥ |1m(z) | ≥ 1m(z)

| z

|

z is always a unimodular complex number if z ≠ 0

|Re(z) | + |1m(z) | ≤ 2 |z|

If z

z+ = a, the greatest and least values of |z| are 1

respectively 2

4 a a+ ²+

and 2

4 a a+ ²+

−

|z1 + z₁²−z²₂ | + |z2 – z₁²−z²₂ | = |z1 + z2| + |z1 – z2| If z1 = z2 ⇔ |z₁| = |z2| or arg z1 = arg z2

|z1 + z2| = |z1 – z2| ⇔ arg (z₁) – arg(z2) = π/2.

If |z1| ≤ 1, |z₂| ≤ 1 then

(i) |z1 + z2|² ≤ (|z₁| – |z2|)² + (arg (z1) – arg (z2))² (ii) |z1 + z2|² ≥ (|z1| + |z2|)² – (arg (z1) – arg(z2))²

|z1 + z2|² = |z1|² + |z2|² + 2||z2| cos(θ₁ – θ₂).

|z1 – z2|² = |z1|² + |z2|² – 2|z1||z2| cos(θ₁ – θ₂).

If z1 and z2 are two complex numbers then |z1 z2| = r1r2; arg(z1z2) = θ₁ + θ₂ and

2 1

z z =

2 1

r r ,

arg 

 





2 1

z

z = θ₁ – θ₂ and where |z1| = r1, |z2| = r2,

arg(z1) = θ₁ and arg(z2) = θ₂.

The area of the triangle whose vertices are z, iz and z + iz is

2 1|z|².

The area of the triangle with vertices z, wz and z + wz is

4 3 |z²|.

If z1, z2, z3 be the vertices of an equilateral triangle and z0 be the circumcentre, then

12

z + z²₂ + z²₃ = 3z²₀.

If z1, z2, z3 ... zn be the vertices of a regular polygon of n sides and z0 be its centroid, then

12

z + z²₂ + ...+ z²_n = nz²₀.

If z1, z2, z3 be the vertices of a triangle, then the triangle is equilateral if

(z1 – z2)² + (z2 – z3)² + (z3 – z1)² = 0 or z₁² + z²₂ + z²₃ = z1z2 + z2z3 + z3z1

or

2

1 z

z 1

− +

3

2 z

z 1

− +

1

3 z

z 1

− = 0

If z1, z2 z3 are the vertices of an isosceles triangle, right angled at z2 then z₁² + z²₂ + z₃² = 2z2(z1 + z2) If z1, z2, z3 are the vertices of a right-angled isosceles

triangle, then (z1 – z2)² = 2(z1 – z3)(z3 – z2).

If z1, z2, z3 be the affixes of the vertices A, B, C respectively of a triangle ABC, then its orthocentre is

C sec c B sec b A sec a

z ) C sec c ( z ) B (sec b z ) A (sec

a _{1 2 3}

+ +

+ +

For any a, b ∈ R

(i) a+ib + a−ib = 2{ a²+b² +a} (ii) a+ib – a−ib = 2{ a²+b² −a}

The sum and product of two complex numbers are real simultaneously if and only if they are conjugate to each other.

If ω and ω² are the complex cube roots of unity, then (i) (aω + bω²)(aω² + bω) = a² + b² – ab

(ii) (a + b (aω + bω²)(aω² + b²ω) = a³ + b³ (iii) (a + bω + cω²)(a + bω² + cω)

= a² + b² + c² – ab – bc – ca (iv) (a + b + c) (a + bω + cω²) (a + bω² + cω)

= a³ + b³ + c³ – 3abc

If three points z1, z2, z3 connected by relation az1 + bz2 + cz3 = 0 where a + b + c = 0, then the three points are collinear.

If three complex numbers are in A.P. then they lie on a straight line in the complex plane.

Progression :

If Tk and Tp of any A.P. are given, then formula for obtaining Tn is

k n

T T_{n k}

−

− = k p

T T_{p k}

−

− .

If pTp = qTq of an A.P., then Tp + q = 0.

If p^th term of an A.P. is q and the q^th term is p, then Tp+q = 0 and Tn = p + q – n.

If the p^th term of an A.P. is 1/q and the q^th term is 1/p, then its pq^th term is 1.

ALGEBRA

Mathematics Fundamentals

MATHS

The common difference of an A.P. is given by d = S2 – 2S1 where S2 is the sum of first two terms and S1 is the sum of first term or the first term.

If sum of n terms Sn is given then general term Tn = Sn – Sn–1, where Sn–1 is sum of (n – 1) terms of A.P.

If for an A.P. sum of p term is q and sum of q terms is p, then sum of (p + q) terms is {–(p + q)}.

If for an A.P., sum of p term is equal to sum of q terms, then sum of (p + q) terms is zero.

If the p^th term of an A.P. is 1/q and q^th term is 1/p, then sum of pq terms is given by Spq =

2

1(pq + 1).

Sum of n A.M.'s between a and b is equal to n times the single A.M. between a and b.

i.e. A1 + A2 + A3 + ... + An = n 



 



 + 2 b

a .

If Tk and Tp of any G.P. are given, then formula for obtaining Tn is ^{n k}

1

k n

T

T ⁻



 



 = ^{p k}

1

k p

T

T ⁻









Product of n G.M.'s between a and b is equal to n^th power of single geometric mean between a and b i.e. G1G2G3 .... Gn = ( ab)ⁿ.

The product of n geometric means between a and 1/a is 1.

If n G.M.'s inserted between a and b then r = ^{n 1}

1

a b +



 





Quadratic Equations and Inequations :

An equation of degree n has n roots, real or imaginary.

If f(α) = 0 and f´(α) = 0, then α is a repeated root of the quadratic equation f(x) = 0 and f(x) = a(x – α)². In fact α = –b/2a.

If α is repeated common root of two quadratic equations f(x) = 0 and φ(x) = 0, then α is also a common root of the equations f´(x) = 0 and φ´(x) = 0.

In the equation ax² + bx + = 0 [a, b, c ∈R], if a + b + c = 0 then the roots are 1, c/a and if a – b + c

= 0, then the roots are –1 and – c/a.

If one root of the quadratic equation ax² + bx + c = 0 is equal to the n^th power of the other, then

( )

^{acn n1+1 +}

( )

^{ancn1+1 + b = 0.}

If one root is k times the other root of the quadratic equation ax² + bx + c = 0, then

k ) 1 k ( + ²

= ac b²

. If an equation has only one change of sign, it has one +ve root and no more.

Permutations and Combinations :

nC0 = ⁿCn = 1, ⁿC1 = n

nCr + ⁿCr–1 = ⁿ⁺¹Cr

nCx = ⁿCy ⇔ x = y or x + y = n n. ^n-1Cr–1 = (n – r + 1)ⁿCr–1

If n is even then the greatest value of ⁿCr is ⁿCn/2. If n is odd then the greatest value of ⁿCr is

2 C_{n 1}

n + or

2 C_{n 1}

n − .

nCr = r

n.^n–1Cr–1.

Number of selection of zero or more things out of n different things is, ⁿC0 + ⁿC1 + ^bC2 + ... + ⁿCn = 2ⁿ.

nC0 + ⁿC2 + ⁿC4 + .... = ⁿC1 + ⁿC3 + ⁿC5 + .... = 2^n–1. Gap method : Suppose 5 moles A, B, C, D, E are arranged in a row as × A × B × C × D × E ×. There will be six gaps between these five. Four in between and two at either end. Now if three females P, Q, R are to be arranged so that no two are together we shall use gap method i.e., arrange them in between these 6 gaps. Hence the answer will be ⁶P3.

Together : Suppose we have to arrange 5 persons in a row which can be done in 5! = 120 ways. But if two particular persons are to be together always, then we tie these two particular persons with a string. Thus we have 5 – 2 + 1 (1 corresponding to these two together) = 3 + 1 = 4 units, which can be arranged in 4! ways. Now we loosen the string and these two particular can be arranged in 2! ways. Thus total arrangements = 24 × 2 = 48.

If we are given n different digits (a, a2, a3 ... an) then sum of the digits in the unit place of all numbers formed without repetition is (n – 1)!(a1 + a2 + a3 + ....

+ an). Sum of the total numbers in this case can be obtatined by applying the formula (n – 1)!(a1 + a2 + a3 + ... + an). (1111 ... n times).

Binomial Theorem & Mathematical Induction :

The number of terms in the expansion of (x + y)ⁿ are (n + 1).

If the coefficients of p^th, q^th terms in the expansion of (1 – x)ⁿ areequal, then p + q = n + 2.

For finding the greatest term in the expansion of (x + y)ⁿ. we rewrite the expansion in this form (x + y)ⁿ = xⁿ

n

x 1 y

 + . Greatest term in (x + y)ⁿ = xⁿ. Greatest term in

n

x 1 y



 

 + .

There are infinite number of terms in the expansion of (1 +x)ⁿ, when n is a negative integer or a fraction.

The number of term in the expansion of (x1 + x2 + .... + x2)ⁿ = ^n+r-1Cr–1.

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