The following topics mostly cover
the mathematics section of the IIT-JEE:

Complex Numbers, Matrices & Determinants,
, Inverse Trigonometric Functions, Trigonometric Equations, Quadratic
Equations, Differential Equations, Circles, Probability, Statistics, P&C,
Progressions, Functions.

In my opinion, one should master
the following concepts for cracking IIT-JEE mathematics paper. These funda are
important in a way that these are tested in different forms and are used in
different topics.

Inverse trigonometric functions:
Generally we neglect the ranges of sin

^{-1}, cos^{-1}etc.,
Range of sin

^{-1}=> [-π/2, π/2]
Range of cos

^{-1}=> [0, π]
Range of tan

^{-1}=> (-π/2, π/2)
Range of cot

^{-1}=>(0, π)
Formulae for expansions of
fractional powers. ex: (1+x)

^{1/2}
Equations of circle, ellipse,
parabola, hyperbola etc., in

**complex plane**
Mixing of concepts (for ex:
complex numbers with conic sections, trigonometry with conic sections)

How these graphs look like, just
an idea: k

^{x}, f(x)^{k}, |x-k| etc.,
Complex number theory approach is a
powerful tool. This can be used to resolve geometry problems on locus,
trigonometric problems etc.

Ex: Try out complex number
approach for the following one:

Q) Find the locus of centre of
circle which touches two circles externally?

Answer) Hyperbola

"AM ≥ GM" and "AM ≥ HM" conditions are
very much useful in solving Maxima-minima problems in 2 or 3 variables. Try out some problems in this angle.

Grip on special functions such as
[x], {x} and their properties.

Expansions of special functions
like e

^{x}, log(1+x), log(1-x) etc.
General equation in ‘x’- relationship
between the coefficients and roots:

x

^{n}+p_{1}x^{n-1}+ p_{2}x^{n-2}+ p_{3}x^{n-3}+..... + p_{n}= 0
If α

_{1}, α_{2},.... α_{n}are the roots of this equation, then the following relations hold good:
S

_{1}= sum of roots taken one at a time = α_{1}+ α_{2}+.... +α_{n}= -p_{1}
S

_{2}= sum of roots taken two at a time = α_{1}α_{2}+ α_{1}α_{3}+..... = p_{2}
S

_{3}= sum of roots taken three at a time = α_{1}α_{2}α_{3}+ α_{1}α_{2}α_{4}+..... = -p_{3}
This logic extends up to n
iterations.

**Important Algebra concept frequently required:**

a

^{3}+b^{3}+c^{3}-3abc = 0
=> (a+b+c)(a

^{2}+b^{2}+c^{2}-ab-bc-ca) = 0
=> a+b+c = 0 (or) a

^{2}+b^{2}+c^{2}-ab-bc-ca = 0
a

^{2}+b^{2}+c^{2}-ab-bc-ca = 0 => 2(a^{2}+b^{2}+c^{2}-ab-bc-ca) = 0
=> (a-b)

^{2}+(b-c)^{2}+(c-a)^{2}= 0 => a-b = b-c = c-a = 0 => a=b=c**If a**

^{3}+b^{3}+c^{3}-3abc = 0, then the possibilities are**(1)**

**a+b+c = 0 (2) a**

^{2}+b^{2}+c^{2}= ab+bc+ca (3) a=b=c**Important Complex numbers concept frequently required:**

e

^{ix}+e^{iy}+e^{iz}= 0
a = e

^{ix};b= e^{iy};c=e^{iz}
a+b+c = 0 =>
cosx+cosy+cosz+i(sinx+siny+sinz) = 0

=> cosx+cosy+cosz =
sinx+siny+sinz = 0

=>
cosx+cosy+cosz-i(sinx+siny+sinz) = 0

=> (cosx-isinx)+ (cosy-isiny)+ (cosz-isinz)
= 0

=>

**1/a + 1/b + 1/c = 0 => ab+bc+ca = 0****=> cos(x+y)+cos(y+z)+cos(x+z) = cos(x+y)+cos(y+z)+cos(x+z)**

a+b+c = 0 => (a+b+c)

^{2}= 0
=> a

^{2}+b^{2}+c^{2}+2(ab+bc+ca)=0
=>

**a**=>^{2}+b^{2}+c^{2}=0**cos2x+cos2y+cos2z = sin2x+sin2y+sin2z = 0**
a+b+c = 0 =>

**a**^{3}+b^{3}+c^{3}= 3abc
=>

**cos3x+cos3y+cos3z = 3cos(x+y+z)****sin3x+sin3y+sin3z = 3sin(x+y+z)**

A useful wikipedia link for inverse
trigonometry formulae, where the formulae are given in tables: Inversetrig formulae

The story does not end here and will be updated from time to time....

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