Sunday, 16 September 2012

Prerequisites to crack IIT-JEE Math section

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: kx, 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 ex, log(1+x), log(1-x) etc.
General equation in ‘x’- relationship between the coefficients and roots:
xn+p1xn-1+ p2xn-2+ p3xn-3+..... + pn = 0
If α1, α2,.... αn are the roots of this equation, then the following relations hold good:
S1 = sum of roots taken one at a time = α1+ α2+....n = -p1
S2 = sum of roots taken two at a time = α1α2+ α1α3+..... = p2
S3 = sum of roots taken three at a time = α1α2α3+ α1α2α4+..... = -p3
This logic extends up to n iterations.

Important Algebra concept frequently required:
a3+b3+c3-3abc = 0
=> (a+b+c)(a2+b2+c2-ab-bc-ca) = 0
=> a+b+c = 0 (or) a2+b2+c2-ab-bc-ca = 0
a2+b2+c2-ab-bc-ca = 0 => 2(a2+b2+c2-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 a3+b3+c3-3abc = 0, then the possibilities are
(1)    a+b+c = 0 (2) a2+b2+c2= ab+bc+ca (3) a=b=c

Important Complex numbers concept frequently required:
eix+eiy+eiz = 0
a = eix ;b= eiy;c=eiz
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
=> a2+b2+c2+2(ab+bc+ca)=0
=> a2+b2+c2=0 => cos2x+cos2y+cos2z = sin2x+sin2y+sin2z = 0
a+b+c = 0 => a3+b3+c3 = 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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