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: 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:

xⁿ+p₁x^n-1+ p₂x^n-2+ p₃x^n-3+..... + p_n = 0

If α₁, α₂,.... α_n are the roots of this equation, then the following relations hold good:

S₁ = sum of roots taken one at a time = α₁+ α₂+.... +α_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:

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

=> 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....