28 Oct 2008 23:27:28 IST

Here are few mathematical shorcuts to solve probs asked in IIT-JEE....

1. (IIT 04)

INCLUDEPICTURE "http://www.goiit.com/equations/2008/10/28/daaebc8a-93c6-4eb1-9413-570416cdb867.png" \* MERGEFORMATINET INCLUDEPICTURE "http://www.goiit.com/equations/2008/10/28/b0eb6005-d9ab-42e4-a96c-c15fe293b599.png" \* MERGEFORMATINET

then the interval in which 'a' lies is(a)a (t+1)^2 - (a-3)t(t+1) + (a-4)t^2 = 0=> (a-5)t - 1 = 0

=> now if a = 6=> t-1 = 0=> x^2 + x = 0=> x(x+1) = 0 and the condition is satisfied ..

if we replace a with 7/8/9 the we'll get 3 other quadratic equations in which the discriminants are < 0 .. hence they dont satisfy the condition..

Complex No. - Assertion Reason

by nikhil_ Tue Mar 08, 2011 8:12 pm

Let z1 and z2 be two complex numbers such that |z1-z2|=|z1+z2|Statement I:If z1, z2 are end point of diameter of circle then it passing through the originStatement II:|arg(z1/z2)| = /2

(A)Both true. II is correct explanation of I.(B)Both true. II is not correct explanation of I.(C)I true, II false.(D)I false, II true.

Both true and II is the correct explanation.

note that the equation of a circle which has z1 and z2 as end points of one of its diameter is |z - (z1 + z2)/2| = |(z1-z2)/2|

z = 0 satisfies this and hence, the circle does pass through the origin.

also, |z|^2 = z * z(bar)put z = z1 - z2 in the above equation to get,z1 * z2(bar) = - z2 * z1(bar)take argument on both the sides

arg(zbar) = - arg(z) and arg(z1 * z2) = arg(z1) + arg(z2) this gives, |arg(z1/z2)| = pi/2