Messages posted by vatika

very nice....srujana.....

if a^2-b^2 is prime then prove that a^2-b^2=a+b if a and b are +ve integers.

Prove that there are infinitely many positive integers 'a' : z = n^4 + a is not a prime for any positive integer n

if a and b are positive integers,then prove that [(2a!)*(2b+1)!]/[(a!)*(b!)*(a+b+1)!] is an integer

aaahhhh what loosers on this site!!!

such pathetically low rankers!!! mera to hai AIR 21

fools pick up ur standards orelse ull all sink . my friends were right when they said this site is not fit for me..neway i am going. getting bored of listenning to such low ranks!!!

shudnt have come!!

excellent

very good job

very good and innovative.keep sending such articles

very good job.keep sending such articles

i feel we require a more of this to make studies systematic

venky honey , do not believe these loosers, who are spreading around wrong rumours....... IIT PATTERN IS NOT GOING TO CHANGE FROM THIS YEAR........

1)on subs y=0 in the equation
x^2+y^2+2gx+2fy+c=0
x^2-[sum of roots] +[product of roots]=0
c=4,g= -29/10
on subs x=0
we get similarly f= -2
centre of circle is given by ( -g,-f)
hence centre is (29/10,2) ans(b)

3)g=k/2,f= (1-k)/2
radius of circle is given by
(g^2+f^2-c)^1/2
on substitution we get
2k^2-2k-119<=0
hence there are no integral soln ans(d)

4)as seen by equation of first circle it passes through the origin hence for second circle to touch the first one it should be a point circle at (0,0) hence c=0 ans (d)

the question has no answer because we can't fing [] of infinity
and infinity+infinity is undefined

1)let (sec[x])^1/2=t
dt/dx= 1/[(sec[x])^(1/2)]sec[x]*tan[x]
on substituting it comes integration 2dt/(t^4 - 1)
usin partial fractions
At+B/(t^2+1) + Ct+D/(t^2-1)
on equating A=0,B=1/2,C=0,D= -1/2
INT dt/(t^2 +1) - INT dt/(t^2-1)

tan inv(sec[x])^1/2 - (1/2)*log [({sec[x]}^(1/2)-1/ {sec[x]}^(1/2) +1]

2) let (tan[x])^(1/2)=p
dp/dx= 1/[2(tan[x])^(1/2)]* (sec[x])^2
on solving it comes
INT 2 (p^2)dp/(1+p^4)
INT p^2 +1/(1+p^4) + INT p^2-1/(1+p^4)
divide by p^2
INT [1+ (1/p^2)]/[(1/p^2)+p^2] + [1 - (1/p^2)]/[(1/p^2)+p^2]
INT [1+ (1/p^2)]/[(1/p -p)^(2) +2]+INT [1-(1/p^2]/[{1/p+p}^(2) -2]
sub 1/p -p =t and 1/p+p=n
you end up getting one of the standard forms
ans=(1/[2]^(1/2)* tan inv( tan[x] -1/[2tan[x] ]^1/2)
+ 1/8^1/2 * log{tan[x] -(2tan[x]^1/2}/[tan[x] + (2tan[x])^1/2}
3)

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info