cross-checking values,

sin^-11=pi/2

tan^-11=pi/4

so sin^-11=2tan^-11

sin^-1-1=-pi/2

tan^-1-1=-pi/4

so sin^-1-1=2tan^-1-1

sin^-10=0

tan^-10=0

hence all 3 are solutions

0.1      0.2       

moles initially present

hence 0.1 moles of each product are formed

so mass o product

162.5 g MW-0.1 moles=16.25 g

233.5 g--0.1 mol=23.35 g

so total=39.6 grams

f(x)=xⁿ/(x-1)=xⁿ-1/x-1+1/x-1

f(x)=1+x+x²+....x^n-1+1/(x-1)

differentiating once...

f'(x)=1+2x+3x²+...(n-1)x^n-2-1/(x-1)²

differentiating again...

f''(x)=2+6x+12x²+....+(n-1)(n-2)x^n-3+2/(x-1)³

aⁿ+bⁿ=(a+b)(a^n-1-a^n-2b+...b^n-1)

hence it is divisible by (a+b)

hence as celestine says divide into pairs of 2 such dat a+b=2008

each pair is divisible by 2008 hence the whole sum is divisble by 2008

Length of tangent from (p,q)==

Also radius=a

tan @=a/

angle =2@

so 2@=2tan^-1(a/)

I want to know about all the decent colleges to which separate applications are to be sent...and if possible the last dates of application!

please provide whatever information u can find!

_{This post should help everyone applying this year...i ran out of time for IICT and didnt even know!!}

a) v=dx/dt, r=acceleration

t=ax²+bx

differentiating wrt t

1=2ax*(dx/dt)+b(dx/dt)

2avx+bv=1

x=(1-bv)/2av

again differentiating wrt t

2a(v(dx/dt)+x(dv/dt))+b(dv/dt)=0

2a(v²+rx)+br=0

2arx+br=-2av²

(2ax+b)r=-2av²

if answer is independant of x

then from above

(2a((1-bv)/2av)+b)r=-2av²

simplify to get answer