IIT JEE , AIEEE Forum - Mathematics , Algebra

jw

1) the no of ways of distributing 8 identical balls in 3 distinct boxes so that none is empty

2) sum of series 1/2! + 1/4! + 1/6!.....is

if lim x - infinity (1+ a/x + b/x^2 ) ^2x = e^2 values of a and b

3) let a , b, c be 3 non collinear vectors such that no 2 of these are collinear . if vector a + 2b is collinear with c and b+2c is collinear with a

then a + 2b + 6c = ????

ans: 0

akhil_o

1) solutions of
a+b+c=8
where a,b,c>0
=7C2
=21

sboosy

ananth_patri

limit prob..... is a=1 and b=R???

sboosy

I think ananth is right

lim (1+a/x+b/x^2)^2x
= (1+(ax+b)/(x^2))^2x
= (1+(ax+b)/(x^2))^{(x^2*(ax+b)*2x)/((ax+b)*x^2)

The previous step is obtained by multiplying and dividing power by
x^2/(ax+b)

It is nothing but e^2(a+b/x)
now this is equal to e^2

Now independent of value of b the b/x term vanishes as xtends to infinity

Thus e^2a = e^2
which gives a = 1 ...and b belongs to R

ananth_patri

nice work dude u r sol is perfect..

sboosy

I think third part ..what is asked to prove is wrong ...
I think it should have been a+2b+4c = 0

Now since a+2b is collinear with c we have
a+2b = kc ...where k is some constant

Since b+2c is collinear with a we have
b+2c = k'a
=> 2b+4c = 2k'(a)

Subtracting the 2 eqns..we get

a-4c = kc - 2k'a
=> (1+2k')a = (k+4)c .....(1)

Now since a,b,c are such that no 2 are collinear ....
a cannot b expressed solely in terms of c
that is a not equal to lambda c

Thus (1) can have solution only if 1+2k'=0 ..that is k' = -1/2
and k+4 = 0 ..that is k = -4

Substituting this value of k in the very first equation we get

a+2b+4c = 0

sweet08

actually ques should be b+3c collinear with a

its a prev. aieee ques

sol:

a+2b=tc -------(i)

b+3c=na

=> b=na-3c

put in (i)

we get

a-6c=tc-2na

comparing,

t=-6 => a+2b=-6c------from(i)

so a+2b+6c=0

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