ISI ME I, 2011 doubts

[Anmol]

Hey can someone plz help me with the following questions from ISI ME I 2011?

Qs 1 (The ans given by Amit Sir is d but I am getting c.. Plz explain)
Qs 8, 10,11, 16, 22, 24, 28 and 29

For qs 28, I am getting the mean as 1, while the ans is different..

[Sinistral]

1) coming out to be sqrt(46). so option d only.

10) satisfies only  if f(x)=1 ∀ x or if f(x) =0 ∀ x .
     so (b)

11) keep n=3. none of the options satisfy . so option (d)

16) replace x by -x and solve the 2 simultaneous equations (in f(x) & f(-x))
    we get f(x) = -(x+2)/(x^2  +4)
    hence option (a)

22)
   f(0) = C ==> C must be an integer
   f(1)= A+B+C , we know C is an integer therefore A+B must also be an integer
   f(2)= 4A+2B+C = 2A + 2(A+B) +C, we already know that A+B & C is an integer therefore 2A has to be integer.
   therefore option c

28)
   integrate f(x) from x_0 to x to get F(x).
   then integrate xf(x) from x_0 to infinity to get E(X). option c

[a]

1. take the first part as a and te next one as b. and solve (a+b)^2
10. replace all x by  y and y by x and solve
11. rationalise

[a]

can u solve ques 6 and 10, me 2, 2013

[Sinistral]

for ques 28 u wont get mean as 1. integrate once again. ans is (c) only

29)
f(x) = k*  nCx,    x=0,1,2,3 ...n
  summing the above from 0 to n and equating it to 1 will give k= 1/2^n

E(X)= Σk*x*nCx
       = k(0+ nC1 + 2*(nC2) + 3(nC3) ... n(nCn) )    ---- (1)
please note (1+x)^n =  nC0 + nC1 x + nC2  x^2 ---  nCn x^n)
                 differentiating both sides:
                 n(1+x)^(n-1) = 0+ nC1  + nC2 (2x) + nC3  (3x^2) --- nCn  (nx^n-1)
                 puttong x=1
                 n2^n-1  =   0+ nC1 +      2*(nC2) +   3(nC3) ...         n(nCn)
                putting the above value in equation (1) gives k(n*2^(n-1))
                 k= 1/2^n, so mean = n/2

make similar adjustments to calculate E(X^2)
                 

[Sinistral]

10) replacing all x by y and all y by x will again give the same equation.

[Anmol]

Thanks for the answers..
Can you plz also solve qs 8, 20 and 24?
Can you explain qs 16? After solving for F(x), how do we determine the number of continuous equations.

[Sinistral]

20)
in the expansion of (1+x)^n, the coefficients of the kth term and the (n-k+2)th term are equal so.
so r-6 = 39-r+6 + 2 = 2r+5, gives r as 14

hence option b

16)
because after solving the simultaneous equation u get a unique value of f(x)
so only one such function

24) vectors padha nahi. (8) i am not confident with my solution.