I looked at the answers for strategy.

(a) obviously not. Take 2 and 1/2 for example. There sum isnt even an integer.

(b) obviously not. Take 1. n=1, n+1/n = 2.

(c) possibly. rexamine later.

(d) obviously not. Same as (a).

Examination of (c)
Let n=1 and a be the number in question. then a = 1 and c holds.

let n=2 and let a be a number greater than 1. this collapses into a completing the square problem.

a + 1/a = (a - 2 + 1/a) + 2
(root(a) - 1/root(a))^2 + 2

And since

(root(a) - 1/root(a))^2 > 0

a + 1/a > 2

and (c) holds

by induction you can see that any n numbers can be broken down into the n = 1 or n = 2 case. for example, n=3 is just n=1 + n=2, n=4 case is just n=2 + n=2, etc.

please don't post fake questions!!!!

harsh.champ wrote:If the product of n positive real numbers is unity, then their sum is necessarily

(A)a multiple of n
(B)equal to n + 1/n
(C)never less than n
(D)a positive integer
(E)None of the above

The OA is C.

A) n+ 1/n is not multiple of n. so it is not true
B) not always true.. since we have a no. called '1' and it would give sum as n+(1/n)+1.
C) Let there be nos. a/b and b/a so its sum = a/b+ b/a = (a^2+b^2)/ab = ((a-b)^2)/2 + 2 >2. if there are more nos. they can be reduced to the form a/b+b/a and hence >2

D) n+1/n is not an integer except for n=1.

Hence C

Gspc.gandhinagar wrote:please don't post fake questions!!!!

Do u mean to say something wrong with this Q ?

Gspc.gandhinagar wrote:please don't post fake questions!!!!

this question seems to be quite correct . what does your comment mean exactly.