How to approach this one

IMO C

ST 1) two ans for x>1 the q is false and for 0<x<1 true
not suff

St 2) diff value for 0<x<1;-1<x<0;x<-1
not suff

together 1 and 2 suff as 0<x<1 any RHS will always be grtr than LHS

OA pls

IMO : A

The given sequence is a GP.

1+X+X^2+X^3+X^4+X^5
Sum to n terms : 1st term * (1-r^n)/(1-r)

1 * (1-x^5)/1-x

Now the question asks

Is (1-x^5)/1-x < 1/1-x ?
or, is 1-x^5 <1
or is x^5 > 0

Statement 1:
------------

If x>0 then x^5 will always be greater than 0

Sufficient.

Statement 2:
------------

x<1
If x=0,then
x^5 = 0
If x is -ve then x^5 < 0

Hence insufficient.

Hence, A

OA please.

IMO A

Ok 1+x+x2+x3+x4 = 1.(x4-1)/(x-1) = (1-x4)/(1-x)

so 1/1-x is common in both so we need to decide the sign and value of 1-x4

for x<1 for x=1/2 it will be positive and less than 1 and LHS less
x=0 it will be positive and equal to 1 so both eul
x=-2 it will be negative and LHS less

for x>0 for x=1/2 it will be positive and less than 1
x=2 it will be negative and so LHS will be less

so as we see stmt x>0 is sufficient to say that LHS will be less and so A

Should be C ..
In reducing the expression (1-x^5)/1-x < 1/1-x to 1-x^5 <1 , you are assuming 1-x > 0 ..else if 1 - x < 0 then the inequality changes to (1-x^5) > 1 ...

So stmnt 1 alone cant suffice , but in conjunction with stmnt 2 it works

sanjana wrote:
Is (1-x^5)/1-x < 1/1-x ? ---- 1
or, is 1-x^5 <1 ---- 2

Is the above correct? I guess if (1-x) is negative, by dividing the inequality with (1-x), the inequality will be reversed. So without knowing whether (1-x) will be positive, we cannot solve.

IMO C.

The answer seems to be A,

From A: X>0

First ex: X =2,

1+2+8 ... < -1

Not true ...

Second ex: X = 1/2

1+1/2+1/8+ .... < 1/1-1/2
=> 1+1/2+1/8+ .... < 1/-1/2
=> 1+1/2+1/8+ .... < - 2

Not true..

A answers the question

From B: x<1

If X = 1/2 - It doesn't answer

If X = -1 - It answers.. The answers are contradicting from B.

viju9162 wrote:The answer seems to be A,


=> 1+1/2+1/8+ .... < 1/-1/2
=> 1+1/2+1/8+ .... < - 2

1-1/2 = 1/2 in which case RHS >LHS

IMO C RHS will always be > LHS providing x>0 and <1

Hello friends, sorry for delayed response OA:C..

I was expecting more of a mathematically inclined solution.Anyways, thanks for your responses.