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Functions

by vishwas.arora » Thu Aug 11, 2011 8:41 am
The function f(x) is defined as,

f(x) = x2 {read it as x squared} for |x|<=1
f(x) = 1/x2 {read it as 1 upon x squared} for |x|>1

Is -0.9<a<0.9?

(1) f(-a)=1/f(b)
(2) a=1/b

OA after discussion.

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by Frankenstein » Thu Aug 11, 2011 8:58 am
Hi,
f(x) = x^2 for |x|<=1. So, 0<=f(x)<=1, for |x|<=1 and 1/f(x) >=1 for |x|<=1
f(x) = 1/x^2 for |x|>1. So, 0<f(x)<1, for |x|>1 and 1/f(x) > 1 for |x|>1
From(1):
f(-a) = 1/f(b)
LHS is at most 1 and RHS is at least 1.
Equality holds only when f(-a) = 1/f(b) = 1
So, a= -1 or +1 and b= -1 or+1
So, a cannot be in the range (-0.9,0.9)
Sufficient

From(2):
a=1/b
We know nothing from this.
Not sufficient

Hence, A
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by HarryPotter » Thu Aug 11, 2011 4:09 pm
This will sound crazy...

consider 2 scenarios:
i. a = 1/2
ii. a = 2

From the given statements :
When a=1/2, f(a) = 1/4
and when a=2 , f(a) = 1/4
Similarly, f(-a) = 1/4


(1) alone doesn't tell anything
(2) alone doesnt tell anything

Let us try to use both (1) and (2)

From (2), we know b= 1/a

Substitute for b in (1)

f(-a) = 1/f(1/a)


When a = 1/2, 1/f(1/a) = 1/f(2) = 4 which is not equal to f(-a) : 1/4
when a = 2, 1/f(1/a) = 1/f(1/2) = 1/4 which is equal to f(-a) : 1/4


Therefore 'a' should be >1

We need [spoiler]both (1) and (2)[/spoiler] to answer the question.
Ans : C
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by Frankenstein » Thu Aug 11, 2011 8:34 pm
when a = 2, 1/f(1/a) = 1/f(1/2) = 1/4 which is equal to f(-a) : 1/4
Hi,
Please check this..
when a = 2, 1/f(1/a) = 1/f(1/2) = 1/(1/4) = 4 but f(-a) = 1/4
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by vishwas.arora » Thu Aug 11, 2011 9:02 pm
Frankenstein wrote:Hi,
f(x) = x^2 for |x|<=1. So, 0<=f(x)<=1, for |x|<=1 and 1/f(x) >=1 for |x|<=1
f(x) = 1/x^2 for |x|>1. So, 0<f(x)<1, for |x|>1 and 1/f(x) > 1 for |x|>1
From(1):
f(-a) = 1/f(b)
LHS is at most 1 and RHS is at least 1.
Equality holds only when f(-a) = 1/f(b) = 1
So, a= -1 or +1 and b= -1 or+1
So, a cannot be in the range (-0.9,0.9)
Sufficient

From(2):
a=1/b
We know nothing from this.
Not sufficient

Hence, A
That's the perfect solution.
OA is A

Cheers Frankenstein !
Jai Hind !
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