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24 Oct 2022, 01:30
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
15% (01:57) correct
85%
(02:22)
wrong
based on 53
sessions
History
Date
Time
Result
Not Attempted Yet
\(f(x) = x^2\) for \(|x| ≤ 1\)
\(f(x) = \frac{1}{x^2}\) for \(x > 1\)
The above equations define how the function f varies with \(x.\) Is \(–0.9 < a < 0.9\)?
(1) \(f(–a) = \frac{1}{f(b)}\)
(2) \(a = \frac{1}{b}\)
Are You Up For the Challenge: 700 Level Questions
\(f(x) = \frac{1}{x^2}\) for \(x > 1\)
The above equations define how the function f varies with \(x.\) Is \(–0.9 < a < 0.9\)?
(1) \(f(–a) = \frac{1}{f(b)}\)
(2) \(a = \frac{1}{b}\)
Are You Up For the Challenge: 700 Level Questions
27 Oct 2022, 22:24
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Bunuel
Two ranges given
A. \(f(x)=x^2\) for -1 ≤ x ≤ 1
B. \(f(x)=\frac{1}{x^2}\) for x>1
There is no function defined for x<-1.
Statement I says \(f(–a) = \frac{1}{f(b)}\)
a) That is, \(f(-a)*f(b)=1……..a^2*b^2=1\)
#Both a^2 and b^2 could be 1 => (a,b) = (1,1) or (-1,1). In both cases answer is NO as -1 or 1 does not lie in range -0.9<a<0.9
# If one of the two is less than 1, the second will have to be greater than 1.
So we come to case b
b) \(f(-a)*f(b)=1……..a^2*\frac{1}{b^2}=1\)
Now, b>1, so \(b^2>1\) and \(\frac{1}{b^2}<1\)
In this scenario, \(a^2*\frac{1}{b^2}\) will always be less than 1. Similar case when -a>1 and b<1.
Not possible.
Only case a is possible where a is either -1 or 1, and in both cases does not fall in the range -0.9 to 0.9
Sufficient
Statement II says that \(a=\frac{1}{b}\)
Clearly insufficient as we cannot work on this further.
Insufficient
A
26 Dec 2025, 09:10
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