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The above equations define how the function f varies with x. Is –0.9 <

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The above equations define how the function f varies with x. Is –0.9 <  [#permalink]

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New post 04 Oct 2019, 07:08
9
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A
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E

Difficulty:

  95% (hard)

Question Stats:

9% (02:13) correct 91% (02:32) wrong based on 54 sessions

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\(f(x) = x^2\) 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) \(a = \frac{1}{b}\)

Source: Nova GMAT

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Re: The above equations define how the function f varies with x. Is –0.9 <  [#permalink]

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New post 02 Dec 2019, 08:28
1
-0.9<a<0.9 means that f(a)= X^2

1. f(-a)=1/f(b)
suppose b is postive that is Greater than 1
Therefore , f(b)= 1/x^2 and
1/f(b) = x^2.
But f(-a)= x^2 since a is negative .So it has to 1/x^2.
Hence f(-a)= 1/f(b) where b more than 1 and a is between -0.9<a<0.9. Only then statement 1 holds true.

Stat 2 - Insuff .

IMO - A

Had initially selected E.

Experts, Views please !!
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The above equations define how the function f varies with x. Is –0.9 <  [#permalink]

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New post 02 Dec 2019, 11:20
2
f(x)= x^2 for x ≤ 1
\(f(x) = \frac{1}{x^2}\) for \(x > 1\)

Range of f(x)- [0, 1]
Range of 1/f(x)- [1, infinity)

Statement 1-
\(f(–a) = \frac{1}{f(b)}\)

As value of LHS is between 0 and 1( both inclusive) and value of RHS is equal to or greater than 1, LHS is equal to RHS if and only both are equal to 1.

f(-a)=1, if a=1 or -1.

Hence, a can never lies in range (-0.9, 0.9)

Sufficient

Statement 2- a*b=1

Case 1- a=2 and b=1/2

Case 2- a=1/2 and b=2

Insufficient

SajjadAhmad wrote:
\(f(x) = x^2\) 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) \(a = \frac{1}{b}\)

Source: Nova GMAT
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The above equations define how the function f varies with x. Is –0.9 <  [#permalink]

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New post 05 Dec 2019, 07:24
2
SajjadAhmad wrote:
\(f(x) = x^2\) 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) \(a = \frac{1}{b}\)

Source: Nova GMAT


Analyzing the question:
One thing to note is that the highest value of the function is 1. For x between -1 and 1 the function value is at it's lowest for x = 0 and highest at |x| = 1. After we go into the x > 1 range the function value starts dropping as \(1/x^2\) is decreasing with x. Hence we are UNABLE to determine where x is according to the value of f(x). There are 3 such values for f(x) = 0.5 for example, two of them within -1 < x < 1 and another with x > 1.

Statement 1:
We noted earlier that the maximum value of this function is 1, so the left side f(-a) must be at most 1. The right side has 1 / f(b), which must exceed 1 unless f(b) = 1. So we must have f(b) = f(-a) = 1. Also note f(-a) = f(a) since the function always squares the input. So f(a) = f(b) = 1. This can only happen when the input is either -1 or 1. Since \(a\) cannot be in range of -0.9 to 0.9, this is sufficient.

Statment 2:
Insufficient.

Ans: A
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The above equations define how the function f varies with x. Is –0.9 <   [#permalink] 05 Dec 2019, 07:24
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