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Bunuel
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 12 Nov 2022, 10:26
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A
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45% (01:46) correct 55% (01:41) wrong based on 76 sessions

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If f(x) = 2^x, is f(k) > f(m) ?

(1) k > m
(2) k/m is an even integer
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A
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 12 Nov 2022, 10:43
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The answer seems to be D. Statement 1 states that K>m which means 2^k>2^m. Hence sufficient.

Statement 2: k/m is an even integer only if k is greater than m. Hence sufficient as shown in statement 1.

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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 12 Nov 2022, 12:50
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the question is asking if 2^k > 2^m
therefore, is k>m (using the exponents rule)

S1: k>m - Sufficient
S2: k/m is an even integer. In any case, the numerator has to be even, and the denominator has to be odd.
And for this statement to be true, k>m. Therefore, Sufficient
Ans: D
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 13 Nov 2022, 03:40
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vijayjd88
The answer seems to be D. Statement 1 states that K>m which means 2^k>2^m. Hence sufficient.

Statement 2: k/m is an even integer only if k is greater than m. Hence sufficient as shown in statement 1.

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vijayjd88 - Zero is an even integer ;) . The part highlighted is incorrect.

  • k can be 0
  • m can be 100

f(k) < f(m)

  • k can be 100
  • m can be 2

f(k) > f(m)

Answer is A
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 14 Nov 2022, 13:35
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@gmatiphobia
Thank you for correcting. Shall keep in mind to test for zero as well unless otherwise specified.

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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 16 Nov 2022, 21:39
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Bunuel
If f(x) = 2^x, is f(k) > f(m) ?

(1) k > m
(2) k/m is an even integer

We need to know whether \(2^k >2^m\), or whether \(k>m\)

(1) k > m

Exactly what we are looking for.

SUFF.

(2) k/m is an even integer

\(k=4,\) \(m =2 \) yes
\(k=-4, m=2\) no

INSUFF.

Note: any integer that is divisible by \(2\) is an even integer, ... \(-4, -2, 0, 2, 4 \)..etc are all even.

Ans A

Hope it's clear.
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 19 Nov 2022, 01:47
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I'm confused about statement A, if K & M are positive fractions such that K=1/2 and M=1/4, doesn't that make 2^M > 2^K?

I thought both statements were insufficient...
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 19 Nov 2022, 06:34
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rlawodnr1997
I'm confused about statement A, if K & M are positive fractions such that K=1/2 and M=1/4, doesn't that make 2^M > 2^K?

I thought both statements were insufficient...

rlawodnr1997 :

How is \(2^{(\frac{1}{4})} > 2^{(\frac{1}{2})} \)

actually \(2^{(\frac{1}{2}) }> 2^{(\frac{1}{4})}\)

Square root of \(2\) is greater than the fourth root of \(2\)

\(\sqrt2 \approx 1.41\)

\(\sqrt[4] {2} \approx 1.18\)

Hope it's clearer now.
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If f(x) = 2^x, is f(k) > f(m) ? (1) k > m (2) k/m is an even integer 19 Nov 2022, 08:29
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Bunuel
If f(x) = 2^x, is f(k) > f(m) ?

(1) k > m
(2) k/m is an even integer


f(k)>f(m) means \(2^k>2^m\)
\(2^{k-m}>1……..2^{k-m}>2^0\)
\(k-m>0\)

(1) k > m
\(k-m>0\)
What we were looking for.
Sufficient

(2) k/m is an even integer
k>m>0……yes as k-m>0
Say k is -4 and m=2, then m>0>k or k-m<0
Insufficient


A
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