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# For which of the following functions is f(−1/2) > f(2)?

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Math Expert
Joined: 02 Sep 2009
Posts: 51230
For which of the following functions is f(−1/2) > f(2)?  [#permalink]

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22 Aug 2018, 01:04
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Difficulty:

5% (low)

Question Stats:

89% (00:59) correct 11% (01:41) wrong based on 37 sessions

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For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2

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Re: For which of the following functions is f(−1/2) > f(2)?  [#permalink]

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22 Aug 2018, 01:12
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2

Since x^2 is in the denominator in option E, it may yield a greater value of f(-1/2).

f(-1/2)=$$\frac{3}{(-1/2)^2}$$=$$3*2^2=12$$
f(2)=$$\frac{3}{2^2}$$=3/4

Ans. (E)
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For which of the following functions is f(−1/2) > f(2)?  [#permalink]

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22 Aug 2018, 01:13
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2

Move through the options -

A. $$f(x) = 3x^2$$ -------------this will lead to positive value but here $$2^2>(-\frac{1}{2})^2$$. Reject

B. $$f(x)= 3x$$ ------------ this will make $$f(-\frac{1}{2})$$ negative and $$f(2)$$ positive. Reject

C. $$f(x)= 3 + x^2$$ ----------- Same reasoning as Option A. Reject

D. $$f(x)= 3 + \frac{1}{x}$$ ------------ Same reasoning as Option B. $$\frac{1}{x}<x$$, hence the value for $$f(-\frac{1}{2})<f(2)$$. Reject

E. $$f(x)= \frac{3}{x^2}$$ ----------- This should be our Answer, we can check it by solving

$$f(x)= \frac{3}{x^2} => f(-\frac{1}{2})=\frac{3}{(-1/2)^2}=>3*4=12$$

and $$f(2)=\frac{3}{2^2}=\frac{3}{4}=0.75$$. Clearly $$f(-\frac{1}{2})>f(2)$$

Option E
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Joined: 22 Feb 2018
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For which of the following functions is f(−1/2) > f(2)?  [#permalink]

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Updated on: 22 Aug 2018, 01:30
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2

OA:E
$$\begin{matrix} & f(-\frac{{1}}{2}) & & f(2) \\ f(x)= 3x^2 & \frac{3}{4} & {<} & 12 \\ f(x)= 3x & -\frac{3}{2} & {<} & 6 \\ f(x)= 3+x^2 & \frac{13}{4} & {<} & 7 \\ f(x)= 3 + 1/x & 1 & {<} & \frac{7}{2} \\ f(x)= 3/x^2 & 12 & {>} & \frac{3}{4} \end{matrix}$$
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Originally posted by Princ on 22 Aug 2018, 01:25.
Last edited by Princ on 22 Aug 2018, 01:30, edited 1 time in total.
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3327
Location: India
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For which of the following functions is f(−1/2) > f(2)?  [#permalink]

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22 Aug 2018, 01:29
1
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2

Of the 5 answer options, you can straight away eliminate Options A, B, and C because all of these
options have an x or x^2 in the numerator. Substituting -$$\frac{1}{2}$$ can never be greater than substituting 2.

Now that we are down to two answer options,
we can substitute the values to check in which case, value of f(-$$\frac{1}{2}$$) > value of f($$2$$)

In Option D, f(-$$\frac{1}{2}$$) = 3 + (-2) = 1 (because $$\frac{1}{\frac{-1}{2}}$$ = -2) but f($$2$$) = $$3 + \frac{1}{2} = \frac{7}{2}$$. Here, f(-$$\frac{1}{2}$$) < f($$2$$)

Therefore, the only answer option we are left with is f(x) = $$\frac{3}{x^2}$$ (Option E) which is our answer!
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For which of the following functions is f(−1/2) > f(2)? &nbs [#permalink] 22 Aug 2018, 01:29
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