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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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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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Bunuel
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)
Bunuel
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
Bunuel
OA:E
\(\begin{matrix}\\
& f(-\frac{{1}}{2}) & & f(2) \\\\
f(x)= 3x^2 & \frac{3}{4} & {<} & 12 \\\\
f(x)= 3x & -\frac{3}{2} & {<} & 6 \\\\
f(x)= 3+x^2 & \frac{13}{4} & {<} & 7 \\\\
f(x)= 3 + 1/x & 1 & {<} & \frac{7}{2} \\\\
f(x)= 3/x^2 & 12 & {>} & \frac{3}{4}\\
\end{matrix}\)
Bunuel
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!
E.
Just put the values of -1/2 and 2.
A faster way could be to see where x is negative or in denominator. That would increase the value of -1/2 and reduce the value of 2.
In E. 3/(-1/2)^2 = 12
And 3/(2^2) = 3/4 =0.75
12>0.75
So E
Posted from my mobile device
Just put the values of -1/2 and 2.
A faster way could be to see where x is negative or in denominator. That would increase the value of -1/2 and reduce the value of 2.
In E. 3/(-1/2)^2 = 12
And 3/(2^2) = 3/4 =0.75
12>0.75
So E
Posted from my mobile device
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Bunuel
only for f(x) = 3/x^2
we get f(-1/2) > f(2)
IMO E
Posts: 90
Students get confused on these type of questions sometimes because we aren't generally used to plugging into answer choices TWICE. But that's what you have to do for each answer choice. You have to plug in -1/2 and then also plug in 2, and see is the result when you plug in -1/2 is greater than the result when you plug in 2. This will only be true for ONE answer choice.
A. f(x) = 3*x^2
If x = -1/2, then f(x) = 3/4.
If x = 2, then f(x) = 12.
Is 3/4 > 12? Nope! Cross this one off.
B. f(x) = 3*x
If x = -1/2, then f(x) = -3/4.
If x = 2, then f(x) = 6.
Is -3/4 > 6? Nope!
C. f(x) = 3 + x^2
If x = -1/2, then f(x) = 3.25
If x = 2, then f(x) = 7
Is 3.25 > 7? No!
D. f(x) = 3 + 1/x
If x = -1/2, then f(x) = -2
If x = 2, then f(x) = 3.5
Is -1/2 > 3.5? Nope.
E. f(x) = 3/x^2
If x = -1/2, then f(x) = 12
If x = 2, then f(x) = 3/4
Is 12 > 3/4? Yes! We finally have an answer!
Even though it's a PS question, it kind of has that Data Sufficiency vibe in which you're testing cases. Here we had to "test out" each of the answer choices and see which one gave us the relationship we were looking for.
A. f(x) = 3*x^2
If x = -1/2, then f(x) = 3/4.
If x = 2, then f(x) = 12.
Is 3/4 > 12? Nope! Cross this one off.
B. f(x) = 3*x
If x = -1/2, then f(x) = -3/4.
If x = 2, then f(x) = 6.
Is -3/4 > 6? Nope!
C. f(x) = 3 + x^2
If x = -1/2, then f(x) = 3.25
If x = 2, then f(x) = 7
Is 3.25 > 7? No!
D. f(x) = 3 + 1/x
If x = -1/2, then f(x) = -2
If x = 2, then f(x) = 3.5
Is -1/2 > 3.5? Nope.
E. f(x) = 3/x^2
If x = -1/2, then f(x) = 12
If x = 2, then f(x) = 3/4
Is 12 > 3/4? Yes! We finally have an answer!
Even though it's a PS question, it kind of has that Data Sufficiency vibe in which you're testing cases. Here we had to "test out" each of the answer choices and see which one gave us the relationship we were looking for.
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Expert reply
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Bunuel
Starting with E, we have:
f(-1/2) = 3/(-1/2)^2 = 3/(1/4) = 12
f(2) = 3/(2^2) = 3/4
We see that f(-1/2) > f(2), so answer is E correct.
Answer: E
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We need to find For which of the following function is f(-1/2)>f(2)
Now f(-1/2) > f(2)
If we look at the numbers inside the bracket then we need to make an expression consisting of -1/2 which has to be greater than the expression consisting of 2
1. We have - sign on one side (for -1/2) and + on the other (+2) and the expression for - sign has to be greater
=> We need to look for even powers in the expression so that - gets converted to +
2. We have fraction on one side (for -1/2) and integer on the other (+2) and the expression for fraction has to be greater
=> We need to look for an expression where x is in the denominator
Only option choice which satisfies these two conditions is f(x)=3/x^2
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
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Now f(-1/2) > f(2)
If we look at the numbers inside the bracket then we need to make an expression consisting of -1/2 which has to be greater than the expression consisting of 2
1. We have - sign on one side (for -1/2) and + on the other (+2) and the expression for - sign has to be greater
=> We need to look for even powers in the expression so that - gets converted to +
2. We have fraction on one side (for -1/2) and integer on the other (+2) and the expression for fraction has to be greater
=> We need to look for an expression where x is in the denominator
Only option choice which satisfies these two conditions is f(x)=3/x^2
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
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