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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:02) correct
43%
(02:02)
wrong
based on 2441
sessions
History
Date
Time
Result
Not Attempted Yet
For each value of \(y\) greater than \(2 \sqrt{3}\), the function \(f(x)\) is such that the equation \(f(x) = y\) has a solution of the form \(x=\frac{y^2+12}{y}\)
Select one value for \(a\) and one value for \(b\) such that the given information implies \(f(a) = b\). Make only two selections, one in each column.
Select one value for \(a\) and one value for \(b\) such that the given information implies \(f(a) = b\). Make only two selections, one in each column.
| \(a\) | \(b\) | Values |
| 1 | ||
| 2 | ||
| 4 | ||
| 6 | ||
| 8 |
Posts: 16,403
Official Explanation
The formula may only be applied for y greater than \(2\sqrt{3}\), therefore, the only choices available for b are 4, 6, and 8.
If f(a) = b and b = 4, then, according to the formula,
\(a=\frac{4^2+12}{4}=\frac{28}{4}=7\)
Since 7 is not an option, b = 4 must not be part of the correct response.
If f(a) = b and b = 6, then, according to the formula,
\(a=\frac{6^2+12}{6}=\frac{48}{6}=8\)
Since 8 is an option, a = 8 and b = 6 must be the correct response.
The correct answer is 8.
The correct answer is 6.
The formula may only be applied for y greater than \(2\sqrt{3}\), therefore, the only choices available for b are 4, 6, and 8.
If f(a) = b and b = 4, then, according to the formula,
\(a=\frac{4^2+12}{4}=\frac{28}{4}=7\)
Since 7 is not an option, b = 4 must not be part of the correct response.
If f(a) = b and b = 6, then, according to the formula,
\(a=\frac{6^2+12}{6}=\frac{48}{6}=8\)
Since 8 is an option, a = 8 and b = 6 must be the correct response.
The correct answer is 8.
The correct answer is 6.
General Discussion
A few things I noted while reading the question:
1. 2√3 is slightly less than 4.
2√4 = 4. So, 2√3 would be slightly less.
2. x will be greater than y.
The given equation can be simplified to x = y + y/12
Since y is positive, x will be greater than y.
3.
The function is written in the form of 'y = <some function of x>', but the equation is then in the form of 'x = 'some expression in terms of y'
I needed a moment to ensure I processed it correctly.
The given equation in terms of x and y is essentially an elaboration of the relation between x and y. f(x) = y tells us that there is a relation between x and y. What that relation is we get from the equation.
4. 'a' maps to 'x' and 'b' maps to 'y'
I realized that I could mix up the two constants and end up getting the answer wrong. So, I spent extra time to ensure I understood what a and b represent.
f(x) = y
maps with
f(a) = b
So, 'a' maps to 'x' and 'b' maps to 'y'.
Since I had already understood that x > y,
I realize a > b.
When I look at the table
1. I can reject 1 and 2 in b's column. Since y > 2√3
2. I can reject 8 for b. Since a > b, so b can't have the highest value.
3. I have 2 options left for b. 4 and 6. I can plug them in.
i. b = 4
x = y + 12/y
if b = 4,
a = 4 + 12/4
a = 4 + 3 = 7.
7 isn't an option for a.
Reject 4 for b.
---
If I were running short on time at this point, I could directly mark b = 6 and a = 8 at this point.
Why a = 8?
Because a > b. So, if b = 6 (which is the only feasible value left for b), a has to be greater than 6. That leaves only one choice for a.
---
ii. b = 6
x = y + 12/y
if b = 6,
a = 6 + 12/6
a = 6 + 2
a = 8.
a = 8, b = 6.
1. 2√3 is slightly less than 4.
2√4 = 4. So, 2√3 would be slightly less.
2. x will be greater than y.
The given equation can be simplified to x = y + y/12
Since y is positive, x will be greater than y.
3.
The function is written in the form of 'y = <some function of x>', but the equation is then in the form of 'x = 'some expression in terms of y'
I needed a moment to ensure I processed it correctly.
The given equation in terms of x and y is essentially an elaboration of the relation between x and y. f(x) = y tells us that there is a relation between x and y. What that relation is we get from the equation.
4. 'a' maps to 'x' and 'b' maps to 'y'
I realized that I could mix up the two constants and end up getting the answer wrong. So, I spent extra time to ensure I understood what a and b represent.
f(x) = y
maps with
f(a) = b
So, 'a' maps to 'x' and 'b' maps to 'y'.
Since I had already understood that x > y,
I realize a > b.
When I look at the table
1. I can reject 1 and 2 in b's column. Since y > 2√3
2. I can reject 8 for b. Since a > b, so b can't have the highest value.
3. I have 2 options left for b. 4 and 6. I can plug them in.
i. b = 4
x = y + 12/y
if b = 4,
a = 4 + 12/4
a = 4 + 3 = 7.
7 isn't an option for a.
Reject 4 for b.
---
If I were running short on time at this point, I could directly mark b = 6 and a = 8 at this point.
Why a = 8?
Because a > b. So, if b = 6 (which is the only feasible value left for b), a has to be greater than 6. That leaves only one choice for a.
---
ii. b = 6
x = y + 12/y
if b = 6,
a = 6 + 12/6
a = 6 + 2
a = 8.
a = 8, b = 6.
scc404
\(x=\frac{y^2+12}{y}\) or \(a=\frac{b^2+12}{b}=b+\frac{12}{b}\)
Surely a>b and \(b>2\sqrt{3}>2\).
Also, both a and b are even.
\(a=b+\frac{12}{b}\) means 12/b is even as even = even + 12/b
12/4 =3...Discard
12/6 = 2..Yes
so \(8=6+(\frac{12}{6})\)
_________________
scc404
Some questions of DI have odd wording but you can figure out what to do by reading what is given and what is asked.
I struggled to understand what this means: the function \(f(x)\) is such that the equation \(f(x) = y\) has a solution of the form \(x=\frac{y^2+12}{y}\)
but I ignored it. I could see that there is a relation between x and y if f(x) = y and that's all I cared about because I was given f(a) = b. This meant that the relation will exist for a and b
\(a=\frac{b^2+12}{b} = b + \frac{12}{b}\)
y needs to be greater than 2*sqrt3 which means that b needs to be greater than 2*1.7 = 3.4 approx.
If b = 4, a = 4 + 3 = 7 (Not available in options)
If b = 6, a = 6 + 2 = 8 - Available
ANSWER: 8, 6
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Earlier I tried plugging in a=1 and rearranged incorrectly.
The correct equation is:
a=b^2+12b
⇒ab=b2+12
⇒b2−ab+12=0.
So if a=1a = 1a=1, the quadratic becomes
b2−b+12=0,
b^2 - b + 12 = 0,
b2−b+12=0,
which has no real solutions because its discriminant is negative:
(−1)2−4(1)(12)=1−48<0.(-1)^2 - 4(1)(12) = 1 - 48 < 0.(−1)2−4(1)(12)=1−48<0.
So b=4b = 4b=4 is not a valid solution — that came from a factorization mistake.
Thanks to Bunuel for pointing this out."
The correct equation is:
a=b^2+12b
⇒ab=b2+12
⇒b2−ab+12=0.
So if a=1a = 1a=1, the quadratic becomes
b2−b+12=0,
b^2 - b + 12 = 0,
b2−b+12=0,
which has no real solutions because its discriminant is negative:
(−1)2−4(1)(12)=1−48<0.(-1)^2 - 4(1)(12) = 1 - 48 < 0.(−1)2−4(1)(12)=1−48<0.
So b=4b = 4b=4 is not a valid solution — that came from a factorization mistake.
Thanks to Bunuel for pointing this out."
MANASH94
(b + 3)(b - 4) = 0 gives b^2 - b - 12 = 0, not b^2 - b + 12 = 0.
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