For each value of y greater than 23, the function f(x) is

Question

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.

Sajjad1994 · Accepted Answer

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.

AnishPassi · Answer

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.

chetan2u · Answer

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})

KarishmaB · Answer

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

MANASH94 · Answer

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."

Bunuel · Answer

(b + 3)(b - 4) = 0 gives  b^2 - b - 12 = 0, not b^2 - b + 12 = 0.

MANASH94 · Answer

Oops, thanks for pointing out my mistake...  Bunuel

For each value of y greater than 23, the function f(x) is

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

\(a\)	\(b\)	Values
		1
		2
		4
		6
		8

GMAT Two-part Analysis (TPA) Questions

For each value of y greater than 23, the function f(x) is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards