GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2019, 21:51 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Retired Moderator D
Joined: 25 Feb 2013
Posts: 1179
Location: India
GPA: 3.82
If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

3
14 00:00

Difficulty:   85% (hard)

Question Stats: 53% (02:14) correct 47% (02:26) wrong based on 226 sessions

HideShow timer Statistics

If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

Originally posted by niks18 on 17 Sep 2017, 08:04.
Last edited by niks18 on 18 Sep 2017, 09:25, edited 1 time in total.
Current Student S
Joined: 22 Apr 2017
Posts: 106
Location: India
GMAT 1: 620 Q46 V30 GMAT 2: 620 Q47 V29 GMAT 3: 630 Q49 V26 GMAT 4: 690 Q48 V35 GPA: 3.7
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

1
1
niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

g(x)= x^2-4bx+9b^2--> (x-2b)^2+5b^2.
Since b is a positive integer, value of g(x) will always be greater than 5b^2 as (x-2b)^2 will alwz be positive.
Hence for min value of b i.e 1, g(x)=5b^2.
Hence E. g(x)>=5b^2
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1179
Location: India
GPA: 3.82
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

1
2
niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

OE

$$g(x) = x^2 – 4bx + 4b^2+5b^2$$

or, $$g(x) = (x – 2b)^2+5b^2$$,

Now, $$(x-2b)^2≥0$$, for all values of $$x$$. Adding $$5b^2$$ to both sides of the inequality we get

$$(x-2b)^2+5b^2≥5b^2$$

Or, $$g(x)≥5b^2$$

Option E
Manager  B
Joined: 06 Aug 2017
Posts: 79
GMAT 1: 570 Q50 V18 GMAT 2: 610 Q49 V24 GMAT 3: 640 Q48 V29 Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

Answer is E as follows.

$$g(x) = x^2 – 4bx + 9b^2$$

Since the above is a quadratic equation, its maxima/minima will be at (-)-4d/2 = 2b. Also since b is a +ve integer 2b will be +ve. Hence as per the rule the quadratic equation will have minimum value at x=2b.
Putting the value of 2b back into the equation will yield g(2b) = $$(2b)^2$$ - 4b*(2b) + 9$$b^2$$ = 5$$b^2$$
_________________
-------------------------------------------------------------------------------
Kudos are the only way to tell whether my post is useful.

GMAT PREP 1: Q50 V34 - 700

Veritas Test 1: Q43 V34 - 630
Veritas Test 2: Q46 V30 - 620
Veritas Test 3: Q45 V29 - 610
Veritas Test 4: Q49 V30 - 650

GMAT PREP 2: Q50 V34 - 700

Veritas Test 5: Q47 V33 - 650
Veritas Test 5: Q46 V33 - 650
Senior Manager  S
Joined: 08 Jun 2015
Posts: 420
Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

+1 for option E. The given expression can be expressed as (x-2b)^2+5b^2. The first part is always positive. The expression then is always greater than 5(b^2). Hence option E it is.
_________________
" The few , the fearless "
Intern  B
Joined: 15 Jan 2016
Posts: 36
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

niks18 wrote:
niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

OE

$$g(x) = x^2 – 4bx + 4b^2+5b^2$$

or, $$g(x) = (x – 2b)^2+5b^2$$,

Now, $$(x-2b)^2≥0$$, for all values of $$x$$. Adding $$5b^2$$ to both sides of the inequality we get

$$(x-2b)^2+5b^2≥5b^2$$

Or, $$g(x)≥5b^2$$

Option E

Can you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated.

Thanks

Posted from my mobile device
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1179
Location: India
GPA: 3.82
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

syedazeem3 wrote:
niks18 wrote:
niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

OE

$$g(x) = x^2 – 4bx + 4b^2+5b^2$$

or, $$g(x) = (x – 2b)^2+5b^2$$,

Now, $$(x-2b)^2≥0$$, for all values of $$x$$. Adding $$5b^2$$ to both sides of the inequality we get

$$(x-2b)^2+5b^2≥5b^2$$

Or, $$g(x)≥5b^2$$

Option E

Can you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated.

Thanks

Posted from my mobile device

hi syedazeem3

will be glad to clarify your doubts. I would like to understand how you deduced option E simply by looking at $$9b^2$$ in the original equation. If instead of $$5b^2$$, option E had mentioned $$4b^2$$, then would you still have chosen option E as correct?
Intern  B
Joined: 15 Jan 2016
Posts: 36
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

niks18 wrote:
syedazeem3 wrote:
niks18 wrote:
[quote="niks18"]If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

OE

$$g(x) = x^2 – 4bx + 4b^2+5b^2$$

or, $$g(x) = (x – 2b)^2+5b^2$$,

Now, $$(x-2b)^2≥0$$, for all values of $$x$$. Adding $$5b^2$$ to both sides of the inequality we get

$$(x-2b)^2+5b^2≥5b^2$$

Or, $$g(x)≥5b^2$$

Option E

Can you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated.

Thanks

Posted from my mobile device

hi syedazeem3

will be glad to clarify your doubts. I would like to understand how you deduced option E simply by looking at $$9b^2$$ in the original equation. If instead of $$5b^2$$, option E had mentioned $$4b^2$$, then would you still have chosen option E as correct?[/quote]

My logic was, regardless of what is before 9b^2, 5b^2 would be smaller - the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2.

Sent from my iPhone using GMAT Club Forum mobile app
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1179
Location: India
GPA: 3.82
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

1
niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

OE

$$g(x) = x^2 – 4bx + 4b^2+5b^2$$

or, $$g(x) = (x – 2b)^2+5b^2$$,

Now, $$(x-2b)^2≥0$$, for all values of $$x$$. Adding $$5b^2$$ to both sides of the inequality we get

$$(x-2b)^2+5b^2≥5b^2$$

Or, $$g(x)≥5b^2$$

Option E

syedazeem3 wrote:
My logic was, regardless of what is before 9b^2, 5b^2 would be smaller - the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2.

Hi syedazeem3

Note this is an algebraic equation where $$x$$ is a variable and $$b$$ is any constant and the question asks which relation for the equation ALWAYS HOLDS TRUE.

By your logic if Option E had been $$8b^2$$, then you would have selected it which is NOT ALWAYS TRUE

Suppose $$x=2$$ & $$b=1$$ then

$$8b^2=8*1^2=8$$, but

$$g(x) = x^2 – 4bx + 9b^2=2^2-4*2*1+9*1^2=5$$

clearly here $$g(x)<8b^2$$ in this case; but $$g(x)=5b^2$$

So until you factories this quadratic equation you will not come to know that the function $$g(x)$$ is always greater than or equal to $$5b^2$$ for any value of $$x$$ & $$b$$
Intern  B
Joined: 15 Jan 2016
Posts: 36
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

Show Tags

niks18 wrote:
niks18 wrote:
If $$g(x) = x^2 – 4bx + 9b^2$$, where $$b$$ is any positive integer, then which of the following must be true

A.) $$g(x) < -b$$

B.) $$g(x) < 0$$

C.) $$0 < g(x) < 4b$$

D.) $$g(x) ≥ 4b$$

E.) $$g(x) ≥ 5b^2$$

OE

$$g(x) = x^2 – 4bx + 4b^2+5b^2$$

or, $$g(x) = (x – 2b)^2+5b^2$$,

Now, $$(x-2b)^2≥0$$, for all values of $$x$$. Adding $$5b^2$$ to both sides of the inequality we get

$$(x-2b)^2+5b^2≥5b^2$$

Or, $$g(x)≥5b^2$$

Option E

syedazeem3 wrote:
My logic was, regardless of what is before 9b^2, 5b^2 would be smaller - the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2.

Hi syedazeem3

Note this is an algebraic equation where $$x$$ is a variable and $$b$$ is any constant and the question asks which relation for the equation ALWAYS HOLDS TRUE.

By your logic if Option E had been $$8b^2$$, then you would have selected it which is NOT ALWAYS TRUE

Suppose $$x=2$$ & $$b=1$$ then

$$8b^2=8*1^2=8$$, but

$$g(x) = x^2 – 4bx + 9b^2=2^2-4*2*1+9*1^2=5$$

clearly here $$g(x)<8b^2$$ in this case; but $$g(x)=5b^2$$

So until you factories this quadratic equation you will not come to know that the function $$g(x)$$ is always greater than or equal to $$5b^2$$ for any value of $$x$$ & $$b$$

Thanks for the help here niks18. Appreciate the insight - finally understood this.

Posted from my mobile device Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then   [#permalink] 08 May 2018, 18:46
Display posts from previous: Sort by

If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  