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

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Joined: 25 Feb 2013
Posts: 1012
Location: India
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If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then [#permalink]

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17 Sep 2017, 08:04
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75% (hard)

Question Stats:

47% (01:32) correct 53% (01:51) wrong based on 104 sessions

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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.) $$g(x) ≥ 5b^2$$
[Reveal] Spoiler: OA

Last edited by niks18 on 18 Sep 2017, 09:25, edited 1 time in total.
Manager
Joined: 22 Apr 2017
Posts: 120
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then [#permalink]

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18 Sep 2017, 09:32
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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
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1012
Location: India
GPA: 3.82
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then [#permalink]

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18 Sep 2017, 11:11
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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
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Joined: 06 Aug 2017
Posts: 79
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GMAT 2: 610 Q49 V24
Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then [#permalink]

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23 Sep 2017, 03:27
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.) $$g(x) ≥ 5b^2$$

$$g(x) = x^2 – 4bx + 9b^2$$

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$$b^2$$
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then   [#permalink] 23 Sep 2017, 03:27
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