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If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

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Bunuel
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If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   22 May 2020, 08:17
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If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

A. 0
B. x
C. x + 2
D. 2x
E. 2x + 1


PS21179
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E
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BrentGMATPrepNow
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Re: If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   22 May 2020, 08:37
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Bunuel wrote:
If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

A. 0
B. x
C. x + 2
D. 2x
E. 2x + 1

PS21179


Given: [x] = x² - 1
So, for example, [5] = 5² - 1 = 25 - 1 = 24
And [7] = 7² - 1 = 49 - 1 = 48
Likewise, [3k] = (3k)² - 1 = 9k² - 1

So......
[x + 1] = (x + 1)² - 1 = (x² + 2x + 1) - 1 = x² + 2x
And [x] = x² - 1 = x² - 1

So, [x + 1] - [x] = (x² + 2x) - (x² - 1) = 2x +1

Answer: E

Cheers,
Brent
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Re: If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   22 May 2020, 09:15
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Bunuel wrote:
If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

A. 0
B. x
C. x + 2
D. 2x
E. 2x + 1


PS21179


[x] = x^2 – 1

Let, x = 2, [x] = x^2 – 1 = 2^2 - 1 = 3

Let, x+1 = 3, [x+1] = (x+1)^2 – 1 = 3^2 - 1 = 8

[x + 1] - [x] = 8-3 = 5



A. 0 WRONG
B. x WRONG
C. x + 2 = 2+2 = 4 WRONG
D. 2x = 2*2 = 4 WRONG
E. 2x + 1 = 2*2+1 = 5 CORRECT

Answer: Option E
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Re: If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   28 Nov 2020, 13:34
Bunuel wrote:
If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

A. 0
B. x
C. x + 2
D. 2x
E. 2x + 1


PS21179


\([x + 1] = (x+1)^2 - 1 = x^2 + 2x + 1 -1\)

\(x^2 + 2x - (x^2 - 1) = x^2 + 2x - x^2 + 1 = 2x + 1\)

Answer is E.
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If [x] is defined as x^2 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   30 Mar 2023, 00:19
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Bunuel wrote:
If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =

A. 0
B. x
C. x + 2
D. 2x
E. 2x + 1


PS21179


We can solve it by plugging in numbers too. Say x = 2, then x + 1 = 3

\([3] - [2] = (3^2 - 1) - (2^2 - 1) = 5\)

Put x = 2 in the options. Only (E) satisfies.
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Re: If [x] is defined as x^2 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   17 Nov 2023, 23:56
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[x] = x^2 – 1

=> [x + 1] - [x] = \((x+1)^2 - 1 - ( x^2 - 1)\) = \(x^2 + 2x + 1 -1 - x^2 + 1\) = 2x + 1

So, Answer will be E
Hope it helps!

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Re: If [x] is defined as x^2 1 for all integers x, then [x + 1] - [x] = [#permalink] If [x] is defined as x^2 – 1 for all integers x, then [x + 1] - [x] =   18 May 2024, 07:45
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­Straightforward function- keep each expression in parentheses to be safe:

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Re: If [x] is defined as x^2 1 for all integers x, then [x + 1] - [x] = [#permalink]
18 May 2024, 07:45
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