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Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)

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Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)  [#permalink]

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New post Updated on: 10 Aug 2018, 01:08
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Is x^3 > 1?

(1) x > -2

(2) 2x – (b – c) < c – (b – 2)

Originally posted by ankita1211 on 08 Aug 2018, 09:10.
Last edited by Bunuel on 10 Aug 2018, 01:08, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)  [#permalink]

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New post 08 Aug 2018, 09:52
ankita1211 wrote:
Is x^3> 1?
(1) x > -2
(2) 2x – (b – c) < c – (b – 2)


OA:B
Rephrasing the question
Subtracting 1 from both sides, we get
\(x^3-1>0\)
\((x-1)(x^2+x+1)>0\) [Using \(a^3 - b^3 = (a - b)(a^2 + b^2 + ab)\)]
\(x^2+x+1\) is always positive, as its lowest value will occurs \(x=-\frac{1}{2}\) i.e at \(\frac{-b}{{2a}}\) given expression is of form \(ax^2+bx+c\)
At \(x=-\frac{1}{2}\);\(x^2+x+1= {(-\frac{1}{2})}^2 -\frac{1}{2}+1=\frac{3}{4}\)

Lowest value of \(x^2+x+1\) is \(\frac{3}{4}\), for all other values of \(x\) , \(x^2+x+1\) would be greater than \(\frac{3}{4}\)

Question reduces: Is \((x-1)>0?\) or Is \(x>1\)

Statement 1 :\(x > -2\)
\(x\) can be\(-1\)
Is \(x>1\) : No
\(x\) can be \(2\)
Is \(x>1\) : yes
Statement 1 alone is not sufficient

Statement 2: \(2x – (b – c) < c – (b – 2)\)
\(2x -b+c< c-b+2\)
Adding \(b-c\) on both sides, we get \(2x<2\) or \(x<1\)
Is \(x>1\) : No always
Statement 2 alone is sufficient
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Re: Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)  [#permalink]

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New post 08 Aug 2018, 12:05
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Re: Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)   [#permalink] 08 Aug 2018, 12:05
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