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

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Intern
Joined: 19 Apr 2018
Posts: 1
Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)  [#permalink]

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Updated on: 10 Aug 2018, 00:08
1
00:00

Difficulty:

35% (medium)

Question Stats:

96% (00:43) correct 4% (01:01) wrong based on 25 sessions

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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, 08:10.
Last edited by Bunuel on 10 Aug 2018, 00:08, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Joined: 22 Feb 2018
Posts: 414
Re: Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2)  [#permalink]

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08 Aug 2018, 08:52
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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08 Aug 2018, 11:05
1
1
This question is based on an important Concept
Attachment:

WhatsApp Image 2018-08-09 at 00.31.35.jpeg [ 111.39 KiB | Viewed 295 times ]

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Re: Is x^3 > 1? (1) x > -2 (2) 2x – (b – c) < c – (b – 2) &nbs [#permalink] 08 Aug 2018, 11:05
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