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then x lies in the interval/set...?

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by RACHVIK » Fri Jan 07, 2011 6:10 am
Nice Ques.

Let us solve first eq.

x^2 - 1 =<0

assume eq as x^2 = 1, we get roots as -1 & 1
Divide number line three parts, namely x<-1, -1=<x<=1, x>1
the second section satisfies the equation.

Similarly divide number line into three parts x=<-1, -1<x<2, x>=2 for equation no.2

If you have a closer look, you will find that [spoiler]x=-1[/spoiler] is the only common point for both equations.

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by anshumishra » Fri Jan 07, 2011 6:24 am
Night reader wrote:x² - 1 < = 0 and x²- x - 2 > = 0, then x lies in the interval/set

A. (-1, 2)
B.(-1, 1)
C. (1, 2)
D. {-1}.
E. (1, -2)
x^2 - 1 <= 0 => (x-1)(x+1) <=0 => -1<=x<=1
x^2-x-2 >= 0 => (x-2)(x+1) >=0 => x>=2 or x<=-1

Hence, D
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by Night reader » Fri Jan 07, 2011 6:28 am
I think the answer is B, though this problem has been posted at other forums without OA and answered D

x^2 - 1 =< 0
|x|=<1, -1 =< x =< 1
intervals
{-infin}___+ve___{-1}___-ve___{1}___+ve___{+infin}

x^2 - x - 2 >= 0
(x-2)(x+1) >= 0
(I omit the individual inequality solutions to save space)
intervals
{-infin}___+ve___{-1}___-ve___{1}___-ve___{2}___+ve___{+infin}

it would be good to have a normal picture for better visualizing :(

So answer is B, because the common solution area is (-1, 1)
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by anshumishra » Fri Jan 07, 2011 6:31 am
Night reader wrote:I think the answer is B, though this problem has been posted at other forums without OA and answered D

x^2 - 1 =< 0
|x|=<1, -1 =< x =< 1
intervals
{-infin}___+ve___{-1}___-ve___{1}___+ve___{+infin}

x^2 - x - 2 >= 0
(x-2)(x+1) >= 0
(I omit the individual inequality solutions to save space)
intervals
{-infin}___+ve___{-1}___-ve___{1}___-ve___{2}___+ve___{+infin}

it would be good to have a normal picture for better visualizing :(

So answer is B, because the common solution area is (-1, 1)
There is a mistake. Check with x=0 (which lies in -1<=x<=1) for both the equations.
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by Night reader » Fri Jan 07, 2011 6:50 am
yes, that's true I overrun the sign in inequality (2), correct answer is D.
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