i f space open vertical bar fraction numerator 2 x minus 1 over denominator x minus 1 end fraction close vertical bar greater than 2 space comma t h e n space x space i s left parenthesis A right parenthesis left parenthesis negative infinity comma negative 3 right square bracket union left parenthesis negative 1 comma 5 right parenthesis space space space space space space space space space space space space left parenthesis B right parenthesis left parenthesis negative 3 comma 0 right square bracket union left parenthesis 1 comma 2 right parenthesis left parenthesis C right parenthesis left parenthesis negative infinity comma negative 2 right square bracket space union space left parenthesis negative 1 comma 4 right parenthesis space space space space space space space space space space left parenthesis D right parenthesis space left square bracket 0 comma 5 right parenthesis
plz ... kindly explain the shortest method to solve this problem....

Asked by araima2001 | 13th Jul, 2016, 09:36: AM

Expert Answer:

If space open vertical bar straight x close vertical bar greater than straight a space left right double arrow straight x less than negative straight a space space or space straight x greater than straight a open vertical bar fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction close vertical bar greater than 2 left right double arrow fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction less than negative 2 space space space space or space space space space fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction greater than 2 space Case space 1 colon fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction less than negative 2 fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction less than negative 2 fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction plus 2 less than 0 fraction numerator 4 straight x minus 3 over denominator straight x minus 1 end fraction less than 0 For space the space above space fraction space to space be space less space than space zero space either space numerator space or space denominator space should space be space negative. All space the space cases space are space expained space in space the space below space table table row blank cell straight x less than 3 over 4 end cell cell straight x equals 3 over 4 end cell cell 3 over 4 less than straight x less than 1 end cell cell straight x equals 1 end cell cell straight x greater than 1 end cell row cell 4 straight x minus 3 end cell Negativee 0 Positive Positive Positive row cell straight x minus 1 end cell Negative Negative Negative 0 Positive row cell fraction numerator 4 straight x minus 3 over denominator straight x minus 1 end fraction end cell Positive 0 Negative Undefined Positive end table From space above space table space we space can space see space that space our space fraction space will space be space less space than space zero space when space 3 over 4 less than straight x less than 1. Case space 2 colon fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction greater than 2 fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction greater than 2 fraction numerator 1 over denominator straight x minus 1 end fraction greater than 0 For space the space above space fraction space to space be space greater space space than space space denominator space should space bepositive. straight x minus 1 greater than 0 straight x greater than 1 So space from space Case space 1 space and space Case space 2 space we space can space see space that open vertical bar fraction numerator 2 straight x minus 1 over denominator straight x minus 1 end fraction close vertical bar greater than 2 when space 3 over 4 less than straight x less than 1 space space space or space space straight x greater than 1 that space is space space open parentheses 3 over 4 comma space 1 close parentheses union open parentheses 1 comma infinity close parentheses

Answered by Vijaykumar Wani | 14th Jul, 2016, 11:31: AM

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