Find the value of k for which f(x)= { (square root of 1 plus k x end root -square root of 1 minus k x end root)divided byx, -1<=x<0
                                                2 x plus 1 divided by x minus 2 , 0<=x<+1  }   is continuous.
 

Asked by Navya Benny | 2nd Sep, 2015, 07:12: PM

Expert Answer:

G i v e n space t h a t space t h e space f u n c t i o n space i s space c o n t i n u o u s. C o n s i d e r space t h e space f u n c t i o n comma space f left parenthesis x right parenthesis. f open parentheses x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator square root of 1 plus k x end root minus square root of 1 minus k x end root over denominator x end fraction comma space minus 1 less or equal than x less than 0 end cell row cell fraction numerator 2 x plus 1 over denominator x minus 2 end fraction comma space 0 less or equal than x less than 1 end cell end table close S i n c e space t h e space f u n c t i o n space i s space c o n t i n u o u s comma space stack l i m with x rightwards arrow 0 to the power of minus below f open parentheses x close parentheses equals stack l i m with x rightwards arrow 0 to the power of plus below f open parentheses x close parentheses  L H L equals stack l i m with x rightwards arrow 0 to the power of minus below f open parentheses x close parentheses space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator square root of 1 plus k x end root minus square root of 1 minus k x end root over denominator x end fraction space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator square root of 1 plus k x end root minus square root of 1 minus k x end root over denominator x end fraction cross times fraction numerator square root of 1 plus k x end root plus square root of 1 minus k x end root over denominator square root of 1 plus k x end root plus square root of 1 minus k x end root end fraction space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator 1 plus k x minus 1 plus k x over denominator x open parentheses square root of 1 plus k x end root plus square root of 1 minus k x end root close parentheses end fraction space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator 2 k x over denominator x open parentheses square root of 1 plus k x end root plus square root of 1 minus k x end root close parentheses end fraction space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator 2 k over denominator square root of 1 plus k x end root plus square root of 1 minus k x end root end fraction space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator 2 k over denominator square root of 1 plus k x end root plus square root of 1 minus k x end root end fraction space space space space space space space equals fraction numerator 2 k over denominator 2 end fraction space space space space space space space equals k R H L equals stack l i m with x rightwards arrow 0 to the power of plus below f open parentheses x close parentheses space space space space space space space equals stack l i m with x rightwards arrow 0 below fraction numerator 2 x plus 1 over denominator x minus 2 end fraction space space space space space space space equals negative 1 half because L H L equals R H L rightwards double arrow k equals negative 1 half

Answered by Vimala Ramamurthy | 3rd Sep, 2015, 11:52: AM