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Question

L t ┬ left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x left parenthesis 1 plus a c o s invisible function application x right parenthesis minus b s i n invisible function application x right parenthesis divided by x cubed equals 1 t h e n a equals comma b equals

  1. left parenthesis negative 5 right parenthesis divided by 2 comma left parenthesis negative 3 right parenthesis divided by 2
  2. space 5 divided by 2 space space 3 divided by 2
  3. 2 divided by 3 comma 5 divided by 2
  4. 3 divided by 2 comma 1 divided by 2

hintHint:

In this question we will use the L'Hôpital's rule to find the values of a space a n d space b.

The correct answer is: left parenthesis negative 5 right parenthesis divided by 2 comma left parenthesis negative 3 right parenthesis divided by 2


    In this question we are given limit as x minus greater than 0 of fraction numerator x left parenthesis 1 plus a cos open parentheses x close parentheses right parenthesis minus b sin open parentheses x close parentheses over denominator x cubed end fraction equals 1 and we have to find the value of a space a n d space b
    Step1: Using L'Hôpital's rule
    By differentiating both numerator and denominator we get,
     
    limit as x minus greater than 0 of fraction numerator left parenthesis 1 plus a cos open parentheses x close parentheses minus a x sin open parentheses x close parentheses minus b cos open parentheses x close parentheses over denominator 3 x squared end fraction equals 1
     
    For the limit to exist 1 plus a minus b equals 0 or b equals a plus 1
     
    Step2: Again using L'Hôpital's rule
     
    Again differentiating both numerator and denominator
     
    => limit as x minus greater than 0 of fraction numerator negative left parenthesis a minus b right parenthesis sin open parentheses x close parentheses minus a sin open parentheses x close parentheses minus a x cos open parentheses x close parentheses over denominator 6 x end fraction equals 1
     
     
     
    We know that limit as x minus greater than a of left parenthesis f left parenthesis x right parenthesis plus g left parenthesis x right parenthesis right parenthesis space equals space limit as x minus greater than a of f left parenthesis x right parenthesis space plus space limit as x minus greater than a of g left parenthesis x right parenthesis
     
     
    => limit as x minus greater than 0 of fraction numerator left parenthesis b minus 2 a right parenthesis sin open parentheses x close parentheses over denominator 6 x end fraction minus space limit as x minus greater than 0 of fraction numerator a cos open parentheses x close parentheses over denominator 6 end fraction equals 1 
     
    => fraction numerator b minus 2 a over denominator 6 end fraction minus a over 6 equals 1 
     
    => By solving b equals a plus 1 and fraction numerator b minus 2 a over denominator 6 end fraction minus a over 6 equals 1 we get,
     
    =>fraction numerator a plus 1 minus 2 a over denominator 6 end fraction minus a over 6 equals 1
     
    => 1 minus 2 a equals 6
    =>a equals 5 over 2 , Similarly we can calculate b equals fraction numerator negative 3 over denominator 2 end fraction
     
    => a plus b equals fraction numerator negative 8 over denominator 2 end fraction equals negative 4

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