Maths-
General
Easy

Question

If f left parenthesis x right parenthesis equals x t a n left parenthesis negative 1 right parenthesis invisible function application x then l i m subscript left parenthesis x rightwards arrow 1 right parenthesis   left parenthesis f left parenthesis x right parenthesis minus f left parenthesis 1 right parenthesis right parenthesis divided by left parenthesis x minus 1 right parenthesis equals

  1. pi divided by 4
  2. left parenthesis pi plus 1 right parenthesis divided by 4
  3. left parenthesis pi plus 2 right parenthesis divided by 4
  4. left parenthesis pi plus 3 right parenthesis divided by 4

hintHint:

In this question we are getting 0 over 0 form. We will use the L'Hôpital's rule to find the value of the limit

The correct answer is: left parenthesis pi plus 2 right parenthesis divided by 4


    In this question we are f left parenthesis x right parenthesis equals tan to the power of negative 1 end exponent open parentheses x close parentheses and we have to find the value of limit as x minus greater than 1 of fraction numerator f left parenthesis x right parenthesis minus f left parenthesis 1 right parenthesis over denominator x minus 1 end fraction
    Step1: Putting the value of limit
    By putting the value of limit we are getting both numerator and denominator as zero. That is 0 over 0 form.
    Step2: Using L'Hôpital's rule
    By differentiating both numerator and denominator we get
    limit as x minus greater than 1 of fraction numerator f apostrophe left parenthesis x right parenthesis over denominator 1 end fraction
    => f apostrophe left parenthesis 1 right parenthesis
    We have to find the value of f apostrophe left parenthesis 1 right parenthesis
    Step3: Finding the value of f apostrophe left parenthesis 1 right parenthesis.
    f left parenthesis x right parenthesis equals tan to the power of negative 1 end exponent open parentheses x close parentheses
    => f apostrophe left parenthesis x right parenthesis equals tan to the power of negative 1 end exponent open parentheses x close parentheses plus fraction numerator x over denominator 1 plus x squared end fraction
    by putting the value of x equals 1,
    f apostrophe left parenthesis 1 right parenthesis space equals space pi over 4 plus 1 half
    =>fraction numerator pi plus 2 over denominator 4 end fraction

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