Maths-
General
Easy

Question

If L t left parenthesis x rightwards arrow 0 right parenthesis left parenthesis left parenthesis c o s invisible function application 4 x plus a c o s invisible function application 2 x plus b right parenthesis divided by x to the power of 4 right parenthesis is finite then the values of a,b are respectively

  1. 5 comma negative 4
  2. negative 5 comma negative 4
  3. negative 4 comma 3
  4. 4,5

hintHint:

In this question for the limit to exist the numerator must also tends to  zero as denominator tends to zero. We will apply L'Hôpital's rule to find the value of a and b

The correct answer is: negative 4 comma 3


    In this question we are given that the limit limit as x minus greater than 0 of open parentheses fraction numerator cos open parentheses 4 x close parentheses plus a cos open parentheses 2 x close parentheses plus b over denominator x to the power of 4 end fraction close parentheses of this expression is finite and we have to find the value of a and b
    Step1: Putting the value of the limit
    By putting the value of limit the denominator of the expression tends to zero. In order for the limit to exist the numerator must also tend to zero.
    limit as x minus greater than 0 of left parenthesis cos open parentheses 4 x close parentheses plus a cos open parentheses 2 x close parentheses plus b right parenthesis space equals 0
    by putting x equals 0 we get,
    1 plus a plus b equals 0
    =>a plus b space equals negative 1
    As we are getting 0 over 0 form. We will use L'Hôpital's rule
    Step2: Using the L'Hôpital's rule
    By differentiating both numerator and denominator.
    => limit as x minus greater than 0 of fraction numerator negative 4 sin open parentheses 4 x close parentheses minus 2 a sin open parentheses 2 x close parentheses over denominator 4 x cubed end fraction
    By putting the value of limit, we are getting both numerator and denominator as zero.
    So, we will again apply L'Hôpital's rule
    limit as x minus greater than 0 of fraction numerator negative 16 cos open parentheses 4 x close parentheses minus 4 a cos open parentheses 2 x close parentheses over denominator 12 x squared end fraction
    By putting the value of limit the denominator tends to  zero. For the limit to exist the numerator must also tend to zero.
    limit as x minus greater than 0 of left parenthesis negative 16 cos open parentheses 4 x close parentheses minus 4 a cos open parentheses 2 x close parentheses right parenthesis equals 0
    =>negative 16 minus 4 a equals 0
    =>a space equals space minus 4
    Then b space equals space 4 minus 1
    b equals 3.
    So, the value of a and b are left parenthesis negative 4 comma 3 right parenthesis

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