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Question

The function f open parentheses x close parentheses equals fraction numerator ln invisible function application left parenthesis pi plus x right parenthesis over denominator ln invisible function application left parenthesis e plus x right parenthesis end fraction comma is

  1. Increasing on left square bracket 0 comma infinity right parenthesis  
  2. Decreasing on left square bracket 0 comma infinity right parenthesis  
  3. Increasing on left square bracket 0 comma pi divided by e right parenthesis and decreasing on left square bracket pi divided by e comma infinity right parenthesis  
  4. Decreasing on left square bracket 0 comma pi divided by e right parenthesis and increasing on left square bracket pi divided by e comma infinity right parenthesis  

The correct answer is: Decreasing on left square bracket 0 comma infinity right parenthesis


    Given, f open parentheses x close parentheses equals cos invisible function application x plus cos invisible function application square root of 2 x
    rightwards double arrow blank f to the power of ´ end exponent open parentheses x close parentheses equals negative sin invisible function application x minus square root of 2 sin invisible function application square root of 2 x

    rightwards double arrow blank f to the power of ´ ´ end exponent open parentheses x close parentheses equals negative cos invisible function application x minus 2 cos invisible function application square root of 2 x

    For extremum value, put f to the power of ´ end exponent open parentheses x close parentheses equals 0

    rightwards double arrow blank minus sin invisible function application x minus square root of 2 sin invisible function application square root of 2 x equals 0 blank rightwards double arrow blank x equals 0

    therefore At x equals 0 comma blank f to the power of ´ ´ end exponent open parentheses x close parentheses less than 0, maxima

    therefore blank f left parenthesis x right parenthesis Is maximum only once.

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