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If f left parenthesis x right parenthesis is a function given by f open parentheses x close parentheses equals open vertical bar table row cell sin invisible function application x end cell cell sin invisible function application a end cell cell sin invisible function application b end cell row cell cos invisible function application x end cell cell cos invisible function application a end cell cell cos invisible function application b end cell row cell tan invisible function application x end cell cell tan invisible function application a end cell cell tan invisible function application b end cell end table close vertical bar comma w h e r e blank 0 less than a less than b less than fraction numerator pi over denominator 2 end fraction
Then the equation f to the power of ´ end exponent open parentheses x close parentheses equals 0

  1. Has at least one root in left parenthesis a comma b right parenthesis  
  2. Has at most one root in left parenthesis a comma b right parenthesis  
  3. Has exactly one root in left parenthesis a comma b right parenthesis  
  4. Has no root in left parenthesis a comma b right parenthesis  

The correct answer is: Has at least one root in left parenthesis a comma b right parenthesis


    We know that sin invisible function application x comma cos invisible function application x and tan invisible function application x are continuous on open square brackets a comma blank b close square brackets and differentiable on open parentheses a comma blank b close parentheses. Therefore, f left parenthesis x right parenthesis is continuous on open square brackets a comma blank b close square brackets and differentiable on left parenthesis a comma blank b right parenthesis
    Also, f open parentheses a close parentheses equals f open parentheses b close parentheses equals 0

    Therefore, by Rolle’s theorem there exists at least one c element of left parenthesis a comma blank b right parenthesis such that f to the power of ´ end exponent open parentheses c close parentheses equals 0

    Hence, f to the power of ´ end exponent open parentheses x close parentheses equals 0 has at least one root in left parenthesis a comma blank b right parenthesis

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