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Application Of Derivatives

Question
CBSEENMA12035458
Wired Faculty App

What is the maximum value of the function sin x + cos x?

Solution

Consider the interval  open square brackets 0 comma space 2 straight pi close square brackets
Let space space straight f left parenthesis straight x right parenthesis space equals space sinx space plus cosx comma space space space space space space therefore space space space space straight f space apostrophe left parenthesis straight x right parenthesis space equals space cosx minus sinx space space
         straight f apostrophe space left parenthesis straight x right parenthesis space equals space 0 space space space space gives space cosx minus sinx space equals space 0 space space space space space rightwards double arrow space space space sinx space equals space space cosx
rightwards double arrow space space space tanx space equals 1 space space space space space space rightwards double arrow space space space space straight x space equals space straight pi over 4 comma space space space fraction numerator 5 straight pi over denominator 4 end fraction
     Now comma space space space straight f left parenthesis 0 right parenthesis space equals space sin space 0 space plus space cos space space 0 space equals space 0 space plus space 1 space equals space 1
               straight f open parentheses straight pi over 4 close parentheses space equals space sin space straight pi over 4 plus space cos space straight pi over 4 space equals space fraction numerator 1 over denominator square root of 2 end fraction plus fraction numerator 1 over denominator square root of 2 end fraction space equals space fraction numerator 2 over denominator square root of 2 end fraction space equals space square root of 2 straight f open parentheses fraction numerator 5 straight pi over denominator 4 end fraction close parentheses space equals space sin space space fraction numerator 5 straight pi over denominator 4 end fraction plus cos space fraction numerator 5 straight pi over denominator 4 end fraction space equals space sin space open parentheses straight pi plus straight pi over 4 close parentheses plus space cos space open parentheses straight pi plus straight pi over 4 close parentheses equals space space minus sin space straight pi over 4 minus cos space straight pi over 4 space space space space space space space space space space space space space space space equals negative fraction numerator 1 over denominator square root of 2 end fraction minus fraction numerator 1 over denominator square root of 2 end fraction space equals space minus fraction numerator 2 over denominator square root of 2 end fraction space equals space minus square root of 2 space space space straight f left parenthesis 2 straight pi right parenthesis space equals space sin space 2 straight pi space plus space cos space 2 straight pi space equals space 0 plus 1 space equals space 1 therefore space space space space maximum space value space space equals space square root of 2

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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