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Application Of Derivatives

Question
CBSEENMA12035475
Wired Faculty App

The maximum value of open square brackets straight x left parenthesis straight x minus 1 right parenthesis space plus space 1 close square brackets to the power of 1 third end exponent comma space space space 0 space less or equal than space straight x space less or equal than space 1 is

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  • 1 half

  • 1

  • 0

Solution

C.

1

 Let space straight y space equals space open square brackets straight x left parenthesis straight x minus 1 right parenthesis plus 1 close square brackets to the power of 1 third end exponent space space equals space left parenthesis straight x squared minus straight x plus 1 right parenthesis to the power of 1 third end exponent
therefore space space space dy over dx space equals space 1 third left parenthesis straight x squared minus straight x plus 1 right parenthesis to the power of negative 2 over 3 end exponent. space left parenthesis 2 straight x minus 1 right parenthesis space equals space fraction numerator 2 straight x minus 1 over denominator 3 left parenthesis straight x squared minus straight x plus 1 right parenthesis to the power of begin display style 2 over 3 end style end exponent end fraction
         space dy over dx space equals space 0 space space space space rightwards double arrow space space space space fraction numerator 2 straight x minus 1 over denominator 3 left parenthesis straight x squared minus straight x plus 1 right parenthesis to the power of begin display style 2 over 3 end style end exponent end fraction space equals space 0 space space space space space rightwards double arrow space space 2 straight x minus 1 space equals space 0 space space space space rightwards double arrow space space space straight x space equals space 1 half

Also,  dy over dx exists at all real values of x
therefore space space space straight x space equals space 1 half space is space the space critical space point. space
Now,  straight y left parenthesis 0 right parenthesis space equals space left parenthesis 0 minus 0 plus 1 right parenthesis to the power of 1 third end exponent space equals space left parenthesis 1 right parenthesis to the power of 1 third end exponent minus 1
     straight y open parentheses 1 half close parentheses space equals space open parentheses 1 fourth minus 1 half plus 1 close parentheses to the power of 1 third end exponent space equals space open parentheses 3 over 4 close parentheses to the power of 1 third end exponent
     straight y left parenthesis 1 right parenthesis space equals space left parenthesis 1 minus 1 plus 1 right parenthesis to the power of 1 third end exponent space minus space left parenthesis 1 right parenthesis to the power of 1 third end exponent space equals 1
therefore space space space space maximum space value space equals space 1 therefore space space space left parenthesis straight C right parenthesis space is space correct space answer.
   

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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