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Application Of Derivatives

Question
CBSEENMA12035466
Wired Faculty App

If y = a log | x | + bx2 + x has its extremum values at x = – 1 and x = 2, find the values of a and b. Prove that these extrema are maximum.

Solution
straight y space equals space straight a space log space open vertical bar straight x close vertical bar space plus space bx squared plus straight x
therefore space space space space dy over dx space equals space straight a fraction numerator 1 over denominator open vertical bar straight x close vertical bar end fraction. space straight d over dx open parentheses open vertical bar straight x close vertical bar close parentheses space plus space 2 bx plus 1
                   equals space straight a. space fraction numerator 1 over denominator open vertical bar straight x close vertical bar end fraction space open parentheses fraction numerator straight x over denominator open vertical bar straight x close vertical bar end fraction. space 1 close parentheses space plus space 2 bx plus 1 space equals space fraction numerator straight a space straight x over denominator straight x squared end fraction plus 2 bx plus 1          open square brackets because space space space open vertical bar straight x close vertical bar squared space equals space straight x squared close square brackets
                    equals straight a over straight x plus 2 bx plus 1
Since y has its extremum values at x = -1 and x = 2
therefore space space space space open square brackets dy over dx close square brackets subscript straight x equals negative 1 end subscript space space equals space 0 comma space space space space space space open square brackets dy over dx close square brackets subscript straight x equals 2 end subscript space space equals 0 therefore space space space minus straight a minus 2 straight b plus 1 space equals 0 comma space space space straight a over 2 plus 4 straight b plus 1 space equals space 0 therefore space space space space space space space straight a plus 2 straight b space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis and space space space space space space straight a plus 8 straight b space equals space minus 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
Subtracting (1) from (2),    6 straight b space equals negative 3 space space space space space space rightwards double arrow space space space straight b space equals space minus 1 half
therefore space space space space space space from space left parenthesis 1 right parenthesis comma space space space straight a minus 1 space equals 1 space space space rightwards double arrow space space space straight a space equals space 2 therefore space space space space space space dy over dx space equals space 2 over straight x minus straight x plus 1 space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus 2 over straight x squared minus 1
At space space straight x space space equals negative 1 comma space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus 2 over 1 minus 1 space equals space minus 3 space less than space 0 At space space space straight x minus 2 comma space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus 1 half minus 1 space equals space minus 3 over 2 less than 0 therefore space space space space both space the space points space straight x space equals negative 1 comma space space space straight x space equals space 2 space of space extrema space are space maxima

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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