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Application Of Derivatives

Question
CBSEENMA12035465

straight y space equals space fraction numerator ax minus straight b over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 4 right parenthesis end fraction has a turning point at P(2, -1). Find the values of a and b and show that y is maximum at P.

Solution
Here space straight y space equals space fraction numerator ax minus straight b over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x minus 4 right parenthesis end fraction space equals space fraction numerator ax minus straight b over denominator straight x squared minus 5 straight x plus 4 end fraction                        ...(1)
              dy over dx space equals space fraction numerator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis. space straight a space minus space left parenthesis ax minus straight b right parenthesis thin space left parenthesis 2 straight x minus 5 right parenthesis over denominator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis squared end fraction      ...(2)
Now,     straight P left parenthesis 2 comma space minus 1 right parenthesis lies on (1)
therefore space space space space space minus 1 space equals space fraction numerator 2 straight a minus straight b over denominator 4 minus 10 plus 4 end fraction space space space rightwards double arrow space space space minus 1 space equals space fraction numerator 2 straight a minus straight b over denominator negative 2 end fraction space space space space rightwards double arrow space space space 2 straight a minus straight b space equals space 2     ...(3)
Since P(2, -1) is a turning point at P(2, -1)
therefore space space space space dy over dx space equals space 0
therefore space space space space space space fraction numerator left parenthesis 4 minus 10 plus 4 right parenthesis. space straight a minus left parenthesis 2 straight a minus straight b right parenthesis thin space left parenthesis 4 minus 5 right parenthesis over denominator left parenthesis 4 minus 10 plus 4 right parenthesis squared end fraction space equals space 0
rightwards double arrow space space space space minus 2 straight a plus left parenthesis 2 straight a minus straight b right parenthesis space equals space 0 space space space space space space rightwards double arrow space space space minus 2 straight a plus left parenthesis 2 right parenthesis space equals space 0 space space space space space space space space space space space space open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets
rightwards double arrow space space space space space space space space space space 2 straight a space equals space 2 space space space space space rightwards double arrow space space space straight a space equals space 1
therefore space space space space from space left parenthesis 3 right parenthesis comma space space space 2 minus straight b space equals space 2 space space space space space space space space space rightwards double arrow space space straight b space equals space 0
therefore space space space space space space dy over dx equals space fraction numerator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis minus straight x left parenthesis 2 straight x minus 5 right parenthesis over denominator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis end fraction space equals fraction numerator negative straight x squared plus 4 over denominator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis squared end fraction
When space straight x space is space slightly space less than space 2 comma space space space dy over dx space equals space plus ve
When space straight x space is space slightly space greater than 2 comma space space space dy over dx space equals space minus ve
therefore space space space at space straight x space equals space 2 comma space space dy over dx space changes space sign space from space plus ve space to space minus ve
therefore space space space straight y space has space maximum space value space at space straight P left parenthesis 2 comma space minus 1 right parenthesis.