Question

If the function y = `(ax + b)/((x - 4)(x - 1))` has an extremum at P(2, –1), then the values of a and b are ______.

Options

• a = 0, b = 1

• a = 0, b = –1

• a = 1, b = 0

• a = –1, b = 0

Answer 1

If the function y = `(ax + b)/((x - 4)(x - 1))` has an extremum at P(2, –1), then the values of a and b are a = 1, b = 0.

Explanation:

y = `(ax + b)/(x^2 - 5x + 4)`

y' = `((ax^2 - 5ax + 4a) - (ax + b)(2x - 5))/((x - 1)^2(x - 4)^2`

= `(-ax^2 + 4a - 2bx + 5b)/((x - 1)^2(x - 4)^2`

= `y_{(2, –1)}^' = 0`

⇒ –4a + 4a – b = 0

⇒ b = 0

and 2a + b = 2  ...(∵ (2, –1) lies on the curve.)

Put b = 0 in the above equation

⇒ a = 1, b = 0