The function $y = \frac{ax + b}{(x - 1) (x - 4)}$ has a turning point or a point of inflexion at P(2, -1). Then, which of the following is(are) true about a and b?

a = 0

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a = 1

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b = 1

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b = 0

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Solution

The correct options are
B a = 1
D b = 0
$y = \frac{ax + b}{(x - 1) (x - 4)} = \frac{ax + b}{x^{2} - 5 x + 4} has a turning point at P (2, - 1) . Thus, P (2, - 1) lies on the curve . ∴ 2 a + b = 2 Also, \frac{dy}{dx} = 0 at P (2, - 1) . Now, \frac{dy}{dx} = \frac{a (x^{2} - 5 x + 4) - (2 x - 5) (ax + b)}{{(x^{2} - 5 x + 4)}^{2}} At P (2, - 1), \frac{dy}{dx} = \frac{- 2 a + 2 a + b}{4} = 0 or, a = 1 and b = 0 .$

Suggest Corrections

Similar questions

y = \frac{ax + b}{(x - 1) (x - 4)}

x^{3} - a x^{2} + b x - a = 0

y = \frac{a x + b}{(x - 1) (x - 4)}

(2, - 1)

a

b

y = {tan}^{- 1} (\frac{a sin x - b cos x}{a cos x + b sin x}),

\frac{d y}{d x} =

y = a x^{3} + b x^{2} + c x + d

x = 1

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