If the function $f(x)=\frac{cx+d}{(x-1)(

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

If the function $f(x)=\frac{cx+d}{(x-1)(x-4)}$ has a turning point at the point (2, –1) then:

Options:

c = 2, d = 0

c = 1, d = 0

c = 1, d = -1

c = 1, d = 1

Correct Answer:

c = 1, d = 0

Explanation:

$f'(x)=\frac{c(x-1)(x-4)-(cx+d)(2x-5)}{(x-1)^2(x-4)^2}$

So, $f'(2)=0⇒\frac{-2c+(2c+d)}{-4}=\frac{d}{4}⇒d=0$

Also, $-1=f(2)=\frac{2c+d}{-2}=-c⇒c=1$