Question

The function \( f(x) = \frac{ax + b}{(x-1)(x-4)} \) has a local maxima at \( (2, -1) \), then

• A. $b=1, a=0$
• B. $a=1, b=0$
• C. $b=-1, a=0$
• D. $a=-1, b=0$

Hint:

The function $f(x)=\fracax+b(x-1)(x-4)}$ has a local maxima at $(2,-1)$ then

Answer 1

The Correct Option is B

Step 1: Concept
A local maximum at a point $(x_0, y_0)$ implies that $f(x_0) = y_0$ and the first derivative $f'(x_0) = 0$.
Step 2: Analysis
Substituting the point $(2, -1)$ into the function:
$-1 = \frac{2a+b}{(2-1)(2-4)} \Rightarrow -1 = \frac{2a+b}{-2} \Rightarrow 2a+b = 2$.
Step 3: Evaluation
Finding the derivative $f'(x)$ and setting $f'(2) = 0$:
The calculated derivative at $x=2$ leads to the condition $b=0$.
Step 4: Conclusion
Substituting $b=0$ back into $2a+b=2$ gives $2a=2$, so $a=1$.
Final Answer: (b)