Applications Of Derivatives Question 52
Question: If the function $ y=\frac{ax+b}{(x-1)(x-4)} $ has turning point at $ P(2,-1) $ , then
Options:
A) $ a=b=1 $
B) $ a=b=0 $
C) $ a=1,b=0 $
D) $ a=b=2 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y=\frac{ax+b}{(x-1)(x-4)}=\frac{ax+b}{x^{2}-5x+4} $ has turning point at $ P(2,-1) $ . Thus, $ P(2,-1) $ lies on the curve. Therefore, $ 2a+b=2…(1) $ Also, $ \frac{dy}{dx}=0 $ at $ P(2,-1) $ . Now, $ \frac{dy}{dx}=\frac{a(x^{2}-5x+4)-(2x-5)(ax+b)}{{{(x^{2}-5x+4)}^{2}}} $ . At $ P(2,-1),\frac{dy}{dx}=\frac{-2a+2a+b}{4}=0 $ or $ b=0 $ or $ a=1 $ [From equation (1)].
BETA
