Differentiation Question 27
Question: If $ \underset{x\to \infty }{\mathop{\lim }}{ \frac{x^{3}+1}{x^{2}+1}-(ax+b) }=2 $ , then
Options:
A) $ a=1,b=1 $
B) $ a=1,b=2 $
C) $ a=1,v=-2 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \underset{x\to \infty }{\mathop{\lim }}( \frac{x^{3}+1}{x^{2}+1}-(ax+b) )=2 $ Or $ \underset{x\to \infty }{\mathop{\lim }}\frac{x^{3}(1-a)-bx^{3}-ax+(1-b)}{x^{2}+1}=2 $ Or $ 1-a=0 $ and $ -b=2 $ Or $ a=1, $
$ b=-2 $
BETA
