Functions Question 307
Question: The value of the constant $ \alpha $ and $ \beta $ such that $ \underset{x\to \infty }{\mathop{\lim }},( \frac{x^{2}+1}{x+1}-\alpha x-\beta )=0 $ are respectively
[Orissa JEE 2005]
Options:
A) (1, 1)
B) (?1, 1)
C) (1, ?1)
D) (0, 1)
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to \infty }{\mathop{\lim }},( \frac{x^{2}+1}{x+1}-2x-\beta )=0 $
Þ $ \underset{x\to \infty }{\mathop{\lim }},\frac{x^{2}(1-\alpha )-x(\alpha +\beta )+1-b}{x+1}=0 $ Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than denominator. \ $ 1-\alpha =0 $ and $ \alpha +\beta =0 $
Þ $ \alpha =1 $ and $ \beta =-1 $ .
BETA
