Differentiation Question 128
Question: Let $ f(x)=x^{2}-1,0<x<2 $ and $ 2x+3,2\le x<3. $ The quadratic equation whose roots are, $ \underset{x\to 2-0}{\mathop{\lim }}f(x) $ And $ \underset{x\to 2+0}{\mathop{\lim }}f(x) $ is
Options:
A) $ x^{2}-6x+9=0 $
B) $ x^{2}-10x+21=0 $
C) $ x^{2}-14x+49=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \underset{x\to 2-0}{\mathop{\lim }}f(x)=\underset{x\to 2-0}{\mathop{\lim }}(x^{2}-1)=2^{2}-1=3 $
$ \underset{x\to 2+0}{\mathop{\lim }}f(x)=\underset{x\to 2+0}{\mathop{\lim }}(2x+3)=2\times 2+3=7 $
$ \therefore $ $ Required quadratic equation is $ x^{2}-10x+21=0 $
BETA
