Limits Continuity And Differentiability Question 113
Question: Consider the function $ f(x)= \begin{cases} ax-2 & for & -2<x<-1 \\ -1 & for & -1\le x\le 1 \\ a+2{{(x-1)}^{2}} & for & 1<x<2 \\ \end{cases} . $ What is the value of a for which f(x) is continuous at x =-1 and x=1?
Options:
A) -1
B) 1
C) 0
D) 2
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)= \begin{cases} ax-2 & -2<x<-1 \\ -1 & -1\le x\le 1 \\ a+2{{(x-1)}^{2}} & 1<x<2 \\ \end{cases} . $ If f(x) is continuous at x = -1 then, $ \underset{x\to -1}{\mathop{\lim }}(ax-2)=\underset{x\to -1}{\mathop{\lim }}(-1) $
$ \Rightarrow a(-1)-2=-1\Rightarrow a=-1 $ If f(x) is continuous at x = 1 then, $ \underset{x\to 1}{\mathop{\lim }}a+2{{(x-1)}^{2}}=\underset{x\to 1}{\mathop{\lim }}-1 $
$ \Rightarrow a+2{{(1-1)}^{2}}=-1\Rightarrow a=-1 $
BETA
