Question

Consider the function
$f\left(x\right)=\{\begin{array}{ccc}ax-2& for& -2<x<-1\\ -1& for& -1\le x\le 1\\ a+2(x-1{)}^2& for& 1<x<2\end{array}$
What is the value of a for which $f\left(x\right)$ is continuous at $x=-1$ and $x=1$?

• A. -1
• B. 1
• C. 0
• D. 2

Answer 1

The correct option is A -1

Solution:

$LHL={lim}_{h\to 0}f(-1-h)={lim}_{h\to 0}a(-1-h)-2=-a-2$

$RHL={lim}_{h\to 0}f(-1+h)={lim}_{h\to 0}(-1)=-1$

At $x=1$, $LHL={lim}_{h\to 0}f(1-h)={lim}_{h\to 0}(-1)=-1$

$RHL={lim}_{h\to 0}f(1+h)={lim}_{h\to 0}(a+2h^2)=a$

Now, for $f$ to be continuous at $x=-1$

we should have $LHL=RHL=f(-1)$

$⟹-a-2=-1$

$⟹a=-1$

Also, for $a=-1$, $f$ is continuous at $x=1$

Hence, $a=-1$

Hence, A is the correct option.