Question

Let a function $f:[0,\infty ]\to R$ be defined as follows.
$f\left(x\right)=\{\begin{array}{cc}|x-a|-1,& 0\le x<2\\ b(x-2{)}^2,& x\ge 2\end{array}$ where $a$ and $b$ are real numbers with $aϵ(0,2)$ and $b\ne 0$. Then

• A. There is exactly one value of $a$ which makes $f$ a continuous function
• B. The continuity of $f$ depends on the values of both $a$ and $b$
• C. $f$ is continuous if an only if $a+b=1$
• D. $f$ is not differentiable at exactly one point in its domain

Answer 1

The correct option is A There is exactly one value of $a$ which makes $f$ a continuous function

If the given function is continuous,

$\therefore$ it is also continuous at$2$

$⟹\underset{x⟶2}{lim}(|x-a|-1)=f\left(0\right)=b\left(0\right)=0$

As $0<a<2$

$⟹2-a>0$,

$⟹2-a=1⟹a=1$

$\therefore$ there is exactly one value of $a$.