Question

If $f\left(x\right)=\{\begin{array}{c}p{x}^2-q,xϵ[0,1)\\ x+1,xϵ(1,2]\end{array}$
and $f\left(1\right)=2$ then the value of the pair $(p,q)$ for which $f\left(x\right)$ cannot be continuous at $x=1$ is

• A. $(2,0)$
• B. $(1,-1)$
• C. $(4,2)$
• D. $(1,1)$

Answer 1

The correct option is D $(1,1)$

$f\left(x\right)$ is continuous at $x=1$ if

$\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}(1-h)=f\left(1\right)=2$

$\underset{h\to o^{+}}{lim}f(1+h)=\underset{h\to 0^{+}}{lim}(1+h+1)=2$

$\underset{h\to 0^{-}}{lim}f(p{(1-h)}^2-q)=p-q$

$\therefore f\left(x\right)$ is not continuous at $x=1$ if $p-q\ne 2$

Hence option (D) is true.