Question

If $f\left(x\right)=\{\begin{array}{cc}ax+3,& x\le 2\\ a^{2}x-1,& x>2\end{array}$, then the values of $a$ for which $f$ is continuous for all $x$ are

• A. $1$ and $-2$
• B. $1$ and $2$
• C. $-1$ and $2$
• D. $-1$ and $-2$
• E. $0$ and $3$

Answer 1

The correct option is C $-1$ and $2$

Given, $f\left(x\right)=\{\begin{array}{cc}ax+3,& x\le 2\\ a^{2}x-1& x>2\end{array}$

Continuity at $x=2$,
$\text{LHL}=\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2}{lim}(ax+3)=2a+3$
$\text{RHL}=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}(a^{2}x-1)=2a^2-1$
Since, $f\left(x\right)$ is continuous for all values of $x$.
Therefore, $\text{LHL}=\text{RHL}$
$\Rightarrow 2a+3=2a^2-1$
$\Rightarrow 2a^2-2a-4=0$
$\Rightarrow a^2-a-2=0$
$\Rightarrow a^2-2a+a-2=0$
$\Rightarrow a(a-2)+1(a-2)=0$
$\Rightarrow (a+1)(a-2)=0$
$\Rightarrow a=-1,2$