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If $f(x)=\left\{\begin{array}{ll}a x+3, & x \leq 2 \\ a^{2} x-1 & x>2\end{array}\right.$, then the values of $a$ for which $f$ is continuous for all $x$ are
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The correct answer is:
$-1$ and 2
Given, $f(x)=\left\{\begin{array}{cc}a x+3, & x \leq 2 \\ a^{2} x-1, & x>2\end{array}\right.$
Continuity at $x=2$,
$$
\begin{aligned}
\mathrm{LHL} &=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(a x+3)=2 a+3 \\
\text { RHL } &=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}\left(a^{2} x-1\right)=2 a^{2}-1
\end{aligned}
$$
Since, $f(x)$ is continuous for all values of $x$.
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow \quad 2 a+3=2 a^{2}-1 \Rightarrow \quad 2 a^{2}-2 a-4=0$
$\Rightarrow \quad a^{2}-a-2=0 \quad \Rightarrow a^{2}-2 a+a-2=0$
$\Rightarrow \quad a(a-2)+1(a-2)=0 \Rightarrow(a+1)(a-2)=0$
$\therefore \quad a=-1,2$
Continuity at $x=2$,
$$
\begin{aligned}
\mathrm{LHL} &=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(a x+3)=2 a+3 \\
\text { RHL } &=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}\left(a^{2} x-1\right)=2 a^{2}-1
\end{aligned}
$$
Since, $f(x)$ is continuous for all values of $x$.
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow \quad 2 a+3=2 a^{2}-1 \Rightarrow \quad 2 a^{2}-2 a-4=0$
$\Rightarrow \quad a^{2}-a-2=0 \quad \Rightarrow a^{2}-2 a+a-2=0$
$\Rightarrow \quad a(a-2)+1(a-2)=0 \Rightarrow(a+1)(a-2)=0$
$\therefore \quad a=-1,2$
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