Find the number a that makes $f(x)$ continuous everywhere?

Question

$$ f(x)= \begin{cases} ax^2-3 & \text{if $x\leqslant 2$,} \\ 2ax+3 & \text{if $x > 2$.} \end{cases} $$ $$\lim_{x\to 2^-} ax^2-3 = 4a-3$$ $$\lim_{x\to 2^+} 2ax+3 =4a+3$$ $$ \begin{align*} \text{limit RHS} &= \text{limit LHS} \\ 4a+3 &= 4a-3 \\ 0 &=6 \end{align*} $$ What mistake have i done here?

Answer 1

You have done no mistake. On computing,$$LHL=RHL=f(2)$$ $$\implies0=6$$ which is impossible and hence there is no such $'a'$ exists or $$a\in\phi$$