Piecewise Functions - 1

Since both $x + 2$ and $x^2 + 5x + 6$ are continous for all real $x$ , the only point of discontinuity possible in $f$ is at $x = a$ .

To make $f$ continuous at $x =a$ , we must set

$\displaystyle\lim_{x\rightarrow a^{-} } f(x) = f(a)= \displaystyle\lim_{x \rightarrow a^{+}} f(x)$ $\displaystyle\lim_{x\rightarrow a^{-} } f(x) = a + 2, \displaystyle\lim_{x \rightarrow a^{+}} f(x) = a^2 + 5a + 6$

$\therefore a + 2 = a^2 + 5a + 6$

$\implies a^2 + 4a + 4 = 0$

$\implies (a + 2)^2 = 0$

$\implies \boxed{a = -2}$