Q.Determine the values of $p$ and $q$ such that the following function is continuous on the entire real number line: $f(x) = x+1$, for $1 < x < 3$, $= x^2+px+q$, for $|x-2| \ge 1$.

$f(x)=x+1$ for $1<x<3$, and $f(x)=x^2+px+q$ for $|x-2|\ge1$, i.e. for $x\le1$ or $x\ge3$. Together these two pieces cover all of $\mathbb{R}$, with junctions at $x=1$ and $x=3$ (both included in the quadratic piece).

At $x=1$: $f(1)=1+p+q$ (quadratic piece, since $x=1$ satisfies $x\le1$). Right-hand limit: $\displaystyle\lim_{x\to1^+}(x+1)=2$. Continuity requires $1+p+q=2\Rightarrow p+q=1.\quad(i)$

At $x=3$: $f(3)=9+3p+q$ (quadratic piece, since $x=3$ satisfies $x\ge3$). Left-hand limit: $\displaystyle\lim_{x\to3^-}(x+1)=4$. Continuity requires $9+3p+q=4\Rightarrow3p+q=-5.\quad(ii)$ …

Identify that the two given pieces meet at exactly two junctions, $x=1$ and $x=3$ (since $|x-2|\ge1$ means $x\le1$ or $x\ge3$), set up one continuity equation at …

Missing that the domain condition $|x-2|\ge1$ actually gives two separate junctions to check, not one — treating this as a single-point …