GATE CS 2000 Q2.4 — Sets, Relations and Functions

We are given $5$ distinct data points for the polynomial $p(x)$:

• $(1, 1)$
• $(2, -1)$
• $(3, 1)$
• $(4, -1)$
• $(5, 1)$

By polynomial interpolation, any $n$ distinct points can always be fitted by a unique polynomial of degree at most $n-1$. With $5$ points, there exists a unique polynomial of degree at most $4$.

To see if a lower degree is possible, suppose the degree of $p(x)$ is at most $3$. Since $p(1) = p(3) = p(5) = 1$, the polynomial $p(x) - 1$ must have roots at $x = 1, 3, 5$. Thus, we can write:

for some constant $c$. Now, we substitute the remaining conditions:

• For $x = 2$: $p(2) - 1 = c(1)(-1)(-3) \implies -2 = 3c \implies c = -\frac{2}{3}$
• For $x = 4$: $p(4) - 1 = c(3)(1)(-1) \implies -2 = -3c \implies c = \frac{2}{3}$

Since we obtain two conflicting values for $c$, no polynomial of degree $\le 3$ can satisfy all the given conditions. Thus, the minimum degree of $p(x)$ is $4$.