GATE CS 2014 Set 2 Q40 — Recursion

The given C function implements the Newton-Raphson method to find the square root of $3$.

The base condition of the recursion is:

$$\lvert x^2 - 3 \rvert < 0.01$$

For $f(q)$ to return $q$ directly without entering any further recursive steps, the base condition must be satisfied for $x = q$. That is:

$$\lvert q^2 - 3 \rvert < 0.01$$

Solving this inequality:

$$-0.01 < q^2 - 3 < 0.01$$

$$2.99 < q^2 < 3.01$$

Taking the positive square root:

$$\sqrt{2.99} < q < \sqrt{3.01}$$

$$1.7291 < q < 1.7349$$

Thus, the only value of $q$ rounded to $2$ decimal places that satisfies this condition is $1.73$ (or $-1.73$ if considering negative roots).

The official GATE answer key accepts any value in the range $1.72$ to $1.74$.