Official PYQ · 2024Set 1 · Q38Programming in CMSQ2 marks

GATE CS 2024 Set 1 Q38 — Programming in C

Consider the following $\mathrm{C}$ function definition.

int f (int x, int y){
for (int i=0 ; i<y ; i++ ) {
x= x + x + y;
}
return x;
}

Which of the following statements is/are TRUE about the above function?

View worked solution

Answer: B;D

To find the value returned by the function f(x, y), let us trace the value of $x$ after each iteration of the loop.

Let $x_i$ denote the value of $x$ after the $i$ -th iteration of the loop, with $x_0 = x$ being the initial value.

The loop runs for $i = 0$ to $y-1$ , which is exactly $y$ iterations. The recurrence relation is:

x_i = 2x_{i-1} + y

Let us expand this recurrence step-by-step:

\begin{aligned} x_1 &= 2x_0 + y \\ x_2 &= 2(2x_0 + y) + y = 2^2 x_0 + (2 + 1)y \\ x_3 &= 2(2^2 x_0 + 3y) + y = 2^3 x_0 + (4 + 2 + 1)y \\ &\vdots \\ x_k &= 2^k x_0 + (2^{k-1} + 2^{k-2} + \dots + 1)y \\ &= 2^k x_0 + (2^k - 1)y \end{aligned}

Since the loop runs for $y$ iterations, the final value returned is $x_y$ :

x_y = 2^y x + (2^y - 1)y = 2^y(x + y) - y

Now, let us evaluate each option using this closed-form expression:

Options A & C: $x = 20, y = 10$ $x_{10} = 2^{10}(20 + 10) - 10 = 30 \cdot 2^{10} - 10 = 30 \cdot 1024 - 10 = 30710$
- Since $30710 < 2^{20}$ (where $2^{20} = 1,048,576$ ), Option A is FALSE.
- Since $30710 > 2^{10}$ (where $2^{10} = 1024$ ), Option C is FALSE.
Option B: $x = 20, y = 20$ $x_{20} = 2^{20}(20 + 20) - 20 = 40 \cdot 2^{20} - 20$ Since $40 \cdot 2^{20} - 20 > 2^{20}$ , Option B is TRUE.
Option D: $x = 10, y = 20$ $x_{20} = 2^{20}(10 + 20) - 20 = 30 \cdot 2^{20} - 20$ Since $30 \cdot 2^{20} - 20 > 2^{20}$ , Option D is TRUE.

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Published 10 Oct 2026

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