ISRO CSE 2014 PYQ

For an ideal R-bit unipolar DAC with codes 0 through $2^R-1$, the maximum code produces V. Therefore one LSB has weight V/($2^R-1)$. The most significant bit has binary weight $2^{R-1}$ LSBs, so its contribution is $2^{R-1$}$\times V/(2^R-1)$. This assumes V is the full-scale output at the maximum digital code.

The original statements execute in order: first y becomes y+1; then z is assigned the current x plus the incremented y; finally x becomes x+1. The expression $z = (x++) + (++y) has$ the same effect: x++ contributes the old x and increments x, while ++y increments y before contributing its new value. The two increments affect different objects, so this expression does not have the unsequenced-modification problem that would arise from modifying the same scalar twice without sequencing.

The pixel coordinates run from 0 through 100 inclusive, so each row contains 101 pixels. In row-major order with zero-based addressing, address(row,col) = row$\times number_of_columns + col$. Thus address(6,10) = 6$\times 101+10 = 616$. One bit per pixel affects storage size but not this element index.

In a singly linked list, deleting the first node is O(1), and swapping or interchanging the first two nodes is also O(1). Appending at the end is O(1) when a valid tail pointer L is maintained. Deleting the last node requires finding its predecessor, and a singly linked list has no backward link; traversal from the head takes O(n).

A function from B into A assigns exactly one element of A to each element of B. There are y elements in B, and each independently has x choices in A. By the multiplication principle, the number of functions is x multiplied by itself y times, or $x^y$. This counts all functions, not only one-to-one or onto functions.