Functions as a special kind of relation

Each of the five elements of A can independently be mapped to either 0 or 1, so the total number of functions is 2⁵ = 32. We need at least two inputs mapped to 1. It is easier to subtract the unwanted cases: no input maps to 1, giving exactly 1 function, and exactly one input maps to 1, giving C(5,1) = 5 functions. Thus the required count is 32 − 1 − 5 = 26. Option D counts every function and therefore includes cases with zero or one occurrence of 1. Options A and C do not follow the complementary-count calculation.

The Cartesian product A × B contains |A||B| = 2 × 3 = 6 ordered pairs. Therefore it has 2⁶ = 64 subsets in total. A function from A to B must contain exactly one ordered pair beginning with each element of A. For each of the two inputs, there are three possible outputs, so the number of functions is 3² = 9. Every such function is a subset of A × B, and all remaining subsets fail the function condition because an input is missing, has multiple outputs, or both. Hence the number of non-function subsets is 64 − 9 = 55. Option D is the total number of subsets, not the requested number.

To test whether a relation from X to Y is a function, every element of the domain X must have exactly one image in Y. Use y² = x − 1. For x = 2, y² = 1, so y = 1 or y = −1; both ordered pairs (2,1) and (2,−1) belong to R. Thus the single input 2 has two images, which violates the defining uniqueness condition for a function. For completeness, x = 5 gives y = 2 or −2, and x = 10 gives y = 3 or −3, so each input actually has two images. Therefore option B states a valid and sufficient reason. Option C is false because 10 does have images, while A and D incorrectly call the relation a function. The relation is many-valued, not a function.

A function from A to B assigns one image in B to each of the four elements of A. Since the range must be exactly {1, 3, 4}, the value 2 cannot be used, while each of 1, 3, and 4 must occur at least once. First count all assignments using only these three values: 3^4 = 81. Subtract assignments omitting one specified value: for each value there are 2^4 assignments, giving 3 × 16 = 48. Add back the three assignments that use only one value, because they were subtracted twice. Thus the required number is 81 − 48 + 3 = 36. Therefore option C is correct; 24 does not account for all valid surjections, 64 counts a different unrestricted situation, and 18 is too small.

The function is not one-one because \(f(-1)=2\) and \(f(2)=2\), although \(-1\ne2\). However, it is onto: for this continuous cubic polynomial, \(f(x)\to-\infty\) as \(x\to-\infty\) and \(f(x)\to\infty\) as \(x\to\infty\). Hence, it takes every real value. Therefore, option B is correct. Exam tip: To test one-one-ness, look for distinct inputs with the same output; for onto-ness, use continuity and end behaviour.

For \(a>b\), both \(a^3>b^3\) and \(a>b\), so \(a^3+a>b^3+b\); hence the function is one-one. As this cubic takes values from \(-\infty\) to \(+\infty\), it is onto \(\mathbb{R}\) too. Exam tip: always check the codomain.