Functions as a special kind of relation

If the domain is (\mathbb{R}), every real (x) must have a real image. Negative (x) values make the rule invalid on the whole domain.

For a mapping f:A→B to be a function, every element of the stated domain A must have exactly one image in B. Here the claimed domain is all real numbers, but the rule f(x)=1/x contains division by x. Division by zero is undefined, so f(0)=1/0 has no real value. Consequently, the rule cannot define a function on all of R; its natural real domain is R extbackslash{0}. For every nonzero real x, the rule gives exactly one real output, so there is no problem at x=1 or x=−1. Option D is also false because no nonzero input has two images. Therefore option A correctly identifies the exceptional domain value and the precise reason the proposed mapping fails.

(0) is not in the domain, and (\frac{1}{x}) is defined for every remaining (x). In exams, changing the domain can change validity as a function.

From (y^2=1), both (y=1) and (y=-1) are obtained. In exams, if one first component has more than one image, it is not a function.

Repeating the same ordered pair does not create a different image for an element. In exams, look for different images, not mere repetition.

Both even and odd elements are present in (A), so both images (0) and (1) occur. In exams, check all types of domain elements.

In option (A), both (1) and (2) have exactly one image. In exams, a subset is a function only when the whole domain is covered.

The graph of a function contains ordered pairs of the form \((x,f(x))\). Here, \(f(1)=2\), \(f(2)=4\), and \(f(3)=6\), so \((1,2)\), \((2,4)\), and \((3,6)\) belong to the graph. For \((3,8)\), the rule would require \(f(3)=8\), but actually \(f(3)=6\). Exam tip: treat the first component as \(x\) and use the given rule to check the second component.

(a) is a preimage of (b) because applying (f) to (a) gives (b). In exams, do not reverse image and preimage.

All domain elements have image (a), so it is a constant function. In exams, identify a constant function when all images are the same.

For (x=1,2,3), the images are (1,2,3). In exams, an extra element of the codomain need not belong to the range.

The function rule is \(f(x)=x+1\). Substituting \(x=0\) gives \(f(0)=0+1=1\), so option B is correct. Option A ignores the \(+1\), while the other values do not follow the given rule. Exam tip: To find a function value, substitute the given input directly into the function rule.

((1,2)) is repeated, but (1) does not have another different image. In exams, distinguish a repeated pair from two different images.

A function must assign exactly one image to every element of its domain. Here, 1, 2, 3 and 4 are assigned the images 1, 2, 3 and 3, respectively, so the relation is a function. Different domain elements may have the same image; therefore, two pre-images of 3 are allowed. Also, a function need not have equally sized domain and codomain, nor must each codomain element have a distinct pre-image. Exam tip: to test a function, check that every input has exactly one output, not that every output has only one input.

In the reversed relation, each of (2,4,6) has exactly one image. In exams, the inverse relation is a function only when no second component repeats with different preimages.

In the reversed relation, (a) as a first component is associated with two different images. In exams, watch repeated second components when reversing.

All elements have image (0), so the range is ({0}). In exams, do not include an unattained codomain element in the range.

In option (A), different domain elements have different images. In exams, check repetition of images while testing one-one behavior.

The range is ({1,4,9}), and (16) is not obtained. In exams, first find the range and then compare it with the codomain.

A function must assign exactly one image to each of the (4) elements of (A), so (4) pairs are needed. In exams, the number of pairs in a function equals the number of domain elements.

Dividing (1,2,3,4) by (3) gives remainders (1,2,0,1), so the range is ({0,1,2}). In exams, write only distinct obtained images in the range.

(f(-1)=1), (f(0)=0), (f(1)=1), and (f(2)=4), so the range is ({0,1,4}). In exams, write repeated images only once.

In every pair, the second component is (1) more than the first. In exams, test one common pattern on all pairs.

Each element of (A) has (2) choices in (B), so total functions are (2^3=8). In exams, use (n(B)^{n(A)}).

(2) is associated with both (q) and (r), so it is not a function. In exams, check different images of a repeated first component.