Functions as a special kind of relation

Correct answer: A. Because (\sqrt{x}) is not real for negative (x)

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.

Correct answer: A. 1/x is not defined at x = 0

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.

Correct answer: A. Because (0) is removed from the domain

(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.

Correct answer: A. Because (x=1) has two images (y=1) and (y=-1)

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.

Correct answer: A. It is a function because the repeated pair does not give a new different image

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

Correct answer: C. ({0,1})

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

Correct answer: A. ({(1,3),(2,4)})

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.

Correct answer: D. \((3,8)\)

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.

Correct answer: B. Preimage

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

Correct answer: A. Constant function

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

Correct answer: A. ({1,2,3})

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.

Correct answer: B. 1

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.

Correct answer: A. Because (1) does not get a different image

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

Correct answer: A. It is a function.

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.

Correct answer: A. Yes because every element of (B) has exactly one image

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.

Correct answer: A. Because (a) has two images (1) and (2)

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

Correct answer: A. ({0})

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

Correct answer: A. ({(1,4),(2,5),(3,6)})

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

Correct answer: D. (16)

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

Correct answer: B. (4)

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.

Correct answer: C. ({0,1,2})

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.

Correct answer: A. ({0,1,4})

(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.

Correct answer: A. (f(x)=x+1)

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

Correct answer: A. (2^3=8)

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

Correct answer: A. Because (2) has two different images

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