Functions as a special kind of relation

In the reversed relation, (p) as a first component is linked to two different images. In exams, watch repeated second components while reversing.

The domain is (\mathbb{R}), but no real image is obtained for (x<1). In exams, check that the expression inside a square root is not negative.

(x=-3) is in the domain but (f(-3)) is not defined. In exams, find values that make the denominator (0).

Now (\frac{1}{x+3}) is defined for every domain element and gives one value. In exams, changing the domain can make a rule valid.

Since \(3\geq 3\), the condition \(x\geq 3\) applies, so \(h(3)=1\). The value 0 is assigned only when \(x<3\), not when \(x=3\). Exam tip: Pay close attention to whether the boundary value is included by symbols such as \(<\) and \(\geq\).

The remainders of (2,3,4,5) are (2,0,1,2), so the range is ({0,1,2}). In exams, write only distinct images.

The image of (3) is (4), which is not in the codomain. In exams, check whether every output lies in the codomain.

For a function f:A→B, every input in the domain A must have exactly one output, and every output must belong to the stated codomain B. Evaluate the rule on all elements of A: f(1)=1²+1=2, f(2)=2²+1=5, and f(3)=3²+1=10. Therefore the image set is {2,5,10}. This set can serve as the codomain, so option A is correct. The set {1,4,9} contains the squares before adding 1, not the actual outputs. The set {2,4,6} misses 5 and 10, while {1,2,3} does not contain all outputs. A larger codomain could also work mathematically, but among the given choices only option A contains exactly the required images.

For every integer (x), (x^2) gives one definite integer. Having the same image does not break the function condition.

Two different elements have the same image, so it is a many-one type function. In exams, the same image is allowed.

In option (A), different domain elements have different images. In exams, images should not repeat for a one-one function.

Every domain element has image (7), so the range is ({7}). In exams, do not write an unattained codomain element in the range.

In the notation f(a)=b, the function takes the input a and produces the output b. The input a is therefore called a preimage, or inverse image, of b under f. The output b is called the image of a. This distinction is based on the direction of the mapping: a→b. Hence option A is correct. Image and preimage are not interchangeable terms. The range is the set of all actual outputs produced by the function from its domain, so it is a set rather than the particular input a. The codomain is the declared target set in which outputs are required to lie; it may contain elements that are not actually attained. Thus neither range nor codomain names a in the equation f(a)=b.

In \(f(5)=12\), 5 is the input and 12 is the output of the function. Therefore, 12 is the image of 5. The preimage is 5, not 12, while the domain is the set of all possible inputs. Exam tip: In \(f(a)=b\), \(a\) is the preimage and \(b\) is the image.

The images are (2,4,6), so (8) is not obtained. In exams, keep the difference between codomain and range clear.

Each domain element can map to one of the (2) elements of (B). In exams, multiply these choices to count total functions.

Substituting \(x=4\) into the rule gives \(f(4)=2(4)-1=8-1=7\). Therefore, option C is correct. Option B results from forgetting to subtract 1 at the end. Exam tip: substitute the given input into the function rule and follow the order of operations carefully.

For an ordered pair \((x,y)\) to belong to the function \(f\), it must satisfy \(y=f(x)=x^2\). Thus, \(f(0)=0\), \(f(1)=1\), and \(f(2)=4\), so options A, B, and C are valid pairs. However, \(f(3)=3^2=9\), not 6; therefore, \((3,6)\) does not belong to \(f\). Exam tip: compare the second component of each pair with the square of its first component.

A function can be represented by ordered pairs (x,f(x)), where the first component is the input from the domain and the second component is its corresponding output in the codomain. Apply f(x)=x³ to each domain element: f(1)=1³=1, f(2)=2³=8, and f(3)=3³=27. Consequently, the function is the set {(1,1),(2,8),(3,27)}, which is option A. Option B uses an unrelated pattern, while option C reverses the input-output roles in the latter pairs. Option D also does not follow the cube rule. Every first component in the correct set belongs to the domain {1,2,3}, and every second component belongs to the codomain {1,8,27}, confirming the representation.

The range is ({2,3,4}), so (5) is not obtained. In exams, every element of the codomain need not be in the range.

In option (A), (1) is associated with both (a) and (b). In exams, look for the same first component with two different second components.

In option (A), there is no pair for (4). In exams, every domain element must be included for a function.

The number 4 is not prime because it has a factor other than 1 and itself, namely 2. Therefore, by the given rule, \(f(4)=0\). Option 1 would apply only if 4 were prime. In an exam, first check whether the input is prime or non-prime.

(2) and (3) give (1), while (1) and (4) give (0). In exams, write distinct obtained values in the range.

A function is a relation in which every domain element is associated with exactly one image. In exams, remember that every relation is not a function.