Functions as a special kind of relation

The number 4 is even, so the given rule gives \(f(4)=0\). Option 1 is incorrect because it corresponds to an odd input, while 4 is even; 4 is the input, not the function value. Exam tip: first check whether the input is even or odd, and then apply the matching rule.

Substituting \(x=3\) into the given rule gives \(f(3)=3^2+1=9+1=10\). Hence, the correct answer is 10. Option 9 results from calculating only \(3^2\) and forgetting to add 1. Exam tip: substitute the input into the complete function rule before simplifying.

A function assigns exactly one image to every element of domain (A), so it will have (3) pairs. In exams, pairs in a function equal the number of domain elements.

The range is ({1,4,9}), so (16) is in the codomain but not in the range. In exams, keep codomain and range separate.

All domain elements have image (a), so it is a constant function. In exams, the same image does not make a function invalid.

(x=2) is in the domain but (f(2)) is not defined. In exams, find values that make the denominator zero.

After removing the value that made the denominator (0), every remaining (x) has one image. In exams, changing the domain can make a rule valid.

(3) has two different images (6) and (4). In exams, if one domain element has two images, the relation is not a function.

(A) has both odd and even elements, so both images (1) and (0) are obtained. In exams, put only obtained values in the range.

(A\times B) has (2\times 3=6) pairs, so the number of relations is (2^6=64). In exams, relations are counted as subsets.

For (2) domain elements, there are (3) codomain choices each, so total functions are (3^2=9). In exams, keep the counts of functions and relations separate.

In the reversed relation, each first component (2,4,6) has exactly one image. In exams, test the inverse relation separately by the function condition.

Substituting (x=0,1,2,3) gives images (1,2,3,4). In exams, apply the rule to every domain element while finding range.

Every element of (A) must have exactly one image in a function, but (2) is missing. In exams, compare first components with the whole domain.

Since \(2\le 2\) is true, the first condition applies, so \(f(2)=1\). The second condition, \(x>2\), does not apply to 2. Exam tip: \(\le\) includes equality, whereas \(<\) does not.

For \(x=2\), the condition \(2<2\) is false, while \(2\ge 2\) is true. Therefore, \(f(2)=2\). Remember that \(\ge\) includes equality, so the boundary point \(x=2\) belongs to the second case.

The direct answer is A: preimage. If f:A→B and f(a)=b, then a is the input that is sent to b, while b is the resulting output. The input a is therefore called the preimage of b, and b is called the image of a. Option A is correct because it names the element before the mapping. Option B, image, is the name for b, not a. Option C, codomain, means the set B into which the function maps all elements; it is a whole set, not the particular input a. Option D, range, means the set of actual output values produced by the function; it is also a set, not the single input a. For example, if f(2)=7, then 2 is the preimage of 7 and 7 is the image of 2. The arrow direction is important: input comes first and output comes second. Memory cue: preimage is the value you start from; image is the value you reach.

(b) is the image of (a) because applying (f) to (a) gives (b). In exams, understand (f(a)) as the output.

The domain is the set of first components, so it is ({1,2,3,4}). In exams, look at the first components of ordered pairs.

The range is the set of distinct second components, so it is ({3,5}). In exams, write a repeated value once in a set.

The range is ({4,9,16}), so (25) is not obtained. In exams, every codomain element need not be in the range.

Every element of \(A\) has exactly one image, so \(R\) is a function. Since both 1 and 3 map to \(a\), it is many-one. Element \(c\) has no preimage, so it is not onto. Exam tip: first check that each input has exactly one output.

The direct answer is A: {(1,4),(2,3),(3,2),(4,1)}. A function is represented by ordered pairs (x,f(x)). Use each domain value in turn and calculate f(x)=5−x. For x=1, f(1)=5−1=4, giving (1,4). For x=2, f(2)=3, giving (2,3). For x=3, f(3)=2, giving (3,2). For x=4, f(4)=1, giving (4,1). Therefore option A is exactly the graph of the function. Option B represents the identity function f(x)=x, not the given rule. Option C incorrectly gives output 5 for every input; that would be a constant function. Option D gives outputs 3,4,1,2, which do not follow 5−x. All outputs are in the stated codomain {1,2,3,4}. Exam cue: substitute every domain element carefully, then write pairs as (input, output), never (output, input).

To find the function value, substitute \(x=6\) in \(f(x)=2x+1\): \(f(6)=2(6)+1=12+1=13\). Option C does not follow the given rule, while option A omits the final addition of 1. Exam tip: substitute the input first, then perform multiplication and addition.

In a constant function, all domain elements map to one chosen element of (B). Therefore the number of constant functions is (4).