Functions as a special kind of relation

Correct answer: A. 0

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.

Correct answer: C. 10

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.

Correct answer: A. (3)

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.

Correct answer: D. (16)

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

Correct answer: A. Constant function

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

Correct answer: A. Because the denominator becomes (0) at (x=2)

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

Correct answer: A. Because (x=2) is removed from the domain

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.

Correct answer: C. (3)

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

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

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

Correct answer: A. (2^6=64)

(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.

Correct answer: B. (3^2=9)

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.

Correct answer: A. Function

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.

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

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

Correct answer: A. Because (2) has no image

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.

Correct answer: A. 1

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.

Correct answer: B. 2

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.

Correct answer: A. Preimage

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.

Correct answer: A. Image

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

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

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.

Correct answer: B. ({3,5})

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

Correct answer: D. (25)

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

Correct answer: B. Many-one function

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.

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

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).

Correct answer: B. 13

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.

Correct answer: B. (4)

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