Functions as a special kind of relation

Each element of (A) has exactly one image in (B). In exams, repetition of the second component does not make a function invalid.

In a function, one domain element cannot have two different images. In exams, check repeated first components carefully.

For a function, every element of (A) must have exactly one image. In exams, identify a missing domain element quickly.

Each of the (2) elements of (A) has (3) choices in (B), so total functions are (3^2=9). In exams, remember the formula (n(B)^{n(A)}).

Here (n(A)=3) and (n(B)=2), so total functions are (2^3=8). In exams, the base is the number of elements in the codomain.

Putting (x=1,2,3) gives images (2,4,6). In exams, the range is only the set of actually obtained images.

The number 3 is odd. According to the given rule, \(f(x)=1\) for every odd \(x\); therefore, \(f(3)=1\). The value 0 applies only when the input is even. In an exam, first check whether the input is even or odd, and then apply the definition.

Every first component has exactly one image and it is (x+1). In exams, identifying the rule from ordered pairs is useful.

In a function, one domain element cannot be associated with two different images. This is the most common exam mistake.

In option (A), every element (1,2,3) of (A) has exactly one image. In exams, match the list of first components with (A).

Substituting \(x=5\) into the given rule gives \(f(5)=5+2=7\). Hence, the correct value is 7. The value 5 is only the input, not the output of the function. Exam tip: To find a function value, substitute the given input directly into the function rule.

(g(0)=1), (g(1)=2), and (g(2)=5), so the range is ({1,2,5}). In exams, apply the rule to every domain element.

For each (x), (y=x^2) gives only one image. Two different (x) values having the same image does not invalidate a function.

For a function, every element of the domain \(A\) must have exactly one image. Here, the first component \(2\) appears in \((2,6)\) and \((2,7)\), so it has two different images and violates the function condition. In contrast, \(1\) and \(3\) each have exactly one image. Exam tip: check the first components of the ordered pairs; a repeated first component with different second components indicates a violation.

The range is the set of distinct obtained images, so it is ({1,2}). In exams, write repeated images only once.

There are (4) choices for each of the (2) elements of (A), so there are (4^2=16) functions. In exams, the exponent is the number of domain elements.

In a constant function, all elements have the same image, chosen from one element of (B). Therefore there are (2) constant functions.

The set of actually obtained images is called the range. In exams, the range is always a subset of the codomain.

The range is ({2,3,4}), so (5) is in the codomain but not in the range. In exams, keep codomain and range separate.

In option (D), (3) has two images (c) and (d), and (4) has no image. In exams, check both conditions.

In every ordered pair, the second component is (2) times the first. In exams, test the rule on two pairs and confirm with the rest.

Since \\(f(x)=x^3\\), we substitute 4 for x: \\(f(4)=4^3=4\times4\times4=64\\). Therefore, option B is correct. The value 16 is \\(4^2\\), not the required cube. Exam tip: apply the function rule directly to the given input.

(A\times B) has (3\times 2=6) elements, so the number of relations is (2^6=64). In exams, counting relations and functions is different.

The number of functions is (n(B)^{n(A)}=2^3=8). In exams, do not confuse it with the number of relations (2^{n(A)n(B)}).

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