Functions as a special kind of relation

There are (3) choices for (f(1)=f(2)) and (3\cdot2) choices for (f(3)\ne f(4)). Total is (3\cdot3\cdot2=18).

For \(x=0\), the equation gives \(y^2=1\), so both \(y=1\) and \(y=-1\) are obtained. Thus, the relation contains both \((0,1)\) and \((0,-1)\). A function must assign exactly one image to each element of its domain, so this relation is not a function. Likewise, \(x=3\) gives \(y=\pm2\), and \(x=8\) gives \(y=\pm3\). Exam tip: to test a relation for being a function, check whether any one domain element is paired with two different elements of the codomain.

The expression (ax^2+4) stays non-negative for all (x\in\mathbb{R}) only when (a\ge0). In square-root functions the radicand must be non-negative.

For each (x), (y=\frac{6}{x}) is unique and lies in the given set. In finite sets check the image of every input separately.

On the given domain (f(x)=x+2), but removing (x=2) removes the value (4). Check how a removed point affects the range.

The values \(f(1),f(2),f(3)\) must be in strictly increasing order and must belong to \(B\). Since \(B\) has exactly the three elements \(0,1,2\), the only possible ordered triple is \((0,1,2)\). Thus \(f(1)=0, f(2)=1, f(3)=2\), giving exactly one function. Exam tip: when all elements of a three-element codomain must appear in strictly increasing order, their assignment is forced.

In option (C), (3) has two images (a) and (c). A function requires both existence and uniqueness.

For (x\in[1,3]), the value is (2), and the maximum (4) occurs at (x=-1). Open moduli in intervals to find range.

From an empty domain to an empty codomain, there is one empty function. Remember this as the function-context meaning of (0^0=1).

A function is a special relation where each input has exactly one image. It is not necessary to use every element of the codomain.

At (x=0), both (y=1) and (y=-1) are possible. Circle-like relations often break uniqueness in function tests.

The denominator must never be zero, so the discriminant of (x^2+ax+1) must satisfy (a^2-4<0). Hence (|a|<2).

Each element of \(A\)—1, 2 and 3—has exactly one image, namely \(a,b,a\), so \(R\) is a function. However, \(c\) has no preimage, so it is not onto. Exam tip: first check that every input has exactly one output; repeated outputs are allowed.

This function pairs values uniquely and the same rule reverses them. Hence the inverse relation is also a function.

For every (n\in\mathbb{N}), (n^2+n+1) is a natural number. The other rules do not give natural values for all natural (n).

A relation is a function only when every input has exactly one output. Here, a pair (x,y) belongs to R when x+y is even. For x=1, the sum is even when y is odd. Both y=1 and y=3 are in B, and 1+1=2 and 1+3=4 are even. Thus the same input x=1 is related to two different outputs, 1 and 3.

This violates the defining rule of a function, which permits only one image for each input. Therefore option A is correct. Notice that x=2 does have images, namely 2, and the set B is not empty. The fact that A and B are equal also has no bearing on whether the relation is a function; the problem is specifically the multiple outputs for x=1.

On the given domain (f(x)=x-1), and removing (x=-1) removes the value (-2). After simplification remove the output of the excluded input.

With exactly (3) pairs, a function needs one image for each of the three elements of (A). Repetition of second components is allowed.

By symmetry the maximum occurs at (x=2) and equals (2\sqrt{2}). In such questions check the balanced point.

The number of relations is (2^{|A||B|}=2^8=256), and the number of functions is (|B|^{|A|}=4^2=16). Remember the two formulas separately.

At the given inputs the values are (0,2,\frac{3}{2}). For a finite domain, direct substitution is the fastest method.

The point (x=1) belongs only to the second part, and (f(1)=2\cdot1-1=1). Read boundary symbols carefully in piecewise functions.

The expression (\sqrt{x-5}) is real only when (x\ge5). A function on all of (\mathbb{R}) needs a value for every real input.

There are (2^4=16) total functions, and the opposite case (f(1)=b) and (f(2)=a) gives (2^2=4) functions. Hence (16-4=12).

For (0\le x\le2), the value is (2), and it increases outside this interval. Since the minimum is (2), the range is ([2,\infty)).