Functions as a special kind of relation

Correct answer: A. (18)

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).

Correct answer: B. It is not a function because \(0\) has two images

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.

Correct answer: A. (\ a\ge0) / (a\ge0)

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.

Correct answer: A. It is a function and range is ({1,2,3,6})

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.

Correct answer: B. (\mathbb{R}-{4})

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

Correct answer: B. 1

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.

Correct answer: C. ({(1,b),(2,c),(3,a),(4,b),(3,c)})

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

Correct answer: A. ([2,4])

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

Correct answer: B. (1)

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

Correct answer: A. For every (a\in A), exactly one (b\in B) satisfies ((a,b)\in f)

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

Correct answer: A. Because (x=0) has two images

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

Correct answer: A. (|a|<2)

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

Correct answer: A. \(R\) is a function from \(A\) to \(B\), but it is not onto.

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.

Correct answer: A. (f^{-1}) is a function and (f^{-1}=f)

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

Correct answer: C. (f(n)=n^2+n+1)

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).

Correct answer: A. Because (x=1) has images (1) and (3)

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.

Correct answer: A. (-2)

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.

Correct answer: A. When each first component (1,2,3) appears exactly once

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

Correct answer: B. (2\sqrt{2})

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

Correct answer: B. (256) and (16)

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.

Correct answer: A. ({0,2,\frac{3}{2}})

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

Correct answer: A. It is a function and (f(1)=1)

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

Correct answer: D. (f(x)=\sqrt{x-5})

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.

Correct answer: C. (12)

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).

Correct answer: A. ([2,\infty))

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)).