Functions as a special kind of relation

Having distinct images for all five elements means that the function must be injective. There are 6 choices for the image of the first element, followed by 5, 4, 3 and 2 choices. Hence, the total number of functions is \(6\times5\times4\times3\times2={}^6P_5=720\). Since the assignment of images to the elements matters, use a permutation rather than a combination.

In option (B), every input has one image, so (R) is a function. In the inverse relation, (5) has two images (1) and (2).

The values look like (5,4,3,2,1,1), but (5) is not in the codomain, so it is not a valid function. The correct check must first test codomain membership.

The direct answer is A: the correct domain is \(\mathbb{R}\setminus\{-2,3\}\). A rational expression is defined only when its denominator is non-zero. Here the denominator is |2x−1|−5, so require |2x−1|−5≠0, or |2x−1|≠5. An absolute value equals 5 when 2x−1=5 or 2x−1=−5. These give 2x=6 and x=3, or 2x=−4 and x=−2. Remove both numbers from all real numbers. Option A is correct. Option B is wrong because 5 is not the excluded value; substituting x=5 gives denominator |9|−5=4, which is non-zero. Option C is wrong because x=1/2 gives denominator |0|−5=−5, also non-zero. Option D is wrong because many numbers greater than or equal to 3 are valid, such as x=4. Memory cue: for a fraction, set only the denominator equal to zero and exclude every resulting x.

A function \(f:A\to B\) assigns each element of \(A\) either to image \(1\) or to image \(0\). We want at least 3 elements mapping to \(1\). If exactly \(k\) elements map to \(1\), then the number of choices is \(\binom{6}{k}\). So the total is
\[\binom{6}{3}+\binom{6}{4}+\binom{6}{5}+\binom{6}{6}
=20+15+6+1=42.\]

The domain includes (-2), but (\frac{3}{-2+2}) is undefined. A function needs a value at every domain element.

There are (2^{12}=4096) total subsets and (4^3=64) functions. Thus the non-function subsets are (4096-64=4032).

To find the range, first rewrite the expression under the square root by completing the square: \\(x^2-10x+29=(x-5)^2+4\\). Since a square is always non-negative, the smallest possible value of the expression inside the root is 4, achieved at x=5.

The square root is therefore at least \\(\sqrt4=2\\). As x moves away from 5, \\( (x-5)^2\\) can take every non-negative value and the function increases without bound. Hence every value from 2 onward occurs, so the range is \\([2,\infty)\\). Option A is correct; 0 is not possible because the radicand never falls below 4.

An absolute value expression assigns one non-negative value to each input. In a function test, check uniqueness from input to output.

For odd (x), (7-x) gives even values, and for even (x), (x) itself is obtained. Hence the obtained values are only ({2,4,6}).

The graph of a function has exactly one ordered pair for each element of the domain. Therefore the number of pairs is (|A|=7).

The expression (\sqrt{x+4}) gives the principal non-negative square root, so each given (x) has one image. The obtained values are ({0,1,2,3}).

The input (x=2) belongs to both parts and gives different values (6) and (3). Two different outputs for one input do not define a function.

At (x=2), (4\cdot2-5=3) and (2+1=3), so there is no conflict. Overlap is valid when both values agree.

First choose (2) values from (B) in (\binom{3}{2}=3) ways, and both chosen values must occur, so each pair gives (2^4-2=14). The total is (3\cdot14=42), which is not in the options.

Here (x^2+2x+2=(x+1)^2+1) and (f(x)=1+\frac{3}{(x+1)^2+1}). The maximum is (\frac{5}{2}), and (1) is never attained.

The function is not one-one because different inputs can produce the same output. For example, \(f(0)=|2|-6=-4\) and \(f\left(-\frac{4}{3}\right)=|-2|-6=-4\). Also, since \(|3x+2|\ge 0\), we have \(f(x)\ge -6\); hence values in \((-\infty,-6)\) are never obtained, so the function is not onto \(\mathbb{R}\). Therefore, option C is correct. Option A is wrong because the function is not one-one, while option B is wrong because it is not onto. Exam tip: For an absolute-value function, test one-one behaviour using symmetric inputs and test onto behaviour by finding its range.

Completing the square gives \(f(x)=x^2-4x+7=(x-2)^2+3\). Since \((x-2)^2\geq 0\), the function has a minimum value of \(3\), attained at \(x=2\). Therefore, its range is \([3,\infty)\). Option C incorrectly treats the constant term 7 as the minimum value, ignoring the effect of the linear term. Exam tip: Complete the square to determine the vertex and range of a quadratic function.