\(\displaystyle f\) has an inverse on a given domain if and only if there are no two distinct values on the domain \(\displaystyle a,b\) such that \(\displaystyle f(a) = f(b)\).

\(\displaystyle f(x) = x^{2} - 12x - 17\) is a quadratic function, so its graph is a parabola. The key is to find the \(\displaystyle x\)-coordinate of the vertex of the parabola, which can be found by completing the square:

\(\displaystyle f(x) = x^{2} - 12x + \left ( \frac{- 12}{2} \right )^{2} - 17 - \left ( \frac{- 12}{2} \right )^{2}\)

\(\displaystyle f(x) = x^{2} - 12x +36 - 17 - 36\)

\(\displaystyle f(x) = \left (x-6 \right )^{2} - 53\)

The vertex occurs at \(\displaystyle x = 6\), so the interval which contains this value will have at least one pair \(\displaystyle a,b\) such that \(\displaystyle f(a) = f(b)\). The correct choice is \(\displaystyle x\in (5, 10)\)..

\(\displaystyle f\) has an inverse on a given domain if and only if there are no two distinct values on the domain \(\displaystyle a,b\) such that \(\displaystyle f(a) = f(b)\).

\(\displaystyle f(x) = \cos \frac{x}{2}\) has a sinusoidal wave as its graph, with period \(\displaystyle \frac{2 \pi }{\frac{1}{3}} =6 \pi\); it begins at a relative maximum of \(\displaystyle (0,1)\) and has a relative maximum or minimum every \(\displaystyle 3\pi\) units. Therefore, any interval containing an integer multiple of \(\displaystyle 3\pi\) will have at least two distinct values \(\displaystyle a,b\) such that \(\displaystyle f(a) = f(b)\).

The only interval among the choices that includes a multiple of \(\displaystyle 3 \pi\) is \(\displaystyle (9,12)\):

\(\displaystyle 3 \pi \approx 3 \cdot 3.14 \approx 9.42 \in (9,12)\) .

This is the correct choice.

We could solve this by actually adding up all terms the from 0 to 30, but it would take way too much time. There is a simple formula to remember for the summations of consecutive terms: \(\displaystyle \frac{n^{2}+n}{2}\) , which gives the sum of all terms from 0 to \(\displaystyle n\).

By substituting the value provided in our problem \(\displaystyle (30)\) into the formula, we can solve for the correct answer.

 \(\displaystyle \frac{30^{2}+30}{2}\)

 \(\displaystyle \frac{900+30}{2}\)

\(\displaystyle \frac{930}{2}\)

\(\displaystyle 465\)

We should notice that since we have a sequence of even numbers, we can factor \(\displaystyle 2\) out, so we can rewrite it as :

\(\displaystyle 2+4+6+...+60=2(1+2+3+... + 30)\)

We can calculate the summation of all numbers from 1 to \(\displaystyle n\) with the formula \(\displaystyle \frac{n^{2}+n}{2}\); so, we simply have to plug in 30 for \(\displaystyle n\) and multiply this formula by two:

\(\displaystyle 2(\frac{30^{n}+30}{2})= 930\)

The formula for the summation of consecutive terms is \(\displaystyle \frac{n^{2}+n}{2}\) , which gives the sum of all terms from 0 to \(\displaystyle n\). We can apply the formula to get the summation of all consecutive terms from 1 to 160. To figure out the summation starting from 120, we simply have to subtract the summation of all terms from 1 to 119. (We don't want to include 120 since we want it in our summation.)

\(\displaystyle \frac{160^{2}+160}{2}-\frac{119^{2}+119}{2} = 12880-7140= 5740\)

We can manipulate summations to make them easier to work with. Here, we are asked for the sum of odd terms from 1 to 59. We can calculate this by subtracting the summation of the even terms from the summation of all numbers from 1 to 59, using the formula \(\displaystyle \frac{n^{2}+n}{2}\) to sum all terms from 1 to \(\displaystyle n\).

In other words, we have to calculate \(\displaystyle \frac{59^{2}+59}{2}-\frac{2\cdot (29^{2}+29)}{2}\), since the even numbers are given by \(\displaystyle 2+4+6...+58\) or \(\displaystyle 2(1+2+3...+29)\).

We obtain the final answer 900.

By definition:

If \(\displaystyle x \geq 0\), then \(\displaystyle |x| = x\) ,and subsequently, \(\displaystyle f (x) = x - x = 0\)

If \(\displaystyle x < 0\), then \(\displaystyle |x| = -x\) ,and subsequently, \(\displaystyle f (x) = x- \left ( - x \right ) = 2x\)

\(\displaystyle \left (g \circ f \right ) (1) = g [f(1)]\)

As seen in the diagram below, the graph of \(\displaystyle f\) includes the point \(\displaystyle (1, -9)\).

Therefore, \(\displaystyle f(1) = -9\), and

\(\displaystyle \left (g \circ f \right ) (1) = g (-9)\).

\(\displaystyle g (x) = x^{2} - 3\), so 

\(\displaystyle g (-9) = (-9)^{2} - 3 = 81 - 3 = 78\).

Therefore, \(\displaystyle \left (g \circ f \right ) (1) = 78\), the correct choice.

\(\displaystyle \left (f \circ g \right ) (1) = f [g(1)]\)

\(\displaystyle g (x) = x^{2} - 3\), so

\(\displaystyle g (1) = 1^{2} - 3 = 1-3 = -2\), so

\(\displaystyle \left (f \circ g \right ) (1) = f (-2)\)

As seen in the diagram below, the graph of \(\displaystyle f\) includes the point \(\displaystyle (-2, -1)\).

Therefore, \(\displaystyle f(-2) = -1\), and \(\displaystyle \left (f \circ g \right ) (1) = -1\), the correct choice.

\(\displaystyle \left (g \circ f^{-1} \right ) (4) = g [f ^{-1} (4)]\)

\(\displaystyle f ^{-1} (4) = c\) such that \(\displaystyle f(c)= 4\).As seen in the diagram below, the graph of \(\displaystyle f\) includes the point \(\displaystyle \left (\frac{1}{2}, 4 \right )\), so \(\displaystyle f ^{-1} (4) = \frac{1}{2}\).

\(\displaystyle \left (g \circ f^{-1} \right ) (4) = g \left ( \frac{1}{2} \right )\)

\(\displaystyle g(x) = x^{2} - 11\), so

\(\displaystyle g\left ( \frac{1}{2} \right )=\left ( \frac{1}{2} \right )^{2} - 11 = \frac{1}{4} -11 = -10 \frac{3}{4}\)

\(\displaystyle \left (g \circ f^{-1} \right ) (4) =-10 \frac{3}{4}\), the correct choice.