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Question 2 NAT · 3.0 marks
Published solution
**1. Find the given slope:** \(x-3y+1=0\) gives \(y=x/3+1/3\), so its slope is 1/3.
**2. Use perpendicularity:** A perpendicular line has slope -3. Through (1,1), its equation is \(y-1=-3(x-1)\), or \(y=-3x+4\).
**3. Find the intercept:** At (0,k), substitute x=0 to obtain \(k=4\).
**Answer: 4.**
**1. Factor the quadratic:** \(x^2-11x+18=(x-2)(x-9)\).
**2. Determine the sign:** The leading coefficient is positive, so the quadratic is negative strictly between its two roots. The largest open interval is \((a,b)=(2,9)\).
**3. Compute its length:** \(b-a=9-2=7\). The endpoints are excluded because the expression equals zero there.
**Answer: 7.**
**1. Apply Prim’s rule:** Choose the least-weight edge crossing from the current tree to a vertex outside it, rather than the globally smallest unused edge.
**2. Identify the current tree:** After DE=6 and EC=12, the included vertices are D,E,C. Edges leaving this set are DB=20, CB=23 and CA=30. CD=15 is internal and cannot add a new vertex.
**3. Choose the crossing minimum:** Edge DB with weight 20 is selected. AB=10 is not eligible yet because neither endpoint belongs to the current tree.
**Answer: 20.**
**1. Check logarithm domains:** Bases x and x-2 must be positive and not one. In particular x>2 and x cannot equal 3.
**2. Rewrite using common logarithms:** Let \(u=\log_x2=\ln2/\ln x\) and \(v=\log_{x-2}2=\ln2/\ln(x-2)\). Since \(\tfrac12\log_x4=u\) and \(\log_{x-2}8=3v\), the equation is \(u+v=3uv\).
**3. Solve:** Dividing by nonzero uv gives \(\log_2x+\log_2(x-2)=3\). Hence \(x(x-2)=8\), or \((x-4)(x+2)=0\). Only x=4 satisfies the domain, so the answer is 4.
**Answer: 4.**
**1. Define an independent set:** It contains vertices with no edges between any two of them. Number rows/columns of the given matrix 1 through 7.
**2. Exhibit five independent vertices:** The set \(\{1,3,5,6,7\}\) has zero adjacency entries between every pair. It therefore proves that the maximum size is at least five.
**3. Prove five is maximal:** The graph contains the disjoint edges (1,2) and (4,5). Any independent set must omit at least one endpoint from each edge, hence omit at least two distinct vertices out of seven. Thus its size is at most \(7-2=5\), matching the constructed set.
**Answer: 5.**
**1. Find the left limit:** The source’s left piece is \(\sin(ax)/x\). Write it as \(a\sin(ax)/(ax)\). Since \(\sin u/u\to1\), its limit as x approaches zero is a.
**2. Evaluate the right value:** The right piece is \(\ln((e+x)^{-b})\), so \(f(0)=\ln(e^{-b})=-b\).
**3. Enforce continuity:** Equality of the left limit and the value gives \(a=-b\). Both parameters are nonzero, so \(a/b=-1\). The individual values are not required.
**Answer: -1.**
The relation is reflexive and symmetric, but not transitive.
6406537368908.
The relation is reflexive and transitive, but not symmetric.
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The relation is symmetric and transitive, but not reflexive.
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The relation is an equivalence relation.
Published solution
**1. Check reflexivity:** Every subset A satisfies \(A\subseteq A\), so the relation is reflexive.
**2. Check transitivity:** If \(A\subseteq B\) and \(B\subseteq C\), then every element of A belongs to C. Thus \(A\subseteq C\).
**3. Disprove symmetry:** For example \(\{0\}\subseteq\{0,1\}\), but \(\{0,1\}\nsubseteq\{0\}\). Therefore it is reflexive and transitive, but not symmetric; it is not an equivalence relation.
**Answer: B — Selected: The relation is reflexive and transitive, but not symmetric.**
A: **Incorrect:** Symmetry fails, while transitivity holds.
B: **Correct:** These are exactly the three established properties.
C: **Incorrect:** It is reflexive and not symmetric, reversing this option.
D: **Incorrect:** Equivalence requires symmetry as well as reflexivity and transitivity.
**1. Test injectivity:** \(\sin0=\sin\pi=0\), although \(0\ne\pi\). Thus the function from the real numbers to the real numbers is not one-to-one.
**2. Test surjectivity:** Sine has range \([-1,1]\), so it never attains real numbers such as 2. Its stated codomain is all real numbers, so it is not onto.
**3. Select:** It is neither one-to-one nor onto. Restricting the domain or codomain could change these properties, but no such restriction is given.
**Answer: D — Selected: none of these.**
A: **Incorrect:** Distinct inputs can have the same sine.
B: **Incorrect:** Real numbers outside [-1,1] have no preimage.
C: **Incorrect:** The not-one-to-one part is true but the onto part is false.
D: **Correct:** None of the other descriptions holds for the specified domain/codomain.
The given data is insufficient to determine the truth values of the given
statements.
Published solution
**1. Interpret adjacency:** \(A[0,2]=0\) means there is no direct edge from 0 to 2, so S1 is false.
**2. Interpret Boolean powers:** A 1 in \(A^k[i,j]\), using Boolean multiplication, establishes a directed walk of length k from i to j. The given \(A^3[0,2]=1\) establishes reachability, despite no length-one or length-two walk.
**3. Conclude:** A walk yields a path after removing repeated-vertex cycles. Here no shorter walk is possible, so a three-edge path exists. S2 is true.
**Answer: D — Selected: S1 is false and S2 is true.**
A: **Incorrect:** S1 is contradicted by A[0,2]=0.
B: **Incorrect:** The cube entry establishes reachability, so S2 is not false.
C: **Incorrect:** Both stated truth values are reversed.
D: **Correct:** S1 is false and S2 is true.
E: **Incorrect:** The supplied entries determine both claims.
Which of the following statements is/are true about directed acyclic graphs (DAG)?
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Every non-empty finite DAG has a vertex with indegree 0.
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Every DAG can be uniquely topologically sorted.
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If a vertex with indegree equal to
is removed from a DAG along with the
edges that go out of it, then the resulting graph is also a DAG.
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The number of topological sorts for a graph with
vertices and no edges is
equal to
.
Published solution
**1. Prove existence of a source:** If every vertex in a finite nonempty directed graph had an incoming edge, repeatedly following predecessors would eventually repeat a vertex and create a cycle. A DAG therefore has a vertex of indegree zero.
**2. Consider deletion:** Removing a vertex and its edges cannot create a new cycle. The remaining graph is still acyclic.
**3. Count possible orders:** A DAG may have several topological orders. With three vertices and no edges, every permutation is valid, giving \(3!=6\), not 3.
**Answer: A, C.**
A: **Correct:** Every finite nonempty DAG has at least one source vertex.
B: **Incorrect:** Two unrelated vertices can appear in either order, so uniqueness is not guaranteed.
C: **Correct:** Vertex/edge deletion preserves acyclicity.
D: **Incorrect:** Three unconstrained vertices admit six permutations.
**1. Find stationary points:** For \(f(x)=x^3-12x\), \(f'(x)=3x^2-12=3(x-2)(x+2)\). Thus stationary points occur at -2 and 2.
**2. Classify with curvature:** \(f''(x)=6x\). At -2 it is -12, giving a local maximum. At 2 it is 12, giving a local minimum.
**3. Match the locations:** Select the maximum at -2 and minimum at 2. The corresponding function values are 16 and -16.
**Answer: A, D.**
A: **Correct:** Negative second derivative at -2 establishes a local maximum.
B: **Incorrect:** -2 is a local maximum, not a minimum.
C: **Incorrect:** 2 is a local minimum, not a maximum.
D: **Correct:** Positive second derivative at 2 establishes a local minimum.
**1. Use the leading term:** For a polynomial of degree n with nonzero leading coefficient, its end behavior agrees with \(a_nx^n\).
**2. Take x toward negative infinity:** An odd power tends to negative infinity; an even power tends to positive infinity.
**3. Apply the coefficient sign:** A positive leading coefficient with odd degree gives negative infinity. A negative leading coefficient with even degree also gives negative infinity. The other two sign/parity combinations give positive infinity.
**Answer: A, D.**
A: **Correct:** Positive coefficient with degree three gives negative infinity on the left.
B: **Incorrect:** Negative coefficient times an odd negative power gives positive infinity.
C: **Incorrect:** Positive coefficient and even degree give positive infinity.
D: **Correct:** Negative coefficient and even degree give negative infinity.
Find the maximum speed attained by the particle in the time interval
.
Published solution
**1. Check the first time interval:** For \(0\le t<1\), speed \(v(t)=t^2\) lies between 0 and 1.
**2. Check the second piece:** For \(1\le t\le1+\pi\), \(v(t)=1+\sin(t-1)\). The sine argument ranges from 0 to pi, with maximum 1 at \(t-1=\pi/2\).
**3. Compare:** The maximum speed is \(1+1=2\), attained at \(t=1+\pi/2\).
**Answer: 2.**
Find the distance traveled by the particle in the time interval
and enter the answer
rounded to two decimal places
(Note that the distance traveled is the integral of the speed function.)
Published solution
**1. Integrate speed piecewise:** Speed is nonnegative, so total distance is
\[d=\int_0^1t^2\,dt+\int_1^{1+\pi}[1+\sin(t-1)]\,dt.\]
**2. Evaluate:** The first integral is 1/3. In the second, put u=t-1 to obtain \(\int_0^\pi(1+\sin u)du=\pi+2\).
**3. Add and round:** \(d=1/3+\pi+2=\pi+7/3\approx5.474926\), giving 5.47 distance units to two decimals. Distance units follow the units used by the supplied speed/time model.
**Answer: 5.47.**
