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Question 2 MCQ · 5.0 marks
Consider the function
defined as follows.
Choose the correct option.
6406536925289.
.
6406536925290.
is not defined for any non-zero
.
6406536925291.
.
6406536925292.
exists for any non-zero
.
Published solution
**1. Set up:** Away from the origin, r=sqrt(x²+y²) is positive and sin(r)/r is a composition of continuous functions.
**2. Work through the reasoning:** Therefore a finite limit exists at every (a,0) with a≠0, equal to sin(|a|)/|a|. At (0,pi/2) it is 2/pi, not zero, and at (pi,0) it is zero, not one.
**3. Check the result:**
\[\lim_{(x,y)\to(a,0)}f(x,y)=\frac{\sin|a|}{|a|}\quad(a\ne0)\]
The assigned special value at the origin does not affect continuity at these nonzero points.
**Answer: D.**
A: **Incorrect:** At radius pi/2 the value is 2/pi, not zero.
B: **Incorrect:** The denominator is nonzero at every such point, so the limit exists.
C: **Incorrect:** sin(pi)/pi is zero.
D: **Correct:** Continuity away from radius zero guarantees these limits.
, which of the following sequences of row
operations can be performed on
so that the resulting matrix has determinant
?
6406536925297.
Interchange two rows of
and then multiply one row by
6406536925298.
Multiply one row of
by
.
6406536925299.
Interchange two rows of
and then multiply one row by
.
6406536925300.
Multiply all rows of
by
.
Published solution
**1. Set up:** Starting from det(A)=-3, interchanging two rows changes its sign to +3.
**2. Work through the reasoning:** Multiplying one row by 1/3 then gives determinant 1.
**3. Check the result:**
\[\det A=-3\longmapsto3\longmapsto\frac13(3)=1\]
Scaling every row would multiply the determinant by the fourth power of the scale because the matrix has order four; that is different from scaling one row.
**Answer: C.**
A: **Incorrect:** After the swap the determinant is 3, and scaling by -1/3 gives -1.
B: **Incorrect:** Scaling -3 by 1/3 gives -1 without a sign-changing swap.
C: **Correct:** The swap and one-row scaling give (-1)(1/3)(-3)=1.
D: **Incorrect:** All-row scaling gives (-1/3)^4(-3)=-1/27, not 1.
**1. Set up:** Convert the representation back to standard coordinates.
**2. Work through the reasoning:** For input (x,y,z), its B1 coordinates are a=(x-y+z)/2,b=(x+y-z)/2,c=(-x+y+z)/2. The displayed matrix gives B2 coordinates (-b+2c,-a+2b). Recombining with B2=((1,1),(1,-1)), the standard output is their sum and difference: (-x+2y,-2x-y+3z). This is B.
**3. Check the result:**
\[T(x,y,z)=(-x+2y,-2x-y+3z)\]
Basis coordinates are not the same as ordinary output coordinates.
**Answer: B.**
A: **Incorrect:** Negates the required output transformation.
B: **Correct:** Matches the result after both input- and output-basis conversions.
C: **Incorrect:** Does not reproduce the given matrix columns on the supplied basis vectors.
D: **Incorrect:** The second component differs from the recombined -2x-y+3z.
**1. Set up:** The bound |f(x,y)|<=|x-y|²<=2(x²+y²) proves continuity at the origin.
**2. Work through the reasoning:** For the x-partial, f(h,0)/h=h sin(1/h)→0. For the y-partial, f(0,h)/h=-h sin(1/h)→0. Thus both partials are zero. To prove differentiability, divide the error from the zero linear map by r=sqrt(x²+y²): |f|/r<=2r→0.
**3. Check the result:**
\[\frac{|f(x,y)|}{\sqrt{x^2+y^2}}\le2\sqrt{x^2+y^2}\longrightarrow0\]
This establishes more than the existence of partial derivatives.
**Answer: A, B, D.**
A: **Correct:** The quadratic bound tends to zero.
B: **Correct:** Both partial derivatives equal 0.
C: **Incorrect:** The two partial derivatives are equal, not unequal.
D: **Correct:** The first-order remainder divided by distance tends to zero.
**1. Set up:** A real subspace is isomorphic to R² iff its dimension is 2.
**2. Work through the reasoning:** W1 has four matrix entries with two independent row-sum constraints, leaving dimension 2. In W2, z=x-y and w=x, so x,y are free and the dimension is 2. W3 has nine entries and three independent column-sum equations, leaving dimension 6.
**3. Check the result:**
\[(\dim W_1,\dim W_2,\dim W_3,\dim W_4)=(2,2,6,2)\]
W4 is the plane x+y=0 in R³; its two independent free coordinates give dimension 2.
**Answer: A, B, D.**
A: **Correct:** Four coordinates minus two independent constraints leaves two degrees of freedom.
B: **Correct:** Two free variables parametrise the space.
C: **Incorrect:** The three column-sum constraints leave six, not two, degrees of freedom.
D: **Correct:** Orthogonality to the nonzero vector (1,1,0) imposes one independent equation on R³.
is an orthogonal matrix of order 5, then find the nullity of the matrix
.
Published solution
**1. Set up:** An orthogonal matrix satisfies A transpose A=I, so it is invertible with inverse A transpose.
**2. Work through the reasoning:** If Ax=0, multiply by A transpose to get x=0. Thus its kernel is the zero subspace and nullity is 0.
**3. Check the result:**
\[\dim\ker A=5-\operatorname{rank}A=5-5=0\]
Equivalently, rank is 5 and rank-nullity gives 5-5=0.
**Answer: 0.**
**1. Set up:** Orthogonality requires the dot product to vanish.
**2. Work through the reasoning:** For u=(2,-1,4,3) and v=(1,c,-2,6), u·v=2-c-8+18=12-c. Therefore 12-c=0 and c=12.
**3. Check the result:**
\[(2,-1,4,3)\cdot(1,c,-2,6)=12-c=0\]
Substituting this gives a zero dot product, as required.
**Answer: 12.**
**1. Set up:** For f=xy+yz+zx+x³+y³+z³, the gradient is (y+z+3x²,x+z+3y²,x+y+3z²).
**2. Work through the reasoning:** At (1,0,1) it is (4,2,4), with norm sqrt(16+4+16)=6. The unit steepest-ascent vector is (a,b,c)=(2/3,1/3,2/3).
**3. Check the result:**
\[\frac{\nabla f(1,0,1)}{\|\nabla f(1,0,1)\|}=\frac{(4,2,4)}6=\left(\frac23,\frac13,\frac23\right)\]
Hence a+2b+c=2/3+2/3+2/3=2.
**Answer: 2.**
**1. Set up:** Solve (1,2,-3)=alpha(1,0,1)+beta(1,1,0)+gamma(0,1,1).
**2. Work through the reasoning:** The coefficients are alpha=-2,beta=3,gamma=-1, as direct substitution checks. Linearity gives T(1,2,-3)=-2(1,0)+3(0,1)-(1,1)=(-3,2).
**3. Check the result:**
\[T(1,2,-3)=-2(1,0)+3(0,1)-(1,1)=(-3,2)\]
Therefore a-b=-3-2=-5.
**Answer: -5.**
.
Based on the above data, answer the given subquestions.
What is the dimension of the subspace corresponding to the affine subspace
?
Published solution
**1. Set up:** For f(x,y)=xy²+yx², both first partial derivatives vanish at (0,0), and f(0,0)=0.
**2. Work through the reasoning:** The tangent plane there is z=0. Its associated direction subspace has basis (1,0,0),(0,1,0), hence dimension 2.
**3. Check the result:**
\[z=f(0,0)+f_x(0,0)x+f_y(0,0)y=0,\qquad\dim V=2\]
This conclusion is independent of the source-key error in the separate tangent-plane point-membership question.
**Answer: 2.**
Based on the above data, answer the given subquestions.
Consider the following statements about the function
.
**S1**: The range of
is the subset
of
.
**S2**: The gradient of
is defined at all points in
.
**S3**: The directional derivative of
in the direction of
at
is
.
What is the number of correct statements? Write 0 if none of the statements is correct.
Published solution
**1. Set up:** The function is exp(x³+y³).
**2. Work through the reasoning:** S1 gives the range as [0,infinity), which is incorrect because exponentials never equal zero; the exact range is (0,infinity), since x³+y³ attains every real number. S2 is true: its gradient exists everywhere in R². S3 is false: at (1,0), the gradient is (3e,0), which has a nonzero dot product with the displayed negative direction; normalising that direction also leaves it nonzero. Only S2 is correct.
**3. Check the result:**
\[\operatorname{range}(e^{x^3+y^3})=(0,\infty)\]
If S1 were read merely as containment inside the nonnegative reals, it would be true, but the source's one-answer count uses the stated set as the proposed exact range.
**Answer: 1.**
Based on the above data, answer the given subquestions.
Suppose function
describes the amount of rain at a point
on a map. If along the unit
direction
the amount of rain remains constant starting at point
, then find
.
Published solution
**1. Set up:** The direction (a,b) is explicitly a unit direction, so by definition a²+b²=1.
**2. Work through the reasoning:** This alone determines the requested value. Under the intended local rate-of-change reading, constancy at (1,1) means the direction is orthogonal to the gradient (3e²,3e²), giving a+b=0 and a,b=±1/sqrt(2) with opposite signs.
**3. Check the result:**
\[\|(a,b)\|=1\quad\Longrightarrow\quad a^2+b^2=1\]
Zero initial directional derivative need not imply constancy along an entire straight line; the unit-vector norm remains 1 in either calculation.
**Answer: 1.**
Based on the above data, answer the given subquestions.
Which of the following options depicts the correct geometric shape of the boundary of
?
6406536925320.
A triangle.
6406536925321.
A rectangle.
6406536925322.
A pentagon.
6406536925323.
A hexagon.
Published solution
**1. Set up:** The domain x>=0,y>=0,x+y<=1 is the filled triangle bounded by the two coordinate axes and the line x+y=1.
**2. Work through the reasoning:** Its three vertices are (0,0),(1,0),(0,1), so the boundary has three line segments.
**3. Check the result:**
\[D=\{(x,y):x\ge0,\ y\ge0,\ x+y\le1\}\]
The displayed calculation establishes the requested result.
**Answer: A — Selected: A triangle.**
A: **Correct:** The three constraints enclose a triangle.
B: **Incorrect:** There is a slanted third side rather than two pairs of parallel sides.
C: **Incorrect:** Only three boundary segments occur, not five.
D: **Incorrect:** Only three boundary segments occur, not six.
Based on the above data, answer the given subquestions.
What is the number of critical points inside the domain
of
? Write zero if there is no critical
point.
Published solution
**1. Set up:** For f=x²-y, the gradient is (2x,-1).
**2. Work through the reasoning:** It can never be the zero vector because its second component is always -1. The function is differentiable everywhere, so there are no nondifferentiable interior critical points either. Thus the number of critical points inside the triangular domain is 0.
**3. Check the result:**
\[\nabla f(x,y)=(2x,-1)\ne(0,0)\]
Boundary extrema are considered separately.
**Answer: 0.**
Based on the above data, answer the given subquestions.
Find the absolute maximum of
over the domain
.
Published solution
**1. Set up:** The domain has 0<=x<=1 and y>=0.
**2. Work through the reasoning:** Hence f=x²-y<=x²<=1. Equality is attained at (x,y)=(1,0), which lies on the domain boundary. Therefore the absolute maximum is 1.
**3. Check the result:**
\[x^2-y\le x^2\le1,\qquad f(1,0)=1\]
The simple upper bound plus a feasible equality point proves global maximality without needing a lengthy boundary search.
**Answer: 1.**
Let and .
Based on the above data, answer the given subquestions.
Find the rank of
.
Published solution
**1. Set up:** For A=[[2,-2,2],[2,-4,3],[5,3,1]], its rows satisfy 2R3=13R1-8R2, so rank is at most 2.
**2. Work through the reasoning:** The upper-left 2×2 minor has determinant 2(-4)-2(-2)=-4≠0, so rank is at least 2. Combining the bounds gives rank(A)=2.
**3. Check the result:**
\[2R_3=13R_1-8R_2,\qquad\det\begin{bmatrix}2&-2\\2&-4\end{bmatrix}=-4\ne0\]
Exhibiting one dependence alone would not exclude rank 1; the nonzero minor does.
**Answer: 2.**
Let and .
Based on the above data, answer the given subquestions.
Find the determinant of
.
Published solution
**1. Set up:** Expand B=[[-1,0,2],[0,1,0],[3,0,7]] along its second row, which contains only one nonzero entry. det(B)=1×det([[-1,2],[3,7]])=(-1)7-2(3)=-13.
**2. Work through the reasoning:** Its nonzero determinant means all three columns are independent.
**3. Check the result:**
\[\det B=\det\begin{bmatrix}-1&2\\3&7\end{bmatrix}=-7-6=-13\]
The displayed calculation establishes the requested result.
**Answer: -13.**
Let and .
Based on the above data, answer the given subquestions.
Choose all the correct options from the following:
6406536925328.
and
are neither equivalent nor similar.
6406536925329.
has a unique solution but
does not have a unique solution.
6406536925330.
The columns of
are linearly dependent and the columns of
are linearly
independent.
6406536925331.
Let
,
and
be the vectors representing the columns of
. The vectors
,
and
lie on the same plane.
Published solution
**1. Set up:** A has rank 2, while B has determinant -13 and rank 3.
**2. Work through the reasoning:** Equivalent matrices must have the same rank, and similar matrices do too, so neither relationship holds. A's homogeneous system has nonzero solutions because its nullity is 1; B's has only the zero solution. A's columns are dependent, while B's span R³ and cannot all lie in a two-dimensional linear plane through the origin.
**3. Check the result:**
\[\operatorname{rank}A=2,\quad\operatorname{rank}B=3,\quad\dim\ker A=1,\quad\dim\ker B=0\]
The course's coplanarity statement concerns column vectors' linear span; any three endpoints as affine points can of course lie in some affine plane.
**Answer: A, C.**
A: **Correct:** The unequal ranks rule out both equivalence and similarity.
B: **Incorrect:** Reverses the homogeneous-solution comparison: B is invertible and A is not.
C: **Correct:** Rank 2 gives dependence for A; nonzero determinant gives independence for B.
D: **Incorrect:** B's three independent column vectors do not share one two-dimensional subspace through the origin.
