Multivariable Calculus Questions and Answers

Q: Consider the function \(f: \mathbb{R}^{2} \to \mathbb{R}\) given by\[ f(x, y) = xy(5x + 3y - 6). \]Which one of the following statements is NOT true?

The statement that is NOT true is \(f\) has two local minima; it has only one.Write \(f = 5x^{2}y + 3xy^{2} - 6xy\). Then\[ \begin{aligned} f_{x} &= y(10x + 3y - 6) \\ f_{y} &= x(5x + 6y - 6) \end{aligned} \]Setting both to zero gives four critical points: \((0,0)\), \(\left(\frac{6}{5}, 0\right)\), \((0, 2)\), and the solution of \(10x + 3y = 6\), \(5x + 6y = 6\), which is \(\left(\frac{2}{5}, \frac{2}{3}\right)\), with \(a + b = \frac{16}{15}\).With \(f_{xx} = 10y\), \(f_{yy} = 6x\), \(f_{xy} = 10x + 6y - 6\), the Hessian determinant \(f_{xx}f_{yy} - f_{xy}^{2}\) is \(-36\) at each of the first three points, so they are saddle points. At \(\left(\frac{2}{5}, \frac{2}{3}\right)\) it is \(\frac{20}{3} \cdot \frac{12}{5} - 2^{2} = 12 \gt 0\) with \(f_{xx} \gt 0\): the only local minimum. So statements A, B and D are true.

Q: Let \(f(x, y) = x^{2} + 5xy^{2}\), then the directional derivative of \(f\) at the point \((-2, 1)\) in the direction of vector \(v = (12, 5)\) is :

The directional derivative is -88/13.The gradient of \(f = x^{2} + 5xy^{2}\) is \(\nabla f = (2x + 5y^{2},\ 10xy)\). At \((-2, 1)\) this is \((-4 + 5,\ -20) = (1, -20)\). The direction must be a unit vector, so divide \(v = (12, 5)\) by its length \(\sqrt{144 + 25} = 13\):\[ \begin{aligned} D_{u}f &= (1, -20) \cdot \tfrac{(12, 5)}{13} \\ &= \tfrac{12 - 100}{13} = -\tfrac{88}{13} \end{aligned} \]\(-88\) is the dot product with \(v\) itself, before normalising, and \(88/13\) drops the sign: the function decreases in this direction because the \(y\)-component of the gradient is large and negative.

Q: Suppose \(f(x, y)\) is of class \(c^{2}\), then under the coordinate transformation \(x = r\cos\theta\), \(y = r\sin\theta\), the expression \(\dfrac{\partial^{2} f}{\partial x^{2}} + \dfrac{\partial^{2} f}{\partial y^{2}}\) transforms to :

The Laplacian becomes \(f_{rr} + \tfrac{1}{r}f_{r} + \tfrac{1}{r^{2}}f_{\theta\theta}\).With \(x = r\cos\theta\), \(y = r\sin\theta\), the chain rule gives \(f_{r} = f_{x}\cos\theta + f_{y}\sin\theta\) and \(f_{\theta} = -rf_{x}\sin\theta + rf_{y}\cos\theta\). Differentiating again and combining, the mixed terms cancel and\[ \begin{aligned} f_{xx} + f_{yy} &= f_{rr} + \frac{1}{r}f_{r} \\ &\quad + \frac{1}{r^{2}}f_{\theta\theta} \end{aligned} \]A check: \(f = r^{2} = x^{2} + y^{2}\) has Laplacian 4, and the formula gives \(2 + \tfrac{1}{r}(2r) + 0 = 4\). Options A and B drop the first-order term \(\tfrac{1}{r}f_{r}\) (B gives 2 for this check), and C has it without the \(1/r\), which is not even dimensionally consistent.

Q: For the function \(f(x, y) = (x - 1)(x^{2} - y^{2})\), the point \(\left(\dfrac{2}{3}, 0\right)\) :

The point \((2/3, 0)\) is a local minimum.Expand \(f = (x - 1)(x^{2} - y^{2})\). Then \(f_{x} = 3x^{2} - 2x - y^{2}\) and \(f_{y} = -2y(x - 1)\). At \((2/3, 0)\): \(f_{x} = 4/3 - 4/3 = 0\) and \(f_{y} = 0\), so it is a critical point.Second derivatives: \(f_{xx} = 6x - 2 = 2\), \(f_{yy} = -2(x - 1) = 2/3\), \(f_{xy} = -2y = 0\). So\[ \begin{aligned} f_{xx}f_{yy} - f_{xy}^{2} &= 2 \cdot \tfrac{2}{3} - 0 \\ &= \tfrac{4}{3} \gt 0 \end{aligned} \]with \(f_{xx} \gt 0\), which is a local minimum. The other critical points, (0, 0), (1, 1) and (1, -1), are where the saddle points of this function lie, which is the trap in option D.

Q: The functions :\(f_{1}(x, y, z) = x + y + z\)\(f_{2}(x, y, z) = xy + yz + zx\)\(f_{3}(x, y, z) = x^{2} + y^{2} + z^{2}\)

The functions are functionally dependent on \(\mathbf{R}^{3}\).They satisfy an identity everywhere:\[ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) \]that is, \(f_{1}^{2} - f_{3} - 2f_{2} = 0\). Equivalently the Jacobian \(\partial(f_{1}, f_{2}, f_{3})/\partial(x, y, z)\) vanishes identically: its rows are \((1, 1, 1)\), \((y + z, x + z, x + y)\) and \((2x, 2y, 2z)\), and the third row equals \(2f_{1}(1, 1, 1) - 2(y + z, x + z, x + y)\), so the rows are linearly dependent at every point.Because the relation holds on all of \(\mathbf{R}^{3}\), the functions are not independent on any subset, so options A, B and C are wrong.

Q: Let \(\alpha : \left(\dfrac{1}{2}, 2\right) \to \mathbf{R}^{3}\) be defined by \(\alpha(t) = (t, -t, t^{3})\). Then the torsion of \(\alpha\) at \(t = 1\) is :

The torsion is 0.For \(\alpha(t) = (t, -t, t^{3})\): \(\alpha' = (1, -1, 3t^{2})\), \(\alpha'' = (0, 0, 6t)\), \(\alpha''' = (0, 0, 6)\). Then \(\alpha' \times \alpha'' = (-6t, -6t, 0)\), and\[ \begin{aligned} \tau &= \frac{(\alpha' \times \alpha'') \cdot \alpha'''}{|\alpha' \times \alpha''|^{2}} \\ &= \frac{0}{72t^{2}} = 0 \end{aligned} \]The geometric reason: \(y = -x\) at every point, so the whole curve lies in the plane \(x + y = 0\), and a plane curve has zero torsion everywhere, not only at \(t = 1\). The values \(\pm 1\) and 2 cannot occur.

Q: The right cylinder over the circle \(x^{2} + y^{2} = 1\) has the parametrization \(\bar{x} : U \to \mathbf{R}^{3}\), where :\[ \begin{aligned} U = \{&(u, v) \in \mathbf{R}^{2} \mid 0 \lt u \lt 2\pi, \\ &-\infty \lt v \lt \infty\}, \end{aligned} \]\(\bar{x}(u, v) = (\cos u, \sin u, v)\)Then the coefficients E, F, G in the first fundamental form of the cylinder are given by :

The coefficients are E = 1, F = 0, G = 1.For \(\bar{x}(u, v) = (\cos u, \sin u, v)\): \(\bar{x}_{u} = (-\sin u, \cos u, 0)\) and \(\bar{x}_{v} = (0, 0, 1)\). So\[ \begin{aligned} E &= \bar{x}_{u} \cdot \bar{x}_{u} = \sin^{2}u + \cos^{2}u = 1 \\ F &= \bar{x}_{u} \cdot \bar{x}_{v} = 0 \\ G &= \bar{x}_{v} \cdot \bar{x}_{v} = 1 \end{aligned} \]This is why the cylinder is locally isometric to the plane: its first fundamental form is \(du^{2} + dv^{2}\). \(\sin^{2}u\) and \(\cos u\) are single components of \(\bar{x}_{u}\), not its squared length, and F = 1 would mean the coordinate curves are not perpendicular.

Q: The function \(f : \mathbf{R}^{2} \to \mathbf{R}^{2}\) defined by \(f(x, y) = (2xy, x^{2} - y^{2})\) is one-one on the set :

\(f\) is one-one on \(\{(x, y) : \tfrac{1}{2} \lt x \lt 1, |y| \lt \tfrac{1}{4}\}\).Write \(w = y + ix\). Then \(w^{2} = (y^{2} - x^{2}) + 2ixy\), so \(f(x, y) = (\mathrm{Im}\,w^{2}, -\mathrm{Re}\,w^{2})\): \(f\) is complex squaring in disguise. Hence \(f(x, y) = f(a, b)\) exactly when \(w^{2} = w'^{2}\) for \(w' = b + ia\), that is \(w' = \pm w\), or \((a, b) = \pm(x, y)\).So \(f\) is one-one on a set precisely when the set contains no pair of opposite points \((x, y)\), \((-x, -y)\). The set in option C lies in the half-plane \(x \gt 0\), which contains no such pair, so \(f\) is one-one there.The others contain opposite points: (1, 0) and \((-1, 0)\) both lie in \(y \ge 0\) and in the unit disc, and both map to (0, 1); \(\mathbf{R}^{2}\) contains every such pair.

Q: Let \(f : \mathbf{R}^{3} \to \mathbf{R}^{2}\) be defined by the equation \(f(x, y, z) = (xy, y + z)\) and \(g : \mathbf{R}^{2} \to \mathbf{R}^{2}\) be defined as \(g(x, y) = (x + y, y)\). Then the derivative \(D(g \circ f)(0, 0, 0) =\)

\(D(g \circ f)(0, 0, 0)\) is \(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\).Compose first: \(g(f(x, y, z)) = (xy + y + z,\ y + z)\). Its Jacobian is\[ \begin{bmatrix} y & x + 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \]which at the origin is \(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\). The chain rule agrees: \(Dg = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) and \(Df(0) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}\), whose product is the same matrix.The \(3 \times 2\) option has the wrong shape for a map from \(\mathbf{R}^{3}\) to \(\mathbf{R}^{2}\), and the other two put non-zero entries in the first column, but nothing depends on \(x\) to first order at the origin.

11.

Let \(S\) be the solid of intersection of two solid spheres \(S_1, S_2\) given by

\(S_1: x^{2} + y^{2} + (z-1)^{2} \leq 4\) and \(S_2: x^{2} + y^{2} + (z+1)^{2} \leq 4\).

Which ONE of the following iterated integrals expresses the volume of \(S\)?

A.\(8\int_0^{\sqrt{3}}\int_0^{\sqrt{3-x^2}} (\sqrt{4-x^2-y^2} - 1)\,dy\,dx\)
B.\(8\int_0^{2}\int_0^{\sqrt{4-x^2}} (\sqrt{4-x^2-y^2} - 1)\,dy\,dx\)
C.\(2\int_0^{2}\int_0^{\sqrt{4-x^2}} (\sqrt{4-x^2-y^2} - 1)\,dy\,dx\)
D.\(2\int_0^{\sqrt{3}}\int_0^{\sqrt{3-x^2}} (\sqrt{4-x^2-y^2} - 1)\,dy\,dx\)

Exam Year(s)

IIT JAM - 2026,

Answer: A

The volume is \(8\int_0^{\sqrt{3}}\int_0^{\sqrt{3-x^2}} (\sqrt{4-x^2-y^2} - 1)\,dy\,dx\).

Subtracting the two sphere equations gives \((z-1)^{2} = (z+1)^{2}\), so the spheres meet in the plane \(z = 0\), on the circle \(x^{2} + y^{2} = 3\). The solid is a lens, symmetric about \(z = 0\).

Above \(z = 0\) the binding surface is the sphere centred at \((0,0,-1)\): \(z \leq \sqrt{4 - x^{2} - y^{2}} - 1\), over the disc \(x^{2} + y^{2} \leq 3\). So

\[ \begin{aligned} V &= 2\iint_{x^2+y^2 \leq 3} (\sqrt{4-x^2-y^2} - 1)\,dA \\ &= 8 \times (\text{first-quadrant part}) \end{aligned} \]

The factor 8 counts the two halves of the lens and the four quadrants of the disc. The option with factor 2 over the same quadrant gives only a quarter of the volume. The two options over the disc of radius 2 include the ring \(\sqrt{3} \lt r \leq 2\), where \(\sqrt{4 - r^{2}} - 1 \lt 0\), so they subtract volume that is not there.

Discuss in Forum

12.

Consider the function \(f: \mathbb{R}^{2} \to \mathbb{R}\) given by

\[ f(x, y) = xy(5x + 3y - 6). \]

Which one of the following statements is NOT true?

A.\(f\) has exactly \(4\) critical points
B.\(f\) has more than one saddle point
C.\(f\) has two local minima
D.\(f\) has a critical point of the form \((a, b)\) such that \(a + b = \frac{16}{15}\)

Exam Year(s)

IIT JAM - 2026,

Answer: C

The statement that is NOT true is \(f\) has two local minima; it has only one.

Write \(f = 5x^{2}y + 3xy^{2} - 6xy\). Then

\[ \begin{aligned} f_{x} &= y(10x + 3y - 6) \\ f_{y} &= x(5x + 6y - 6) \end{aligned} \]

Setting both to zero gives four critical points: \((0,0)\), \(\left(\frac{6}{5}, 0\right)\), \((0, 2)\), and the solution of \(10x + 3y = 6\), \(5x + 6y = 6\), which is \(\left(\frac{2}{5}, \frac{2}{3}\right)\), with \(a + b = \frac{16}{15}\).

With \(f_{xx} = 10y\), \(f_{yy} = 6x\), \(f_{xy} = 10x + 6y - 6\), the Hessian determinant \(f_{xx}f_{yy} - f_{xy}^{2}\) is \(-36\) at each of the first three points, so they are saddle points. At \(\left(\frac{2}{5}, \frac{2}{3}\right)\) it is \(\frac{20}{3} \cdot \frac{12}{5} - 2^{2} = 12 \gt 0\) with \(f_{xx} \gt 0\): the only local minimum. So statements A, B and D are true.

Discuss in Forum

13.

Let \(f(x, y) = x^{2} + 5xy^{2}\), then the directional derivative of \(f\) at the point \((-2, 1)\) in the direction of vector \(v = (12, 5)\) is :

A.88/13
B.-88/13
C.78/7
D.-88

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The directional derivative is -88/13.

The gradient of \(f = x^{2} + 5xy^{2}\) is \(\nabla f = (2x + 5y^{2},\ 10xy)\). At \((-2, 1)\) this is \((-4 + 5,\ -20) = (1, -20)\). The direction must be a unit vector, so divide \(v = (12, 5)\) by its length \(\sqrt{144 + 25} = 13\):

\[ \begin{aligned} D_{u}f &= (1, -20) \cdot \tfrac{(12, 5)}{13} \\ &= \tfrac{12 - 100}{13} = -\tfrac{88}{13} \end{aligned} \]

\(-88\) is the dot product with \(v\) itself, before normalising, and \(88/13\) drops the sign: the function decreases in this direction because the \(y\)-component of the gradient is large and negative.

Discuss in Forum

14.

Suppose \(f(x, y)\) is of class \(c^{2}\), then under the coordinate transformation \(x = r\cos\theta\), \(y = r\sin\theta\), the expression \(\dfrac{\partial^{2} f}{\partial x^{2}} + \dfrac{\partial^{2} f}{\partial y^{2}}\) transforms to :

A.\(\dfrac{\partial^{2} f}{\partial r^{2}}\) + \(\dfrac{\partial^{2} f}{\partial \theta^{2}}\)
B.\(\dfrac{\partial^{2} f}{\partial r^{2}}\) + \(\dfrac{1}{r^{2}}\dfrac{\partial^{2} f}{\partial \theta^{2}}\)
C.\(\dfrac{\partial^{2} f}{\partial r^{2}}\) + \(\dfrac{\partial f}{\partial r}\) + \(\dfrac{1}{r^{2}}\dfrac{\partial^{2} f}{\partial \theta^{2}}\)
D.\(\dfrac{\partial^{2} f}{\partial r^{2}}\) + \(\dfrac{1}{r}\dfrac{\partial f}{\partial r}\) + \(\dfrac{1}{r^{2}}\dfrac{\partial^{2} f}{\partial \theta^{2}}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The Laplacian becomes \(f_{rr} + \tfrac{1}{r}f_{r} + \tfrac{1}{r^{2}}f_{\theta\theta}\).

With \(x = r\cos\theta\), \(y = r\sin\theta\), the chain rule gives \(f_{r} = f_{x}\cos\theta + f_{y}\sin\theta\) and \(f_{\theta} = -rf_{x}\sin\theta + rf_{y}\cos\theta\). Differentiating again and combining, the mixed terms cancel and

\[ \begin{aligned} f_{xx} + f_{yy} &= f_{rr} + \frac{1}{r}f_{r} \\ &\quad + \frac{1}{r^{2}}f_{\theta\theta} \end{aligned} \]

A check: \(f = r^{2} = x^{2} + y^{2}\) has Laplacian 4, and the formula gives \(2 + \tfrac{1}{r}(2r) + 0 = 4\). Options A and B drop the first-order term \(\tfrac{1}{r}f_{r}\) (B gives 2 for this check), and C has it without the \(1/r\), which is not even dimensionally consistent.

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15.

For the function \(f(x, y) = (x - 1)(x^{2} - y^{2})\), the point \(\left(\dfrac{2}{3}, 0\right)\) :

A.is not a critical point
B.is a local maximum
C.is a local minimum
D.is a saddle point

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The point \((2/3, 0)\) is a local minimum.

Expand \(f = (x - 1)(x^{2} - y^{2})\). Then \(f_{x} = 3x^{2} - 2x - y^{2}\) and \(f_{y} = -2y(x - 1)\). At \((2/3, 0)\): \(f_{x} = 4/3 - 4/3 = 0\) and \(f_{y} = 0\), so it is a critical point.

Second derivatives: \(f_{xx} = 6x - 2 = 2\), \(f_{yy} = -2(x - 1) = 2/3\), \(f_{xy} = -2y = 0\). So

\[ \begin{aligned} f_{xx}f_{yy} - f_{xy}^{2} &= 2 \cdot \tfrac{2}{3} - 0 \\ &= \tfrac{4}{3} \gt 0 \end{aligned} \]

with \(f_{xx} \gt 0\), which is a local minimum. The other critical points, (0, 0), (1, 1) and (1, -1), are where the saddle points of this function lie, which is the trap in option D.

Discuss in Forum

16.

The functions :

\(f_{1}(x, y, z) = x + y + z\)
\(f_{2}(x, y, z) = xy + yz + zx\)
\(f_{3}(x, y, z) = x^{2} + y^{2} + z^{2}\)

A.are functionally independent on \(\mathbf{R}^{3}/\{x\text{-axis}\}\)
B.are functionally independent on the set \(\{(x, y, z) \in \mathbf{R}^{3}/x \gt 0\}\)
C.are functionally independent on the set \(\{(x, y, z) \in \mathbf{R}^{3}/x \gt 0, y \gt 0, z \gt 0\}\)
D.are functionally dependent on \(\mathbf{R}^{3}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The functions are functionally dependent on \(\mathbf{R}^{3}\).

They satisfy an identity everywhere:

\[ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) \]

that is, \(f_{1}^{2} - f_{3} - 2f_{2} = 0\). Equivalently the Jacobian \(\partial(f_{1}, f_{2}, f_{3})/\partial(x, y, z)\) vanishes identically: its rows are \((1, 1, 1)\), \((y + z, x + z, x + y)\) and \((2x, 2y, 2z)\), and the third row equals \(2f_{1}(1, 1, 1) - 2(y + z, x + z, x + y)\), so the rows are linearly dependent at every point.

Because the relation holds on all of \(\mathbf{R}^{3}\), the functions are not independent on any subset, so options A, B and C are wrong.

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17.

Let \(\alpha : \left(\dfrac{1}{2}, 2\right) \to \mathbf{R}^{3}\) be defined by \(\alpha(t) = (t, -t, t^{3})\). Then the torsion of \(\alpha\) at \(t = 1\) is :

A.1
B.-1
C.0
D.2

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The torsion is 0.

For \(\alpha(t) = (t, -t, t^{3})\): \(\alpha' = (1, -1, 3t^{2})\), \(\alpha'' = (0, 0, 6t)\), \(\alpha''' = (0, 0, 6)\). Then \(\alpha' \times \alpha'' = (-6t, -6t, 0)\), and

\[ \begin{aligned} \tau &= \frac{(\alpha' \times \alpha'') \cdot \alpha'''}{|\alpha' \times \alpha''|^{2}} \\ &= \frac{0}{72t^{2}} = 0 \end{aligned} \]

The geometric reason: \(y = -x\) at every point, so the whole curve lies in the plane \(x + y = 0\), and a plane curve has zero torsion everywhere, not only at \(t = 1\). The values \(\pm 1\) and 2 cannot occur.

Discuss in Forum

18.

The right cylinder over the circle \(x^{2} + y^{2} = 1\) has the parametrization \(\bar{x} : U \to \mathbf{R}^{3}\), where :

\[ \begin{aligned} U = \{&(u, v) \in \mathbf{R}^{2} \mid 0 \lt u \lt 2\pi, \\ &-\infty \lt v \lt \infty\}, \end{aligned} \]

\(\bar{x}(u, v) = (\cos u, \sin u, v)\)

Then the coefficients E, F, G in the first fundamental form of the cylinder are given by :

A.E = 1, F = 1, G = 1
B.E = 1, F = 0, G = 1
C.E = \(\sin^{2} u\), F = 0, G = 1
D.E = \(\cos u\), F = 0, G = 1

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The coefficients are E = 1, F = 0, G = 1.

For \(\bar{x}(u, v) = (\cos u, \sin u, v)\): \(\bar{x}_{u} = (-\sin u, \cos u, 0)\) and \(\bar{x}_{v} = (0, 0, 1)\). So

\[ \begin{aligned} E &= \bar{x}_{u} \cdot \bar{x}_{u} = \sin^{2}u + \cos^{2}u = 1 \\ F &= \bar{x}_{u} \cdot \bar{x}_{v} = 0 \\ G &= \bar{x}_{v} \cdot \bar{x}_{v} = 1 \end{aligned} \]

This is why the cylinder is locally isometric to the plane: its first fundamental form is \(du^{2} + dv^{2}\). \(\sin^{2}u\) and \(\cos u\) are single components of \(\bar{x}_{u}\), not its squared length, and F = 1 would mean the coordinate curves are not perpendicular.

Discuss in Forum

19.

The function \(f : \mathbf{R}^{2} \to \mathbf{R}^{2}\) defined by \(f(x, y) = (2xy, x^{2} - y^{2})\) is one-one on the set :

A.\(\{(x, y) \in \mathbf{R}^{2} / y \ge 0\}\)
B.\(\{(x, y) \in \mathbf{R}^{2} / x^{2} + y^{2} \le 1\}\)
C.\(\{(x, y) \in \mathbf{R}^{2} / \tfrac{1}{2} \lt x \lt 1, |y| \lt \tfrac{1}{4}\}\)
D.\(\mathbf{R}^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

\(f\) is one-one on \(\{(x, y) : \tfrac{1}{2} \lt x \lt 1, |y| \lt \tfrac{1}{4}\}\).

Write \(w = y + ix\). Then \(w^{2} = (y^{2} - x^{2}) + 2ixy\), so \(f(x, y) = (\mathrm{Im}\,w^{2}, -\mathrm{Re}\,w^{2})\): \(f\) is complex squaring in disguise. Hence \(f(x, y) = f(a, b)\) exactly when \(w^{2} = w'^{2}\) for \(w' = b + ia\), that is \(w' = \pm w\), or \((a, b) = \pm(x, y)\).

So \(f\) is one-one on a set precisely when the set contains no pair of opposite points \((x, y)\), \((-x, -y)\). The set in option C lies in the half-plane \(x \gt 0\), which contains no such pair, so \(f\) is one-one there.

The others contain opposite points: (1, 0) and \((-1, 0)\) both lie in \(y \ge 0\) and in the unit disc, and both map to (0, 1); \(\mathbf{R}^{2}\) contains every such pair.

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20.

Let \(f : \mathbf{R}^{3} \to \mathbf{R}^{2}\) be defined by the equation \(f(x, y, z) = (xy, y + z)\) and \(g : \mathbf{R}^{2} \to \mathbf{R}^{2}\) be defined as \(g(x, y) = (x + y, y)\). Then the derivative \(D(g \circ f)(0, 0, 0) =\)

A.\(\begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\)
B.\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ 1 & 1 \end{bmatrix}\)
C.\(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\)
D.\(\begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

\(D(g \circ f)(0, 0, 0)\) is \(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\).

Compose first: \(g(f(x, y, z)) = (xy + y + z,\ y + z)\). Its Jacobian is

\[ \begin{bmatrix} y & x + 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \]

which at the origin is \(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\). The chain rule agrees: \(Dg = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) and \(Df(0) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}\), whose product is the same matrix.

The \(3 \times 2\) option has the wrong shape for a map from \(\mathbf{R}^{3}\) to \(\mathbf{R}^{2}\), and the other two put non-zero entries in the first column, but nothing depends on \(x\) to first order at the origin.

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