Multivariable Calculus Questions and Answers

Q: Let \(f, g : \mathbb{R}^{2} \to \mathbb{R}\) be defined by\[ f(x,y) = x^{2} - \frac{3}{2}xy^{2} \]and\[ g(x,y) = 4x^{4} - 5x^{2}y + y^{2} \]for all \((x,y) \in \mathbb{R}^{2}\).Consider the following statements:P: f has a saddle point at (0,0).Q: g has a saddle point at (0,0).Then

Both P and Q are TRUE.Both functions have zero gradient at the origin, and their Hessians there are only semi-definite, so the second derivative test is inconclusive; look at the sign of each function near (0, 0).f: on the x-axis \(f = x^{2} \gt 0\). On the curve \(x = y^{2}\), \(f = y^{4} - \tfrac{3}{2}y^{4} = -\tfrac{1}{2}y^{4} \lt 0\). So f takes both signs in every neighbourhood of the origin, where f = 0: a saddle point.g factorises:\[ g = (y - x^{2})(y - 4x^{2}) \]It is negative between the parabolas \(y = x^{2}\) and \(y = 4x^{2}\) (for example at \(y = 2x^{2}\), \(g = -2x^{4}\)) and positive elsewhere near the origin, such as on the x-axis. So g also has a saddle point, although it has a minimum along every straight line through the origin.

Q: Let \(g : M_{2}(\mathbb{R}) \to \mathbb{R}\) be given by \(g(A) = \text{Trace}(A^{2})\). Let 0 be the \(2 \times 2\) zero matrix. The space \(M_{2}(\mathbb{R})\) may be identified with \(\mathbb{R}^{4}\) in the usual manner. Which one of the following is correct?

0 is a saddle point of g.For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(\text{Trace}(A^{2}) = a^{2} + 2bc + d^{2}\). This quadratic form has gradient zero at the origin, so 0 is a critical point.Along \(b = c = t\) (others zero) \(g = 2t^{2} \gt 0\), and along \(b = t, c = -t\) it is \(-2t^{2} \lt 0\). The form takes both signs in every neighbourhood of 0, so 0 is neither a local minimum nor a local maximum: it is a saddle point.

Q: Let \(g(x,y) = f(x,y)e^{2x+3y}\) be defined in \(\mathbb{R}^{2}\), where \(f(x,y)\) is a continuously differentiable non-zero homogeneous function of degree 4. Then, \(x\dfrac{\partial g}{\partial x} + y\dfrac{\partial g}{\partial y} = 0\) holds for

The relation holds on the line 2x + 3y + 4 = 0.By Euler's theorem for a homogeneous function of degree 4, \(xf_{x} + yf_{y} = 4f\). Then\[ \begin{aligned} xg_{x} + yg_{y} &= e^{2x+3y}\,[\,xf_{x} + yf_{y} + f(2x + 3y)\,] \\ &= f\,e^{2x+3y}\,(4 + 2x + 3y) \end{aligned} \]The exponential never vanishes, so the expression is zero wherever \(2x + 3y + 4 = 0\), whatever the value of f. It does not vanish on all of \(\mathbb{R}^{2}\), and the option excluding that line describes the opposite set. The line \(2x + 3y = 0\) would come from forgetting the degree 4.

Q: Consider the function \(F: \mathbb{R}^{2} \to \mathbb{R}^{2}\) given by\[ F(x,y) = (x^{3} - 3xy^{2} - 3x,\ 3x^{2}y - y^{3} - 3y). \]Then, for the function F, the inverse function theorem is

The inverse function theorem is not applicable at exactly two points.With z = x + iy, the two components of F are the real and imaginary parts of \(h(z) = z^{3} - 3z\), since \(z^{3} = (x^{3} - 3xy^{2}) + i(3x^{2}y - y^{3})\). For a holomorphic map the Jacobian determinant is \(|h'(z)|^{2}\):\[ \det J_{F} = |3z^{2} - 3|^{2} \]which vanishes only at \(z = \pm 1\), the points (1, 0) and (-1, 0). Everywhere else the Jacobian is invertible and the theorem applies. So it fails at exactly two points; 'all points' ignores them, and one or three points miscount the roots of \(z^{2} = 1\).

Q: Let \(f: \mathbb{R}^{2} - \{(0,0)\} \to \mathbb{R}\) be a function defined by\[ \begin{aligned} f(x,y) &= \left(\frac{x^{2} - y^{2}}{x^{2} + y^{2}}\right) \\ &\quad + x\sin\left(\frac{1}{x^{2} + y^{2}}\right). \end{aligned} \]Consider the following three statements:S1: \(\lim_{x \to 0} \lim_{y \to 0} f(x,y)\) exists.S2: \(\lim_{y \to 0} \lim_{x \to 0} f(x,y)\) exists.S3: \(\lim_{(x,y) \to (0,0)} f(x,y)\) exists.Then, which of the following is/are correct?

The correct option is S1 and S2 are TRUE and S3 is FALSE.S1: for fixed \(x \neq 0\), letting \(y \to 0\) gives \(f(x, 0) = 1 + x\sin(1/x^{2})\). Then as \(x \to 0\), \(|x\sin(1/x^{2})| \le |x| \to 0\), so the repeated limit is 1.S2: for fixed \(y \neq 0\), letting \(x \to 0\) gives \(f(0, y) = -1\), so the repeated limit is -1.S3: the second term tends to 0 in any manner, but the first term equals \(\cos 2\theta\) in polar form and takes every value between -1 and 1 near the origin, so the double limit does not exist. The two repeated limits differing (1 and -1) already shows this. The paper allows several answers, and the key gives only this one.

Q: Let C be the curve of intersection of the surfaces \(z^{2} = x^{2} + y^{2}\) and \(4x + z = 7\). If P is a point on C at a minimum distance from the xy-plane, then the distance of P from the origin is

The distance is \(\dfrac{7\sqrt{2}}{5}\).On the curve, \(z = 7 - 4x\) and \((7 - 4x)^{2} = x^{2} + y^{2}\), so\[ \begin{aligned} y^{2} &= 15x^{2} - 56x + 49 \\ &= 15\left(x - \tfrac{7}{5}\right)\left(x - \tfrac{7}{3}\right) \ge 0 \end{aligned} \]which forces \(x \le 7/5\) or \(x \ge 7/3\). The distance from the xy-plane is \(|z| = |7 - 4x|\): on the first branch \(z \ge 7/5\), and on the second \(z \le -7/3\). The minimum is 7/5, at \(P = (7/5, 0, 7/5)\), and\[ \begin{aligned} |OP| &= \sqrt{\left(\tfrac{7}{5}\right)^{2} + \left(\tfrac{7}{5}\right)^{2}} \\ &= \frac{7\sqrt{2}}{5} \end{aligned} \]7/5 is the distance from the plane, not from the origin; 14/5 and \(14\sqrt{2}/5\) double the coordinates.

Q: Let \(f : \mathbb{R}^2 \to \mathbb{R}^2\) be defined as\[ f(x, y) = (e^x \cos y,\ e^x \sin y). \]Which one of the following is TRUE?

The true statement is f is locally invertible.The Jacobian determinant is\[ \det \begin{pmatrix} e^x \cos y & -e^x \sin y \\ e^x \sin y & e^x \cos y \end{pmatrix} = e^{2x} \gt 0 \]at every point, so by the inverse function theorem f has a smooth local inverse around each point.It is not one-to-one, because f is \(2\pi\)-periodic in y: \(f(x, y) = f(x, y + 2\pi)\). So it cannot be invertible on the whole plane (it also never takes the value (0,0)). The Jacobian is positive, not negative. This map is the complex exponential \(e^{x+iy}\) written in real coordinates.

Let \(f, g : \mathbb{R}^{2} \to \mathbb{R}\) be defined by

\[ f(x,y) = x^{2} - \frac{3}{2}xy^{2} \]

and

\[ g(x,y) = 4x^{4} - 5x^{2}y + y^{2} \]

for all \((x,y) \in \mathbb{R}^{2}\).

Consider the following statements:

P: f has a saddle point at (0,0).
Q: g has a saddle point at (0,0).

Then

A.both P and Q are TRUE
B.P is FALSE but Q is TRUE
C.P is TRUE but Q is FALSE
D.both P and Q are FALSE

Exam Year(s)

GATE Mathematics - 2023,

Answer: A

Both P and Q are TRUE.

Both functions have zero gradient at the origin, and their Hessians there are only semi-definite, so the second derivative test is inconclusive; look at the sign of each function near (0, 0).

f: on the x-axis \(f = x^{2} \gt 0\). On the curve \(x = y^{2}\), \(f = y^{4} - \tfrac{3}{2}y^{4} = -\tfrac{1}{2}y^{4} \lt 0\). So f takes both signs in every neighbourhood of the origin, where f = 0: a saddle point.

g factorises:

\[ g = (y - x^{2})(y - 4x^{2}) \]

It is negative between the parabolas \(y = x^{2}\) and \(y = 4x^{2}\) (for example at \(y = 2x^{2}\), \(g = -2x^{4}\)) and positive elsewhere near the origin, such as on the x-axis. So g also has a saddle point, although it has a minimum along every straight line through the origin.

Discuss in Forum

Let \(g : M_{2}(\mathbb{R}) \to \mathbb{R}\) be given by \(g(A) = \text{Trace}(A^{2})\). Let 0 be the \(2 \times 2\) zero matrix. The space \(M_{2}(\mathbb{R})\) may be identified with \(\mathbb{R}^{4}\) in the usual manner. Which one of the following is correct?

A.0 is a point of local minimum of g
B.0 is a point of local maximum of g
C.0 is a saddle point of g
D.0 is not a critical point of g

Exam Year(s)

GATE Mathematics - 2024,

Answer: C

0 is a saddle point of g.

For \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(\text{Trace}(A^{2}) = a^{2} + 2bc + d^{2}\). This quadratic form has gradient zero at the origin, so 0 is a critical point.

Along \(b = c = t\) (others zero) \(g = 2t^{2} \gt 0\), and along \(b = t, c = -t\) it is \(-2t^{2} \lt 0\). The form takes both signs in every neighbourhood of 0, so 0 is neither a local minimum nor a local maximum: it is a saddle point.

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Let \(g(x,y) = f(x,y)e^{2x+3y}\) be defined in \(\mathbb{R}^{2}\), where \(f(x,y)\) is a continuously differentiable non-zero homogeneous function of degree 4. Then, \(x\dfrac{\partial g}{\partial x} + y\dfrac{\partial g}{\partial y} = 0\) holds for

A.all points (x, y) in \(\mathbb{R}^{2}\)
B.all points (x, y) on the line given by \(2x + 3y + 4 = 0\)
C.all points (x, y) in the region of \(\mathbb{R}^{2}\) except on the line given by \(2x + 3y + 4 = 0\)
D.all points (x, y) on the line given by \(2x + 3y = 0\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

The relation holds on the line 2x + 3y + 4 = 0.

By Euler's theorem for a homogeneous function of degree 4, \(xf_{x} + yf_{y} = 4f\). Then

\[ \begin{aligned} xg_{x} + yg_{y} &= e^{2x+3y}\,[\,xf_{x} + yf_{y} + f(2x + 3y)\,] \\ &= f\,e^{2x+3y}\,(4 + 2x + 3y) \end{aligned} \]

The exponential never vanishes, so the expression is zero wherever \(2x + 3y + 4 = 0\), whatever the value of f. It does not vanish on all of \(\mathbb{R}^{2}\), and the option excluding that line describes the opposite set. The line \(2x + 3y = 0\) would come from forgetting the degree 4.

Discuss in Forum

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

\[ F(x,y) = (x^{3} - 3xy^{2} - 3x,\ 3x^{2}y - y^{3} - 3y). \]

Then, for the function F, the inverse function theorem is

A.applicable at all points of \(\mathbb{R}^{2}\)
B.not applicable at exactly one point of \(\mathbb{R}^{2}\)
C.not applicable at exactly two points of \(\mathbb{R}^{2}\)
D.not applicable at exactly three points of \(\mathbb{R}^{2}\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: C

The inverse function theorem is not applicable at exactly two points.

With z = x + iy, the two components of F are the real and imaginary parts of \(h(z) = z^{3} - 3z\), since \(z^{3} = (x^{3} - 3xy^{2}) + i(3x^{2}y - y^{3})\). For a holomorphic map the Jacobian determinant is \(|h'(z)|^{2}\):

\[ \det J_{F} = |3z^{2} - 3|^{2} \]

which vanishes only at \(z = \pm 1\), the points (1, 0) and (-1, 0). Everywhere else the Jacobian is invertible and the theorem applies. So it fails at exactly two points; 'all points' ignores them, and one or three points miscount the roots of \(z^{2} = 1\).

Discuss in Forum

Let \(f: \mathbb{R}^{2} - \{(0,0)\} \to \mathbb{R}\) be a function defined by

\[ \begin{aligned} f(x,y) &= \left(\frac{x^{2} - y^{2}}{x^{2} + y^{2}}\right) \\ &\quad + x\sin\left(\frac{1}{x^{2} + y^{2}}\right). \end{aligned} \]

Consider the following three statements:

S1: \(\lim_{x \to 0} \lim_{y \to 0} f(x,y)\) exists.
S2: \(\lim_{y \to 0} \lim_{x \to 0} f(x,y)\) exists.
S3: \(\lim_{(x,y) \to (0,0)} f(x,y)\) exists.

Then, which of the following is/are correct?

A.S2 and S3 are TRUE and S1 is FALSE
B.S1 and S2 are TRUE and S3 is FALSE
C.S1 and S3 are TRUE and S2 is FALSE
D.S1, S2 and S3 all are TRUE

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

The correct option is S1 and S2 are TRUE and S3 is FALSE.

S1: for fixed \(x \neq 0\), letting \(y \to 0\) gives \(f(x, 0) = 1 + x\sin(1/x^{2})\). Then as \(x \to 0\), \(|x\sin(1/x^{2})| \le |x| \to 0\), so the repeated limit is 1.

S2: for fixed \(y \neq 0\), letting \(x \to 0\) gives \(f(0, y) = -1\), so the repeated limit is -1.

S3: the second term tends to 0 in any manner, but the first term equals \(\cos 2\theta\) in polar form and takes every value between -1 and 1 near the origin, so the double limit does not exist. The two repeated limits differing (1 and -1) already shows this. The paper allows several answers, and the key gives only this one.

Discuss in Forum

Let C be the curve of intersection of the surfaces \(z^{2} = x^{2} + y^{2}\) and \(4x + z = 7\). If P is a point on C at a minimum distance from the xy-plane, then the distance of P from the origin is

A.\(\dfrac{7}{5}\)
B.\(\dfrac{7\sqrt{2}}{5}\)
C.\(\dfrac{14}{5}\)
D.\(\dfrac{14\sqrt{2}}{5}\)

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

The distance is \(\dfrac{7\sqrt{2}}{5}\).

On the curve, \(z = 7 - 4x\) and \((7 - 4x)^{2} = x^{2} + y^{2}\), so

\[ \begin{aligned} y^{2} &= 15x^{2} - 56x + 49 \\ &= 15\left(x - \tfrac{7}{5}\right)\left(x - \tfrac{7}{3}\right) \ge 0 \end{aligned} \]

which forces \(x \le 7/5\) or \(x \ge 7/3\). The distance from the xy-plane is \(|z| = |7 - 4x|\): on the first branch \(z \ge 7/5\), and on the second \(z \le -7/3\). The minimum is 7/5, at \(P = (7/5, 0, 7/5)\), and

\[ \begin{aligned} |OP| &= \sqrt{\left(\tfrac{7}{5}\right)^{2} + \left(\tfrac{7}{5}\right)^{2}} \\ &= \frac{7\sqrt{2}}{5} \end{aligned} \]

7/5 is the distance from the plane, not from the origin; 14/5 and \(14\sqrt{2}/5\) double the coordinates.

Discuss in Forum

Let \(f : \mathbb{R}^2 \to \mathbb{R}^2\) be defined as

\[ f(x, y) = (e^x \cos y,\ e^x \sin y). \]

Which one of the following is TRUE?

A.f is one-to-one.
B.The Jacobian of f is negative.
C.f is locally invertible.
D.f is invertible on \(\mathbb{R}^2\).

Exam Year(s)

GATE Mathematics - 2026,

Answer: C

The true statement is f is locally invertible.

The Jacobian determinant is

\[ \det \begin{pmatrix} e^x \cos y & -e^x \sin y \\ e^x \sin y & e^x \cos y \end{pmatrix} = e^{2x} \gt 0 \]

at every point, so by the inverse function theorem f has a smooth local inverse around each point.

It is not one-to-one, because f is \(2\pi\)-periodic in y: \(f(x, y) = f(x, y + 2\pi)\). So it cannot be invertible on the whole plane (it also never takes the value (0,0)). The Jacobian is positive, not negative. This map is the complex exponential \(e^{x+iy}\) written in real coordinates.

Discuss in Forum

Let \(f: \mathbb{R}^{3} \to \mathbb{R}\) be a continuous function. Then

\(\int_{0}^{2}\int_{0}^{x}\int_{0}^{y} f(x,y,z)\,dz\,dy\,dx =\) _____.

A.\(\int_{0}^{2}\int_{0}^{y}\int_{y}^{2} f(x,y,z)\,dx\,dz\,dy\)
B.\(\int_{0}^{2}\int_{0}^{x}\int_{z}^{2} f(x,y,z)\,dy\,dz\,dx\)
C.\(\int_{0}^{2}\int_{y}^{2}\int_{0}^{x} f(x,y,z)\,dz\,dx\,dy\)
D.\(\int_{0}^{2}\int_{z}^{2}\int_{0}^{x} f(x,y,z)\,dy\,dx\,dz\)

Exam Year(s)

IIT JAM - 2026,

Answer: A

The equal integral is \(\int_{0}^{2}\int_{0}^{y}\int_{y}^{2} f(x,y,z)\,dx\,dz\,dy\).

Read the region from the given limits: \(z\) runs from 0 to \(y\), \(y\) from 0 to \(x\), and \(x\) from 0 to 2, so the region is

\[ 0 \leq z \leq y \leq x \leq 2. \]

Describe it again with \(y\) outermost: \(y\) runs over \([0,2]\); for fixed \(y\), \(z\) runs over \([0,y]\) and, independently, \(x\) over \([y,2]\). That is exactly the first option, with \(dx\) innermost, then \(dz\), then \(dy\).

The others each lose one inequality. The \(dy\,dz\,dx\) option lets \(y\) run up to 2 instead of stopping at \(x\). The \(dz\,dx\,dy\) option lets \(z\) run up to \(x\) instead of \(y\). The \(dy\,dx\,dz\) option lets \(y\) start at 0, so \(y\) can fall below \(z\).

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

\(f(x,y) = (x^{2} - 1)^{2} + (y^{2} - 1)^{2}\) for all \((x,y) \in \mathbb{R}^{2}\).

Which ONE of the following statements is TRUE?

A.\(f\) has local maxima at exactly two points.
B.\(f\) has local minima at exactly two points.
C.\(f\) has local minima at exactly three points.
D.\(f\) has exactly four saddle points.

Exam Year(s)

IIT JAM - 2026,

Answer: D

The TRUE statement is: \(f\) has exactly four saddle points.

Write \(f(x,y) = g(x) + g(y)\) with \(g(t) = (t^{2} - 1)^{2}\). Since \(g'(t) = 4t(t^{2} - 1)\), \(g\) has local minima at \(t = \pm 1\) (value 0) and a local maximum at \(t = 0\) (value 1). The critical points of \(f\) are the nine points with \(x, y \in \{-1, 0, 1\}\), and their type follows from the two coordinates:

\((\pm 1, \pm 1)\): a minimum in both directions, so 4 local minima, where \(f = 0\).
\((0,0)\): a maximum in both directions, so 1 local maximum, where \(f = 2\).
\((0, \pm 1)\) and \((\pm 1, 0)\): a maximum in one direction and a minimum in the other, so 4 saddle points.

So there is only one local maximum, not two, and there are four local minima, not two or three.

Discuss in Forum

10.

Let \(f: \mathbb{R}^{2} \to \mathbb{R}\) be the function defined by

\[ f(x,y) = \begin{cases} \frac{5x^{2}y - 4y^{3}}{x^{2} + y^{2}}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases} \]

Which ONE of the following statements is FALSE?

A.\(f\) is continuous at \((0,0)\)
B.\(\frac{\partial f}{\partial x}\) at (0,1) = 0
C.\(\frac{\partial f}{\partial y}\) at (1,0) = 5
D.\(\frac{\partial f}{\partial y}\) is continuous at \((0,0)\)

Exam Year(s)

IIT JAM - 2026,

Answer: D

The FALSE statement is: \(\frac{\partial f}{\partial y}\) is continuous at \((0,0)\).

At the origin, \(f(0,y) = \frac{-4y^{3}}{y^{2}} = -4y\), so \(\frac{\partial f}{\partial y}(0,0) = -4\). At a point \((x,0)\) with \(x \neq 0\), the linear term in \(y\) is \(\frac{5x^{2}y}{x^{2}} = 5y\), so \(\frac{\partial f}{\partial y}(x,0) = 5\). Approaching the origin along the \(x\)-axis the partial derivative stays at 5, not \(-4\), so it is not continuous there.

The other statements are true. Continuity: \(|f(x,y)| \leq \frac{5x^{2}|y| + 4|y|^{3}}{x^{2} + y^{2}} \leq 5|y| \to 0\). For \(\frac{\partial f}{\partial x}\) at \((0,1)\), \(f(x,1) = \frac{5x^{2} - 4}{x^{2} + 1}\) is even in \(x\), so its derivative at \(x = 0\) is 0. For \(\frac{\partial f}{\partial y}\) at \((1,0)\), \(f(1,y) = \frac{5y - 4y^{3}}{1 + y^{2}}\) has derivative 5 at \(y = 0\), the same value found above.

Multivariable Calculus Questions and Answers

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Multivariable Calculus Questions and Answers

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