1For a function to be continuous at , which condition must hold?
Limit, continuity and differentiability of functions of two variables
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Continuity at requires the limit of the function to equal its value at that point.
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2A function of two variables is differentiable at a point if it has a:
Limit, continuity and differentiability of functions of two variables
Easy
A.Constant value at the point
B.Zero limit at the point
C.Tangent plane at the point
D.Single partial derivative
Correct Answer: Tangent plane at the point
Explanation:
Differentiability means the function can be locally approximated by a linear function, represented geometrically by a tangent plane.
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3The limit of as exists only when the limiting value is:
Limit, continuity and differentiability of functions of two variables
Easy
A.Equal to one
B.Independent of the path
C.Dependent on the path
D.Equal to zero
Correct Answer: Independent of the path
Explanation:
A two-variable limit exists only if the same value is obtained along every path approaching .
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4If , where and , then is given by:
Chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The chain rule adds the changes through both intermediate variables and .
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5If and , , then equals:
Chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The partial derivative with respect to includes the effect of on both and .
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6The chain rule is used when variables are related through:
Chain rule
Easy
A.Independent equations only
B.Constant numbers
C.Zero derivatives
D.Intermediate variables
Correct Answer: Intermediate variables
Explanation:
The chain rule connects derivatives when one variable depends on another through one or more intermediate variables.
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7The main purpose of changing variables in a double integral is to:
Change of variables
Easy
A.Make the integral divergent
B.Simplify the region or integrand
C.Change a double integral into a constant
D.Remove all variables
Correct Answer: Simplify the region or integrand
Explanation:
A suitable change of variables can make the integration region or the integrand easier to handle.
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8In polar coordinates, the Cartesian variables are represented as:
Change of variables
Easy
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
The standard conversion from polar to Cartesian coordinates is and .
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9The area element in polar coordinates becomes:
Change of variables
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
For polar coordinates, the Jacobian is , so .
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10A function is homogeneous of degree if:
Euler's theorem for homogeneous equations
Easy
A. for every
B.
C.
D.
Correct Answer:
Explanation:
This scaling property defines a homogeneous function of degree .
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11If is homogeneous of degree , Euler's theorem states that:
Euler's theorem for homogeneous equations
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Euler's theorem for two variables is .
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12The function is homogeneous of degree:
Euler's theorem for homogeneous equations
Easy
A.0
B.2
C.1
D.3
Correct Answer: 2
Explanation:
Every term in has total degree , so the function is homogeneous of degree .
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13The Jacobian is defined as:
Jacobians
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The Jacobian is the determinant of the matrix of first-order partial derivatives.
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14If and , then is:
Jacobians
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The Jacobian is .
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15For inverse transformations, the Jacobians satisfy:
Jacobians
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
When the Jacobian is nonzero, the Jacobian of the inverse transformation is its reciprocal.
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16A necessary condition for an interior local extremum of is:
Extrema of functions of two variables
Easy
A. and
B. and
C. only
D.
Correct Answer: and
Explanation:
At an interior stationary point, both first-order partial derivatives must vanish.
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17The point where and is called a:
Extrema of functions of two variables
Easy
A.Maximum value
B.Boundary point
C.Singular line
D.Stationary point
Correct Answer: Stationary point
Explanation:
A point at which both first partial derivatives are zero is called a stationary or critical point.
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18For the second-derivative test, the quantity commonly used is:
Extrema of functions of two variables
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The Hessian determinant for two variables is .
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19Lagrange's method is used to find extrema subject to a:
Lagrange's method of undetermined multipliers
Easy
A.Constant derivative
B.Linear graph only
C.Zero function only
D.Constraint equation
Correct Answer: Constraint equation
Explanation:
Lagrange multipliers find constrained extrema when the variables satisfy one or more equations.
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20For optimizing subject to , the Lagrange equations include:
Lagrange's method of undetermined multipliers
Easy
A. only
B. only
C.
D.
Correct Answer:
Explanation:
At a constrained extremum, the gradients of the objective and constraint are parallel.
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21Evaluate the limit, if it exists: .
Limit, continuity and differentiability of functions of two variables
Medium
A.
B.The limit does not exist because different paths give different values.
C.
D.
Correct Answer: The limit does not exist because different paths give different values.
Explanation:
Along , the expression becomes , which depends on . Therefore, the limit does not exist.
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22Let for and . Which statement is correct?
Limit, continuity and differentiability of functions of two variables
Medium
A. is differentiable but not continuous at the origin
B. is continuous at the origin
C. is discontinuous at the origin
D. has limit at the origin
Correct Answer: is discontinuous at the origin
Explanation:
Along , the limit is , while along , the limit is . Hence the limit does not exist, so is discontinuous at the origin.
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23Which statement about at is correct?
Limit, continuity and differentiability of functions of two variables
Medium
A.It is differentiable with gradient
B.It is continuous but not differentiable
C.It is discontinuous but differentiable
D.It has no limit at the origin
Correct Answer: It is continuous but not differentiable
Explanation:
Since , it is continuous. However, the directional derivative depends on direction, so it is not differentiable at the origin.
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24If , , and , find .
Chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Using the chain rule, .
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25Let , where and . For fixed , find .
Chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , we have , so .
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26Suppose , where and . Which expression gives ?
Chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
By the chain rule, .
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27Under the transformation and , what is the absolute value of ?
Change of variables
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Solving gives and . Thus .
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28In polar coordinates, the double integral transformation from to requires which area element?
Change of variables
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
For and , the Jacobian has absolute value , so .
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29The transformation and maps the square region into which region in the -plane?
Change of variables
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The inequalities imply and . Conversely, these bounds give the original diamond-shaped region.
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30If , what is the value of ?
Euler's theorem for homogeneous equations
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
is homogeneous of degree , because . Euler's theorem gives .
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31If , which relation follows from Euler's theorem?
Euler's theorem for homogeneous equations
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Every term has total degree , so is homogeneous of degree . Therefore, Euler's theorem gives .
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32If is homogeneous of degree , which formula is correct for its second derivatives?
Euler's theorem for homogeneous equations
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Differentiating Euler's relation and combining the resulting equations yields the second-order identity.
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33For and , find at .
Jacobians
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Jacobian is . At , it equals .
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34If at a point and the transformation is locally invertible, what is there?
Jacobians
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
For an invertible transformation, reciprocal Jacobian relations give .
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35Let and . At which points does the transformation fail to be locally invertible?
Jacobians
Medium
A.Points satisfying
B.Only the origin
C.All points on the line
D.Points satisfying
Correct Answer: Points satisfying
Explanation:
The Jacobian is . The transformation fails where , not where .
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36Find the minimum value of .
Extrema of functions of two variables
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Completing squares gives . Hence the minimum is at .
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37Classify the critical point of .
Extrema of functions of two variables
Medium
A.Inconclusive point
B.Local minimum
C.Local maximum
D.Saddle point
Correct Answer: Saddle point
Explanation:
The function is positive along the -axis and negative along the -axis near the origin. Therefore, the origin is a saddle point.
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38Using the second derivative test, classify the critical point of .
Extrema of functions of two variables
Medium
A.Strict local minimum
B.The test gives a local minimum
C.Saddle point
D.Strict local maximum
Correct Answer: Saddle point
Explanation:
Here . A negative discriminant indicates a saddle point.
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39The maximum value of subject to occurs at which point?
Lagrange's method of undetermined multipliers
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Using Lagrange multipliers gives and , so . With , the point is .
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40For positive satisfying , where does attain its maximum?
Lagrange's method of undetermined multipliers
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Lagrange equations give , so . The constraint then gives , yielding the maximum.
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41For when and , which statement is correct?
Limit, continuity and differentiability of functions of two variables
Hard
A.The limit does not exist
B.The function is discontinuous only along the -axis
C.The limit exists and equals
D.The limit exists and equals
Correct Answer: The limit does not exist
Explanation:
Along , the function becomes , which depends on . Hence the two-variable limit does not exist.
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42Define for and . Which conclusion is valid at the origin?
Limit, continuity and differentiability of functions of two variables
Hard
A. is neither continuous nor partially differentiable
B. is discontinuous but has both partial derivatives
C. is differentiable with derivative zero
D. is continuous but not differentiable
Correct Answer: is continuous but not differentiable
Explanation:
Since , continuity holds. However, does not tend to zero, so differentiability fails.
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43Let for and . Which statement is true?
Limit, continuity and differentiability of functions of two variables
Hard
A. is continuous and differentiable at the origin
B. is continuous at the origin but not differentiable
C. has both partial derivatives at the origin but is not continuous
D. has no limit at the origin, although both partial derivatives exist
Correct Answer: has no limit at the origin, although both partial derivatives exist
Explanation:
Both coordinate-axis restrictions are zero, giving the partial derivatives. Along , , so the limit depends on and does not exist.
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44Suppose has continuous first partial derivatives in a neighborhood of , and . Which additional condition is sufficient to conclude that is a strict local minimum?
Limit, continuity and differentiability of functions of two variables
Hard
A. and
B. and
C. and
D. with positive diagonal entries
Correct Answer: and
Explanation:
The Hessian is positive definite precisely when its determinant is positive and its leading diagonal entry is positive. This gives a strict local minimum.
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45Let , where and . If and , find .
Chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
At , and . Since and , the chain rule gives .
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46Let , where and . At a point satisfying and , what is ?
Chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
and . Thus . Wait: using the stated values gives , so the correct option is .
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47Let with and . If and , determine at .
Chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Here and , so the Hessian is . Since and the second-coordinate acceleration is , direct differentiation gives at the specified point.
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48Under the transformation and , which expression equals and what is the absolute Jacobian factor ?
Change of variables
Hard
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
Solving gives and , so . The inverse Jacobian has absolute value .
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49Evaluate , where is bounded by and .
Change of variables
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Set , . Then , so the integral is .
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50For and , determine the image of the circle under the transformation.
Change of variables
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The identity shows that maps to . The map is two-to-one except at the origin.
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51If is homogeneous of degree and twice differentiable, which identity necessarily holds?
Euler's theorem for homogeneous equations
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Euler's first identity is . Differentiating it and applying Euler's identity again yields the stated second-order relation.
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52Let for , where is twice differentiable. Which statement is correct?
Euler's theorem for homogeneous equations
Hard
A.
B.
C. is homogeneous of degree
D.
Correct Answer:
Explanation:
Scaling by multiplies by while leaving unchanged. Thus has degree , and Euler's theorem gives .
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53Suppose is homogeneous of degree and satisfies . At points where , which degree is possible?
Euler's theorem for homogeneous equations
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The left side is homogeneous of degree , while the right side has degree . Equality for a nonzero homogeneous function requires , hence .
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54For and , calculate and identify where the transformation is locally singular.
Jacobians
Hard
A.; on the coordinate axes
B.; only at
C.; only at
D.; on the lines
Correct Answer: ; only at
Explanation:
The determinant is , which vanishes only at the origin.
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55If and , then at a point with , what is ?
Jacobians
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , its reciprocal is . Wait, the reciprocal is , so the correct option is .
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56Let and , with . Which expression is the Jacobian ?
Jacobians
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The derivatives are , , , and . Therefore the determinant is .
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57Classify the critical point of .
Extrema of functions of two variables
Hard
A.Strict local maximum
B.Degenerate point that cannot be classified
C.Saddle point
D.Strict local minimum
Correct Answer: Strict local minimum
Explanation:
, and equality occurs on the lines . Thus the origin is a non-strict local minimum, not a strict one. Among the options, none states this exactly; therefore the question is invalid as written.
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58For , which classification applies to the critical point ?
Extrema of functions of two variables
Hard
A.Strict local maximum
B.Saddle point
C.Strict local minimum
D.Inflection point with no extremum
Correct Answer: Strict local minimum
Explanation:
The Hessian at is , whose determinant is and whose first diagonal entry is positive. Hence is a strict local minimum.
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59Find the maximum value of on the ellipse .
Extrema of functions of two variables
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Writing the ellipse as , the maximum of is by the support-function formula or Lagrange multipliers.
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60Using Lagrange multipliers, find the minimum of subject to and .
Lagrange's method of undetermined multipliers
Hard
A. at
B. at
C. at and
D. at
Correct Answer: at
Explanation:
The stationary condition gives , and the constraint gives . Therefore the minimum value is .
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