Since is continuous, substitute and to obtain the limit .
Incorrect! Try again.
2A function is continuous at if which condition holds?
Limits and continuity
Easy
A.
B. is always an integer
C. is always positive
D.
Correct Answer:
Explanation:
Continuity at requires the limit of the function to exist and equal its value at that point.
Incorrect! Try again.
3Which function is continuous at every point in ?
Limits and continuity
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
A polynomial in and is continuous at every point in .
Incorrect! Try again.
4For , find .
Partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Treating as a constant, the derivative of with respect to is .
Incorrect! Try again.
5For , find .
Partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Treating as a constant, differentiating with respect to gives .
Incorrect! Try again.
6If , which expression gives its total differential?
Partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The total differential combines the changes in both variables: .
Incorrect! Try again.
7For , what is the mixed partial derivative ?
Partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The correct option follows directly from the given concept and definitions.
Incorrect! Try again.
8If , , and , what is ?
Chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Substitution gives , so .
Incorrect! Try again.
9If , where and , which formula represents the chain rule?
Chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The chain rule adds the contributions caused by the changes in and .
Incorrect! Try again.
10Let , , and . What is ?
Chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , differentiation gives .
Incorrect! Try again.
11A function is homogeneous of degree if which relation holds?
Euler's theorem for homogeneous functions
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
By definition, scaling all variables by scales a degree- homogeneous function by .
Incorrect! Try again.
12What is the degree of the homogeneous function ?
Euler's theorem for homogeneous functions
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Every term has total degree , so the function is homogeneous of degree .
Incorrect! Try again.
13If is homogeneous of degree , Euler's theorem states that:
Euler's theorem for homogeneous functions
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Euler's theorem for a degree- homogeneous function is .
Incorrect! Try again.
14At an interior stationary point of a differentiable function , which conditions usually hold?
Maxima and minima for a function of two variables
Easy
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
At an interior stationary point, both first-order partial derivatives are zero.
Incorrect! Try again.
15Which point is the minimum point of ?
Maxima and minima for a function of two variables
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Because and equals at , the minimum occurs there.
Incorrect! Try again.
16For the second derivative test, the discriminant at a stationary point is:
Maxima and minima for a function of two variables
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The discriminant used in the two-variable second derivative test is .
Incorrect! Try again.
17If and at a stationary point, the function has a:
Maxima and minima for a function of two variables
Easy
A.Local minimum
B.Local maximum
C.Saddle point
D.Discontinuity
Correct Answer: Local minimum
Explanation:
The conditions and identify a local minimum.
Incorrect! Try again.
18The Lagrange multiplier method is mainly used to find:
Lagrange method of multiplier
Easy
A.Partial derivatives
B.Constrained extrema
C.Ordinary limits
D.Taylor coefficients
Correct Answer: Constrained extrema
Explanation:
The Lagrange multiplier method finds maxima or minima subject to one or more constraints.
Incorrect! Try again.
19To optimize subject to , the basic Lagrange condition is:
Lagrange method of multiplier
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
At a constrained extremum, the gradients of the objective and constraint are parallel.
Incorrect! Try again.
20For the constraint , which function may be used as in the equation ?
Lagrange method of multiplier
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Rearranging the constraint into zero form gives .
Incorrect! Try again.
21Evaluate the limit
Limits and continuity
Medium
A.The limit does not exist because different paths approaching the origin produce different values
B.The limit is
C.The limit is
D.The limit is
Correct Answer: The limit does not exist because different paths approaching the origin produce different values
Explanation:
Along , the expression is , while along , it is . Since the path limits differ, the limit does not exist.
Incorrect! Try again.
22Let For which value of is continuous at the origin?
Limits and continuity
Medium
A.
B.
C.No value of
D.
Correct Answer:
Explanation:
Since , the limit at the origin is . Thus continuity requires .
Incorrect! Try again.
23Find
Limits and continuity
Medium
A.
B.
C.The limit does not exist
D.
Correct Answer:
Explanation:
Set . Then and the expression becomes , whose limit is .
Incorrect! Try again.
24For , find the total differential at .
Partial derivatives and total derivative
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
At , and . Therefore, .
Incorrect! Try again.
25If , what is ?
Partial derivatives and total derivative
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Here . Substituting gives .
Incorrect! Try again.
26Find the tangent plane to at the point .
Partial derivatives and total derivative
Medium
A.
B.
C., since both partial derivatives equal the coordinates at the given point
D.
Correct Answer:
Explanation:
Since and at , the tangent plane is .
Incorrect! Try again.
27For , determine the mixed partial derivative .
Partial derivatives and total derivative
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The correct option follows directly from the given concept and definitions.
Incorrect! Try again.
28Let , where and . Find .
Chain rule
Medium
A., obtained by differentiating both transformed variables without combining terms
B.
C.
D.
Correct Answer:
Explanation:
Substitution gives . Hence .
Incorrect! Try again.
29If , , and , find at .
Chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Using the chain rule, . At , this equals .
Incorrect! Try again.
30Suppose , where and . Which expression equals ?
Chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The chain rule gives and . Therefore, .
Incorrect! Try again.
31Let . According to Euler's theorem, what is ?
Euler's theorem for homogeneous functions
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The function is homogeneous of degree . Euler's theorem therefore gives .
Incorrect! Try again.
32For , where defined, evaluate .
Euler's theorem for homogeneous functions
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The ratio is unchanged by scaling, while contributes degree . Thus is homogeneous of degree , so .
Incorrect! Try again.
33If is twice differentiable and homogeneous of degree , what is
Euler's theorem for homogeneous functions
Medium
A.
B., after applying Euler's theorem separately to every second derivative
C.
D.
Correct Answer:
Explanation:
For a homogeneous function of degree , the second-order Euler relation is . With , this is .
Incorrect! Try again.
34Classify the stationary point of .
Maxima and minima for a function of two variables
Medium
A.An inconclusive stationary point
B.A saddle point
C.A strict local minimum
D.A strict local maximum
Correct Answer: A strict local minimum
Explanation:
At the origin, , , and . Since and , it is a strict local minimum.
Incorrect! Try again.
35For , which statement about the stationary point is correct?
Maxima and minima for a function of two variables
Medium
A.It is a saddle point with value
B.It is a local minimum with value
C.It is a local maximum with value
D.It is a local minimum with value
Correct Answer: It is a local minimum with value
Explanation:
At , the Hessian has , , and , so with . Also, .
Incorrect! Try again.
36Determine the nature of the stationary point of .
Maxima and minima for a function of two variables
Medium
A.A local minimum at
B.A saddle point at
C.A saddle point at
D.A local maximum at
Correct Answer: A saddle point at
Explanation:
Solving and gives . Since , it is a saddle point.
Incorrect! Try again.
37Find the minimum value of .
Maxima and minima for a function of two variables
Medium
A. at
B. at
C. at
D. at
Correct Answer: at
Explanation:
Completing squares gives . Hence the minimum value is at .
Incorrect! Try again.
38Using Lagrange multipliers, find the maximum value of subject to .
Lagrange method of multiplier
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Lagrange equations imply . The maximum occurs when and have the same sign, giving or and .
Incorrect! Try again.
39Find the minimum value of subject to the constraint .
Lagrange method of multiplier
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Lagrange equations give and , so . The constraint then gives and the minimum value is .
Incorrect! Try again.
40For and , determine the constrained minimum of subject to .
Lagrange method of multiplier
Medium
A.The minimum is at
B.The minimum is at
C.The minimum is approached as one variable tends to zero and the other increases without bound
D.The minimum is at
Correct Answer: The minimum is at
Explanation:
The Lagrange equations give and , so . With and positivity, , yielding the minimum .
Incorrect! Try again.
41Evaluate the limit
Limits and continuity
Hard
A.The limit does not exist
B.
C.The function is unbounded near
D.
Correct Answer: The limit does not exist
Explanation:
Along the expression is , while along it equals . Since the path limits differ, the limit does not exist.
Incorrect! Try again.
42For both iterated limits at are zero. Which statement about the two-variable limit is correct?
Limits and continuity
Hard
A.The limit exists and is zero, but continuity fails only because the function is undefined at the origin
B.The limit does not exist because different straight-line paths give different values
C.The limit is because both iterated limits agree
D.The limit is because the numerator and denominator have equal degree
Correct Answer: The limit does not exist because different straight-line paths give different values
Explanation:
Along , the value is , which depends on . Agreement of the two iterated limits is therefore insufficient.
Incorrect! Try again.
43Define and, for , For which real values of is continuous at the origin?
Limits and continuity
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Writing and gives . This tends uniformly to zero exactly when .
Incorrect! Try again.
44Let Which statement is correct at ?
Partial derivatives and total derivative
Hard
A. is continuous, but neither partial derivative exists
B. is differentiable with total derivative
C.Only the partial derivative with respect to exists
D.Both partial derivatives exist, but is not differentiable
Correct Answer: Both partial derivatives exist, but is not differentiable
Explanation:
At the origin, and . The proposed linear part is , but along the normalized remainder does not approach zero, so total differentiability fails.
Incorrect! Try again.
45For with , what is the total derivative applied to an increment ?
Partial derivatives and total derivative
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The Jacobian is . At it becomes , giving .
Incorrect! Try again.
46Define Which assertion is correct?
Partial derivatives and total derivative
Hard
A.The mixed partial derivatives fail to exist because their defining one-variable limits approach unequal finite values
B. is twice differentiable with zero Hessian at the origin
C.All second-order partial derivatives exist at the origin, but is not twice differentiable there
D. is not differentiable at the origin
Correct Answer: All second-order partial derivatives exist at the origin, but is not twice differentiable there
Explanation:
All second-order partial derivatives at the origin are zero. However, along , , which is not ; hence no second-order Fréchet expansion with the zero Hessian exists.
Incorrect! Try again.
47Let , where is twice continuously differentiable. If and , what is ?
Chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
For and , the chain rule gives . At this is .
Incorrect! Try again.
48Let . Suppose . Find the directional derivative of at in the direction of the unit vector .
Chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The chain rule gives and at . Thus .
Incorrect! Try again.
49Let , where is twice differentiable. If and , what is ?
Chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
At , the velocity is and the acceleration is . Therefore .
Incorrect! Try again.
50If is twice continuously differentiable and homogeneous of degree , which identity must hold?
Euler's theorem for homogeneous functions
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Differentiate Euler's identity and combine the resulting equations after multiplying by and .
Incorrect! Try again.
51Let on a region where this expression is twice differentiable and nonzero. What is
Euler's theorem for homogeneous functions
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The function is homogeneous of degree . The second-order Euler identity therefore gives .
Incorrect! Try again.
52Suppose is twice differentiable and homogeneous of degree , and let . Assuming is differentiable, which identity is satisfied by ?
Euler's theorem for homogeneous functions
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Each second derivative of is homogeneous of degree . Hence the Hessian determinant is homogeneous of degree , and Euler's theorem gives the stated identity.
Incorrect! Try again.
53For , which classification of its stationary points is correct?
Maxima and minima for a function of two variables
Hard
A.Both and are local minima
B. is a local maximum and is a saddle point
C. is inconclusive and is a local maximum
D. is a saddle point and is a local minimum
Correct Answer: is a saddle point and is a local minimum
Explanation:
The stationary equations give only and . The Hessian determinant is negative at and positive with at .
Incorrect! Try again.
54At the stationary point of , the second derivative test has determinant zero. What is the correct classification?
Maxima and minima for a function of two variables
Hard
A.Saddle point
B.Strict local minimum
C.Non-strict local minimum
D.Strict local maximum
Correct Answer: Saddle point
Explanation:
Along , for , while along , for . Thus every neighborhood contains values of both signs.
Incorrect! Try again.
55On the domain , determine the global minimum of
Maxima and minima for a function of two variables
Hard
A.The minimum is , attained at
B.No global minimum exists because the domain has four disconnected components
C.The infimum is , but it is not attained in the domain
D.The minimum is , attained at all four points
Correct Answer: The minimum is , attained at all four points
Explanation:
Set and . The stationary equations for give , and the function diverges at the domain boundary and at infinity. Thus the global minimum is .
Incorrect! Try again.
56For , which statement describes all stationary points?
Maxima and minima for a function of two variables
Hard
A. is a local minimum, while and are saddles
B. is a saddle, while and are global minima
C.The origin is the only stationary point because the quartic terms dominate the mixed term near infinity
D.All three stationary points are local minima with different function values
Correct Answer: is a saddle, while and are global minima
Explanation:
The stationary points are and with matching signs. The Hessian classifies the origin as a saddle and the other two as minima; coercivity makes those minima global.
Incorrect! Try again.
57Using the constraint , what are the maximum and minimum values of ?
Lagrange method of multiplier
Hard
A.Maximum and minimum
B.Maximum and minimum
C.Maximum and minimum
D.Maximum and minimum
Correct Answer: Maximum and minimum
Explanation:
The multiplier equations give and , leading to . Combining this with the constraint yields , , so .
Incorrect! Try again.
58Consider optimizing subject to . Which conclusion is correct?
Lagrange method of multiplier
Hard
A.There is a maximum of at and , but no minimum
B.The multiplier equations locate every extremum, and because they have one solution the function has exactly one constrained maximum
C.There is a minimum of at , but no maximum
D.There are both a minimum of and a maximum of
Correct Answer: There is a minimum of at , but no maximum
Explanation:
The multiplier equations give , which is the constrained minimum. Along the line , the function tends to infinity as , so no maximum exists.
Incorrect! Try again.
59Minimize subject to . Which statement correctly describes the role of Lagrange multipliers?
Lagrange method of multiplier
Hard
A.The minimum is at , but no multiplier satisfies there
B.Every real multiplier works because both gradients vanish at the feasible point
C.No minimum exists because the multiplier equations have no solution
D.The minimum is at with the unique multiplier
Correct Answer: The minimum is at , but no multiplier satisfies there
Explanation:
The constraint set contains only , so it is necessarily the minimum. However, while , so the regularity condition fails and no multiplier exists.
Incorrect! Try again.
60Find the squared minimum distance from to the parabola .
Lagrange method of multiplier
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
On the parabola the squared distance is . Its global minima occur when , giving .
Incorrect! Try again.
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