2A function is continuous at if which condition holds?
limits and continuity
Easy
A. is always constant
B.
C. is always positive
D.
Correct Answer:
Explanation:
Continuity requires the limit at the point to exist and equal the function value there.
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3Find .
limits and continuity
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Both and approach zero, so their sum approaches zero.
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4If , what is ?
partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Treating as constant, the derivative of with respect to is .
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5If , what is ?
partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Treating as constant, differentiating with respect to gives .
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6For , which expression gives the total differential ?
partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The total differential is the sum of the changes contributed by each independent variable.
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7If , what is its total differential?
partial derivatives and total derivative
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Here and .
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8If , , and , what is ?
chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Substitution gives , whose derivative is .
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9If , where and , which formula represents ?
chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The multivariable chain rule adds the contributions through and .
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10Let , where and . Find .
chain rule
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , differentiation gives .
Incorrect! Try again.
11If is homogeneous of degree , what does Euler's theorem state?
Euler's theorem for homogeneous functions
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Euler's theorem relates a homogeneous function to its first partial derivatives.
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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 .
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13For , what is ?
Euler's theorem for homogeneous functions
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The function has degree , so Euler's theorem gives .
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14At an interior stationary point of a differentiable function , which conditions normally 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 partial derivatives vanish.
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15What type of point is for ?
maxima and minima for a function of two variables
Easy
A.Local maximum
B.Local minimum
C.Saddle point
D.Nonstationary point
Correct Answer: Local minimum
Explanation:
Because and equals zero at , the point is a minimum.
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16For the second derivative test, let . If and , what is the stationary point?
maxima and minima for a function of two variables
Easy
A.Saddle point
B.Local maximum
C.Inconclusive point
D.Local minimum
Correct Answer: Local minimum
Explanation:
The conditions and identify a local minimum.
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17If at a stationary point, how is the point classified?
maxima and minima for a function of two variables
Easy
A.Saddle point
B.Local maximum
C.Constant point
D.Local minimum
Correct Answer: Saddle point
Explanation:
A negative value of means the surface curves in opposite directions, producing a saddle point.
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18To find extrema of subject to , which Lagrange multiplier equation is used?
Lagrange method of multiplier
Easy
A.
B. only
C.
D. only
Correct Answer:
Explanation:
At a constrained extremum, the gradients of the objective and constraint are parallel.
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19In the Lagrange method, what does represent?
Lagrange method of multiplier
Easy
A.A second derivative
B.A Lagrange multiplier
C.A constraint variable
D.A partial derivative
Correct Answer: A Lagrange multiplier
Explanation:
The symbol is an auxiliary multiplier used to incorporate the constraint.
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20Subject to , at which point is minimized?
Lagrange method of multiplier
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The sum of the squares is smallest when the fixed total is divided equally, giving .
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21Evaluate
limits and continuity
Medium
A.
B.
C.The limit does not exist
D.
Correct Answer: The limit does not exist
Explanation:
Along , the expression is . Along , it equals . Since the path limits differ, the limit does not exist.
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22Let Which statement is correct?
limits and continuity
Medium
A. is continuous at
B. has no limit at the origin
C. has limit at the origin
D. is discontinuous along
Correct Answer: is continuous at
Explanation:
Since , we have . This equals .
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23For which pair of paths proves that the limit at the origin does not exist?
limits and continuity
Medium
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
Along , , while along , . The unequal path limits show that the two-variable limit does not exist.
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24If , which pair gives and ?
partial derivatives and total derivative
Medium
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
Differentiate while holding the other variable constant. The chain rule gives and .
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25For , what is the total differential at ?
partial derivatives and total derivative
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Here and . At , these equal and , so .
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26Use linearization at to approximate for .
partial derivatives and total derivative
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
At , , , and . Thus .
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27For , estimate the change at when and .
partial derivatives and total derivative
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since and at , .
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28If , where and , find at .
chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Substitution gives . Therefore , which equals at .
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29Let , where and . Find at .
chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , . Hence at .
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30Given , , and , find at .
chain rule
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
In polar form, . Thus , giving at .
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31For , evaluate .
Euler's theorem for homogeneous functions
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Each term of has total degree , so is homogeneous of degree . Euler's theorem gives .
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32Let . What is ?
Euler's theorem for homogeneous functions
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Scaling and by scales by , so is homogeneous of degree . Euler's theorem therefore gives .
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33If is twice differentiable and homogeneous of degree , which identity is valid?
Euler's theorem for homogeneous functions
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Differentiating Euler's identity and combining the resulting equations yields the second-order Euler identity with factor .
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34Find the minimum value of .
maxima and minima for a function of two variables
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Completing squares gives . Therefore the minimum value is , attained at .
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35Classify the stationary point of .
maxima and minima for a function of two variables
Medium
A.A local minimum at
B.A local maximum at
C.A saddle point at
D.A saddle point at
Correct Answer: A saddle point at
Explanation:
The stationary point is . Since , , and , the Hessian determinant is , indicating a saddle point.
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36For , how is the stationary point classified?
maxima and minima for a function of two variables
Medium
A.An inconclusive point
B.A saddle point
C.A local maximum
D.A local minimum
Correct Answer: A local minimum
Explanation:
At , , , and . The Hessian determinant is , and , so it is a local minimum.
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37Find the global maximum of on .
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 global maximum is at .
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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 or , giving .
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39Find the point on the line that is closest to the origin.
Lagrange method of multiplier
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Minimize subject to . The Lagrange equations give , so .
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40Find the maximum of subject to .
Lagrange method of multiplier
Medium
A. at
B. at
C. at
D. at
Correct Answer: at
Explanation:
The objective vector is parallel to the maximizing radius. Since lies on the circle, the maximum is .
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41Consider Which statement about the limit as is correct?
limits and continuity
Hard
A.The limit diverges to infinity along every parabolic path.
B.The limit does not exist because along it equals .
C.The limit is because that value occurs along .
D.The limit is because it is along every straight line.
Correct Answer: The limit does not exist because along it equals .
Explanation:
Straight-line paths all give , but substituting gives , which depends on . Hence the joint limit does not exist.
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42For real , define For which values of is continuous at the origin?
limits and continuity
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Since , the magnitude is bounded by , which tends to exactly when . At , the path gives .
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43Evaluate the behavior of as .
limits and continuity
Hard
A.The joint limit exists and equals .
B.The joint limit exists and equals .
C.Both iterated limits fail to exist.
D.The joint limit fails, although both iterated limits equal .
Correct Answer: The joint limit fails, although both iterated limits equal .
Explanation:
Along , the expression tends to , so the joint limit is path-dependent. Taking either variable to zero first gives an iterated limit of .
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44Define Which statement is correct at the origin?
partial derivatives and total derivative
Hard
A. is differentiable with total derivative .
B. is continuous and has every directional derivative, but is not differentiable.
C. is continuous, but its directional derivative along does not exist.
D. is discontinuous, although both partial derivatives exist.
Correct Answer: is continuous and has every directional derivative, but is not differentiable.
Explanation:
For , . This dependence is not linear in , so no total derivative exists, even though is continuous and all directional derivatives exist.
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45Near , the equation defines as a function of and . What is its total differential at ?
partial derivatives and total derivative
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Differentiating implicitly gives . At this reduces to .
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46Let At the origin, what are respectively?
partial derivatives and total derivative
Hard
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
Directly from the definitions, and . Differentiating these gives and , showing that mixed partials need not agree without suitable continuity.
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47Let , where , , and has continuous second derivatives. Which identity is correct?
chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The gradients of and are orthogonal and each has squared norm . Also, and are harmonic, so their Laplacian terms vanish.
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48Let , where has continuous second derivatives. Which expression equals , with all derivatives of evaluated at ?
chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
At , , , , and . Substitution into the second-order chain rule leaves .
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49The variables are transformed successively by What is the Jacobian determinant ?
chain rule
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The two Jacobian determinants are and . Since , their product is .
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50If is twice continuously differentiable and homogeneous of degree , which second-order identity follows from Euler's theorem?
Euler's theorem for homogeneous functions
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Applying the radial operator to yields the stated second-order identity.
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51Suppose is homogeneous of degree on a cone excluding the origin, and define Which relation must satisfy?
Euler's theorem for homogeneous functions
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Both numerator and denominator are homogeneous of degree , so is homogeneous of degree . Euler's theorem therefore gives .
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52Let be twice continuously differentiable and homogeneous of degree . What must be true of its Hessian at every point ?
Euler's theorem for homogeneous functions
Hard
A.Its determinant must equal .
B.It must be the zero matrix.
C.Its determinant must equal .
D.Its trace must equal .
Correct Answer: Its determinant must equal .
Explanation:
Differentiating Euler's identity gives . Thus the nonzero radial vector lies in the Hessian's null space, making the Hessian singular.
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53For which classification of its critical points is correct?
maxima and minima for a function of two variables
Hard
A.All three critical points are non-strict local minima.
B.The origin is a local minimum, while and are saddles.
C.The origin is a local maximum, while and are minima.
D.The origin is a saddle, while and are global minima.
Correct Answer: The origin is a saddle, while and are global minima.
Explanation:
The critical points are , , and . The Hessian is indefinite at the origin and positive definite at the other two points, where . Coercivity makes these minima global.
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54For the parameterized function how is the origin classified?
maxima and minima for a function of two variables
Hard
A.It is a strict local minimum for and a saddle otherwise.
B.It is a non-strict local minimum for and a strict minimum for .
C.It is a strict local minimum for , non-strict for , and a saddle for .
D.It is a strict local minimum for every real .
Correct Answer: It is a strict local minimum for , non-strict for , and a saddle for .
Explanation:
At , , giving a non-strict minimum. For the quartic form is positive away from the origin; for it is negative along but positive along either axis.
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55The function has a critical point at the origin, where the Hessian test is inconclusive. What is the correct classification?
maxima and minima for a function of two variables
Hard
A.A saddle point
B.A strict local maximum
C.A non-strict local minimum
D.A strict local minimum
Correct Answer: A saddle point
Explanation:
Along , , which takes both positive and negative values arbitrarily close to the origin. Hence the origin is a saddle despite the zero Hessian.
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56Classify the origin for
maxima and minima for a function of two variables
Hard
A.It is a strict local maximum.
B.It is a saddle point.
C.It is a non-strict local minimum.
D.It is a strict local minimum.
Correct Answer: It is a saddle point.
Explanation:
Along , , while along , . Thus both signs occur arbitrarily close to the origin.
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57Find the constrained extrema of subject to .
Lagrange method of multiplier
Hard
A.The maximum is and the minimum is .
B.The maximum is and the minimum is .
C.The maximum is and the minimum is .
D.The maximum is and the minimum is .
Correct Answer: The maximum is and the minimum is .
Explanation:
Writing gives and . Since , the extrema are , attained when .
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58Minimize subject to the cusp constraint . Which statement is correct at the origin?
Lagrange method of multiplier
Hard
A.The origin is not an extremum because no Lagrange multiplier exists.
B.The origin is a strict constrained minimum, but the standard multiplier equation fails there.
C.The origin is a strict constrained maximum detected by a zero multiplier.
D.The origin is a non-strict constrained minimum detected by a unique multiplier.
Correct Answer: The origin is a strict constrained minimum, but the standard multiplier equation fails there.
Explanation:
The constraint can be parameterized as , so has a strict minimum at . However, , so cannot hold.
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59Determine the maximum and minimum of subject to
Lagrange method of multiplier
Hard
A.The maximum is at , and the minimum is at .
B.The maximum is at , and the minimum is at .
C.The maximum is at , and the minimum is at .
D.The maximum is at , and the minimum is at .
Correct Answer: The maximum is at , and the minimum is at .
Explanation:
The multiplier equations force at an extremum. The constraint then gives , so or , yielding values and .
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60Find the global minimum of subject to .
Lagrange method of multiplier
Hard
A., attained at
B., attained at
C., attained at only
D., attained at
Correct Answer: , attained at
Explanation:
Using reduces the objective to . Its minimum occurs when , giving , , and minimum value .
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