Three of these four appear in the second-order group of a two-variable Taylor expansion. The exception is
Answer: A
The second-order group is \( \tfrac{1}{2}\left[h^{2}f_{xx} + 2hk\,f_{xy} + k^{2}f_{yy}\right] \) with \( h = x-a \) and \( k = y-b \). A product of two *first* derivatives never appears: the expansion is built from single derivatives of increasing order, not from products of lower ones
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Match the failure of continuity with its description. 1. Removable 2. Jump 3. Infinite P. The one-sided limits exist and differ Q. The limit exists but the value assigned differs R. The function grows without bound at the point
Answer: D
A removable discontinuity is repaired by reassigning one value, the limit already existing. A jump has two one-sided limits that disagree, so no reassignment helps. An infinite discontinuity has no finite limit on at least one side, the function running away as the point is approached
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Match the curve with the number of its asymptotes parallel to the axes. 1. \( y = 1/x \) 2. \( y = 1/(x^{2}-1) \) 3. \( y = x + 1/x \) P. Two vertical and one horizontal Q. One vertical and one horizontal R. One vertical and none horizontal
Answer: C
The first has \( x = 0 \) and \( y = 0 \). The second vanishes in the denominator at \( x = \pm 1 \) and tends to zero at infinity, so two vertical and one horizontal. The third has \( x = 0 \) vertical but grows like \( x \), so its remaining asymptote is the oblique line \( y = x \) rather than a horizontal one
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Just one of these four is a legitimate use of L'Hopital's rule. Which?
Answer: B
Repeated application is standard and often necessary, as in \( (1-\cos x)/x^{2} \). The rule needs an indeterminate form, so a determinate quotient is out; it replaces \( f/g \) by \( f'/g' \), not by the derivative of the quotient; and if only the numerator vanishes the limit is simply zero
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True or false: the mixed partial derivatives of every function of two variables are equal.
Answer: A
Clairaut's theorem gives equality only when the mixed derivatives are continuous at the point. Counterexamples exist in which \( f_{xy} \) and \( f_{yx} \) differ at the origin, both being discontinuous there. Mere existence of the first derivatives, or a domain covering the plane, is not enough
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True or false: a function of two variables with a negative discriminant at a stationary point has a maximum there.
Answer: D
A negative discriminant \( f_{xx}f_{yy} - f_{xy}^{2} \) means the surface curves upward in one direction and downward in another, which is a saddle and not an extremum of any kind. A maximum needs a positive discriminant together with \( f_{xx} < 0 \), so both conditions carry weight
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