Differential Calculus Questions and Answers

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

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

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

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

Answer: C

The tangent is y - 1 = 2(x - 1).

The slope of \(y = x^{2}\) is \(dy/dx = 2x\), which is 2 at x = 1. The point-slope form through (1,1) is therefore y - 1 = 2(x - 1), that is, y = 2x - 1.

y = 2x - 2 has the right slope but passes through (1, 0), not (1, 1). x = 2y - 1 and x - 1 = 2(y - 1) have slope 1/2, which comes from swapping x and y, the tangent to the inverse curve.

Answer: C

The height is \(\frac{4}{3}R\).

Put the cone's apex on the sphere and its base circle on the sphere too. If the height is h, the base lies a distance h - R from the centre, so the base radius r satisfies \(r^{2} = R^{2} - (h - R)^{2} = 2Rh - h^{2}\).

\[ \begin{aligned} V &= \frac{\pi}{3}(2Rh^{2} - h^{3}) \\ \frac{dV}{dh} &= \frac{\pi}{3}(4Rh - 3h^{2}) = 0 \\ h &= \frac{4R}{3} \end{aligned} \]

The second derivative, \(4R - 6h\), is negative there, so this is a maximum. h = R puts the base through the centre (a cone of volume \(\pi R^{3}/3\), smaller than the optimum \(32\pi R^{3}/81\)), and 3R/2 and 5R/4 are other heights with smaller volume.

Answer: D

Rolle's theorem needs \(f(0) = f(\pi)\) and a point \(c\) in \((0, \pi)\) where \(f'(c) = 0\). Here \(f(0) = \cos 0 = 1\) and \(f(\pi) = \cos 4\pi = 1\), so the theorem applies.

\[ f'(x) = -4\sin 4x = 0 \]

\[ 4x = k\pi \Rightarrow x = \frac{k\pi}{4} \]

Inside the open interval this gives \(\pi/4\), \(\pi/2\) and \(3\pi/4\). Of the values offered, only \(\pi/4\) is one of them. The values \(0\) and \(\pi\) are the end points, not interior points, and \(\pi/3\) gives \(f'(\pi/3) = -4\sin(4\pi/3) \neq 0\). The answer is \(\pi/4\), option D.

Answer: B

Use the sum to product identity, or simply note that the first term is fixed because \(x + y = 6\).

\[ z = \cos 6 + \cos(x - y) \]

The only term that varies is \(\cos(x - y)\), whose largest value is \(1\) (reached when \(x - y = 0\), that is \(x = y = 3\)). So the maximum is \(\cos 6 + 1 = 1 + \cos 6\). The option \(1 - \cos 6\) has the wrong sign on the fixed term, \(1 + \sin 2\) and \(1 + \cos 3\) do not equal \(1 + \cos 6\), since \(\cos 3 \neq \cos 6\). The answer is option B.

Answer: B

The statement that is not true is that every continuous function is differentiable. The function \(f(x) = |x|\) is continuous at \(0\) but has a corner there, so its derivative does not exist at \(0\). The opposite statement, that every differentiable function is continuous, is a standard theorem and is true.

The function \(|x| + \sin|x-3|\) is a sum of continuous functions, so it is continuous, and it has corners at \(x = 0\) and \(x = 3\) where the absolute values change sign, so it is not differentiable there. Both of those statements are true, which leaves option B as the false one.

Answer: A

Take natural logarithms of \(x^y y^x = C\) to get \(y\ln x + x\ln y = \ln C\) and differentiate implicitly:

The negative sign is essential because the two sides of the constraint trade off against each other. The unsigned version in the second option and the options with x and y swapped (the third and fourth) come from errors in collecting the terms with y' or in inverting the fraction.