Differential Calculus Questions and Answers

Answer: A

The integral equals \(f(\xi)(b - a)\).

This is the Mean Value Theorem for integrals: a continuous function on [a, b] takes its average value somewhere in the interval. Since the average value is \(\frac{1}{b - a}\int_{a}^{b} f(x)dx\), there is a \(\xi\) with

\[ \int_{a}^{b} f(x)dx = f(\xi)(b - a) \]

Geometrically, the area under the curve equals a rectangle of width b - a and height \(f(\xi)\). The forms with \(\xi - a\) or \(b - \xi\) use only part of the interval, and 0 is true only for special functions, not in general.

Answer: A

The values are \(\alpha = -3\pi\) and \(p = \pi\).

At \(x = \pi\) the denominator is \(\pi - \pi + 2\sin\pi = 0\). For the limit to be finite, the numerator must vanish there too:

\[ \begin{aligned} \pi^{2} + \alpha\pi + 2\pi^{2} &= 0 \\ \alpha &= -3\pi \end{aligned} \]

The limit is then of the form 0/0, and L'Hôpital's rule gives

\[ \begin{aligned} p &= \lim_{x \to \pi}\frac{2x + \alpha}{1 + 2\cos x} \\ &= \frac{2\pi - 3\pi}{1 - 2} = \pi \end{aligned} \]

Any other \(\alpha\) leaves a non-zero numerator over a zero denominator, so the limit would be infinite; that rules out the other three options.

Answer: C

At x = 0, \(|x|\) is continuous and non-differentiable.

It is continuous because both one-sided limits equal the value of the function: \(\lim_{x \to 0}|x| = 0 = f(0)\).

It is not differentiable because the two one-sided derivatives differ:

\[ \begin{aligned} f'(0^{+}) &= \lim_{h \to 0^{+}}\frac{h}{h} = 1 \\ f'(0^{-}) &= \lim_{h \to 0^{-}}\frac{-h}{h} = -1 \end{aligned} \]

The graph has a sharp corner at the origin. 'Non-continuous and differentiable' is impossible, since differentiability implies continuity, and the function is plainly continuous, which rules out the remaining two options.

Answer: B

The limit is 1/2.

Both numerator and denominator tend to 0, so apply L'Hospital's rule twice:

\[ \begin{aligned} \lim_{x \to 0}\frac{1-\cos x}{x^{2}} &= \lim_{x \to 0}\frac{\sin x}{2x} \\ &= \lim_{x \to 0}\frac{\cos x}{2} = \frac{1}{2} \end{aligned} \]

The same result follows from the series \(\cos x = 1 - x^{2}/2 + \dots\), which makes the numerator \(x^{2}/2\) for small x.

1 is the value of \(\lim \sin x / x\), a different standard limit; 1/4 and 2 come from mishandling the factor of 2.

Answer: D

At x = 0, \(x^{3}+1\) has a point of inflection.

The first derivative \(f'(x) = 3x^{2}\) is zero at x = 0, so it is a stationary point. But \(f'(x)\) is positive on both sides, so the function keeps increasing through x = 0 and there is no maximum or minimum. The second derivative \(f''(x)=6x\) changes sign from negative to positive there, so the curve changes from concave down to concave up: a stationary point of inflection.

A singularity would need the function or its derivative to blow up, but a polynomial is smooth everywhere.

Answer: B

The true statement is \(\nabla f(\mathbf{x}_{0})^{T}\mathbf{p} = 0\) and \(\mathbf{p}^{T}\nabla^{2}f(\boldsymbol{\psi})\mathbf{p} \gt 0\).

At a local minimum of a differentiable function the gradient is zero, so the first-order term \(\nabla f(\mathbf{x}_{0})^{T}\mathbf{p}\) vanishes for every direction. A strict minimum means \(f(\mathbf{x}_{0} + \mathbf{p}) \gt f(\mathbf{x}_{0})\) for every small non-zero p. With the first-order term gone, the expansion leaves

\[ \begin{aligned} &f(\mathbf{x}_{0} + \mathbf{p}) - f(\mathbf{x}_{0}) \\ &= \tfrac{1}{2}\mathbf{p}^{T}\nabla^{2}f(\boldsymbol{\psi})\mathbf{p} \end{aligned} \]

so the quadratic term must be positive.

A positive gradient term (option A) contradicts the zero gradient at a minimum. A zero quadratic term would leave f unchanged, so the minimum would not be strict, and a negative one would make x₀ a point where f decreases, a maximum or saddle.

Answer: A

The minimum value is 0.

Find the critical points: \(f'(x) = 6x^{2} - 18x + 12 = 6(x - 1)(x - 2)\), so \(x = 1\) and \(x = 2\). Evaluate at these and at the end points:

\[ \begin{aligned} f(0) &= 0,\quad f(1) = 5 \\ f(2) &= 4,\quad f(3) = 9 \end{aligned} \]

The smallest is \(f(0) = 0\), at the left end point. 4 is only the local minimum at \(x = 2\), and 5 and 9 are the local maximum and the value at \(x = 3\). On a closed interval the end points must always be checked.

Answer: B

The derivative is \(-\dfrac{x}{y^{2}} \ln v - \dfrac{u}{v}\).

By the chain rule, with \(\partial u/\partial y = -x/y^{2}\) and \(\partial v/\partial y = -1\):

\[ \begin{aligned} \frac{\partial F}{\partial y} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} \\ &= \ln v \left(-\frac{x}{y^{2}}\right) + \frac{u}{v}(-1) \\ &= -\frac{x}{y^{2}}\ln v - \frac{u}{v} \end{aligned} \]

The first option has the right first term but a plus sign on the second, which drops the -1 from \(\partial v/\partial y\). The third and fourth lose the minus sign from differentiating x/y with respect to y.

Answer: D

The correct statement is that \(g(x)\) is differentiable at \(x = 0\).

Near 0, \(\sin x \approx x\), so \(g(x) \approx |x|x\), and

\[ \begin{aligned} g'(0) &= \lim_{h \to 0} \frac{|h|\sin h}{h} \\ &= \lim_{h \to 0} |h|\frac{\sin h}{h} = 0 \end{aligned} \]

from both sides, so the derivative exists and equals 0.

The others fail. \(|f(x)| = |x||\sin x|\), which differs from g wherever \(\sin x \lt 0\). g is odd, the product of the even |x| and the odd sin x. And since \(g = -f\) for \(x \lt 0\), the maxima of g there sit at the minima of f, so the maxima do not coincide.

Answer: A

The true statement is \(f\) is NOT continuous at (0,0).

Partial derivatives first. On the \(x\)-axis, \(f(x,0) = 0\), and on the \(y\)-axis, \(f(0,y) = 0\), so both difference quotients are 0 and \(f_{x}(0,0) = f_{y}(0,0) = 0\). That rules out every option giving a partial derivative the value 1.

Now approach the origin along a path close to the excluded curve \(y = -|x|\), where the denominator vanishes. Take \(x \gt 0\) and \(y = -x + x^{2}\):

\[ \begin{aligned} |x| + y &= x^{2} \\ xy &= -x^{2} + x^{3} \\ f &= -1 + x \to -1 \end{aligned} \]

Along the axes \(f \to 0\), but along this path \(f \to -1\), so the limit at (0,0) does not exist and \(f\) is not continuous there, even though both partial derivatives exist.