Differential Calculus Questions and Answers

Answer: A

The limit is \(-\frac{1}{3}\).

Expand numerator and denominator in powers of \(x\) near 0, using \(\cos x = 1 - x^{2}/2 + \cdots\) and \(\sin x = x - x^{3}/6 + \cdots\):

\[ \begin{aligned} x\cos x - \sin x &= x - \tfrac{x^{3}}{2} - x + \tfrac{x^{3}}{6} + \cdots \\ &= -\tfrac{x^{3}}{3} + \cdots \\ x^{2}\sin x &= x^{3} + \cdots \end{aligned} \]

The ratio therefore tends to \(-\frac{1}{3}\). The same result follows from L'Hopital's rule applied repeatedly to the 0/0 form.

\(\frac{1}{3}\) drops the sign: the numerator is negative for small positive \(x\), because \(x\cos x \lt \sin x\) there. 3 and \(-3\) invert the ratio.

Answer: C

The value is \(\frac{1}{4u}\).

Every term of \(u = x^{4} + y^{4} + 3x^{2}y^{2}\) has total degree 4, so \(u\) is a homogeneous function of degree 4. By Euler's theorem on homogeneous functions,

\[ \begin{aligned} &x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} \\ &= 4u \end{aligned} \]

Directly: \(x u_x = 4x^{4} + 6x^{2}y^{2}\) and \(y u_y = 4y^{4} + 6x^{2}y^{2}\), which add to \(4x^{4} + 4y^{4} + 12x^{2}y^{2} = 4u\). The question asks for the reciprocal, \((4u)^{-1} = \frac{1}{4u}\).

\(4u\) is the bracket itself, before the power \(-1\) is applied. \(2u\) uses the wrong degree, and 0 would hold only for a function of degree zero.

Answer: D

\(f\) is not differentiable at \(\{-1, 0, 1\}\).

The curves \(y = x\) and \(y = x^{3}\) cross where \(x^{3} = x\), at x = −1, 0 and 1. Between the crossings the larger one switches:

• \(x \lt -1\): \(x \gt x^{3}\), so f = x
• \(-1 \lt x \lt 0\): f = \(x^{3}\)
• \(0 \lt x \lt 1\): f = x
• \(x \gt 1\): f = \(x^{3}\)

At each switch the two pieces have different slopes (1 against \(3x^{2}\), which is 3 at x = ±1 and 0 at x = 0), so each is a corner. The options containing 2 or −2 name points where f is smooth, and \(\{0, 1\}\) misses the corner at −1.

Answer: D

The limit is -1.

Split the fraction:

\[ \begin{aligned} \frac{x^{3} - \sin x}{x} &= x^{2} - \frac{\sin x}{x} \\ &\to 0 - 1 = -1 \end{aligned} \]

using the standard limit \(\sin x / x \to 1\). L'Hospital's rule gives the same: \((3x^{2} - \cos x)/1 \to -1\). 1 drops the minus sign, 0 treats the whole numerator as negligible, and 3 comes from differentiating \(x^{3}\) but forgetting the sine term.

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.