Limits, Continuity and Differentiability Questions and Answers

Answer: B

\(\frac{dy}{dx} = \) \(\frac{2x - 1}{1 + (x^{2} - x)^{2}}\).

With \(u = x^{2} - x\), \(\frac{d}{dx}\tan^{-1}u = \frac{1}{1 + u^{2}}\cdot\frac{du}{dx}\) and \(\frac{du}{dx} = 2x - 1\).

Option A differentiates \(x^{2} - x\) as \(2x\), dropping the \(-1\); option C uses \(1 - u^{2}\), which is not the denominator in the derivative of \(\tan^{-1}u\); option D has the sign reversed, as for \(\cot^{-1}\); option E multiplies by \(1 + u^{2}\) instead of dividing.

Answer: E

\(f\) is strictly decreasing in \(\left(0, \frac{4}{3}\right)\).

\[ \begin{aligned} f(x) &= x^{3} - 2x^{2} \\ f'(x) &= 3x^{2} - 4x = x(3x - 4) \end{aligned} \]

\(f'(x) \lt 0\) between the roots 0 and \(\frac{4}{3}\), so \(f\) decreases there and increases outside. (1,2) runs past \(\frac{4}{3}\) into the increasing part; (-1,0) and (-1,1) include negative \(x\), where \(f' \gt 0\); \(\left(\frac{4}{3}, \infty\right)\) is where \(f\) increases.

Answer: C

The volume increases at \(\frac{40}{3}\) cm³/s.

A solid hemisphere has total surface area \(S = 2\pi r^{2} + \pi r^{2} = 3\pi r^{2}\) (curved surface plus flat base) and volume \(V = \frac{2}{3}\pi r^{3}\).

\[ \begin{aligned} \frac{dS}{dt} &= 6\pi r\frac{dr}{dt} = 8 \Rightarrow \frac{dr}{dt} = \frac{8}{30\pi} \\ \frac{dV}{dt} &= 2\pi r^{2}\frac{dr}{dt} = 2\pi(25)\frac{8}{30\pi} = \frac{40}{3} \end{aligned} \]

A shortcut: \(\frac{dV}{dS} = \frac{2\pi r^{2}}{6\pi r} = \frac{r}{3}\), so \(\frac{dV}{dt} = \frac{5}{3} \times 8\). Using only the curved surface \(2\pi r^{2}\) gives \(\frac{r}{2} \times 8 = 20\), not offered, and \(\frac{20}{3}\) or \(\frac{80}{3}\) halve or double the answer.

Answer: B

\(f\) is increasing in \((-\infty, -2] \cup [3, \infty)\).

\[ \begin{aligned} f'(x) &= 6x^{2} - 6x - 36 \\ &= 6(x^{2} - x - 6) = 6(x - 3)(x + 2) \end{aligned} \]

\(f'(x) \geq 0\) outside the roots, for \(x \leq -2\) or \(x \geq 3\). The other options shift one of the critical points: -1, -5, 5 and 8 are not roots of \(x^{2} - x - 6\).

Answer: A

\(f(0) = \) 10.

A local minimum at \(x = 2\) needs \(f'(2) = 0\), and the point (2, 6) lies on the curve:

\[ \begin{aligned} f'(2) &= 4 + \alpha = 0 \Rightarrow \alpha = -4 \\ f(2) &= 4 - 8 + \beta = 6 \Rightarrow \beta = 10 \end{aligned} \]

So \(f(x) = x^{2} - 4x + 10 = (x - 2)^{2} + 6\) and \(f(0) = \beta = 10\). 6 is the minimum value itself, not \(f(0)\); -6 and the other signs come from slips in \(\alpha\).

Answer: D

The limit is 2.

Use \(1 - \cos 4x = 2\sin^{2}2x\):

\[ \begin{aligned} \frac{1 - \cos 4x}{\tan^{2}2x} \\ &= \frac{2\sin^{2}2x}{\tan^{2}2x} \\ &= 2\cos^{2}2x \to 2 \end{aligned} \]

as \(x \to 0\). Equivalently, \(1 - \cos 4x \approx 8x^{2}\) and \(\tan^{2}2x \approx 4x^{2}\), whose ratio is 2. \(\frac{1}{2}\) inverts the ratio, and 0 or 3 come from mismatched small-angle approximations.

Answer: A

\(\alpha =\) 2.

For continuity at 0 the left-hand limit must equal \(f(0)\):

\[ \begin{aligned} \lim_{x \to 0^{-}}(x + \alpha) &= \alpha \\ f(0) &= \max(2\cos 0, 2\sin 0) = \max(2, 0) = 2 \end{aligned} \]

so \(\alpha = 2\). Taking \(2\sin 0 = 0\) as the maximum would give 0, and the other values leave a jump at 0.

For \(x \ge 0\) near 0 the maximum is \(2\cos x\), since \(\cos x \gt \sin x\) there, so the right-hand piece starts at 2 and the left-hand piece must meet it.

Answer: B

The limit is \(-1\).

For \(x\) slightly less than 0, say \(-0.1\), the greatest integer \([x]\) is \(-1\), and \(|x| = -x\), so \(\frac{x}{|x|} = -1\):

\[ \begin{aligned} \lim_{x \to 0^{-}}\left(2[x] - \frac{x}{|x|}\right) \\ &= 2(-1) - (-1) \\ &= -1 \end{aligned} \]

Using \([x] = 0\) (the value at 0 itself) gives 1, and taking \(\frac{x}{|x|} = 1\) gives \(-3\); both are the right-hand or wrong-sign readings.

Answer: D

\(\frac{dy}{dx} =\) \(\frac{\sqrt{3}}{2}\).

If \(\theta = \tan^{-1}x\), then \(\sec\theta = \sqrt{1 + x^{2}}\), so \(y = \sqrt{1 + x^{2}}\) and

\[ \begin{aligned} \frac{dy}{dx} &= \frac{x}{\sqrt{1 + x^{2}}} \\ &= \frac{\sqrt{3}}{\sqrt{4}} \\ &= \frac{\sqrt{3}}{2} \end{aligned} \]

at \(x = \sqrt{3}\). \(\frac{1}{2}\) is \(\frac{1}{\sqrt{1 + x^{2}}}\), and 2 is the value of \(y\) itself; the other options do not follow.

Answer: B

\(f'(27) =\) \(\frac{2}{9}\).

Simplify first: \(f(x) = \frac{x^{4/3}}{x^{2/3}} = x^{2/3}\). Then

\[ \begin{aligned} f'(x) &= \frac{2}{3}x^{-1/3} \\ f'(27) &= \frac{2}{3}\cdot\frac{1}{3} = \frac{2}{9} \end{aligned} \]

since \(27^{1/3} = 3\). \(\frac{1}{9}\) drops the factor 2, \(\frac{4}{9}\) doubles it, and \(\frac{1}{3}\) and \(\frac{5}{9}\) come from wrong exponents.

The trap is to differentiate numerator and denominator separately; simplifying the radicals to a single power first avoids the quotient rule entirely.