Limits, Continuity and Differentiability Questions and Answers

Answer: E

The limit is 1.

Add and subtract \(f(1)\) in the numerator:

\[ \begin{aligned} \frac{xf(1) - f(x)}{x - 1} &= \frac{(x - 1)f(1) - (f(x) - f(1))}{x - 1} \\ &= f(1) - \frac{f(x) - f(1)}{x - 1} \\ &\to f(1) - f'(1) = 2 - 1 = 1 \end{aligned} \]

The same result follows from L'Hospital's rule: the derivative of the numerator is \(f(1) - f'(x)\), which is 1 at \(x = 1\). A sign slip gives \(-1\).

Answer: C

\(f(x) - g(x) = 5\) at \(x = \frac{3}{2}\).

Let \(h = f - g\). Then \(h'' = 0\), so \(h\) is linear: \(h(x) = px + q\).

\[ \begin{aligned} h'(1) &= 4 - 2 = 2 = p \\ h(2) &= 9 - 3 = 6 = 2p + q, \quad q = 2 \\ h\left(\tfrac{3}{2}\right) &= 2 \times \tfrac{3}{2} + 2 = 5 \end{aligned} \]

6 is the value at \(x = 2\) (not offered) and 2 is the slope; neither is the value at \(\frac{3}{2}\).

Answer: B

\(\frac{dy}{dx} = \frac{3}{2(x - 1)(x + 2)}\).

Simplify with log laws before differentiating:

\[ \begin{aligned} y &= \tfrac{1}{2}[\log(x - 1) - \log(x + 2)] \\ \frac{dy}{dx} &= \frac{1}{2}\left[\frac{1}{x - 1} - \frac{1}{x + 2}\right] \\ &= \frac{1}{2} \cdot \frac{3}{(x - 1)(x + 2)} \end{aligned} \]

Forgetting the \(\frac{1}{2}\) from the square root gives \(\frac{3}{(x - 1)(x + 2)}\), and forgetting the 3 from combining the fractions gives \(\frac{1}{2(x - 1)(x + 2)}\).

Answer: E

\(\frac{d^{2}x}{dy^{2}} = \frac{\sin x - e^{x}}{(\cos x + e^{x})^{3}}\).

Here \(\frac{dy}{dx} = \cos x + e^{x}\) and \(\frac{d^{2}y}{dx^{2}} = -\sin x + e^{x}\). For the inverse function,

\[ \begin{aligned} \frac{dx}{dy} &= \frac{1}{y'} \\ \frac{d^{2}x}{dy^{2}} &= \frac{d}{dx}\left(\frac{1}{y'}\right)\frac{dx}{dy} = -\frac{y''}{(y')^{3}} \end{aligned} \]

\[ \begin{aligned} \frac{d^{2}x}{dy^{2}} &= -\frac{e^{x} - \sin x}{(\cos x + e^{x})^{3}} \\ &= \frac{\sin x - e^{x}}{(\cos x + e^{x})^{3}} \end{aligned} \]

The options with a square in the denominator forget the chain-rule factor \(\frac{dx}{dy}\), and \(\frac{e^{x} - \sin x}{(\cos x + e^{x})^{3}}\) drops the minus sign.

Answer: E

\(\frac{dy}{dx} = \frac{1}{x}\left[\frac{1}{\log_{e}10} + 1\right]\).

Change the base: \(\log_{10}x = \frac{\log_{e}x}{\log_{e}10}\). Then

\[ \begin{aligned} \frac{dy}{dx} &= \frac{1}{x\log_{e}10} + \frac{1}{x} \\ &= \frac{1}{x}\left[\frac{1}{\log_{e}10} + 1\right] \end{aligned} \]

Since \(\frac{1}{\log_{e}10} = \log_{10}e\), this also equals \(\frac{1 + \log_{10}e}{x}\), which is not among the options: option B has \(\log_{e}10\) where \(\log_{10}e\) is needed, and option A has the wrong sign. The options without \(\frac{1}{x}\) are not derivatives of logarithms at all.

Answer: C

\(f\) is strictly increasing on [-1, 0] and [1, \(\infty\)).

\[ f'(x) = 4x^{3} - 4x = 4x(x - 1)(x + 1) \]

The critical points are \(-1\), 0 and 1. The sign of \(f'\) is negative for \(x \lt -1\), positive on \((-1, 0)\), negative on \((0, 1)\) and positive for \(x \gt 1\). A function is strictly increasing on a closed interval where \(f' \gt 0\) inside and vanishes only at isolated endpoints, so it is strictly increasing on \([-1, 0]\) and \([1, \infty)\).

Intervals starting at \(-2\) include part of \((-\infty, -1)\), where \(f\) decreases, and \([0, \infty)\) includes \((0, 1)\), where it also decreases.

Answer: D

The area increases at \(200\pi\) cm²/sec.

With \(A = \pi r^{2}\), differentiate with respect to time:

\[ \begin{aligned} \frac{dA}{dt} &= 2\pi r\frac{dr}{dt} \\ &= 2\pi \times 20 \times 5 = 200\pi \end{aligned} \]

\(100\pi\) forgets the factor 2 from differentiating \(r^{2}\), and \(400\pi\) is \(\pi r^{2}\) itself at \(r = 20\), the area rather than its rate of change.

Answer: A

\(a\) must be greater than or equal to -2.

\(f'(x) = 2x + a\). For \(f\) to be increasing on [1, 2], \(f'(x) \ge 0\) throughout the interval. Since \(2x + a\) is smallest at \(x = 1\), the condition is

\[ 2(1) + a \ge 0 \quad \Rightarrow \quad a \ge -2 \]

Using the other end, \(x = 2\), would give \(a \ge -4\), which is not enough: with \(a = -3\), for example, \(f'(1) = -1 \lt 0\) and \(f\) decreases at the start of the interval.

Answer: E

The absolute maximum is 4.

\(f'(x) = 3x^{2} - 3 = 0\) gives \(x = \pm 1\); only \(x = 1\) lies in [0, 2]. Compare the critical point with the endpoints:

\[ \begin{aligned} f(0) &= 2 \\ f(1) &= 1 - 3 + 2 = 0 \\ f(2) &= 8 - 6 + 2 = 4 \end{aligned} \]

The largest is 4, at the endpoint \(x = 2\). 0 is the absolute minimum, and 2 is the value at the left endpoint. Checking only the critical point misses the maximum, which is at an end of the interval.

Answer: E

The limit is 0.

As \(x \to 2^{+}\), x is slightly greater than 2, so \([x] = 2\) and the numerator is \(2 - 2 = 0\) for every such x.

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

The denominator is small but not zero, so the quotient is exactly 0 throughout. From the left the numerator would be \(1 - 2 = -1\) and the quotient would tend to infinity, which is why only the right-hand limit is asked.