Limits, Continuity and Differentiability Questions and Answers

Answer: B

The limit is 1.

The numerator is the expansion of a sine of a difference.

\[ \sin x\cos 2 - \cos x\sin 2 = \sin(x - 2) \]

So the expression is \(\frac{\sin(x - 2)}{x - 2}\), and with t = x - 2 tending to 0 it tends to \(\frac{\sin t}{t} \to 1\). The value 0 is what direct substitution gives in the numerator alone, and 2 or 4 bring the point x = 2 into the answer.

Answer: E

There are 5 discontinuities.

The greatest integer function jumps by 1 at every integer and is constant between consecutive integers. In the open interval (0, 6) the integers are 1, 2, 3, 4 and 5.

The end points 0 and 6 are not in the interval, so they are not counted. Counting them would give 7; the smaller options miss some of the five interior jumps.

Answer: D

f is not differentiable at x = 5.

\(|x - 5|\) has a corner where \(x - 5 = 0\): its slope is -1 to the left and +1 to the right. So f has slopes +1 and -1 on the two sides of x = 5, and no single derivative exists there.

Everywhere else f is linear, so it is differentiable at x = 10, 15, -5 and -15. x = 15 and x = -5 are where f(x) = 0, which has nothing to do with differentiability.

Answer: E

The value is \(g'(0) = cos(log3)\).

By the chain rule \(g'(x) = f'(f(x))\cdot f'(x)\), and \(f'(x) = -\cos x\). Also \(f(0) = \log 3 - 0 = \log 3\).

\[ \begin{aligned} g'(0) &= f'(\log 3)\cdot f'(0) \\ &= (-\cos(\log 3))(-\cos 0) = \cos(\log 3) \end{aligned} \]

\(-\cos(\log 3)\) forgets the second factor \(f'(0) = -1\), and the sine options differentiate \(\sin\) wrongly. \(2\cos(\log 3)\) adds the two factors instead of multiplying them.

Answer: A

The value is \(\frac{-3}{5}\).

Differentiate \(y = \cos x\cos y\) implicitly, using the product rule.

\[ \begin{aligned} y' &= -\sin x\cos y - \cos x\sin y\, y' \\ y' &= \frac{-\sin x\cos y}{1 + \cos x\sin y} \end{aligned} \]

At \(x = \frac{\pi}{3}, y = \frac{\pi}{6}\): \(\sin x\cos y = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{3}{4}\) and \(\cos x\sin y = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}\), so

\[ y' = \frac{-3/4}{1 + 1/4} = -\frac{3}{5} \]

Dropping the minus sign gives \(\frac{3}{5}\), and inverting the fraction gives \(\pm\frac{5}{3}\).

Answer: B

The value is \(f'''(3) = 6\).

The numbers \(f'(1)\), \(f''(2)\) and \(f'''(3)\) are constants, so f is a cubic \(x^{3} + ax^{2} + bx + c\) with leading coefficient 1.

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

The third derivative of a cubic with leading coefficient 1 is the constant 3! = 6, whatever a, b and c are, so \(f'''(3) = 6\). 3 and 9 come from stopping at a lower derivative or multiplying by the point 3, and \(f''(2)\) is one of the coefficients, not the answer.

Answer: B

The value is \(2\sqrt{2}\).

Simplify u first. Using \(\sec^{-1}(-y) = \pi - \sec^{-1}y\) and \(\sec^{-1}(\sec 2\theta) = 2\theta\) for \(2\theta\) in the principal range,

\[ \begin{aligned} u = \pi - 2\theta &\Rightarrow \frac{du}{d\theta} = -2 \\ v = \cos\theta &\Rightarrow \frac{dv}{d\theta} = -\sin\theta \end{aligned} \]

\[ \begin{aligned} \frac{du}{dv} &= \frac{-2}{-\sin\theta} = \frac{2}{\sin\theta} \\ &= \frac{2}{1/\sqrt{2}} = 2\sqrt{2} \end{aligned} \]

(At exactly \(\theta = \frac{\pi}{4}\), \(\sec 2\theta\) is undefined; the value is that of the derivative formula, which is the same on both sides of that point.) \(\sqrt{2}\) loses the factor 2 from \(2\theta\), and \(-\sqrt{2}\) also loses a sign.

Answer: E

f is increasing on \((0, \infty)\).

A differentiable function increases where its derivative is positive.

\[ f'(x) = e^{x} - 1 \gt 0 \iff e^{x} \gt 1 \iff x \gt 0 \]

So f increases on the whole of \((0, \infty)\) and decreases on \((-\infty, 0)\), with its minimum value 1 at x = 0. \((-\infty, 0)\) and \((-1, 0)\) are where it decreases, and \((-1, 1)\) straddles the minimum. \((0, 4)\) lies inside the interval of increase, so f increases there too, but it is only part of it; the question asks for the interval in which f increases, which is all of \((0, \infty)\).

Answer: B

The maximum occurs at x = 3.

\(|x - 3| \ge 0\), with equality only at x = 3. So \(f(x) = 10 - |x - 3| \le 10\), and the largest value, 10, is reached at x = 3.

At every other option the subtracted term is positive: f(0) = 7, f(-3) = 4, f(10) = 3 and f(1) = 8. x = 10 confuses the maximum value with the point where it occurs.

Answer: D

The velocity is zero at t = 6.

Velocity is the rate of change of distance.

\[ v = \frac{ds}{dt} = t - 6 = 0 \Rightarrow t = 6\ \text{s} \]

t = 8 confuses the constant term of s with a time, and t = 3 halves the root, as if differentiating \(\frac{t^{2}}{2}\) gave \(2t\). t = 1 and t = 4 leave a non-zero velocity of -5 and -2 m/s.