Limits, Continuity and Differentiability Questions and Answers

Answer: D

The minimum occurs at \(x = -2\).

\(|x + 2| \ge 0\) for all x, with equality only when \(x + 2 = 0\). So the minimum value 0 is attained at \(x = -2\).

At 0, 1, 2 and −1 the function takes the values 2, 3, 4 and 1, all larger than 0.

Answer: E

g is increasing in \(\left[\frac{1}{2}, \infty\right)\).

\(g'(x) = 2x - 1\), which is non-negative for \(x \ge \frac{1}{2}\), so g increases on \(\left[\frac{1}{2}, \infty\right)\). It decreases on \(\left(-\infty, \frac{1}{2}\right]\), with its minimum at \(x = \frac{1}{2}\).

\((-\infty, \infty)\) and (−5, 5) include part of the decreasing stretch, and \((-\infty, 0)\) lies wholly in it. '\((0, -\infty)\)' is not a valid interval.

A quadratic \(ax^{2} + bx + c\) with \(a \gt 0\) always decreases up to its vertex at \(x = -\frac{b}{2a}\) and increases after it; here the vertex is at \(x = \frac{1}{2}\).

Answer: D

The particle is at rest when t = 3.

The velocity is the derivative of displacement:

\[ v = \frac{ds}{dt} = 2t - 6 = 0 \Rightarrow t = 3 \text{ s} \]

Setting s itself to zero has no real solution here, and the other times give non-zero velocities (for example \(v = 2\) at \(t = 4\)).

Answer: D

The maximum value is \(3\sqrt{3}\).

The domain is \(0 \le x \le 4\). Maximise \(f^{2} = x^{2}(4x - x^{2}) = 4x^{3} - x^{4}\):

\[ \begin{aligned} \frac{d}{dx}(4x^{3} - x^{4}) &= 12x^{2} - 4x^{3} = 4x^{2}(3 - x) \\ x = 3: \ f &= 3\sqrt{12 - 9} = 3\sqrt{3} \end{aligned} \]

The function is 0 at both ends of the domain, so this critical point gives the maximum. \(4\sqrt{3}\) and \(6\sqrt{3}\) evaluate the root at the wrong point, and \(\sqrt{3}\) forgets the factor x.

Answer: D

The limit is \(-\frac{1}{6}\).

The limit is the definition of the derivative \(f'(1)\):

\[ \begin{aligned} f'(x) &= \frac{-1}{2\sqrt{10 - x}} \\ f'(1) &= \frac{-1}{2\sqrt{9}} = -\frac{1}{6} \end{aligned} \]

The minus sign comes from the chain rule, since the derivative of \(10 - x\) is -1. \(\frac{1}{6}\) drops that sign, and 3 is \(f(1)\), the value of the function rather than its rate of change.

Answer: D

The limit is -1.

Split the \(x^{2}\) between the two factors and use \(\frac{\sin kx}{x} \to k\):

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

The product of the limits is \((-1)(1) = -1\). Taking \(\sin 2x\) as x instead of 2x near 0 would make the first factor 0 and give the trap answer 0. The values -2 and 2 come from multiplying the wrong pieces.

Answer: E

\(\lambda\) is 12.

For continuity at x = 2, \(\lambda\) must equal the limit of the first expression:

\[ \begin{aligned} \frac{3x^{2} - 12}{x - 2} &= \frac{3(x - 2)(x + 2)}{x - 2} = 3(x + 2) \\ \lim_{x \to 2} 3(x + 2) &= 12 \end{aligned} \]

Everywhere else the function is a rational expression with a non-zero denominator, so x = 2 is the only point to check. 4 is the limit of \(x + 2\) without the factor 3, and 0 is the value of the numerator at x = 2, which says nothing about the limit by itself.

Answer: E

The function is discontinuous at {-2, 2}.

A rational function is continuous wherever its denominator is non-zero. Here \(x^{2} - 4 = 0\) at x = 2 and x = -2, where f is not defined, so these are the points of discontinuity.

At x = 0 the numerator is zero but the denominator is -4, so f(0) = 0 and f is continuous there; any option including 0 is wrong. {2, 4} and {0, 2} miss the root -2.

Answer: A

\(\frac{dy}{dx}\) at x = 1 is 0.

Take logarithms of \((xe)^{y} = e^{x}\): \(y(\ln x + 1) = x\), so \(y = \frac{x}{1 + \ln x}\). Differentiate by the quotient rule:

\[ \begin{aligned} \frac{dy}{dx} &= \frac{(1 + \ln x) - x \cdot \frac{1}{x}}{(1 + \ln x)^{2}} \\ &= \frac{\ln x}{(1 + \ln x)^{2}} \end{aligned} \]

At x = 1, \(\ln 1 = 0\), so the derivative is 0: y has a stationary point there. The other values come from forgetting that the derivative of \(\ln x\) is \(\frac{1}{x}\).

Answer: E

\(\frac{dy}{dt}\) is \(2(1 - 3t^{2})\).

Write y in terms of t. Since \(\sin 2x = 2\sin x\cos x\),

\[ \begin{aligned} y &= 2\sin^{2}x\cos x = 2(1 - t^{2})t \\ &= 2t - 2t^{3} \\ \frac{dy}{dt} &= 2 - 6t^{2} = 2(1 - 3t^{2}) \end{aligned} \]

\(1 - 3t^{2}\) and \(\frac{1}{2}(1 - 3t^{2})\) lose the factor 2 from the double angle formula, and the options with \(3t^{2} - 1\) have the wrong sign.