Limits, Continuity and Differentiability Questions and Answers

Answer: B

The derivative at \(x = \frac{1}{2}\) is \(\frac{-35}{4}\).

By the product rule, \(y'\) is a sum of four terms, each with one factor differentiated. At \(x = \frac{1}{2}\) the factor \(2x - 1\) is zero, so every term that keeps \(2x - 1\) vanishes, and only the term in which \(2x - 1\) is differentiated (giving 2) survives:

\[ \begin{aligned} y'\left(\tfrac{1}{2}\right) &= (x - 1) \cdot 2 \cdot (3 - x)(4 - x) \\ &= \left(-\tfrac{1}{2}\right)(2) \\ &\quad \times \left(\tfrac{5}{2}\right)\left(\tfrac{7}{2}\right) \\ &= -\frac{35}{4} \end{aligned} \]

\(\frac{35}{4}\) drops the minus sign from \(x - 1 = -\frac{1}{2}\). \(\frac{-35}{2}\) is twice the answer, and 35 ignores the halves altogether.

Answer: A

\(f'\left(\frac{\pi}{6}\right) = \frac{-(\sqrt{3} + 1)}{2}\).

At \(x = \frac{\pi}{6}\), \(\cos x - \sin x = \frac{\sqrt{3}}{2} - \frac{1}{2} \gt 0\), and it stays positive nearby, so the modulus can be dropped there: \(f(x) = \cos x - \sin x\).

\[ \begin{aligned} f'(x) &= -\sin x - \cos x \\ f'\left(\tfrac{\pi}{6}\right) &= -\tfrac{1}{2} - \tfrac{\sqrt{3}}{2} \\ &= -\frac{\sqrt{3} + 1}{2} \end{aligned} \]

The positive option \(\frac{\sqrt{3} + 1}{2}\) is what you get by treating the inside as negative and flipping the sign. \(\frac{\sqrt{3}}{2}\) is \(\cos\frac{\pi}{6}\) alone. The others do not arise from differentiating the function.

Answer: A

\(f'\left(\frac{3}{2}\right) = 3\).

At \(x = \frac{3}{2}\), \(x^{2} - 1 = \frac{5}{4} \gt 0\), so near that point \(f(x) = x^{2} - 1\) without the modulus:

\[ \begin{aligned} f'(x) &= 2x \\ f'\left(\tfrac{3}{2}\right) &= 3 \end{aligned} \]

The modulus only changes the formula where \(x^{2} - 1 \lt 0\), that is for \(|x| \lt 1\). \(\frac{3}{2}\) is the point itself, not the derivative, and 2 and 4 come from \(2x\) at other points (\(x = 1\) and \(x = 2\)).

Answer: A

A critical point is \(\left(1, \frac{-11}{3}\right)\).

Critical points are where \(f'(x) = 0\):

\[ \begin{aligned} f'(x) &= x^{2} + 6x - 7 = (x + 7)(x - 1) \\ x &= 1 \text{ or } x = -7 \\ f(1) &= \tfrac{1}{3} + 3 - 7 = -\tfrac{11}{3} \end{aligned} \]

The other critical point, \(x = -7\), is not offered. The options at \(x = 0\), \(-1\), 2 and \(-2\) are not zeros of \(f'\): for example \(f'(2) = 4 + 12 - 7 = 9\).

Answer: C

\(f\) is decreasing on \((-2, -1)\).

A function decreases where its derivative is negative:

\[ \begin{aligned} f'(x) &= 6x^{2} + 18x + 12 \\ &= 6(x^{2} + 3x + 2) = 6(x + 1)(x + 2) \end{aligned} \]

The product \((x + 1)(x + 2)\) is negative only between its roots, so \(f'(x) \lt 0\) for \(-2 \lt x \lt -1\). Outside that interval \(f'\) is positive and \(f\) increases.

Intervals such as \((-1, 1)\) and \((-1, 3)\) lie where \(f' \gt 0\); for instance \(f'(0) = 12\). \([-2, 1]\) and \((-3, 1)\) include points of increase as well.

Answer: E

The volume changes at \(48\pi\) cm³/s.

Differentiate \(V = \pi r^{2}h\) with respect to time by the product rule, with \(\frac{dr}{dt} = 2\) and \(\frac{dh}{dt} = -3\) (the height is decreasing):

\[ \begin{aligned} \frac{dV}{dt} &= \pi\left(2rh\frac{dr}{dt} + r^{2}\frac{dh}{dt}\right) \\ &= \pi(2 \cdot 4 \cdot 6 \cdot 2 + 16 \cdot (-3)) \\ &= \pi(96 - 48) = 48\pi \end{aligned} \]

The positive result means the volume is still increasing: the widening outweighs the shortening at this instant. Ignoring the falling height gives \(96\pi\), and dropping the factor 2 in \(2rh\) gives 0; neither is offered, and \(24\pi\) to \(44\pi\) do not follow from the formula.

Answer: D

The numbers are 6 and 6.

Let the numbers be \(x\) and \(12 - x\). Minimise the sum of squares:

\[ \begin{aligned} S &= x^{2} + (12 - x)^{2} \\ \frac{dS}{dx} &= 2x - 2(12 - x) = 0, \quad x = 6 \\ \frac{d^{2}S}{dx^{2}} &= 4 \gt 0 \end{aligned} \]

The positive second derivative confirms a minimum: \(S = 72\). The other pairs give larger sums: 3,9 gives 90, 4,8 gives 80, 5,7 gives 74 and 2,10 gives 104.

Answer: E

The limit is \(-1\).

Divide the numerator and denominator by \(x\) and use \(\frac{\tan kx}{x} \to k\) and \(\frac{\sin kx}{x} \to k\) as \(x \to 0\):

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

The option \(-\frac{3}{2}\) ignores the \(x\) in the numerator, and \(-2\) forgets to divide by the 2 from \(\sin 2x\).

Answer: B

\(\alpha = 3\).

Use \(1 - \sec^{2}\theta = -\tan^{2}\theta\):

\[ \begin{aligned} \frac{1 - \sec^{2}(\alpha x)}{\alpha x^{2}} &= -\frac{\tan^{2}(\alpha x)}{\alpha x^{2}} \\ &= -\alpha\left(\frac{\tan \alpha x}{\alpha x}\right)^{2} \to -\alpha \end{aligned} \]

For continuity at 0 the limit must equal \(f(0) = -3\), so \(-\alpha = -3\) and \(\alpha = 3\). The option \(-3\) forgets the minus sign that comes from \(1 - \sec^{2}\).

Answer: D

The limit is 1.

Factor the denominator as a difference of squares: \(x - 1 = (\sqrt{x} - 1)(\sqrt{x} + 1)\).

\[ \begin{aligned} f(x) &= \frac{\sqrt{x + 3}\,(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} \\ &= \frac{\sqrt{x + 3}}{\sqrt{x} + 1} \\ \lim_{x \to 1} f(x) &= \frac{\sqrt{4}}{2} = 1 \end{aligned} \]

Note that the first factor is \(\sqrt{x + 3}\), the square root of \(x + 3\). Reading it as \(\sqrt{x} + 3\) would give \(\frac{4}{2} = 2\), the first option.