Limits, Continuity and Differentiability Questions and Answers

Answer: D

\(\frac{dy}{dx} = \frac{3}{4}\) at P.

First find \(x\) at P: \(4\ln 2 = \ln 16\), so

\[ x^{3} + 24 = 16 \Rightarrow x^{3} = -8 \Rightarrow x = -2 \]

\[ \begin{aligned} \frac{dy}{dx} &= \frac{3x^{2}}{x^{3} + 24} \\ &= \frac{3 \times 4}{16} = \frac{3}{4} \end{aligned} \]

Since \(x^{2}\) is positive whether \(x\) is 2 or \(-2\), the derivative is positive; \(-\frac{3}{4}\) has the wrong sign, and \(\frac{3}{8}\) halves the numerator.

Answer: A

\(\alpha = -3\).

Differentiate twice:

\[ \begin{aligned} y' &= -4e^{-x} + 4e^{-2x} + 3e^{-3x} \\ y'' &= 4e^{-x} - 8e^{-2x} - 9e^{-3x} \\ &= e^{-3x}(4e^{2x} - 8e^{x} - 9) \end{aligned} \]

Factoring out \(e^{-3x}\) turns \(e^{-x}\) into \(e^{2x}\) and \(e^{-2x}\) into \(e^{x}\), matching the bracket given. So \(\alpha = -3\). The option 3 has the sign of the exponent reversed.

Answer: C

\(\frac{d^{2}y}{dx^{2}} = \frac{-8}{y^{2}}\frac{dy}{dx}\).

\[ \begin{aligned} y &= 4x^{1/2}, \quad y' = 2x^{-1/2} \\ y'' &= -x^{-3/2} \end{aligned} \]

Since \(y^{2} = 16x\):

\[ \begin{aligned} \frac{-8}{y^{2}}\cdot y' &= \frac{-8}{16x}\cdot\frac{2}{\sqrt{x}} \\ &= -\frac{1}{x^{3/2}} \\ &= y'' \end{aligned} \]

The coefficient must make \(\frac{k}{16x}\cdot\frac{2}{\sqrt{x}} = -\frac{1}{x\sqrt{x}}\), which forces \(k = -8\); the options with \(-4\) or \(-2\) give half or a quarter of \(y''\), and the positive options have the wrong sign, since \(y''\) is negative.

Answer: C

\(f'\left(\frac{\pi}{8}\right) = 2\sqrt{2}\).

Recognise the identity \(\frac{1 + \tan^{2}x}{1 - \tan^{2}x} = \frac{1}{\cos 2x} = \sec 2x\), since \(\cos 2x = \frac{1 - \tan^{2}x}{1 + \tan^{2}x}\).

\[ \begin{aligned} f'(x) &= 2\sec 2x\tan 2x \\ f'\left(\tfrac{\pi}{8}\right) &= 2\sec\tfrac{\pi}{4}\tan\tfrac{\pi}{4} \\ &= 2 \times \sqrt{2} \times 1 = 2\sqrt{2} \end{aligned} \]

The option \(\sqrt{2}\) forgets the factor 2 from the chain rule.

Answer: A

\(\alpha = -3\).

Complete the square:

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

The maximum, at \(x = -2\), is \(\alpha + 4\). Setting \(\alpha + 4 = 1\) gives \(\alpha = -3\). Check: \(f(-2) = -3 + 8 - 4 = 1\).

Answer: D

The volume increases at \(\frac{1}{4\sqrt{2}}\) cm³/s.

With edge \(a\): \(S = 6a^{2}\) and \(V = a^{3}\). When \(S = 12\), \(a^{2} = 2\) and \(a = \sqrt{2}\).

\[ \begin{aligned} \frac{dS}{dt} &= 12a\frac{da}{dt} = 0.5 \\ \frac{da}{dt} &= \frac{0.5}{12\sqrt{2}} = \frac{1}{24\sqrt{2}} \\ \frac{dV}{dt} &= 3a^{2}\frac{da}{dt} = \frac{6}{24\sqrt{2}} = \frac{1}{4\sqrt{2}} \end{aligned} \]

A shortcut: \(\frac{dV}{dS} = \frac{3a^{2}}{12a} = \frac{a}{4}\), so \(\frac{dV}{dt} = \frac{\sqrt{2}}{4} \times 0.5 = \frac{1}{4\sqrt{2}}\).

Answer: E

\(f\) is decreasing in (0,2).

Write \(f(x) = \frac{2}{3}x^{-3/2} - x^{-1/2}\) and differentiate:

\[ \begin{aligned} f'(x) &= -x^{-5/2} + \tfrac{1}{2}x^{-3/2} \\ &= \tfrac{1}{2}x^{-5/2}(x - 2) \end{aligned} \]

For \(x \gt 0\), \(x^{-5/2} \gt 0\), so \(f'(x) \lt 0\) exactly when \(x \lt 2\). Hence \(f\) decreases on \((0, 2)\) and increases for \(x \gt 2\). Every other interval contains points beyond 2, where \(f\) is increasing.

Answer: B

The minimum value is \(-\frac{1}{4}\).

Use \(2\sin A\sin B = \cos(A - B) - \cos(A + B)\):

\[ \begin{aligned} f(x) &= \tfrac{1}{2}\left[\cos\tfrac{\pi}{3} - \cos\left(2x + \tfrac{\pi}{3}\right)\right] \\ &= \tfrac{1}{4} - \tfrac{1}{2}\cos\left(2x + \tfrac{\pi}{3}\right) \end{aligned} \]

The cosine reaches 1, so the minimum is \(\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\), and the maximum is \(\frac{3}{4}\). The option \(\frac{3}{4}\) is therefore the maximum, not the minimum.

Answer: A

The range is [5,12].

Area and perimeter in terms of x:

\[ \begin{aligned} A &= (x + 2)(x - 2) = x^{2} - 4 \le 140 \\ &\Rightarrow x^{2} \le 144,\ x \le 12 \\ P &= 2(2x) = 4x \ge 20 \\ &\Rightarrow x \ge 5 \end{aligned} \]

Both conditions together give \(5 \le x \le 12\); the breadth x − 2 is then positive, as it must be. [2,12] ignores the perimeter condition, and [5,10] and [6,12] misplace one end.

Answer: B

The limit is \(\frac{8}{3}\).

Expand the sines to the \(x^{3}\) term, using \(\sin u = u - \frac{u^{3}}{6} + \dots\):

\[ \begin{aligned} 6\sin x &= 6x - x^{3} + \dots \\ 2\sin 3x &= 6x - 9x^{3} + \dots \\ \text{numerator} &= 8x^{3} + \dots \end{aligned} \]

\[ \lim_{x \to 0}\frac{8x^{3}}{3x^{3}} = \frac{8}{3} \]

Equivalently, \(3\sin x - \sin 3x = 4\sin^{3}x\), so the numerator is \(8\sin^{3}x\). The negative options reverse the sign of the numerator.