Limits, Continuity and Differentiability Questions and Answers

Answer: C

The second derivative is \(-y\).

Differentiate twice, using \(\frac{d}{dx}\cos x = -\sin x\) and \(\frac{d}{dx}\sin x = \cos x\):

\[ \begin{aligned} y &= 15\cos x - 13\sin x \\ \frac{dy}{dx} &= -15\sin x - 13\cos x \\ \frac{d^{2}y}{dx^{2}} &= -15\cos x + 13\sin x \\ &= -(15\cos x - 13\sin x) = -y \end{aligned} \]

Any combination of \(\sin x\) and \(\cos x\) satisfies \(y'' = -y\), which is the equation of simple harmonic motion. The constants 15 and 13 do not matter. The result is not a constant such as 2, and it contains no \(\pi\) or \(x\) in a denominator, so the other three options are ruled out.

Answer: D

The derivative is \(1 + \log_{e}(x)\).

Use the product rule with \(u = x\) and \(v = \log_{e}x\):

\[ \begin{aligned} \frac{d}{dx}(x\log_{e}x) &= 1\cdot\log_{e}x + x\cdot\frac{1}{x} \\ &= \log_{e}x + 1 \end{aligned} \]

The answer 1 keeps only the \(x\cdot\frac{1}{x}\) term and forgets to differentiate \(x\). The answers 0 and e are constants, but the derivative changes with \(x\): at \(x = 1\) it is 1 and at \(x = e\) it is 2.

Answer: D

The true statement is that \(f\) has neither a local maximum nor a local minimum in \((0, \infty)\).

Differentiate and put everything over a common denominator:

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

Call the numerator \(g(x) = \ln x + 2 - 2\sqrt{x}\). Then \(g'(x) = \frac{1}{x} - \frac{1}{\sqrt{x}} = \frac{1 - \sqrt{x}}{x}\), so \(g\) rises on \((0, 1)\), falls after 1, and its largest value is \(g(1) = 0\). Hence \(f'(x) \le 0\) everywhere, with equality only at \(x = 1\): \(f\) is strictly decreasing and has no local extremum, which rules out the maximum and minimum options.

The first option fails too. Differentiating again gives \(f''(x) = -\frac{\ln x}{4x^{3/2}}\), which is positive on \((0, 1)\), so \(f'\) is increasing there, climbing from \(-\infty\) up to 0 at \(x = 1\).

Answer: A

The limit is \(4\sqrt{2}\).

Rationalise the denominator:

\[ \begin{aligned} \sqrt{2} - \sqrt{1 + \cos x} &= \frac{1 - \cos x}{\sqrt{2} + \sqrt{1 + \cos x}} \\ \frac{x^{2}}{\sqrt{2} - \sqrt{1 + \cos x}} &= \frac{x^{2}}{1 - \cos x}\left(\sqrt{2} + \sqrt{1 + \cos x}\right) \end{aligned} \]

As \(x \to 0\), \(\frac{x^{2}}{1 - \cos x} \to 2\) and the bracket tends to \(2\sqrt{2}\), so the limit is \(2 \times 2\sqrt{2} = 4\sqrt{2}\). \(2\sqrt{2}\) forgets the factor 2 from \(1 - \cos x \approx \frac{x^{2}}{2}\).

Answer: C

\(\alpha + \beta = \) 4.

Continuity at 0 means the limit of the first branch equals \(f(0) = 5\). Using \(\lim\limits_{x \to 0}\frac{\tan kx}{x} = k\):

\[ \begin{aligned} \lim_{x \to 0}\frac{\tan\alpha x}{x} + (\beta + 1)\frac{\tan x}{x} &= \alpha + \beta + 1 \\ \alpha + \beta + 1 &= 5 \Rightarrow \alpha + \beta = 4 \end{aligned} \]

5 is the answer obtained by forgetting the +1 attached to \(\beta\); 6 adds it on the wrong side.

Answer: B

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

Rationalise the numerator. The difference of the radicands is \(-\sin x\):

\[ \begin{aligned} &\frac{\sqrt{u} - \sqrt{u + \sin x}}{x} \\ &= \frac{-\sin x}{x\left(\sqrt{u} + \sqrt{u + \sin x}\right)} \end{aligned} \]

with \(u = \cos^{2}x + 3 \to 4\). As \(x \to 0\), \(\frac{\sin x}{x} \to 1\) and the bracket tends to \(2 + 2 = 4\), so the limit is \(-\frac{1}{4}\). \(\frac{1}{4}\) loses the sign, and \(\frac{-1}{2}\) forgets that two square roots of 4 are added in the denominator.

Answer: D

\(a^{2} + b^{2} = \) 13.

The modulus of a polynomial has a corner where the polynomial crosses zero with a simple root. Here

\[ x^{2} + x - 6 = (x + 3)(x - 2) \]

so \(f\) is not differentiable at \(x = -3\) and \(x = 2\), giving \(a^{2} + b^{2} = 9 + 4 = 13\). Using the roots 3 and -2 (the wrong signs) gives the same 13, but 11, 12, 14 and 16 do not come from \(\{2, -3\}\).

Answer: A

\(f'\left(\frac{\pi}{4}\right) = \) \(-3\sqrt{2}\).

For \(\frac{\pi}{6} \leq x \leq \frac{\pi}{3}\), \(3x\) lies between \(\frac{\pi}{2}\) and \(\pi\), where \(\sin 3x \geq 0\) and \(\cos 3x \leq 0\). So the moduli open as

\[ \begin{aligned} f(x) &= \sin 3x + \cos 3x \\ f'(x) &= 3\cos 3x - 3\sin 3x \end{aligned} \]

At \(x = \frac{\pi}{4}\), \(3x = \frac{3\pi}{4}\), with \(\cos = -\frac{1}{\sqrt{2}}\) and \(\sin = \frac{1}{\sqrt{2}}\):

\[ \begin{aligned} f'\left(\tfrac{\pi}{4}\right) &= 3\left(-\tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}}\right) \\ &= -\tfrac{6}{\sqrt{2}} = -3\sqrt{2} \end{aligned} \]

Ignoring the moduli (\(f = \sin 3x - \cos 3x\)) gives 0 at this point; \(\frac{\pm 3}{\sqrt{2}}\) drops one of the two terms.

Answer: B

\(h'(3) = \) 3.

By the chain rule, \(h'(x) = f'\left(\sqrt{g(x)}\right)\cdot\frac{g'(x)}{2\sqrt{g(x)}}\). At \(x = 3\), \(\sqrt{g(3)} = 3\):

\[ \begin{aligned} h'(3) &= f'(3)\cdot\frac{g'(3)}{2 \times 3} \\ &= 6 \times \frac{3}{6} = 3 \end{aligned} \]

18 multiplies \(f'(3)\) by \(g'(3)\) and forgets the derivative of the square root; 9 is \(g(3)\); 6 is \(f'(3)\) alone.

Answer: C

\(a = \) \(\frac{-3}{4}\).

By the product rule:

\[ \begin{aligned} f'(x) &= -2\cos x\sin x\,(a + \cos x) - \cos^{2}x\sin x \\ &= -\sin x\cos x\,(2a + 3\cos x) \end{aligned} \]

At \(\frac{\pi}{3}\), \(\sin x\cos x \neq 0\), so \(2a + 3\cos\frac{\pi}{3} = 2a + \frac{3}{2} = 0\), giving \(a = -\frac{3}{4}\). \(\frac{-3}{2}\) forgets to divide by 2, and \(\frac{3}{4}\) has the sign wrong.