Limits, Continuity and Differentiability Questions and Answers

Answer: E

\(a =\) \(-16\).

Differentiability at 2 needs the two pieces to have the same value and the same slope there:

\[ \begin{aligned} \text{value: } 20 + 2a + 16 &= 4 \;\Rightarrow\; a = -16 \\ \text{slope: } 10(2) + a &= 2(2) \;\Rightarrow\; a = -16 \end{aligned} \]

Both conditions agree, so \(a = -16\). \(16\) and \(\pm 32\) come from sign or factor-of-2 slips, and \(-18\) satisfies neither condition.

Answer: C

\(f'\left(\frac{\pi}{6}\right) =\) \(\frac{4}{3}\).

Since \(1 + \cos 2x = 2\cos^{2}x\), near \(\frac{\pi}{6}\) (where \(\cos x \gt 0\)):

\[ \begin{aligned} f(x) &= \frac{\sqrt{2}\sin x}{\sqrt{2}\cos x} \\ &= \tan x \\ f'(x) \\ &= \sec^{2}x \end{aligned} \]

so \(f'\left(\frac{\pi}{6}\right) = \frac{1}{\cos^{2}(\pi/6)}\) \(= \frac{1}{3/4} = \frac{4}{3}\). \(\frac{3}{4}\) is \(\cos^{2}\frac{\pi}{6}\) itself, and the other options do not follow from \(\sec^{2}x\).

Answer: A

\(\frac{du}{dx} =\) \(-36x^{-7}\).

With \(f(x) = x^{-2}\):

\[ \begin{aligned} f''(x) &= 6x^{-4} \\ u &= x^{-2}\cdot 6x^{-4} = 6x^{-6} \\ \frac{du}{dx} &= -36x^{-7} \end{aligned} \]

\(36x^{-7}\) loses the minus sign from differentiating a negative power, and \(\pm 42x^{-7}\) and \(-30x^{-7}\) come from wrong exponents.

A general check: if \(f = x^{-2}\) then \(f f''\) is \(6x^{-6}\), and any power \(x^{k}\) differentiates to \(kx^{k-1}\), here \(-6 \times 6 = -36\).

Answer: B

The function is decreasing in \((0, \pi)\).

\[ f'(x) = -2\sin x - 1 \]

On \((0, \pi)\), \(\sin x \gt 0\), so \(f'(x) \lt -1 \lt 0\) throughout. The function decreases on the whole interval, with no turning point, so the options that switch between increasing and decreasing at \(\frac{\pi}{4}\) or \(\frac{\pi}{2}\) are wrong, as is 'increasing'.

Since \(f'(x) \le -1\) everywhere on the interval, \(f\) falls by at least \(\pi\) between the endpoints, from 5 to \(1 - \pi\).

Answer: D

\(\frac{d^{2}y}{dx^{2}}\) at \(x = 3\) is \(\frac{1}{2}\).

By the product rule:

\[ \begin{aligned} \frac{dy}{dx} &= \log_{e}(x - 1) + 1 \\ \frac{d^{2}y}{dx^{2}} &= \frac{1}{x - 1} = \frac{1}{2} \end{aligned} \]

at \(x = 3\). \(\frac{1}{4}\) uses \(\frac{1}{(x - 1)^{2}}\), and \(e\), \(e^{2}\) and 3 confuse the derivative with the function's values.

The derivative of \((x - 1)\log(x - 1)\) is a standard product-rule result: the '+1' comes from \((x - 1)\cdot\frac{1}{x - 1}\).

Answer: A

\(a =\) \(-9\).

\[ f'(x) = e^{x}[2x^{2} + (a + 4)x + 2] \]

A local minimum at 2 needs \(f'(2) = 0\): \(8 + 2(a + 4) + 2 = 0\), so \(a = -9\). Then \(f'(x) = e^{x}(2x^{2} - 5x + 2) = e^{x}(2x - 1)(x - 2)\), which changes from negative to positive at \(x = 2\), confirming a minimum.

The other values do not make \(f'(2) = 0\).

The other root of \(f'(x) = 0\), at \(x = \frac{1}{2}\), is a local maximum, since \(f'\) changes from positive to negative there.

Answer: D

\(-\frac{dy}{dx} =\) 6.

Differentiate \(e^{x}(y + 2\sqrt{1 + x}) = 5\) with the product rule:

\[ \begin{aligned} e^{x}(y + 2\sqrt{1 + x}) + e^{x}\left(y' + \frac{1}{\sqrt{1 + x}}\right) \\ &= 0 \end{aligned} \]

At \((0, 3)\): \(5 + y' + 1 = 0\), so \(y' = -6\) and \(-\frac{dy}{dx} = 6\). \(-6\) is \(\frac{dy}{dx}\) itself, the trap in the question's minus sign.

Implicit differentiation is the efficient route here, because solving for \(y\) first would introduce \(e^{-x}\) terms to differentiate.

Answer: E

The limit is \(\frac{1}{2}\).

Use \(\lim_{t \to 0}\frac{\sin t}{t} = 1\) for both sines, writing each over its own angle.

\[ \begin{aligned} \frac{\sin x}{2\sqrt{2}\sin\frac{x}{\sqrt{2}}} &= \frac{\frac{\sin x}{x} \cdot x}{2\sqrt{2}\cdot\frac{\sin(x/\sqrt{2})}{x/\sqrt{2}}\cdot\frac{x}{\sqrt{2}}} \\ &\to \frac{x}{2x} = \frac{1}{2} \end{aligned} \]

The \(\sqrt{2}\) in the angle cancels the \(\sqrt{2}\) in the coefficient, leaving \(\frac{1}{2}\). \(\frac{1}{2\sqrt{2}}\) ignores the factor \(\frac{1}{\sqrt{2}}\) in the angle, and \(\frac{1}{\sqrt{2}}\) uses it twice.

Answer: A

The limit is 1.

Use \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(1 - \cos 2\theta = 2\sin^{2}\theta\).

\[ \begin{aligned} \frac{\theta\cdot 2\sin\theta\cos\theta}{2\sin^{2}\theta} &= \frac{\theta}{\sin\theta}\cdot\cos\theta \\ &\to 1 \cdot 1 = 1 \end{aligned} \]

\(\frac{1}{2}\) comes from expanding \(1 - \cos 2\theta\) as \(2\theta^{2}\) correctly but forgetting the 2 in \(\sin 2\theta \approx 2\theta\). Both numerator and denominator are positive for small \(\theta\), so negative answers are impossible, and the limit is not 0 because both behave like \(2\theta^{2}\).

Answer: D

The function is continuous on \([-1, \infty)\).

\(\sqrt{x + 2}\) needs \(x \ge -2\) and \(\sqrt{x + 1}\) needs \(x \ge -1\); both must be defined, so the domain is \(x \ge -1\). On that interval f is a product and sum of continuous functions, so it is continuous there.

\([-3, \infty)\) and \((-\infty, \infty)\) include points where a square root is of a negative number, and \((-\infty, 1]\) does too. \([4, \infty)\) lies inside the domain, so f is continuous there too, but it is only part of the interval of continuity; the question asks for the whole of it, which is \([-1, \infty)\).