Limits, Continuity and Differentiability Questions and Answers

Answer: B

The possible values are 1 and -4.

The limit at x = 1 exists when the left and right limits are equal.

\[ \begin{aligned} a(1) + 3 &= \frac{4(1)}{a} \\ a^{2} + 3a - 4 &= 0 \\ (a - 1)(a + 4) &= 0 \end{aligned} \]

So a = 1 or a = -4. Check: for a = -4 both sides equal -1, and for a = 1 both equal 4. The other pairs are not roots of this quadratic. For a limit to exist the function need not be defined at x = 1 itself; only the two one-sided limits must agree.

Answer: E

The limit is 3.

Rationalise the numerator.

\[ \begin{aligned} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x} &= \frac{(1 + 3x) - (1 - 3x)}{x(\sqrt{1 + 3x} + \sqrt{1 - 3x})} \\ &= \frac{6}{\sqrt{1 + 3x} + \sqrt{1 - 3x}} \to \frac{6}{2} = 3 \end{aligned} \]

Substituting x = 0 directly gives the indeterminate \(\frac{0}{0}\), not 0. \(\frac{1}{2}\) and \(\frac{1}{3}\) come from inverting part of the result.

Answer: A

The points of discontinuity are 2 and \(\frac{9}{4}\).

\(f(f(x))\) is undefined where f(x) itself is undefined (x = 2), and where f(x) equals the bad input 2.

\[ \begin{aligned} f(x) = 2 &\Rightarrow \frac{1}{2x - 4} = 2 \\ 2x - 4 &= \tfrac{1}{2} \Rightarrow x = \tfrac{9}{4} \end{aligned} \]

Algebraically, \(f(f(x)) = \frac{2x - 4}{18 - 8x}\), which fails at x = \(\frac{9}{4}\); the point x = 2 is lost from the simplified form but is still excluded because the inner function is undefined there. Options without 2 miss this.

Answer: E

The limit equals 6.

Multiply and divide by 10 to make the definition of the derivative appear.

\[ \begin{aligned} &\lim_{h \to 0} \frac{f(5 + 10h) - f(5)}{h} \\ &= 10 \lim_{h \to 0} \frac{f(5 + 10h) - f(5)}{10h} \\ &= 10f'(5) \end{aligned} \]

So the value is \(10 \times \frac{3}{5} = 6\). Forgetting the factor 10 would give \(\frac{3}{5}\), which is not offered. The general result is \(\lim_{h \to 0}\frac{f(a + kh) - f(a)}{h} = kf'(a)\).

Answer: D

\(f'(2)\) is \(-\frac{1}{4}\).

Near x = 2, x is positive, so \(|x| = x\) and \(f(x) = \frac{x}{x^{2}} = \frac{1}{x}\).

\[ f'(x) = -\frac{1}{x^{2}} \Rightarrow f'(2) = -\frac{1}{4} \]

\(\frac{1}{4}\) drops the minus sign, and \(-\frac{1}{2}\) is \(-\frac{1}{x}\) instead of \(-\frac{1}{x^{2}}\). The modulus matters only for negative x, where \(f(x) = -\frac{1}{x}\) and the derivative would be \(+\frac{1}{x^{2}}\); at x = 2 that branch is not used.

Answer: D

\(f'(1)\) is -1.

Divide the numerator and denominator by \(5\cos x\) and let \(\tan\alpha = \frac{3}{5}\).

\[ \begin{aligned} \frac{3\cos x - 5\sin x}{5\cos x + 3\sin x} &= \frac{\frac{3}{5} - \tan x}{1 + \frac{3}{5}\tan x} \\ &= \tan(\alpha - x) \end{aligned} \]

So \(f(x) = \alpha - x\) (on the principal branch), a straight line with slope -1, and \(f'(x) = -1\) at every point, including x = 1. The positive options would need f to increase with x.

Answer: E

The value is \(-125e^{5}\).

Differentiate with the product rule.

\[ \begin{aligned} f'(x) &= 2e^{5x} + 5(2x + 3)e^{5x} = (10x + 17)e^{5x} \\ f''(x) &= 10e^{5x} + 5(10x + 17)e^{5x} = (50x + 95)e^{5x} \\ f''(1) - 10f'(1) &= 145e^{5} - 270e^{5} = -125e^{5} \end{aligned} \]

The positive \(125e^{5}\) reverses the sign. Note that \(f'' - 10f' = -25(2x + 3)e^{5x}\) for every x, which at x = 1 gives the same \(-125e^{5}\).

Answer: A

f is strictly decreasing on (2, 3).

\[ \begin{aligned} f'(x) &= 6x^{2} - 30x + 36 \\ &= 6(x - 2)(x - 3) \end{aligned} \]

The derivative is negative between its roots, so f decreases exactly on (2, 3) and increases outside it. Each other interval contains points beyond 3 or below 2, where \(f'(x) \gt 0\). A sign check at x = 2.5: \(f'(2.5) = 6(0.5)(-0.5) = -1.5 \lt 0\), confirming the decrease, while at x = 1 or x = 4 the derivative is positive.

Answer: A

The velocity is \(e^{t}(19\cos 3t - 7\sin 3t)\).

Velocity is \(\frac{ds}{dt}\). Use the product rule.

\[ \begin{aligned} \frac{ds}{dt} &= e^{t}(4\cos 3t + 5\sin 3t) \\ &\quad + e^{t}(-12\sin 3t + 15\cos 3t) \\ &= e^{t}(19\cos 3t - 7\sin 3t) \end{aligned} \]

The option with \(+7\sin 3t\) gets the sign of the derivative of \(\cos 3t\) wrong (\(\frac{d}{dt}\cos 3t = -3\sin 3t\)). The others combine the coefficients 4, 5, 12 and 15 incorrectly; only 4 + 15 = 19 and 5 - 12 = -7 are right.

Answer: C

\(\lambda\) is 12.

At a local minimum inside the interval, \(f'(2) = 0\).

\[ \begin{aligned} f'(x) &= 6x^{2} - 6x - \lambda \\ f'(2) &= 24 - 12 - \lambda = 0 \Rightarrow \lambda = 12 \end{aligned} \]

Check: \(f''(x) = 12x - 6\), and \(f''(2) = 18 \gt 0\), so x = 2 is a minimum. -12 comes from a sign slip with \(-\lambda\), and 36 from using \(f'(3)\). The other stationary point, from \(6x^{2} - 6x - 12 = 0\), is at x = -1, outside the interval [0, 3].