Limits, Continuity and Differentiability Questions and Answers

Answer: A

The value is a = 5.

Put b = 10 - a, so \(ab = a(10 - a) = 10a - a^{2}\).

\[ \frac{d}{da}(10a - a^{2}) = 10 - 2a = 0 \Rightarrow a = 5 \]

The second derivative is -2, so this is a maximum: ab = 25. That makes 25 the maximum product, not the value of a. a = 3 or 6 gives products 21 and 24, and a = 10 gives 0.

Answer: B

The limit is \(\pi\).

As \(x \to 0\), \(\pi\sin^{2}x \to 0\), so \(\sin(\pi\sin^{2}x) \approx \pi\sin^{2}x\):

\[ \begin{aligned} \lim_{x \to 0} \frac{\sin(\pi\sin^{2}x)}{\pi\sin^{2}x} \cdot \frac{\pi\sin^{2}x}{x^{2}} &= 1 \cdot \pi \cdot 1 \\ &= \pi \end{aligned} \]

\(\pi^{2}\) and \(\frac{\pi}{2}\) come from mishandling the factor \(\pi\) or the square.

Answer: B

The limit is sin 1.

For \(-1 \le x \lt 0\), \([x] = -1\). So as \(x \to 0^{-}\) the expression is constant:

\[ \frac{\sin(-1)}{-1} = \frac{-\sin 1}{-1} = \sin 1 \]

The trap is option A: the standard limit \(\frac{\sin t}{t} \to 1\) needs \(t \to 0\), but here \([x]\) stays at −1. −sin 1 forgets that the denominator is also negative.

Answer: C

The limit is 12.

Factor the numerator and denominator: \(x^{3} - 8 = (x - 2)(x^{2} + 2x + 4)\) and \(x^{2} - 4x + 4 = (x - 2)^{2}\). Then

\[ \begin{aligned} \frac{(x^{3} - 8)\sin(x - 2)}{(x - 2)^{2}} &= (x^{2} + 2x + 4)\frac{\sin(x - 2)}{x - 2} \\ &\to 12 \times 1 = 12 \end{aligned} \]

8 and 4 use only part of \(x^{2} + 2x + 4\) at \(x = 2\), and the negative values come from a sign slip.

Answer: E

The limit is 1.

\[ \begin{aligned} \lim_{x \to 0} \frac{x\cos^{2}x}{\sin x} &= \lim_{x \to 0} \frac{x}{\sin x} \cdot \cos^{2}x \\ &= 1 \times 1 \\ &= 1 \end{aligned} \]

0 would come from substituting \(x = 0\) in the numerator only, ignoring that the denominator also tends to 0, and 2 or 4 have no basis in the standard limit.

Answer: C

The limit is 3.

Use \(t - 1 \lt [t] \le t\). With \(t = \frac{1}{x}\) and \(t = \frac{2}{x}\) for \(x \gt 0\):

\[ \begin{aligned} x\left(\frac{3}{x} - 2\right) &\lt x\left\{\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right]\right\} \\ &\le x \cdot \frac{3}{x} \end{aligned} \]

The left side is \(3 - 2x \to 3\) and the right side is 3, so by the squeeze theorem the limit is 3. Taking only one of the two terms gives 1 or 2, and 0 treats the brackets as bounded.

Answer: B

It is not differentiable at \(x = 1\) and \(x = 2\).

\(x^{2} - 3x + 2 = (x - 1)(x - 2)\). The modulus of a function has a corner wherever the function crosses zero with a non-zero slope. Here the derivative \(2x - 3\) is −1 at \(x = 1\) and 1 at \(x = 2\), so the graph folds sharply at both roots and the left and right derivatives differ.

At every other point, including 3, 4, 5, −1 and −2, the quadratic is non-zero and \(|f|\) is smooth.

A quick way to see it is to sketch the parabola and reflect the part below the x-axis upwards; the reflected arc meets the rest of the curve in sharp corners at the two roots.

Answer: E

\(\frac{dy}{dx}\) at (0, 1) is \(\frac{-1}{e}\).

Differentiate implicitly:

\[ e^{y}y' + 2xy + x^{2}y' + y^{2} + 2xyy' = 0 \]

At (0, 1) every term with a factor x vanishes, leaving \(e\,y' + 1 = 0\), so \(y' = -\frac{1}{e}\).

\(\frac{1}{e}\) drops the sign, and ±e invert the factor.

Answer: D

\(f'(-10)\) equals 20.

For \(x \lt 0\), \(|x| = -x\), so \(f(x) = -x^{2}\) and \(f'(x) = -2x\). At \(x = -10\):

\[ f'(-10) = -2(-10) = 20 \]

In general \(f'(x) = 2|x|\), which is never negative, so the negative options are impossible; 40 doubles the derivative.

Answer: A

\(\frac{1}{y}\frac{dy}{dx} = \log(\tan x) + 2x\, cosec(2x)\).

Take logarithms and differentiate:

\[ \begin{aligned} \log y &= x\log(\tan x) \\ \frac{1}{y}\frac{dy}{dx} &= \log(\tan x) + x\frac{\sec^{2}x}{\tan x} \end{aligned} \]

and \(\frac{\sec^{2}x}{\tan x} = \frac{1}{\sin x\cos x}\) \(= \frac{2}{\sin 2x} = 2\,cosec\,2x\). So the second term is \(2x\,cosec\,2x\).

Options B and E lose the factor 2 or halve it, and options C and D multiply \(\log(\tan x)\) by an extra x, which the product rule does not produce.