Limits, Continuity and Differentiability Questions and Answers

Answer: C

\(\frac{dy}{dx}\) at x = 0 is 2.

Put \(x = \sin\theta\). Then \(2x\sqrt{1 - x^{2}} = 2\sin\theta\cos\theta = \sin 2\theta\). Near x = 0, \(\theta\) is small, so \(2\theta\) lies within \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and

\[ \begin{aligned} y &= 2\theta = 2\sin^{-1}x \\ \frac{dy}{dx} &= \frac{2}{\sqrt{1 - x^{2}}} = 2 \text{ at } x = 0 \end{aligned} \]

1 is the derivative of \(\sin^{-1}x\) alone, forgetting the factor 2. The simplification holds for \(|x| \le \frac{1}{\sqrt{2}}\), which includes x = 0.

Answer: C

\(\frac{f'(nx)}{f'(x)}\) equals \(\frac{f(nx)}{f(x)}\).

Differentiate both sides of \((f(x))^{n} = f(nx)\) with respect to x, using the chain rule on each side:

\[ \begin{aligned} n(f(x))^{n-1}f'(x) &= n f'(nx) \\ \frac{f'(nx)}{f'(x)} &= (f(x))^{n-1} \end{aligned} \]

Finally \((f(x))^{n-1} = \frac{(f(x))^{n}}{f(x)} = \frac{f(nx)}{f(x)}\), using the given relation. \((f(x))^{n}\) has one power too many, and \(\frac{f(x)}{f(nx)}\) is the reciprocal.

Answer: B

\(x\frac{dy}{dx}\) is \(\frac{y^{6} - 1}{y^{5}}\).

Differentiate each relation with respect to \(\theta\): \(3x^{2}\,dx = \cos\theta\,d\theta = y^{3}\,d\theta\) and \(3y^{2}\,dy = -\sin\theta\,d\theta = -x^{3}\,d\theta\). Dividing,

\[ \begin{aligned} \frac{dy}{dx} &= -\frac{x^{3}}{y^{2}} \cdot \frac{x^{2}}{y^{3}} = -\frac{x^{5}}{y^{5}} \\ x\frac{dy}{dx} &= -\frac{x^{6}}{y^{5}} \end{aligned} \]

Now \(x^{6} + y^{6} = \sin^{2}\theta + \cos^{2}\theta = 1\), so \(-x^{6} = y^{6} - 1\), and \(x\frac{dy}{dx} = \frac{y^{6} - 1}{y^{5}}\). The options with \(y^{6}\) or \(y^{3}\) in the denominator lose track of the powers.

Answer: A

\(\frac{d^{2}y}{dx^{2}}\) is \(e^{x}(5x + 8)\).

Differentiate the product twice; each time the derivative of \((ax + b)e^{x}\) is \((ax + b + a)e^{x}\):

\[ \begin{aligned} y' &= 5e^{x} + (5x - 2)e^{x} = (5x + 3)e^{x} \\ y'' &= 5e^{x} + (5x + 3)e^{x} = (5x + 8)e^{x} \end{aligned} \]

\(e^{x}(5x + 3)\) is the first derivative, one step short. \(e^{x}(5x - 3)\) and \(e^{x}(5x - 5)\) subtract the 5 instead of adding it.

Answer: E

The value of a is 2.

At a maximum or minimum the derivative is zero:

\[ \begin{aligned} f'(x) &= 3ax^{2} - 18x + 6a \\ f'(1) &= 9a - 18 = 0 \Rightarrow a = 2 \end{aligned} \]

Check the other point: \(f'(2) = 12a - 36 + 6a = 18a - 36 = 0\) for a = 2 as well. With a = 2, \(f''(x) = 12x - 18\) is -6 at x = 1 (maximum) and 6 at x = 2 (minimum), as stated. The other values make \(f'(1)\) non-zero.

Answer: C

The function is decreasing on \(\frac{\pi}{4} \le x \le \frac{5\pi}{4}\).

The paper writes the function in \(\theta\) but the interval in x; read it as \(f(x) = \sin x + \cos x\). Then

\[ \begin{aligned} f'(x) &= \cos x - \sin x \\ &= \sqrt{2}\cos\left(x + \frac{\pi}{4}\right) \end{aligned} \]

\(f'(x) \le 0\) when \(\frac{\pi}{2} \le x + \frac{\pi}{4} \le \frac{3\pi}{2}\), that is \(\frac{\pi}{4} \le x \le \frac{5\pi}{4}\). On \(\left[0, \frac{\pi}{4}\right]\) and \(\left[\frac{5\pi}{4}, 2\pi\right]\) the function increases, and the other intervals mix increasing and decreasing parts.

Answer: B

The maximum value is 50.

For a fixed sum, the product xy is greatest when x = y. With x + y = 50, that is x = y = 25, and \(xy = 625\). (By AM-GM, \(xy \le \left(\frac{x + y}{2}\right)^{2} = 625\).)

\[ \sqrt{4xy} \le \sqrt{4 \times 625} = \sqrt{2500} = 50 \]

625 is the maximum of xy and 2500 the maximum of 4xy, both before the square root is taken. 25 is the value of x or y at the maximum.

Answer: E

The limit is 15.

At \(x = 1\) the numerator is \(5 - 5 = 0\), so the limit is the derivative of \(g(x) = x + x^{2} + x^{3} + x^{4} + x^{5}\) at 1, since \(g(1) = 5\):

\[ \begin{aligned} g'(x) &= 1 + 2x + 3x^{2} + 4x^{3} + 5x^{4} \\ g'(1) &= 1 + 2 + 3 + 4 + 5 = 15 \end{aligned} \]

Term by term, \(\lim_{x \to 1}\frac{x^{n} - 1}{x - 1} = n\), and the five terms add to 15.

5 counts the terms rather than adding their powers; 14 and 12 drop one of the terms; 0 forgets that the denominator also vanishes.

Answer: C

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

Divide top and bottom by \(x\) and use \(\lim_{x \to 0}\frac{\sin kx}{x} = k\):

\[ \begin{aligned} \lim_{x \to 0}\frac{\frac{\sin 2x}{x} + 3}{4 + \frac{\sin 6x}{x}} \\ &= \frac{2 + 3}{4 + 6} \\ &= \frac{5}{10} \\ &= \frac{1}{2} \end{aligned} \]

Near 0 each sine behaves like its argument, so the expression behaves like \(\frac{5x}{10x}\).

Keeping only the sine terms would give \(\frac{2}{6}\), and only the linear terms \(\frac{3}{4}\); neither is offered, and the full ratio \(\frac{5}{10}\) is the only consistent value. 1, 2, 3 and \(\frac{1}{4}\) do not come from it.

Answer: A

\(\frac{dy}{dx} = 2^{x}\log_{e}2\).

Convert the logarithm to exponential form first: \(\log_{2}y = x\) means \(y = 2^{x}\). The derivative of \(a^{x}\) is \(a^{x}\log_{e}a\):

\[ \begin{aligned} y &= 2^{x} = e^{x\log_{e}2} \\ \frac{dy}{dx} &= \log_{e}2 \cdot e^{x\log_{e}2} \\ &= 2^{x}\log_{e}2 \end{aligned} \]

\(2^{x}\) alone treats \(2^{x}\) like \(e^{x}\), which is its own derivative only for base \(e\). \(2x\) and \(x^{2}\) confuse \(2^{x}\) with \(x^{2}\). The last option divides by \(\log_{e}y\) instead of multiplying by \(\log_{e}2\).