Limits, Continuity and Differentiability Questions and Answers

Answer: C

\(h(2026) - h(2025)\) is less than or equal to 0.

As \(x\) increases, \(g(x)\) decreases (or stays level), and since \(f\) is increasing, \(f(g(x))\) then decreases or stays level. By the chain rule,

\[ h'(x) = f'(g(x))\,g'(x) \leq 0 \]

because \(f' \geq 0\) and \(g' \leq 0\). So \(h\) is non-increasing and \(h(2026) \leq h(2025)\).

'Greater than or equal to 0' would hold if both functions were increasing or both decreasing. The large positive bounds cannot follow from monotonicity alone.

Answer: D

\(g(x)\) is greater than 0 for every \(x\).

With \(f'(x) = 20x + a\) and \(f''(x) = 20\):

\[ \begin{aligned} g(x) &= 10x^{2} + ax + 20x + a + 20 \\ &= 10x^{2} + (a + 20)x + (a + 20) \end{aligned} \]

This quadratic has a positive leading coefficient. Its discriminant is

\[ \begin{aligned} D &= (a + 20)^{2} - 40(a + 20) \\ &= (a + 20)(a - 20) = a^{2} - 400 \lt 0 \end{aligned} \]

A quadratic with positive leading coefficient and negative discriminant is positive for all real \(x\). It is unbounded above, so 'less than 10', 'less than 1' and the bounded ranges are all false.

Answer: C

The minimum is -1.

Rewrite the function:

\[ \begin{aligned} f(x) &= \frac{x^{100} + 1 - 2}{x^{100} + 1} \\ &= 1 - \frac{2}{x^{100} + 1} \end{aligned} \]

Since \(x^{100} \geq 0\), the denominator is at least 1, so \(\frac{2}{x^{100} + 1} \leq 2\) and \(f(x) \geq -1\), with equality at \(x = 0\): \(f(0) = \frac{-1}{1} = -1\).

Values below -1, such as -1.5, -2, -3 or -5, are never reached, because \(f\) always lies in \([-1, 1)\).

Answer: A

\(f(x)\) is increasing on \((0, \infty)\).

Differentiate, using \(\frac{d}{dx}\log(x + \sqrt{x^{2} + 1}) = \frac{1}{\sqrt{x^{2} + 1}}\):

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

For \(x \gt 0\), \(x + \sqrt{x^{2} + 1} \gt 1\), so its logarithm is positive and \(f'(x) \gt 0\). Hence \(f\) increases on the whole of \((0, \infty)\).

The options limiting the increase to \((10, \infty)\) or \((0, e)\), or claiming a decrease, contradict \(f' \gt 0\) everywhere on \((0, \infty)\).

Answer: D

The limit is \(\dfrac{1}{4\sqrt{3}}\).

The limit is the definition of \(f'(1)\). Differentiating:

\[ \begin{aligned} f'(x) &= -\frac{-2x}{2\sqrt{49 - x^{2}}} = \frac{x}{\sqrt{49 - x^{2}}} \\ f'(1) &= \frac{1}{\sqrt{48}} = \frac{1}{4\sqrt{3}} \end{aligned} \]

The minus sign in front of the root and the minus from the chain rule cancel, so the derivative is positive. \(\sqrt{48} = 4\sqrt{3}\); the other options use a different root.

Answer: D

The value of \(a\) is 3.

Both pieces are polynomials, so the only point to check is \(x = 1\), where the left-hand limit must equal \(f(1)\):

\[ \begin{aligned} \lim_{x \to 1^{-}} \frac{x^{2}}{2} &= \frac{1}{2} \\ f(1) &= 2 - a + 1.5 = 3.5 - a \\ 3.5 - a &= 0.5 \Rightarrow a = 3 \end{aligned} \]

Sign errors in solving \(3.5 - a = 0.5\) give the negative options.

Answer: C

The limit is 0.

Write both terms over \(\cos x\):

\[ \tan x - \sec x = \frac{\sin x - 1}{\cos x} \]

Multiply top and bottom by \(\sin x + 1\):

\[ \begin{aligned} \frac{\sin^{2}x - 1}{\cos x(\sin x + 1)} &= \frac{-\cos x}{\sin x + 1} \\ &\to \frac{0}{2} \\ &= 0 \end{aligned} \]

Each of \(\tan x\) and \(\sec x\) grows without bound as \(x \to \frac{\pi}{2}^{-}\), but their difference tends to 0; this is an \(\infty - \infty\) form, not one that can be read off term by term.

Answer: E

The statement that is not true is \(f(x)\) is continuous at \(x\) = 0.

\(f(x) = xe^{-2/x}\) contains \(\frac{2}{x}\), so it is not defined at \(x = 0\), and a function cannot be continuous at a point where it is not defined.

Everywhere else it is a product of continuous functions (\(x\), and \(e\) raised to a continuous power), so it is continuous at 1, -1, 2 and -2. (As \(x \to 0^{-}\), \(-\frac{2}{x} \to +\infty\) and the function blows up, so no value at 0 could make it continuous either.)

Answer: D

The statement that is not true is \(f(x)\) is not differentiable at \(x\) = 1.

At \(x = 1\) both pieces give 0, so \(f\) is continuous there. Compare the one-sided derivatives:

\[ \begin{aligned} \text{left: } \tfrac{d}{dx}(1 - x) &= -1 \\ \text{right: } \tfrac{d}{dx}(x^{2} - 3x + 2) &= 2x - 3 = -1 \end{aligned} \]

They are equal, so \(f\) is differentiable at 1, and the claim that it is not is false. The other statements are true: \(1 - x\) is differentiable everywhere left of 1 (including -2026 and the interval (-1, 1)), and at 1.5 the quadratic piece is smooth.

Answer: D

\(g'''(0)\) is -320.

Only the first row depends on \(x\), so differentiating the determinant three times differentiates that row three times:

\[ \begin{aligned} \tfrac{d^{3}}{dx^{3}}x^{4} &= 24x \to 0 \\ \tfrac{d^{3}}{dx^{3}}(-\cos x) &= -\sin x \to 0 \\ \tfrac{d^{3}}{dx^{3}}(-\sin 2x) &= 8\cos 2x \to 8 \end{aligned} \]

At \(x = 0\) the first row becomes (0, 0, 8), so

\[ g'''(0) = 8\begin{vmatrix} -8 & 4 \\ 2 & 4 \end{vmatrix} = 8(-32 - 8) = -320 \]

A sign slip in the third derivative of \(-\sin 2x\) gives +320.