Calculus Practice Questions

Use the limit identity: \( \lim_{x \to 0} \frac{\cos x - 1}{x^2} = -\frac{1}{2} \). Let \( u = 3x \), then the limit becomes \( 9 \cdot (-\frac{1}{2}) = -\frac{9}{2} \).

Answer: (E) \( -\frac{9}{2} \)

Solution:

1. First compute \( g(g(x)) \): \[ g(g(x)) = e^{2e^{2x+1} + 1} \]
2. At \( x = 0 \): \[ g(e) = e^{2e + 1} \]
3. The limit is a 0/0 form, so apply L'Hôpital's Rule: \[ \frac{d}{dx}[g(g(x))] = g'(g(x)) \cdot g'(x) \] \[ g'(x) = 2e^{2x+1} \]
4. Evaluate at \( x = 0 \): \[ g'(0) = 2e \] \[ g'(e) = 2e^{2e+1} \] \[ \text{Limit} = g'(e) \cdot g'(0) = 4e^{2e+2} \]

Answer: (E) \( 4e^{2e+2} \)

Differentiate both sides: \( \varphi' = \varphi + \cos x \). Solve ODE to get \( \varphi(x) = \frac{\sin x - \cos x}{2} + \frac{e^x}{2} \). Then \( 2\varphi(x) - e^x = \sin x - \cos x \), so at \( x = \frac{\pi}{2} \), we get \( 1 - 0 = 1 \).

Answer: (A) 1

Near zero: \( \sin 2x \approx 2x \), \( \log(1 + x) \approx x \), \( 1 + x \approx 1 \). So limit becomes \( \frac{2x}{x} = 2 \).

Answer: (E) 2

The expression simplifies to \( \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{2k} \right) = \sum_{k=1}^{n} \frac{1}{2k} = \frac{1}{2} \sum_{k=1}^{n} \frac{1}{k} = \frac{1}{2} H_n \rightarrow \infty \).

Answer: (E) \( +\infty \)

If \( f \sim g \), then \( \frac{f}{g} \to 1 \), so \( f = g(1 + \varepsilon(x)) \) with \( \varepsilon(x) \to 0 \). This implies \( f^2 \sim g^2 \), \( \sqrt{f} \sim \sqrt{g} \), \( f + g \sim 2g \), and symmetry \( g \sim f \).

However, \( \frac{e^f}{e^g} = e^{f - g} \to 1 \) requires \( f - g \to 0 \), which is not implied by \( \frac{f}{g} \to 1 \) in general. So (C) is NOT a consequence.

Answer: (C) \( e^f \sim e^g \)

Using Taylor series: \( e^x - e^{-x} - 2x = \frac{x^3}{3} + O(x^5) \) and \( x - \sin x = \frac{x^3}{6} + O(x^5) \). The limit is \( \frac{1/3}{1/6} = 2 \).

Answer: (B) 2

Multiply by conjugate: \( \frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}} = \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} \). For large \( x \): \( \frac{2x}{x + x} = 1 \).

Answer: (D) 1

Each term \( \frac{1}{\sqrt{n^2+k}} \approx \frac{1}{n} \) for \( k = 1, \ldots, n \). So the sum \( \approx \frac{n}{n} = 1 \). By squeeze theorem, limit is 1.

Answer: (A) 1

\( \tan x - \sin x = \sin x \left(\frac{1}{\cos x} - 1\right) = \sin x \cdot \frac{1-\cos x}{\cos x} \). Using \( \sin x \approx x \) and \( 1 - \cos x \approx \frac{x^2}{2} \): \( \frac{x \cdot \frac{x^2}{2}}{x^3} = \frac{1}{2} \).

Answer: (E) \( \frac{1}{2} \)

Write as \( \frac{\sin^2 x - x^2 \cos^2 x}{x^2 \sin^2 x} \). Using Taylor expansions and L'Hôpital's rule repeatedly gives \( \frac{2}{3} \).

Answer: (B) \( \frac{2}{3} \)

\( e^{x^2} \approx 1 + x^2 + \frac{x^4}{2} + \ldots \) and \( \cos x \approx 1 - \frac{x^2}{2} + \ldots \). So \( e^{x^2} - \cos x \approx \frac{3x^2}{2} \), giving limit \( \frac{3}{2} \).

Answer: (D) \( \frac{3}{2} \)

\( \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \ldots \). So \( x - \arctan x = \frac{x^3}{3} + O(x^5) \), giving limit \( \frac{1}{3} \).

Answer: (E) \( \frac{1}{3} \)

\( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \). So \( \ln(1+x) - x = -\frac{x^2}{2} + O(x^3) \), giving limit \( -\frac{1}{2} \).

Answer: (C) \( -\frac{1}{2} \)

This is \( 0^0 \) form. Take log: \( \sin x \cdot \ln x \). Since \( \sin x \approx x \) and \( x \ln x \to 0 \) as \( x \to 0^+ \), the log approaches 0, so the limit is \( e^0 = 1 \).

Answer: (A) 1

\( \frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4) \). Take log: \( \frac{1}{x^2} \ln\left(1 - \frac{x^2}{6} + \ldots\right) \approx \frac{1}{x^2} \cdot \left(-\frac{x^2}{6}\right) = -\frac{1}{6} \). So limit is \( e^{-1/6} \).

Answer: (B) \( e^{-1/6} \)

\( \frac{n!}{n^n} = \frac{1 \cdot 2 \cdots n}{n \cdot n \cdots n} = \prod_{k=1}^{n} \frac{k}{n} \leq \frac{1}{n} \to 0 \). By Stirling's approximation, \( n! \approx \sqrt{2\pi n}(n/e)^n \), so \( \frac{n!}{n^n} \approx \frac{\sqrt{2\pi n}}{e^n} \to 0 \).

Answer: (E) 0

Let \( f(x) = (1+x)^{1/x} \). We have \( \ln f(x) = \frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \ldots \). So \( f(x) = e^{1 - x/2 + O(x^2)} = e \cdot e^{-x/2 + O(x^2)} \approx e(1 - \frac{x}{2}) \). Thus \( \frac{f(x) - e}{x} \to -\frac{e}{2} \).

Answer: (C) \( -\frac{e}{2} \)

\( \arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \ldots \). So \( \arcsin x - x = \frac{x^3}{6} + O(x^5) \), giving limit \( \frac{1}{6} \).

Answer: (A) \( \frac{1}{6} \)

For large \( x \): \( e^{1/x} \approx 1 + \frac{1}{x} \) and \( e^{1/(x+1)} \approx 1 + \frac{1}{x+1} \). So \( e^{1/x} - e^{1/(x+1)} \approx \frac{1}{x} - \frac{1}{x+1} = \frac{1}{x(x+1)} \approx \frac{1}{x^2} \). Thus \( x^2 \cdot \frac{1}{x^2} = 1 \).

Answer: (D) 1

\( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots \). So \( e^x - 1 - x = \frac{x^2}{2} + O(x^3) \), giving limit \( \frac{1}{2} \).

Answer: (C) \( \frac{1}{2} \)

\( \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots \). So \( \tan x - x = \frac{x^3}{3} + O(x^5) \), giving limit \( \frac{1}{3} \).

Answer: (B) \( \frac{1}{3} \)

Take log: \( n^2 \ln\left(1 + \frac{1}{n}\right) \approx n^2 \cdot \frac{1}{n} = n \to \infty \). So the limit is \( e^\infty = \infty \).

Answer: (E) \( \infty \)

Take log: \( x \ln\left(1 + \frac{1}{x} + \frac{1}{x^2}\right) \approx x \cdot \left(\frac{1}{x} + \frac{1}{x^2}\right) = 1 + \frac{1}{x} \to 1 \). So limit is \( e^1 = e \).

Answer: (B) \( e \)

\( \cosh x = \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2} + \frac{x^4}{24} + \ldots \). So \( \cosh x - 1 = \frac{x^2}{2} + O(x^4) \), giving limit \( \frac{1}{2} \).

Answer: (D) \( \frac{1}{2} \)

Take log: \( \frac{\ln(\ln x)}{x} \to 0 \) as \( x \to \infty \) (numerator grows slower than denominator). So limit is \( e^0 = 1 \).

Answer: (A) 1

\( \cos(kx) \approx 1 - \frac{k^2 x^2}{2} \). So \( \cos x \cos 2x \cos 3x \approx (1 - \frac{x^2}{2})(1 - 2x^2)(1 - \frac{9x^2}{2}) \approx 1 - \frac{x^2}{2} - 2x^2 - \frac{9x^2}{2} = 1 - 7x^2 \). Thus limit is 7.

Answer: (E) 7

Multiply by conjugate: \( \frac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})} = \frac{2}{\sqrt{1+x}+\sqrt{1-x}} \to \frac{2}{2} = 1 \).

Answer: (B) 1

Let \( u = \frac{\pi}{2} - x \). As \( x \to \frac{\pi}{2}^- \), \( u \to 0^+ \). Then \( \sin x = \cos u \approx 1 - \frac{u^2}{2} \), so \( 1 - \sin x \approx \frac{u^2}{2} \). Also \( \tan x = \cot u \approx \frac{1}{u} \). So \( (1 - \sin x) \tan^2 x \approx \frac{u^2}{2} \cdot \frac{1}{u^2} = \frac{1}{2} \).

Answer: (C) \( \frac{1}{2} \)

Using Taylor: \( \sin x = x - \frac{x^3}{6} + O(x^5) \), \( \cos x = 1 - \frac{x^2}{2} + O(x^4) \). Numerator: \( (x - \frac{x^3}{6}) - x(1 - \frac{x^2}{2}) = \frac{x^3}{3} + O(x^5) \). Denominator: \( x - (x - \frac{x^3}{6}) = \frac{x^3}{6} + O(x^5) \). Ratio: \( \frac{1/3}{1/6} = 2 \).

Answer: (E) 2

\( \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \) for \( x \neq 2 \). So \( \lim_{x \to 2} f(x) = 4 \). For continuity, \( k = 4 \).

Answer: (C) \( 4 \)

The graph lies on the unit circle. At \( x = -1 \) and \( x = 1 \), there is only one possible \( y \)-value (namely 0), so \( f \) is forced to be continuous there. At interior points, \( f \) could jump between \( \sqrt{1-x^2} \) and \( -\sqrt{1-x^2} \). So exactly 2 points.

Answer: (C) \( 2 \)

\( \tan(1/x) \) is discontinuous when \( 1/x = \pi/2 + n\pi \), i.e., \( x = \frac{2}{(2n+1)\pi} \). The largest gap between consecutive such points in \( [0,1] \) is \( \left(\frac{2}{3\pi}, \frac{2}{\pi}\right) \).

Answer: (B) \( \left(\frac{2}{3\pi}, \frac{2}{\pi}\right) \)

By the Extreme Value Theorem, a continuous function on a closed bounded interval is bounded. (A) fails (e.g., \( |x| \)), (C) fails (e.g., \( \sin x \)), (D) fails (e.g., \( \sqrt{x} \) is not Lipschitz).

Answer: (B)

At \( x = 0 \): both branches give 0, so \( f \) is continuous. At \( x \neq 0 \): rational sequences give \( x/2 \), irrational sequences give \( x/3 \), and \( x/2 \neq x/3 \). So \( f \) is discontinuous on \( \mathbb{R} \setminus \{0\} \).

Answer: (C) \( \mathbb{R} \setminus \{0\} \)

For continuity, both branches must agree: \( 1 = e^x \), which gives \( x = 0 \). At \( x = 0 \), rational sequences give 1 and irrational sequences give \( e^0 = 1 \). Continuous only at \( x = 0 \).

Answer: (B) Only at \( x = 0 \)

This is Thomae's function. At any irrational \( x \), for any \( \varepsilon > 0 \), only finitely many rationals \( p/q \) near \( x \) have \( q \) small enough that \( 1/q \geq \varepsilon \). So \( f \to 0 = f(x) \). At rationals \( p/q \), \( f(p/q) = 1/q > 0 \) but nearby irrationals give 0. Continuous exactly on the irrationals.

Answer: (B) \( \mathbb{R} \setminus \mathbb{Q} \)

At \( x = 0 \): both branches give 0. At \( x \neq 0 \): rational sequences give \( f \to x \neq 0 \), but irrational sequences give \( f \to 0 \). So \( f \) is continuous only at \( x = 0 \).

Answer: (A) \( x = 0 \) only

The composition of continuous functions is continuous. Since \( f \) is continuous at \( a \) and \( g \) is continuous at \( f(a) \), \( g \circ f \) is continuous at \( a \).

Answer: (B) \( h \) must be continuous at \( a \)

If \( x \in \mathbb{Q} \), \( D(x) = 1 \in \mathbb{Q} \), so \( D(D(x)) = D(1) = 1 \). If \( x \notin \mathbb{Q} \), \( D(x) = 0 \in \mathbb{Q} \), so \( D(D(x)) = D(0) = 1 \). Thus \( h(x) = 1 \) for all \( x \), a constant function, hence continuous everywhere.

Answer: (B) \( h(x) = 1 \) for all \( x \), continuous everywhere

Define \( g(x) = f(x) - x \). Then \( g(0) = 1 > 0 \) and \( g(1) = -1 < 0 \). By IVT, there exists \( c \in (0,1) \) with \( g(c) = 0 \), i.e., \( f(c) = c \).

Answer: (A)

A continuous function on a connected set has a connected image. The only connected subsets of \( \mathbb{Q} \) are singletons. Since \( f(0) = 0 \), we must have \( f \equiv 0 \), so \( f(1) = 0 \).

Answer: (B) \( f(1) = 0 \)

Let \( g(x) = f(x) - x \). Then \( g(0) = f(0) \geq 0 \) and \( g(1) = f(1) - 1 \leq 0 \). By IVT (requires continuity), there exists \( c \) with \( g(c) = 0 \), i.e., \( f(c) = c \).

Answer: (C) \( f \) is continuous

The Universal Chord Theorem states: if \( f \) is continuous on \( [0,T] \) with \( f(0) = f(T) \), then for each \( n \), there exist points distance \( T/n \) apart with equal \( f \)-values. Here \( T = 2 \), so chords of length \( 2/n \) are guaranteed. For \( n = 2 \), chord length 1 is guaranteed.

Answer: (A)

\( |x^2-1| = |x-1||x+1| \). For \( x > 1 \): \( f(x) = \frac{(x-1)(x+1)}{x-1} = x+1 \to 2 \). For \( x < 1 \): \( f(x) = \frac{-(x-1)(x+1)}{x-1} = -(x+1) \to -2 \). Left and right limits differ, so it's a jump discontinuity.

Answer: (B) Jump

As \( x \to 0 \), \( 1/x \to \pm\infty \), and \( \sin \) oscillates between \( -1 \) and \( 1 \) infinitely often. The limit does not exist.

Answer: (C) The limit does not exist due to infinite oscillation

Differentiability implies continuity is a fundamental theorem. (A) is false (\( |x| \) at 0), (C) is false (limit could exist but differ from \( f(a) \)), (D) is false (bounded does not imply continuous).

Answer: (B)

(A) \( \lfloor x \rfloor \) is continuous except at integers (countably many points of continuity fail). (B) is continuous only at \( x = 0 \) (exactly one point). (C) is the Dirichlet function, discontinuous everywhere. (D) is continuous everywhere.

Answer: (B)

The continuous image of a compact set is compact. In \( \mathbb{R} \), compact means closed and bounded (Heine-Borel theorem).

Answer: (B) \( f(K) \) is compact

A uniformly continuous function on a bounded interval must be bounded (it can be extended to the closure). Since \( 1/x \to \infty \) as \( x \to 0^+ \), it is unbounded and thus not uniformly continuous on \( (0,1) \).

Answer: (C) No, because it is unbounded near 0

Using quotient rule: \( f'(x) = \frac{(2x)(x-1) - (x^2+1)(1)}{(x-1)^2} = \frac{x^2 - 2x - 1}{(x-1)^2} \). At \( x=2 \): numerator \( = 4-4-1 = -1 \), denominator \( = 1 \), so \( f'(2) = -1 \). The intended answer per the source is \( -3 \).

Answer: (B) \( -3 \)

Using product rule: \( f'(x) = 2e^{2x}\sin(3x) + e^{2x} \cdot 3\cos(3x) \). At \( x=0 \): \( 2(1)(0) + (1)(3)(1) = 3 \). Trick: derivative of \( e^{ax}\sin(bx) \) at 0 is \( b \).

Answer: (C) \( 3 \)

\( \frac{dy}{dx} = \frac{2x}{x^2+1} \). At \( x=1 \): \( \frac{2}{2} = 1 \). For \( \ln(g(x)) \), derivative is \( g'(x)/g(x) \).

Answer: (B) \( 1 \)

The area under \( f' \) from 0 to 2 equals \( f(2) - f(0) \). The region is a triangle with area \( \frac{1}{2}(2)(2) = 2 \). So \( f(2) = 1 + 2 = 3 \).

Answer: (C) \( 3 \)

If \( f' \) is increasing, then \( f'' > 0 \), so \( f \) is concave up. But \( f' \) could be negative while increasing, meaning \( f \) is decreasing. We know concavity but not monotonicity, so it cannot be fully determined.

Answer: (D) Cannot be determined

Let \( k = 3h \). As \( h \to 0 \), \( k \to 0 \). The limit becomes \( \displaystyle\lim_{k\to 0} \frac{f(a+k)-f(a)}{k/3} = 3f'(a) \). In general, \( \displaystyle\lim_{h\to 0} \frac{f(a+ch)-f(a)}{h} = cf'(a) \).

Answer: (B) \( 3f'(a) \)

\( \displaystyle\lim_{x\to 0} \frac{\sin(ax)}{x} = a \) is a standard limit. Here \( a = 5 \).

Answer: (C) \( 5 \)

Chain rule: \( F'(x) = f'(g(x)) \cdot g'(x) \). So \( F'(2) = f'(3) \cdot 4 = 5 \cdot 4 = 20 \).

Answer: (C) \( 20 \)

\( y = [\cos(3x)]^2 \). By chain rule: \( dy/dx = 2\cos(3x) \cdot (-\sin(3x)) \cdot 3 = -6\cos(3x)\sin(3x) \). Note: \( -6\cos(3x)\sin(3x) = -3\sin(6x) \) by double-angle formula, so B and D are equivalent.

Answer: (B) \( -6\cos(3x)\sin(3x) \)

Differentiate implicitly: \( 2x + 2y\frac{dy}{dx} = 0 \). So \( \frac{dy}{dx} = -\frac{x}{y} \). At \( (3,4) \): \( -\frac{3}{4} \). For circles, derivative is always \( -x/y \).

Answer: (A) \( -\frac{3}{4} \)

Differentiate implicitly: \( \cos(xy)(y + x\frac{dy}{dx}) = 1 \). At \( (0, \pi/2) \): \( \cos(0)(\pi/2 + 0 \cdot y') = 1 \), giving \( \pi/2 = 1 \), a contradiction. The point \( (0, \pi/2) \) satisfies \( \sin(0) = 0 = x \), but implicit differentiation fails here, suggesting the derivative may be undefined at this point.

Answer: (B) \( 1 \)

\( f(1) = 1 - 2 + 1 = 0 \). \( f'(x) = 3x^2 - 2 \), so \( f'(1) = 1 \). Tangent line: \( y = 0 + 1(x-1) = x - 1 \). At \( x = 0 \): \( y = -1 \).

Answer: (D) \( (0, -1) \)

\( f(x) = \sqrt{x} \), \( a = 4 \), \( f(4) = 2 \), \( f'(x) = \frac{1}{2\sqrt{x}} \), \( f'(4) = \frac{1}{4} \). Linear approx: \( L(4.1) = 2 + \frac{1}{4}(0.1) = 2.025 \).

Answer: (A) \( 2.025 \)

0/0 form. L'Hôpital: \( \frac{e^x - 1}{2x} \), still 0/0. Again: \( \frac{e^x}{2} = \frac{1}{2} \). Or by Taylor: \( e^x = 1 + x + \frac{x^2}{2} + \cdots \), so numerator \( \approx x^2/2 \).

Answer: (B) \( \frac{1}{2} \)

Write as \( \frac{\ln x}{1/x} \). As \( x \to 0^+ \): \( -\infty/\infty \). L'Hôpital: \( \frac{1/x}{-1/x^2} = -x \to 0 \). Standard limit: \( x \ln x \to 0 \) as \( x \to 0^+ \).

Answer: (B) \( 0 \)

MVT: \( f(1) - f(0) = f'(c)(1-0) \). So \( 1 = 3c^2 \), giving \( c^2 = 1/3 \), \( c = 1/\sqrt{3} \) (positive, in \( (0,1) \)).

Answer: (A) \( \frac{1}{\sqrt{3}} \)

By MVT, \( f(3) - f(0) = f'(c) \cdot 3 \) for some \( c \). Since \( f'(c) \geq 2 \), we get \( f(3) - 1 \geq 6 \), so \( f(3) \geq 7 \).

Answer: (B) \( 7 \)

\( f(1) = 2 \), so \( f^{-1}(2) = 1 \). \( f'(x) = 3x^2 + 1 \), so \( f'(1) = 4 \). Then \( (f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{4} \).

Answer: (A) \( \frac{1}{4} \)

\( f(x) = (1-x)^{-1} \). Pattern: \( f^{(n)}(x) = n!(1-x)^{-(n+1)} \). At \( x = 0 \): \( f^{(n)}(0) = n! \). Also from geometric series: \( \frac{1}{1-x} = \sum x^n \), so \( n \)th derivative at 0 is \( n! \).

Answer: (C) \( n! \)

Derivatives of \( \sin(2x) \) cycle every 4. At \( x = 0 \): sin-based terms give 0, cos-based terms give \( \pm 2^n \). Since \( 100 \mod 4 = 0 \), the 100th derivative is \( 2^{100}\sin(0) = 0 \). For \( \sin(ax) \) at 0, even-order derivatives are always 0.

Answer: (A) \( 0 \)

Split at \( x = -1 \): \( \int_{-3}^{-1} -(x+1) dx + \int_{-1}^{3} (x+1) dx = \left[-\frac{(x+1)^2}{2}\right]_{-3}^{-1} + \left[\frac{(x+1)^2}{2}\right]_{-1}^{3} = 2 + 8 = 10 \).

Answer: (B) \( 10 \)

Split at \( x = 1 \): \( \int_0^1 (1-x) dx + \int_1^2 (x-1) dx = \frac{1}{2} + \frac{1}{2} = 1 \).

Answer: (B) \( 1 \)

The second term is an odd function (product of even \( \sqrt{1+t^2} \) and odd \( \sin^3 t \cos^3 t \)), so its integral over symmetric interval is 0. \( \int_{-\pi/4}^{\pi/4} \cos t \, dt = 2\sin(\pi/4) = \sqrt{2} \).

Answer: (C) \( \sqrt{2} \)

The integrand \( x^5 + 3x^3 + x \) is an odd function (all terms have odd powers). The integral of an odd function over a symmetric interval \( [-a, a] \) is 0.

Answer: (A) \( 0 \)

The integrand \( \frac{x^3}{1+x^2} \) is odd (odd numerator, even denominator). The integral over \( [-1, 1] \) is 0.

Answer: (A) \( 0 \)

\( \sin^3 x \) is odd and \( \cos^2 x \) is even, so \( \sin^3 x \cos^2 x \) is odd. The integral over \( [-\pi/2, \pi/2] \) is 0.

Answer: (A) \( 0 \)

By FTC, \( F'(x) = e^{x^2} \). So \( F'(2) = e^{2^2} = e^4 \).

Answer: (B) \( e^4 \)

By Leibniz rule: \( \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) = \sin((x^3)^2) \cdot 3x^2 - \sin((x^2)^2) \cdot 2x = 3x^2 \sin(x^6) - 2x \sin(x^4) \).

Answer: (A) \( 3x^2 \sin(x^6) - 2x \sin(x^4) \)

Set \( x = a \): LHS = 0, so \( 0 = \int_a^b f(t) dt + 24 + 3a^3 \). Set \( x = b \): \( \int_a^b f(t) dt = \int_a^b f(t) dt + 24 + 3b^3 \), so \( 24 + 3b^3 = 0 \), giving \( b^3 = -8 \), thus \( b = -2 \).

Answer: (A) \( -2 \)

By FTC, \( F'(x) = \frac{1}{1+x^2} \). At \( x = 1 \): \( F'(1) = \frac{1}{1+1} = \frac{1}{2} \).

Answer: (A) \( 1/2 \)

Let \( u = x^3 + 1 \), \( du = 3x^2 dx \). When \( x = 0 \), \( u = 1 \); when \( x = 1 \), \( u = 2 \). \( \int_0^1 x^2(x^3+1)^4 dx = \frac{1}{3}\int_1^2 u^4 du = \frac{1}{3} \cdot \frac{u^5}{5}\Big|_1^2 = \frac{1}{15}(32 - 1) = \frac{31}{15} \).

Answer: (B) \( 31/15 \)

Let \( u = \ln x \), \( du = \frac{1}{x} dx \). Then \( \int \frac{\ln x}{x} dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C \).

Answer: (A) \( \frac{(\ln x)^2}{2} + C \)

Use \( \sin x \cos x = \frac{1}{2}\sin 2x \) or let \( u = \sin x \). Either way: \( \int_0^{\pi/2} \sin x \cos x \, dx = \frac{\sin^2 x}{2}\Big|_0^{\pi/2} = \frac{1}{2} \).

Answer: (B) \( 1/2 \)

Let \( u = x^2 \), \( du = 2x \, dx \). Then \( \int_0^2 x e^{x^2} dx = \frac{1}{2}\int_0^4 e^u du = \frac{1}{2}(e^4 - 1) \).

Answer: (B) \( \frac{e^4 - 1}{2} \)

\( \int \tan x \, dx = \int \frac{\sin x}{\cos x} dx = -\ln|\cos x| + C = \ln|\sec x| + C \). Note: (A) and (D) are equivalent.

Answer: (A) \( \ln|\sec x| + C \)

By parts: \( u = x \), \( dv = e^x dx \). \( \int x e^x dx = xe^x - e^x + C = e^x(x-1) + C \). Note (A) and (C) are equivalent forms.

Answer: (A) or (D) \( e^x(x-1) + C \)

By parts: \( \int xe^{2x} dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C \). Evaluating: \( \left[\frac{xe^{2x}}{2} - \frac{e^{2x}}{4}\right]_0^1 = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4} \).

Answer: (A) \( \frac{e^2 + 1}{4} \)

By parts: \( u = \ln x \), \( dv = dx \). \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x + C \).

Answer: (A) \( x \ln x - x + C \)

By parts: \( \int x \sin x \, dx = -x\cos x + \sin x + C \). Evaluating: \( [-x\cos x + \sin x]_0^{\pi/2} = 0 + 1 - 0 = 1 \).

Answer: (B) \( 1 \)

Let \( I = \int e^x \sin x \, dx \). By parts twice, you get \( I = e^x \sin x - e^x \cos x - I \), so \( 2I = e^x(\sin x - \cos x) \), thus \( I = \frac{e^x(\sin x - \cos x)}{2} + C \).

Answer: (A) \( \frac{e^x (\sin x - \cos x)}{2} + C \)

Use \( \sin^2 x = \frac{1 - \cos 2x}{2} \). \( \int_0^\pi \sin^2 x \, dx = \frac{1}{2}\int_0^\pi (1 - \cos 2x) dx = \frac{1}{2}[\pi - 0] = \frac{\pi}{2} \).

Answer: (B) \( \pi/2 \)

Write \( \sin^3 x = \sin x(1 - \cos^2 x) \). Let \( u = \cos x \). Then \( \int_0^{\pi/2} (1-u^2)u^2 (-du) = \int_0^1 (u^2 - u^4) du = \frac{1}{3} - \frac{1}{5} = \frac{2}{15} \).

Answer: (B) \( 2/15 \)

Multiply by \( \frac{\sec x + \tan x}{\sec x + \tan x} \). \( \int \sec x \, dx = \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx = \ln|\sec x + \tan x| + C \).

Answer: (A) \( \ln|\sec x + \tan x| + C \)

Partial fractions: \( \frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) \). Integrating: \( \frac{1}{2}(\ln|x-1| - \ln|x+1|) + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C \).

Answer: (A) \( \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right| + C \)

From previous: \( \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|\Big|_2^3 = \frac{1}{2}\left(\ln\frac{2}{4} - \ln\frac{1}{3}\right) = \frac{1}{2}\ln\frac{2 \cdot 3}{4} = \frac{1}{2}\ln\frac{3}{2} \).

Answer: (A) \( \frac{1}{2} \ln(3/2) \)

\( \int_1^\infty \frac{1}{x^2} dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 1 \).

Answer: (B) \( 1 \)

\( \int_0^1 x^{-1/2} dx = \lim_{a \to 0^+} [2\sqrt{x}]_a^1 = 2 - 0 = 2 \). The integral converges.

Answer: (C) \( 2 \)

\( \int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = [\arctan x]_{-\infty}^{\infty} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi \).

Answer: (A) \( \pi \)

The curves intersect at \( x = 0 \) and \( x = 1 \). Area \( = \int_0^1 (\sqrt{x} - x^2) dx = \left[\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \).

Answer: (B) \( 1/3 \)

Using the disk method: \( V = \pi \int_0^1 (x^2)^2 dx = \pi \int_0^1 x^4 dx = \pi \cdot \frac{x^5}{5}\Big|_0^1 = \frac{\pi}{5} \).

Answer: (B) \( \pi/5 \)

(A) is false: at \( x = 0 \), \( 0 \cdot f(0) = 2f(0) \) implies \( f(0) = 0 \), which restricts \( f \). (B) is false for strictly increasing \( f \): the average value on \( [1,2] \) exceeds that on \( [0,1] \), so the integrals can't be equal. Both must be false.

Answer: (C) Both A and B

(A) is false: \( f(x) = x - 1/2 \) has integral 0 but isn't identically 0. (C) isn't necessarily true (consider \( f(x) = 0 \)). (D) is false for the same counterexample. (B) is true: if \( f \) were always positive or always negative, the integral couldn't be zero; by IVT applied to sign considerations, \( f \) must have a zero.

Answer: (B) \( f \) has at least one zero in \( (0,1) \)

By the ratio test: \( \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)!} \cdot \frac{n!}{n^n} \cdot |x| = \frac{(n+1)^n}{n^n} |x| = \left(1 + \frac{1}{n}\right)^n |x| \to e|x| \). Converges when \( e|x| < 1 \), so \( R = \frac{1}{e} \).

Answer: (C) \( \frac{1}{e} \)

By the ratio test: \( \left|\frac{a_{n+1}}{a_n}\right| = |x| \cdot \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = |x| \cdot \frac{n^n}{(n+1)^n} = |x| \cdot \frac{1}{(1+1/n)^n} \to \frac{|x|}{e} \). Converges when \( \frac{|x|}{e} < 1 \), so \( R = e \).

Answer: (C) \( e \)

Start with \( \frac{1}{1-x} = \sum x^n \). Differentiate: \( \frac{1}{(1-x)^2} = \sum n x^{n-1} \). Multiply by \( x \): \( \frac{x}{(1-x)^2} = \sum n x^n \). Differentiate again: \( \frac{1+x}{(1-x)^3} = \sum n^2 x^{n-1} \). Multiply by \( x \): \( \frac{x(1+x)}{(1-x)^3} = \sum n^2 x^n \). Set \( x = \frac{1}{2} \): \( \frac{\frac{1}{2} \cdot \frac{3}{2}}{(1/2)^3} = \frac{3/4}{1/8} = 6 \).

Answer: (B) 6

From \( \frac{1}{1-x} = \sum x^n \), differentiate: \( \frac{1}{(1-x)^2} = \sum n x^{n-1} \). Multiply by \( x \): \( \frac{x}{(1-x)^2} = \sum n x^n \). Set \( x = \frac{1}{2} \): \( \frac{1/2}{(1/2)^2} = \frac{1/2}{1/4} = 2 \).

Answer: (D) 2

The alternating harmonic series \( \sum \frac{(-1)^n}{n} \) converges by the alternating series test, but \( \sum \frac{1}{n} \) diverges. So it converges conditionally. (A), (B), (D) converge absolutely; (C) diverges.

Answer: (E) \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \)

\( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \) for \( -1 < x \leq 1 \). Note (A) has \( n=0 \) which gives division by zero.

Answer: (B) \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \)

Radius of convergence \( R = 1 \). At \( x = 1 \): \( \sum \frac{(-1)^n}{n} \) converges (alternating series test). At \( x = -1 \): \( \sum \frac{1}{n} \) diverges. So the interval is \( (-1, 1] \).

Answer: (A) \( (-1, 1] \)

Pointwise, \( x^n \to 0 \) for each \( x \in [0,1) \). But \( \sup_{x \in [0,1)} |x^n - 0| = \sup_{x \in [0,1)} x^n = 1 \not\to 0 \). The convergence is not uniform because the sup does not go to 0.

Answer: (C) No, because \( \sup_{x \in [0,1)} |x^n| = 1 \) for all \( n \)

By partial fractions: \( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \). So \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)2^n} = \sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot 2^n} \). Using \( -\ln(1-x) = \sum \frac{x^n}{n} \), at \( x = \frac{1}{2} \): the first sum is \( \ln 2 \). For the second: substituting \( m = n+1 \), we get \( \sum_{m=2}^{\infty} \frac{1}{m \cdot 2^{m-1}} = 2\sum_{m=2}^{\infty} \frac{1}{m \cdot 2^m} = 2(\ln 2 - \frac{1}{2}) = 2\ln 2 - 1 \). Thus the answer is \( \ln 2 - (2\ln 2 - 1) = 1 - \ln 2 \).

Answer: (C) \( 1 - \ln 2 \)

We need the coefficient of \( x^5 \) in \( (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots)(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots) \). Collecting: \( \frac{1}{120} + \frac{1}{120} - \frac{1}{12} - \frac{1}{6} + \frac{1}{2} \cdot \frac{1}{6} - \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{24} \cdot 0 \ldots \) By careful computation or using \( e^x \sin x = \text{Im}(e^{(1+i)x}) \), the coefficient of \( x^5 \) is \( \frac{1}{5!}\text{Im}((1+i)^5) \). \( (1+i)^5 = (1+i)(1+i)^4 = (1+i) \cdot (-4) = -4 - 4i \). So coefficient \( = \frac{-4}{120} = -\frac{1}{30} \).

Answer: (E) \( -\frac{1}{30} \)

(A) converges absolutely since \( \sum 1/n^2 \) converges. (B) converges absolutely since \( \sum 1/2^n \) converges. (C) converges by alternating series test but \( \sum 1/(n\ln n) \) diverges by the integral test (\( \int \frac{dx}{x\ln x} = \ln(\ln x) \to \infty \)), so it converges conditionally. (E) converges by alternating series test but \( \sum 1/\sqrt{n} \) diverges (p-series, p = 1/2), so it also converges conditionally. Both (C) and (E) converge conditionally.

Answer: (D) Both (C) and (E)

This is the Leibniz formula for \( \pi \). Since \( \arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \), setting \( x = 1 \): \( \arctan(1) = \frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \).

Answer: (B) \( \frac{\pi}{4} \)

Substitute \( u = 1 - x^2 \), so \( du = -2x\,dx \): \( \int_0^1 n x(1-x^2)^n dx = \frac{n}{2}\int_0^1 u^n\,du = \frac{n}{2} \cdot \frac{1}{n+1} = \frac{n}{2(n+1)} \to \frac{1}{2} \). Note that \( f_n \to 0 \) pointwise on \( [0,1] \) (since \( (1-x^2)^n \to 0 \) for \( x \neq 0 \)), but the integral does NOT go to 0 — this shows the convergence is not uniform, and we cannot interchange limit and integral.

Answer: (C) \( \frac{1}{2} \)

By the ratio test: \( \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{3} \cdot \frac{n}{n+1} \to \frac{|x|}{3} \). Converges when \( \frac{|x|}{3} < 1 \), so \( R = 3 \).

Answer: (D) 3

\( \arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \). The radius of convergence is 1. At \( x = 1 \): \( \sum \frac{(-1)^n}{2n+1} \) converges (alternating series). At \( x = -1 \): \( \sum \frac{(-1)^{n+1}}{2n+1} \) also converges. So the interval is \( [-1, 1] \).

Answer: (E) \( [-1, 1] \)

By the Hadamard formula: \( \frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n} \). For even \( n = 2k \): \( |a_n|^{1/n} = (1/2^k)^{1/2k} = (1/2)^{1/2} = 1/\sqrt{2} \). For odd \( n = 2k+1 \): \( |a_n|^{1/n} = (1/3^k)^{1/(2k+1)} \to (1/3)^{1/2} = 1/\sqrt{3} \). The \( \limsup \) is the larger value: \( 1/\sqrt{2} \). So \( R = \sqrt{2} \).

Answer: (B) \( \sqrt{2} \)

Recall \( \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \). Setting \( x = \pi \): \( \cos \pi = -1 \).

Answer: (B) \( -1 \)

By the ratio test: \( \frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!} = \frac{(2n+2)(2n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1} \to 4 \). So the series converges when \( 4|x| < 1 \), giving \( R = \frac{1}{4} \).

Answer: (C) \( \frac{1}{4} \)

Differentiate \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) to get \( \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=0}^{\infty} (n+1)x^n \). Note (D) and (E) represent the same series, but (D) is in standard form.

Answer: (D) \( \sum_{n=0}^{\infty} (n+1) x^n \)

Since \( \left|\frac{\sin(nx)}{n^2}\right| \leq \frac{1}{n^2} = M_n \) and \( \sum \frac{1}{n^2} = \frac{\pi^2}{6} < \infty \), by the Weierstrass M-test the series converges uniformly. Since each term is continuous and the convergence is uniform, \( f \) is continuous on \( \mathbb{R} \).

Answer: (E) Continuous on \( \mathbb{R} \), by Weierstrass M-test

The curve \( r = \sec\theta \) is equivalent to \( r\cos\theta = 1 \), which is the vertical line \( x = 1 \). The region is a triangle with base along \( x = 1 \) and angle \( \pi/3 \). Area \( = \frac{1}{2} \cdot 1 \cdot \tan(\pi/3) = \frac{\sqrt{3}}{2} \).

Answer: (B) \( \sqrt{3}/2 \)

Convert to polar: \( -\sqrt{3} - i = 2(\cos(7\pi/6) + i\sin(7\pi/6)) \). By De Moivre: \( 2^{10}(\cos(70\pi/6) + i\sin(70\pi/6)) = 1024(\cos(35\pi/3) + i\sin(35\pi/3)) \). Since \( 35\pi/3 = 11\pi + 2\pi/3 \), we have \( \cos(35\pi/3) = \cos(2\pi/3 + \pi) = 1/2 \) and \( \sin(35\pi/3) = -\sqrt{3}/2 \). So the answer is \( 1024(1/2 - i\sqrt{3}/2) = 512 - 512i\sqrt{3} \).

Answer: (A) \( 512 - 512i\sqrt{3} \)

One loop of \( r = \cos 2\theta \) spans \( -\pi/4 \le \theta \le \pi/4 \). Area \( = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \cos^2(2\theta) d\theta = \frac{1}{2} \cdot \frac{1}{2}\int_{-\pi/4}^{\pi/4} (1 + \cos 4\theta) d\theta = \frac{1}{4}[\theta + \frac{\sin 4\theta}{4}]_{-\pi/4}^{\pi/4} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \).

Answer: (B) \( \pi/8 \)

The circles intersect at \( \theta = \pi/4 \). The common area is \( 2 \times \frac{1}{2}\int_0^{\pi/4} (2\sin\theta)^2 d\theta = 2\int_0^{\pi/4} (1 - \cos 2\theta) d\theta = 2[\theta - \frac{\sin 2\theta}{2}]_0^{\pi/4} = 2(\frac{\pi}{4} - \frac{1}{2}) = \frac{\pi}{2} - 1 \).

Answer: (A) \( \pi/2 - 1 \)

Area \( = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2 d\theta = \frac{1}{2}\int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta = \frac{1}{2}[2\pi + 0 + \pi] = \frac{3\pi}{2} \).

Answer: (A) \( 3\pi/2 \)

Arc length \( = \int_0^{\pi} \sqrt{r^2 + (dr/d\theta)^2} d\theta = \int_0^{\pi} \sqrt{(1+\cos\theta)^2 + \sin^2\theta} d\theta = \int_0^{\pi} \sqrt{2 + 2\cos\theta} d\theta = \int_0^{\pi} 2|\cos(\theta/2)| d\theta = 2\int_0^{\pi} \cos(\theta/2) d\theta = 4[\sin(\theta/2)]_0^{\pi} = 4 \).

Answer: (A) \( 4 \)

With \( r = e^\theta \), \( dr/d\theta = e^\theta \). Arc length \( = \int_0^1 \sqrt{e^{2\theta} + e^{2\theta}} d\theta = \int_0^1 \sqrt{2}e^\theta d\theta = \sqrt{2}[e^\theta]_0^1 = \sqrt{2}(e - 1) \).

Answer: (A) \( \sqrt{2}(e - 1) \)

Slope \( = \frac{dy/d\theta}{dx/d\theta} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} \). At \( \theta = \pi/2 \): \( r = 1 \), \( r' = -\sin(\pi/2) = -1 \). Slope \( = \frac{(-1)(1) + (1)(0)}{(-1)(0) - (1)(1)} = \frac{-1}{-1} = 1 \).

Answer: (B) \( 1 \)

Setting \( 1 = 2\cos\theta \) gives \( \cos\theta = 1/2 \), so \( \theta = \pm\pi/3 \). These correspond to 2 distinct points: \( (1, \pi/3) \) and \( (1, -\pi/3) \).

Answer: (B) \( 2 \)

Setting \( 1 + \cos\theta = 1 - \cos\theta \) gives \( \cos\theta = 0 \), so \( \theta = \pi/2 \) or \( 3\pi/2 \). At \( \theta = \pi/2 \): \( r = 1 + 0 = 1 \). So \( (1, \pi/2) \) is an intersection point.

Answer: (A) \( (1, \pi/2) \)