Answer: Degree 6

Set up the error inequality: We need $\frac{(0.3)^{n+1}}{(n+1)!} < 10^{-6}$ where $M = 1$ for all derivatives of $\cos x$. Test degree 4 ($n = 4$): $\frac{(0.3)^5}{5!} = \frac{0.00243}{120} = 2.025 \times 10^{-5} > 10^{-6}$. Not sufficient. Test degree 6 ($n = 6$): $\frac{(0.3)^7}{7!} = \frac{0.0002187}{5040} \approx 4.34 \times 10^{-8} < 10^{-6}$. ✓ Sufficient. Why distractors fail: Degree 4 gives error $\approx 2 \times 10^{-5}$, too large. Degree 2 is far too crude. Degree 8 works but is more terms than necessary.

Answer: $\sum_{n=0}^{\infty} (-1)^n x^{2n}$

Substitute into the geometric series: $\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n$. Replacing $u$ with $-x^2$: $\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}$. Why Option B is correct: The substitution $u = -x^2$ yields $\sum_{n=0}^{\infty} (-1)^n x^{2n}$. Why distractors fail: Option A uses $x^n$ instead of $x^{2n}$ — it corresponds to $\frac{1}{1+x}$, not $\frac{1}{1+x^2}$. Option C omits the alternating sign. Option D has odd powers $x^{2n+1}$.

Answer: $1 - \frac{1}{3} + \frac{1}{10}$

Write the series and integrate term by term: $e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2}$. Integrating: $\int_0^1 (1 - x^2 + \frac{x^4}{2})\,dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10}\right]_0^1 = 1 - \frac{1}{3} + \frac{1}{10}$. Why Option A is correct: Term-by-term integration of the three-term approximation yields $1 - \frac{1}{3} + \frac{1}{10} = \frac{30 - 10 + 3}{30} = \frac{23}{30} \approx 0.7667$. Why distractors fail: Option B integrates incorrectly (divides by wrong powers). Option C gets $\frac{1}{12}$ for the last term instead of $\frac{1}{10}$. Option D has the wrong sign on the $\frac{1}{3}$ term.

Answer: Integrate the Maclaurin series $\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}$ term by term.

Identify the relationship: Since $\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$, integrating the known series for $\frac{1}{1+x^2}$ term by term gives the series for $\arctan x$. Why Option B is correct: Integrating $\sum_{n=0}^{\infty} (-1)^n x^{2n}$ gives $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$, which is the well-known series for $\arctan x$. Why distractors fail: Option A suggests differentiating $\tan x$, but the series for $\tan x$ is harder to compute and differentiation gives $\sec^2 x$, not $\arctan x$. Option C is an unclear and incorrect substitution. Option D is valid in principle but extremely tedious for higher-order derivatives.

Answer: $1 - x^2 + \frac{x^4}{2}$

Substitute into the known series for $e^u$: Start with $e^u = 1 + u + \frac{u^2}{2!} + \cdots$ and substitute $u = -x^2$: $e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \cdots = 1 - x^2 + \frac{x^4}{2} - \cdots$. Why Option B is correct: The substitution yields $1 - x^2 + \frac{x^4}{2}$ as the first three nonzero terms. Why distractors fail: Option A has the wrong sign on the $x^2$ term. Option C uses $\frac{x^4}{4}$ instead of $\frac{x^4}{2!} = \frac{x^4}{2}$. Option D has incorrect coefficients from a computational error.

Answer: $1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040}$

Divide the $\sin x$ series by $x$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$. Dividing by $x$: $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$. Why Option A is correct: Each power of $x$ in $\sin x$ is reduced by 1, giving even powers: $1, -x^2/6, x^4/120, -x^6/5040$. Why distractors fail: Option B is just the series for $\sin x$ itself, not divided by $x$. Option C divides by $x$ incorrectly (reducing the power in the denominator rather than in $x$). Option D has the wrong denominators — those match $\cos x$ coefficients.

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

Derive the series from the geometric series: Since $\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n$, integrating term by term gives $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$. Why Option B is correct: It correctly starts at $n=1$, includes the alternating sign $(-1)^{n+1}$, and has $n$ (not $n!$) in the denominator. Why distractors fail: Option A starts at $n=0$, which would cause division by zero. Option C lacks the alternating sign. Option D uses $(n+1)!$ instead of $n$ in the denominator.

Answer: $\sum_{k=0}^{n} a_k b_{n-k}$

Apply the Cauchy product formula: When multiplying two power series, the coefficient of $x^n$ in the product is $\sum_{k=0}^{n} a_k b_{n-k}$, which is the discrete convolution of the coefficient sequences. Why Option B is correct: This is the standard Cauchy product formula for multiplying power series. Why distractors fail: Option A multiplies corresponding coefficients, which is not how series multiplication works. Option C adds corresponding coefficients, which corresponds to $f(x) + g(x)$, not $f(x) \cdot g(x)$. Option D introduces an unjustified $n!$.

Answer: $\frac{1}{4}$

Extract the coefficient of $(x-3)^2$: The $n=2$ term is $\frac{(-1)^2}{2^2 \cdot 2!}(x-3)^2 = \frac{1}{4 \cdot 2}(x-3)^2 = \frac{1}{8}(x-3)^2$. Relate to $f''(3)$: In a Taylor series, the coefficient of $(x-a)^n$ equals $\frac{f^{(n)}(a)}{n!}$. So $\frac{f''(3)}{2!} = \frac{1}{8}$, giving $f''(3) = 2 \cdot \frac{1}{8} = \frac{1}{4}$. Why distractors fail: Option B ($-1/2$) has the wrong sign and magnitude. Option C ($1/2$) arises from forgetting the $2^n$ factor and using only $\frac{1}{n!}$, which gives $\frac{f''(3)}{2!} = \frac{1}{2}$ and $f''(3) = 1$, then further miscomputing. Option D ($-1/4$) has the wrong sign.

Answer: Because $\cos x$ is an even function, so all odd-order derivatives vanish at $x = 0$.

Recall even/odd function properties: $\cos x$ is an even function: $\cos(-x) = \cos x$. Even functions have the property that all odd-order derivatives at the origin are zero. Why Option B is correct: Since $\cos x$ is even, $f'(0) = 0$, $f'''(0) = 0$, etc., so only even-power terms survive in the Maclaurin expansion. Why distractors fail: Option A incorrectly calls $\cos x$ an odd function. Option C attributes the absence of odd powers to the factorial, which is not the reason. Option D incorrectly invokes the radius of convergence.

Answer: It normalizes the $n$th derivative to account for the repeated differentiation of $(x-a)^n$.

Understand why $n!$ appears: When you differentiate $(x-a)^n$ a total of $n$ times, you get $n!$. The $\frac{1}{n!}$ factor cancels this, so that $\frac{f^{(n)}(a)}{n!}$ correctly recovers the coefficient of each term. Why Option B is correct: The factorial normalizes the derivative to produce the correct coefficient, matching the polynomial whose $n$th derivative at $a$ equals $f^{(n)}(a)$. Why distractors fail: Option A confuses a consequence of convergence with the structural role of $n!$. Option C invokes Fourier series, which are unrelated. Option D is incorrect because the radius of convergence depends on the function, not the factorial.

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

Differentiate the geometric series: $\frac{d}{dx}\left[\frac{1}{1-x}\right] = \frac{1}{(1-x)^2}$. Differentiating $\sum_{n=0}^{\infty} x^n$ term by term: $\sum_{n=1}^{\infty} n x^{n-1}$. Re-index: Letting $m = n-1$: $\sum_{m=0}^{\infty} (m+1) x^m = \sum_{n=0}^{\infty} (n+1) x^n$. Why distractors fail: Option A is correct before re-indexing but is not in standard form (the $n=0$ term contributes $0 \cdot x^{-1}$, which is problematic notation). Option C has coefficient $n$ instead of $n+1$. Option D shifts the power by 1.

Answer: $-\frac{1}{6}$

Write the series for $\sin x$ and multiply by $x^2$: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$, so $x^2 \sin x = x^3 - \frac{x^5}{6} + \frac{x^7}{120} - \cdots$. Why Option B is correct: The $x^5$ term comes from $x^2 \cdot (-\frac{x^3}{6}) = -\frac{x^5}{6}$, so the coefficient is $-\frac{1}{6}$. Why distractors fail: Option A has the wrong sign. Option C uses $\frac{1}{5!} = \frac{1}{120}$, which is the coefficient of $x^5$ in $\sin x$ itself, not after multiplying by $x^2$. Option D confuses the factorial as $4! = 24$.

Answer: $a_n = \frac{f^{(n)}(0)}{n!}$

Apply the Maclaurin series definition: The Maclaurin series is $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$, so $a_n = \frac{f^{(n)}(0)}{n!}$. Why Option B is correct: It directly follows from the definition of the Maclaurin series. Why distractors fail: Option A omits the $n!$ denominator. Option C uses $(n+1)!$ instead of $n!$. Option D uses the wrong derivative index.

Answer: 3 terms (up to $x^4$)

Set up the error bound: For $\cos x$, all derivatives are bounded by 1 in absolute value. After using $n$ nonzero terms of the $\cos x$ series (up to degree $2n-2$), the error bound is $|R| \leq \frac{(0.1)^{2n}}{(2n)!}$ (since the next term has degree $2n$, approximately). Check with 3 terms (up to $x^4$, polynomial of degree 4): The next term after degree 4 is degree 6. Error bound: $\frac{(0.1)^6}{6!} = \frac{10^{-6}}{720} \approx 1.39 \times 10^{-9} < 10^{-8}$. ✓ Check with 2 terms (up to $x^2$): Error bound: $\frac{(0.1)^4}{4!} = \frac{10^{-4}}{24} \approx 4.17 \times 10^{-6}$, which is much larger than $10^{-8}$. ✗ Why distractors fail: Option A (2 terms) gives error $\approx 4.17 \times 10^{-6}$, too large. Options C and D use more terms than necessary.

Answer: $\frac{1}{3}$

Compute derivatives of $\ln x$ at $a = 1$: $f'(x) = 1/x$, $f''(x) = -1/x^2$, $f'''(x) = 2/x^3$. At $x = 1$: $f'''(1) = 2$. Find the coefficient: The coefficient of $(x-1)^3$ is $\frac{f'''(1)}{3!} = \frac{2}{6} = \frac{1}{3}$. Why distractors fail: Option B has the wrong sign; $f'''(1) = 2 > 0$. Option C uses $\frac{1}{3!} = \frac{1}{6}$, forgetting to multiply by $f'''(1) = 2$. Option D uses $\frac{f'''(1)}{3} = \frac{2}{3}$ without the factorial.

Answer: No — the terms going to zero is necessary but not sufficient; however, the series does converge for all $x$ by the ratio test.

Evaluate the student's reasoning: The divergence test states that if terms do not approach zero, the series diverges. However, terms going to zero does not guarantee convergence (e.g., the harmonic series). So the student's reasoning is flawed. Why the series still converges everywhere: Applying the ratio test: $\lim_{n\to\infty} \left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \lim_{n\to\infty} \frac{|x|}{n+1} = 0 < 1$ for all $x$. This properly establishes convergence for all $x$. Why distractors fail: Option A accepts flawed reasoning. Option C claims a finite radius of convergence, which is false for $e^x$. Option D reaches the right conclusion but with incomplete logic — the rate of factorial growth alone doesn't constitute a valid convergence argument.

Answer: $x - \frac{x^3}{18} + \frac{x^5}{600}$

Start from $\frac{\sin t}{t}$: $\frac{\sin t}{t} = 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \cdots = 1 - \frac{t^2}{6} + \frac{t^4}{120} - \cdots$. Integrate term by term: $\int_0^x \frac{\sin t}{t}\,dt = x - \frac{x^3}{3 \cdot 6} + \frac{x^5}{5 \cdot 120} - \cdots = x - \frac{x^3}{18} + \frac{x^5}{600} - \cdots$. Why distractors fail: Option B gives the series for $\frac{\sin x}{x}$ divided by wrong factors (not integrated). Option C is the series for $\frac{\sin x}{x}$ itself, not its integral. Option D uses $\frac{1}{9}$ instead of $\frac{1}{18}$ for the $x^3$ coefficient.

Answer: $\sum_{n=0}^{\infty} \frac{x^n}{n!}$

Recall the Maclaurin series for $e^x$: Since every derivative of $e^x$ is $e^x$, and $e^0 = 1$, the Maclaurin series is $\sum_{n=0}^{\infty} \frac{x^n}{n!}$. Why Option C is correct: It correctly begins at $n=0$ and uses $\frac{x^n}{n!}$, matching the standard memorized expansion. Why distractors fail: Option A uses $(n+1)!$ instead of $n!$. Option B starts at $n=1$, missing the constant term $1$. Option D has $x^{n+1}$ instead of $x^n$, shifting all powers by one.

Answer: $\sum_{n=0}^{\infty} (-1)^n x^{3n+1}$

Start with the geometric series substitution: $\frac{1}{1+x^3} = \frac{1}{1-(-x^3)} = \sum_{n=0}^{\infty} (-x^3)^n = \sum_{n=0}^{\infty} (-1)^n x^{3n}$. Multiply by $x$: $\frac{x}{1+x^3} = x \cdot \sum_{n=0}^{\infty} (-1)^n x^{3n} = \sum_{n=0}^{\infty} (-1)^n x^{3n+1}$. Why distractors fail: Option A is the series for $\frac{1}{1+x^3}$ without the factor of $x$. Option C adds 3 to the exponent instead of 1 (as if multiplying by $x^3$). Option D re-indexes incorrectly.

Answer: $\sum_{n=0}^{\infty} x^n$, valid for $|x| < 1$

Recall the geometric series: $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ is the geometric series, converging for $|x| < 1$. Why Option A is correct: It matches both the series formula (starting at $n=0$ with no alternating sign) and the correct interval of convergence $|x| < 1$. Why distractors fail: Option B omits the $n=0$ term (which is 1). Option C has $(-1)^n$, which corresponds to $\frac{1}{1+x}$. Option D claims convergence for all $x$, but the geometric series diverges for $|x| \geq 1$.

Answer: It provides a guaranteed upper bound on the magnitude of the approximation error after truncating the series at degree $n$.

Interpret the Lagrange error bound: The Lagrange error bound states that the absolute error $|R_n(x)|$ when using a degree-$n$ Taylor polynomial is at most $\frac{M}{(n+1)!}|x-a|^{n+1}$. It provides an upper bound, not the exact error. Why Option B is correct: It accurately describes the bound as a guaranteed worst-case upper limit on the approximation error. Why distractors fail: Option A is wrong because the bound is an inequality, not an exact value. Option C confuses the error bound with the radius of convergence. Option D conflates the bound with convergence criteria.

Answer: $|R_n(x)| \leq \frac{M}{(n+1)!} |x - a|^{n+1}$ where $M = \max|f^{(n+1)}(z)|$

State the Lagrange error bound: The Lagrange error bound says $|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}$ where $M$ is the maximum of $|f^{(n+1)}(z)|$ for $z$ between $a$ and $x$. Why Option B is correct: It correctly pairs the $(n+1)$th derivative with $(n+1)!$ and $|x-a|^{n+1}$. Why distractors fail: Option A uses $n!$ and $|x-a|^n$ with the $n$th derivative — all indices are off by one. Option C has the correct derivative and factorial but uses $|x-a|^n$ instead of $|x-a|^{n+1}$. Option D pairs $n!$ with the $n$th derivative instead of $(n+1)!$ with the $(n+1)$th derivative.

Answer: $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$

Recall the Taylor series definition: The Taylor series of $f(x)$ centered at $x = a$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$, where $f^{(n)}(a)$ denotes the $n$th derivative of $f$ evaluated at $a$. Why Option A is correct: It matches the standard definition exactly: summation starts at $n=0$, uses $n!$ in the denominator, and evaluates derivatives at the center $a$. Why distractors fail: Option B incorrectly evaluates derivatives at $x$ instead of the center $a$. Option C starts the sum at $n=1$, omitting the constant term $f(a)$. Option D uses $(n+1)!$ instead of $n!$ in the denominator.

Answer: $\frac{e^{0.5} \cdot (0.5)^4}{4!} \approx 0.00429$

Set up the Lagrange error bound: For the degree-3 polynomial, the error satisfies $|R_3(0.5)| \leq \frac{M}{4!}(0.5)^4$, where $M = \max_{z \in [0, 0.5]}|f^{(4)}(z)| = e^{0.5}$ since $f^{(4)}(x) = e^x$ is increasing. Compute the bound: $\frac{e^{0.5}(0.5)^4}{24} = \frac{1.6487 \times 0.0625}{24} \approx \frac{0.1030}{24} \approx 0.00429$. Why distractors fail: Option A uses $M = 1$ (i.e., $e^0$) instead of $e^{0.5}$, underestimating the bound. Option B uses the wrong degree: it applies a degree-2 error formula with $(0.5)^3/3!$. Option D omits the $M = e^{0.5}$ factor and uses the wrong power and factorial.

Answer: $2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2$

Compute derivatives at $a = 4$: $f(x)=x^{1/2}$, so $f(4)=2$. $f'(x)=\frac{1}{2}x^{-1/2}$, so $f'(4)=\frac{1}{4}$. $f''(x)=-\frac{1}{4}x^{-3/2}$, so $f''(4)=-\frac{1}{32}$. Assemble the polynomial: $P_2(x) = 2 + \frac{1}{4}(x-4) + \frac{-1/32}{2!}(x-4)^2 = 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2$. Why distractors fail: Option B has the wrong sign on the quadratic term. Option C uses $f'(4)=\frac{1}{2}$ and $f''(4)/2 = -\frac{1}{16}$, both incorrect. Option D has the constant term as 4 instead of $\sqrt{4} = 2$.

Answer: $x^3 - \frac{x^9}{6}$

Substitute $x^3$ into the $\sin$ series: $\sin u = u - \frac{u^3}{3!} + \frac{u^5}{5!} - \cdots$. Set $u = x^3$: $\sin(x^3) = x^3 - \frac{(x^3)^3}{6} + \cdots = x^3 - \frac{x^9}{6} + \cdots$. Why Option B is correct: Up to the $x^9$ term, only $x^3$ and $-\frac{x^9}{6}$ appear; the next term would be $\frac{x^{15}}{120}$, which exceeds $x^9$. Why distractors fail: Option A includes a spurious $x^6$ term, which does not arise from the substitution. Option C has the wrong sign on the $x^9$ term. Option D introduces $x^6$ and $x^9$ terms with incorrect groupings.

Answer: Yes, because the next nonzero term after degree 3 is degree 5, so the error is bounded by $\frac{(0.2)^5}{5!}$.

Analyze the polynomial degree: The fourth-degree Maclaurin polynomial for $\sin x$ is $P_4(x) = x - \frac{x^3}{6}$ (since the coefficient of $x^4$ is 0 for $\sin x$, as $f^{(4)}(0) = \sin(0) = 0$). Determine the error bound: Since $P_4(x)$ matches the Taylor series through degree 4, the remainder involves the 5th derivative. For $\sin x$, $|f^{(5)}(z)| \leq 1$ for all $z$, so $|R_4(0.2)| \leq \frac{1 \cdot (0.2)^5}{5!}$. Why distractors fail: Option A incorrectly uses $4!$. Option C claims the polynomial is only degree 3, but since the $x^4$ coefficient is zero, $P_4(x) = P_3(x)$, yet the error bound for a degree-4 polynomial still uses the 5th derivative. Option D is wrong because for $\sin x$, all derivatives are bounded by $M = 1$.

Answer: $2x + \frac{2x^3}{3} + \frac{2x^5}{5}$

Split using logarithm properties: $\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)$. Use known series: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots$ and $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \cdots$. Subtract: $\ln(1+x) - \ln(1-x) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots$. All even-power terms cancel. Why distractors fail: Option B has the wrong sign on $x^3$. Option C is missing the factor of 2. Option D has 2 on the first term but not on the others.

Answer: $-\frac{1}{6}$

Multiply the two series up to degree 4: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$ and $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots$. Collect $x^4$ terms from the product: The $x^4$ contributions come from pairing terms whose degrees sum to 4: $(1)(\frac{x^4}{24}) + (\frac{x^2}{2})(-\frac{x^2}{2}) + (\frac{x^4}{24})(1) = \frac{1}{24} - \frac{1}{4} + \frac{1}{24} = \frac{2}{24} - \frac{6}{24} = -\frac{4}{24} = -\frac{1}{6}$. Why distractors fail: Option B ($-1/3$) doubles the correct answer, a common error from mishandling cross terms. Option C ($1/6$) has the wrong sign. Option D ($-1/4$) comes from using only the middle cross term $\frac{x^2}{2} \cdot (-\frac{x^2}{2})$.

Answer: $\frac{1}{2}$

Expand $e^x$ and simplify: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$, so $e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots$. Divide by $x^2$: $\frac{e^x - 1 - x}{x^2} = \frac{1}{2} + \frac{x}{6} + \cdots \to \frac{1}{2}$ as $x \to 0$. Why distractors fail: Option A would result if the numerator were $e^x - 1 - x - x^2/2$. Option C would require the leading term to be $x^2$. Option D is wrong because the limit clearly exists.

Answer: $\frac{e^{1.5}(0.5)^3}{3!} \approx 0.0934$

Apply the Lagrange error bound for degree 2: $|R_2(1.5)| \leq \frac{M}{3!}|1.5 - 1|^3$, where $M = \max_{z \in [1, 1.5]} |f'''(z)| = e^{1.5}$ since $f'''(x) = e^x$ is increasing. Compute: $\frac{e^{1.5}(0.5)^3}{6} = \frac{4.4817 \times 0.125}{6} = \frac{0.5602}{6} \approx 0.0934$. Why distractors fail: Option B uses $M = e$ (the value of $f'''$ at the center $a = 1$) instead of $e^{1.5}$ (the maximum on $[1, 1.5]$), underestimating the bound. Option C uses the wrong power $(0.5)^2$ and factorial $2!$, corresponding to the degree-1 error formula rather than degree-2. Option D uses the correct derivative and power but divides by $2!$ instead of $3!$, misapplying the Lagrange formula.

Answer: $\frac{1}{3}$

Use long division of series: Dividing $\sin x = x - x^3/6 + \cdots$ by $\cos x = 1 - x^2/2 + \cdots$, or equivalently computing $f'''(0)/3!$ for $f(x) = \tan x$: $f'(x) = \sec^2 x$, $f''(x) = 2\sec^2 x \tan x$, $f'''(x)$ evaluated at 0 gives $f'''(0) = 2$. So the coefficient is $\frac{2}{3!} = \frac{1}{3}$. Why Option B is correct: Both the series division method and the direct derivative computation yield $\frac{1}{3}$ as the coefficient of $x^3$. Why distractors fail: Option A ($1/6$) is the coefficient of $x^3$ in $\sin x$, not $\tan x$. Option C ($1/2$) confuses the coefficient with $f''(0)/2!$. Option D has the wrong sign.

Answer: The second student is correct: the Lagrange bound gives $\frac{M \cdot |1|^6}{6!}$ with $M = 1$.

Apply the Lagrange error bound precisely: For a degree-$n$ Taylor polynomial, $|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}$. With $n = 5$ and $x = 1$, $a = 0$: $|R_5(1)| \leq \frac{1}{6!} \cdot 1^6 = \frac{1}{720}$. Address the first student's error: The first student jumps to the 7th-degree bound. While the $x^6$ coefficient of $\sin x$ is zero (so $P_5 = P_6$ for $\sin x$), the strict Lagrange bound for $P_5$ uses $(n+1) = 6$. The sharper bound using $P_6$ would give $\frac{1}{7!}$, but the question specifies the degree-5 polynomial. Why distractors fail: Option A confuses the degree-5 bound with the degree-6 bound (which would be valid if we explicitly stated we were using $P_6$). Option C is numerically false: $1/6! = 1/720 \neq 1/5040 = 1/7!$. Option D uses the wrong formula entirely.

Answer: $a = 0$

Define the Maclaurin series: A Maclaurin series is simply a Taylor series with the center $a = 0$, so it takes the form $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$. Why Option C is correct: By definition, a Maclaurin series is centered at $a = 0$. Why distractors fail: Option A ($a=1$) and Option B ($a=-1$) are valid centers for Taylor series but do not define a Maclaurin series. Option D describes Taylor series in general, not the specific Maclaurin case.

Answer: $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

Identify the pattern of sin x derivatives at 0: The derivatives of $\sin x$ at $0$ cycle through $0, 1, 0, -1, \ldots$, so only odd powers appear with alternating signs. Why Option B is correct: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$, which matches Option B. Why distractors fail: Option A uses even powers $x^{2n}/(2n)!$, which is the series for $\cos x$. Option C lacks the alternating sign $(-1)^n$. Option D uses even powers and starts at $n=1$.

Answer: Both methods are valid, but multiplying known series is generally more efficient.

Evaluate both methods: Computing derivatives of $e^x \sin x$ requires the product rule repeatedly and becomes tedious. Multiplying the series $e^x = \sum \frac{x^n}{n!}$ and $\sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ and collecting like terms is typically faster. Why Option B is correct: Both approaches are mathematically valid, but series multiplication avoids repeated application of the product rule and is generally more efficient. Why distractors fail: Option A incorrectly dismisses series multiplication. Option C incorrectly dismisses the derivative method. Option D proposes integration by parts, which doesn't directly produce a Maclaurin series.