Differential Calculus Questions and Answers

Answer: A

Rolle's theorem asks for \( f(a) = f(b) \), and concludes that \( f'(c) = 0 \) for some \( c \) between them. Nothing is asked of the sign of the derivative, of monotonicity, or of where the interval sits: dropping the endpoint condition gives Lagrange's theorem instead, whose conclusion is a slope rather than a zero

Answer: D

The endpoint values agree, \( f(1) = f(3) = 0 \), so the theorem applies. Setting the derivative to zero,

\[ f'(x) \;=\; 2x - 4 \;=\; 0 \quad \Longrightarrow \quad c \;=\; 2, \]

which lies inside the interval as the theorem promises. It is the vertex of the parabola, the one place where the tangent is horizontal

Answer: C

The theorem equates the derivative at \( c \) to the average slope:

\[ f'(c) \;=\; \frac{f(4)-f(1)}{4-1} \;=\; \frac{16-1}{3} \;=\; 5. \]

Since \( f'(x) = 2x \), this gives \( 2c = 5 \) and \( c = 2.5 \). For a quadratic the point always falls at the midpoint of the interval, which is why the answer is the mean of one and four

Answer: B

Cauchy's form states that \( [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c) \) for some interior \( c \), so two functions are compared at once. Taking \( g(x) = x \) recovers Lagrange's theorem. No second derivative is needed, monotonicity is not required, and the vanishing derivative is Rolle's conclusion

Answer: A

The theorem says \( f'(c) = [f(b)-f(a)]/(b-a) \): a tangent somewhere inside has the same slope as the chord, so the two are parallel. A horizontal tangent is Rolle's special case, and neither an interior crossing nor a curve staying above its chord follows, the latter being a statement about convexity instead

Answer: D

The derivatives at the origin cycle through \( 1, 0, -1, 0 \), so only odd powers survive and the signs alternate:

\[ \sin x \;=\; x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \;=\; x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots \]

The even-power series with leading one is \( \cos x \), and the all-positive version is \( \sinh x \)

Answer: C

Writing \( f(x) = \sum c_{n}(x-a)^{n} \) and differentiating \( n \) times at \( x = a \) leaves only the \( n \)th term, giving \( f^{(n)}(a) = n!\,c_{n} \) and so

\[ c_{n} \;=\; \frac{f^{(n)}(a)}{n!}. \]

The factorial divisor is what makes the series converge for well-behaved functions; omitting it produces a series that diverges for all but the smallest arguments

Answer: B

The binomial expansion for small \( t \) is \( (1+t)^{1/2} \approx 1 + t/2 \). With \( t = 0.02 \),

\[ \sqrt{1.02} \;\approx\; 1 + \frac{0.02}{2} \;=\; 1.01. \]

The next term is \( -t^{2}/8 = -0.00005 \), so the true value is 1.00995 and the two-term estimate is high by five parts in a million

Answer: A

The series \( x - x^{2}/2 + x^{3}/3 - \cdots \) has radius of convergence one. At \( x = 1 \) it becomes the alternating harmonic series, which converges to \( \ln 2 \); at \( x = -1 \) it is the harmonic series with a minus sign and diverges, matching \( \ln 0 \). So the interval is closed at the right and open at the left

Answer: D

The rule replaces the quotient by \( f'/g' \) and requires the original to be \( 0/0 \) or \( \infty/\infty \). The other indeterminate forms must first be rearranged into one of those two, a product by inverting one factor and a power by taking logarithms. Applying it to a determinate quotient gives a wrong answer