Differential Calculus Questions and Answers

Answer: A

The nth derivative is \(\dfrac{(-1)^{n}n!}{(x + 3)^{n+1}}\).

Write \(y = (x + 3)^{-1}\) and differentiate repeatedly:

\[ \begin{aligned} y' &= -(x + 3)^{-2} \\ y'' &= 2(x + 3)^{-3} \\ y''' &= -6(x + 3)^{-4} \end{aligned} \]

Each step multiplies by the current power (giving \(1, 2, 6 = n!\)), flips the sign and raises the power of \((x+3)\) in the denominator by one, so \(y^{(n)} = (-1)^{n}n!/(x + 3)^{n+1}\).

The form with \((-1)^{n+1}\) has the wrong sign (it gives \(+1\) for the first derivative), and the forms with \((x + 3)^{n}\) or \((n + 1)!\) fail at \(n = 1\).

Answer: B

The roots lie on an elliptical curve.

The roots are the points where \(f(x, y) = 0\):

\[ \begin{aligned} \frac{x^{2}}{2} + \frac{y^{2}}{3} &= 5 \\ \frac{x^{2}}{10} + \frac{y^{2}}{15} &= 1 \end{aligned} \]

This is an ellipse in the xy-plane with semi-axes \(\sqrt{10}\) along x and \(\sqrt{15}\) along y.

An ellipse has infinitely many points, so the roots are not a finite set. A sphere would need three variables, and the function has only two. A hyperbola would need the two squared terms to have opposite signs, but both are positive here.

Answer: B

The minimum is at \(x = -\frac{1}{2}\).

Expand and differentiate:

\[ \begin{aligned} y &= (x + 3)(x - 2) = x^{2} + x - 6 \\ y' &= 2x + 1 = 0 \Rightarrow x = -\tfrac{1}{2} \\ y'' &= 2 \gt 0 \end{aligned} \]

The positive second derivative confirms a minimum, and \(-\tfrac{1}{2}\) lies inside \(-4 \lt x \lt 4\). A quicker route: the parabola opens upward and is symmetric about the midpoint of its roots \(-3\) and \(2\), which is \(-\tfrac{1}{2}\). The minimum value is \(y = -6.25\).

The value \(\tfrac{1}{2}\) has the sign wrong, and \(\pm\tfrac{3}{2}\) do not make the derivative zero.

Answer: C

\(A = 3\).

For continuity at x = 2, the left-hand limit must equal the value of the function there. The value and right-hand limit come from \(1 + x^{2}\), and the left-hand limit from \(A + x\):

\[ \begin{aligned} A + 2 &= 1 + 2^{2} = 5 \\ A &= 3 \end{aligned} \]

A = 2 would leave a jump of 1 at x = 2, and 2.5 and 3.5 leave jumps of 0.5. Only A = 3 makes the two pieces meet.

Answer: A

The tangent meets the x axis at \(\left(x_{0} - y_{0}/f'(x_{0}),\ 0\right)\).

The tangent at \((x_{0}, y_{0})\) has slope \(f'(x_{0})\):

\[ \begin{aligned} y - y_{0} &= f'(x_{0})(x - x_{0}) \\ \text{at } y = 0:\ x &= x_{0} - \frac{y_{0}}{f'(x_{0})} \end{aligned} \]

The condition \(f'(x) \neq 0\) is what guarantees the tangent is not horizontal and does cut the axis. This is the step used in Newton-Raphson iteration. The option with a plus sign has the sign of the correction wrong, and the two options with \(f'(x_{0})/y_{0}\) invert the ratio.

Answer: D

\(f(1)\) lies between 6 and (5 + e).

By the Mean Value Theorem there is some c with 0 < c < 1 such that

\[ \begin{aligned} f(1) - f(0) &= f'(c)(1 - 0) = e^{c} \\ 1 &\lt e^{c} \lt e \\ 6 &\lt f(1) \lt 5 + e \end{aligned} \]

Indeed \(f(x) = e^{x} + 4\) fits both conditions and gives \(f(1) = 4 + e \approx 6.72\), inside that interval. The other options forget to add \(f(0) = 5\), or add the wrong constant, so their bounds do not contain the value.

Answer: A

The function has a minimum at x = 1/2.

Set the derivative to zero and check the second derivative:

\[ \begin{aligned} f'(x) &= 2x - 1 = 0 \Rightarrow x = \tfrac{1}{2} \\ f''(x) &= 2 \gt 0 \end{aligned} \]

A positive second derivative means a minimum; the parabola opens upwards because the coefficient of \(x^{2}\) is positive, so it has no maximum at all. The minimum value is \(f(1/2) = -6.25\). The options at \(x = -1/2\) come from a sign slip in solving \(2x - 1 = 0\).

Answer: B

The interval is \(\left(\frac{\pi}{2}, \pi\right)\).

On \([0, 2\pi]\), \(\sin x\) decreases on \(\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)\), where its derivative \(\cos x\) is negative, and \(\cos x\) decreases on \((0, \pi)\), where its derivative \(-\sin x\) is negative. Both decrease only on the overlap, \(\left(\frac{\pi}{2}, \pi\right)\), the second quadrant.

On \(\left(0, \frac{\pi}{2}\right)\) the sine is increasing; on \(\left(\pi, \frac{3\pi}{2}\right)\) the cosine is increasing; on \(\left(\frac{3\pi}{2}, 2\pi\right)\) both are increasing.

Answer: C

The function has minima at -2 and +2.

Write it as \(f(x) = (x^{2} - 4)^{2}\). As a square it is never negative, and it equals 0 exactly at \(x = \pm 2\), so both points are minima (the least possible value). Differentiating confirms the turning points:

\[ \begin{aligned} f'(x) &= 4x(x^{2} - 4) = 0 \\ x &= 0,\ \pm 2 \\ f''(x) &= 12x^{2} - 16 \end{aligned} \]

At \(x = \pm 2\), \(f'' = 32 \gt 0\), so both are minima; at \(x = 0\), \(f'' = -16 \lt 0\), a local maximum of value 16.

The function is even, \(f(-x) = f(x)\), so whatever happens at +2 happens at -2 too. That rules out the two options giving a minimum at one of them and a maximum at the other, and neither point can be a maximum since the function is zero there and positive nearby.

Answer: B

The limit is 1/2.

Substituting x = 0 gives 0/0, so rationalise the numerator by multiplying above and below by \(\sqrt{1+x} + 1\):

\[ \begin{aligned} \frac{\sqrt{1+x} - 1}{x} &= \frac{(1 + x) - 1}{x(\sqrt{1+x} + 1)} \\ &= \frac{1}{\sqrt{1+x} + 1} \to \frac{1}{2} \end{aligned} \]

The same value is the derivative of \(\sqrt{t}\) at t = 1, or the first term of the binomial expansion \(\sqrt{1+x} \approx 1 + x/2\). 0 and 1 come from mishandling the indeterminate form, and 2 from inverting the result.