Differential Calculus Questions and Answers

Answer: B

The limit is \(\frac{1}{2}\).

Expand both sides of \(f(x + h) - f(x) = hf'(x + \theta h)\) in powers of h, using Taylor's theorem. For the cubic \(f = x^{2} + x^{3}\), \(f''' = 6\) is constant:

\[ \begin{aligned} \text{LHS} &= hf' + \tfrac{h^{2}}{2}f'' + \tfrac{h^{3}}{6}f''' \\ \text{RHS} &= hf' + \theta h^{2}f'' + \tfrac{\theta^{2}h^{3}}{2}f''' \end{aligned} \]

Cancelling \(hf'\) and dividing by \(h^{2}\): \(\tfrac{1}{2}f'' + \tfrac{h}{6}f''' = \theta f'' + \tfrac{\theta^{2}h}{2}f'''\). As \(h \to 0\) this leaves \(\theta f'' = \tfrac{1}{2}f''\), so \(\theta \to \tfrac{1}{2}\) wherever \(f''(x) = 2 + 6x \neq 0\).

The general result is that \(\theta \to 1/2\) for any twice-differentiable f with \(f'' \neq 0\), so 1/3, 1/4 and 4/5 are ruled out.

Answer: D

\((\frac{3}{2}, \frac{3}{4})\) is a saddle point of f.

At \((\frac{3}{2}, \frac{3}{4})\) both first derivatives vanish:

\[ \begin{aligned} f_{x} &= 2xy+2y^{2}-4xy^{2} = 2.25+1.125-3.375 = 0 \\ f_{y} &= x^{2}+4xy-4x^{2}y = 2.25+4.5-6.75 = 0 \end{aligned} \]

The second derivatives there are:

\[ \begin{aligned} f_{xx} &= 2y-4y^{2} = -0.75 \\ f_{yy} &= 4x-4x^{2} = -3 \\ f_{xy} &= 2x+4y-8xy = -3 \end{aligned} \]

\(f_{xx}f_{yy}-f_{xy}^{2} = 2.25 - 9 \lt 0\), so the point is a saddle, not a maximum. At \((\frac{3}{2}, 0)\) and \((0, \frac{3}{4})\) the gradient is not zero (\(f_{y} = 2.25\) and \(f_{x} = 1.125\) respectively), so they are not extreme points at all.

Answer: A, B

The function is continuous for all \(x\) and differentiable for all \(x\): A and B.

The only point that might cause trouble is \(x = 0\), where \(|x|\) has a corner. Everywhere else \(\sin^{2}|x|\) is a composition of smooth functions.

• Continuity (A): \(|x|\) is continuous and so is \(\sin^{2}\), so the composition is continuous for all \(x\). Option C is false.
• Differentiability (B): since \(\sin^{2}|x| = \sin^{2}x\) (sine is odd and is squared), the function equals \(\sin^{2}x\) for every \(x\), which is differentiable everywhere, with derivative \(\sin 2x\), which is 0 at \(x = 0\) from both sides. Option D is false.

Answer: A, B, C

The official key accepts A, B and C.

At \(x = 0\) the left-hand limit of \(x^{2}\) is 0 and \(f(0) = 0\) from the second rule, so the function is continuous there, and it is continuous everywhere else because each piece is. A is true, D false.

The derivative is \(2x\) for \(x \lt 0\) and 1 for \(x \geq 0\). As \(x\) approaches 0 from the left it tends to 0, but from the right it is 1, so the derivative jumps at \(x = 0\) (B true). At \(x = 1\) the derivative is 1 on both sides, so it is continuous (C true).

Answer: B, D

The official key accepts B and D: \(d^{2}y/dx^{2} = 0\) and \(d^{3}y/dx^{3} \neq 0\).

At a point of inflection the curve changes from concave to convex or the reverse, so the second derivative must change sign, and being continuous it must be zero there. That makes B necessary. The usual textbook test then asks for the first higher derivative that does not vanish to be of odd order; with \(y'' = 0\), a non-zero third derivative guarantees that \(y''\) changes sign, so D is the condition the key accepts.

A is wrong because \(y'' \neq 0\) means the curve is strictly bending one way and cannot be at an inflection. C is not accepted: for \(y = x^{3}\), the simplest inflection at \(x = 0\), the third derivative is 6, so \(y''' = 0\) does not hold there. (A function such as \(x^{5}\) has \(y''' = 0\) at its inflection, which is why D is the standard sufficient test and not a strict requirement; the key follows the textbook form.)

Answer: A, C, D

The true statements are A, C and D.

Work out \(f\) piece by piece. For \(x \ge 0\), \(|x| = x\): \(f = (\tfrac{x}{2} - x)(x - \tfrac{x}{2}) = -\tfrac{x^{2}}{4}\). For \(x \lt 0\), \(|x| = -x\): \(f = (-\tfrac{3x}{2})(\tfrac{3x}{2}) = -\tfrac{9x^{2}}{4}\).

\[ f'(x) = \begin{cases} -x/2, & x \gt 0 \\ -9x/2, & x \lt 0 \end{cases} \]

A: \(f \le 0 = f(0)\), so 0 is a local (indeed global) maximum. B: no minimum exists, since \(f\) falls without bound on both sides. C: both pieces of \(f'\) tend to 0 at 0 and \(f'(0) = 0\), so \(f'\) is continuous. D: the slopes of \(f'\) are \(-1/2\) on the right and \(-9/2\) on the left, which differ, so \(f'\) has a corner at 0 and is not differentiable over \(\mathbb{R}\).

Answer: B, D

Options B and D are correct; the key accepts both.

The derivative is

\[ \begin{aligned} f'(x) &= 3x^{2} + 30x - 33 \\ &= 3(x^{2} + 10x - 11) \\ &= 3(x + 11)(x - 1), \end{aligned} \]

with critical points \(x = -11\) and \(x = 1\). The second derivative \(f''(x) = 6x + 30\) is \(-36\) at \(-11\), so that is a local maximum, and \(36\) at \(1\), so that is a local minimum. A cubic with two distinct critical points always has one of each, so the statements that deny them (A and C) are false.

Answer: A

Write the quotient so that the standard limit appears with a matching argument:

\[ \frac{\sin 5x}{3x} \;=\; \frac{5}{3}\cdot\frac{\sin 5x}{5x}. \]

Since \( \sin\theta/\theta \to 1 \) as \( \theta \to 0 \), the second factor tends to one and

\[ \lim_{x \to 0} \frac{\sin 5x}{3x} \;=\; \frac{5}{3}. \]

The reciprocal \( 3/5 \) comes of matching the arguments the wrong way round

Answer: D

Use the half-angle identity \( 1 - \cos x = 2\sin^{2}(x/2) \):

\[ \frac{1-\cos x}{x^{2}} \;=\; \frac{2\sin^{2}(x/2)}{x^{2}} \;=\; \frac{1}{2}\left(\frac{\sin (x/2)}{x/2}\right)^{2}. \]

The bracket tends to one, so the limit is \( 1/2 \). Two applications of L'Hopital's rule give the same value, the numerator differentiating to \( \sin x \) and then to \( \cos x \)

Answer: C

Take logarithms and let \( L \) be the limit:

\[ \ln L \;=\; \lim_{x \to \infty} 2x \ln\!\left(1+\frac{3}{x}\right) \;=\; \lim_{x \to \infty} 2x \cdot \frac{3}{x} \;=\; 6, \]

using \( \ln(1+t) \approx t \) for small \( t \). Hence \( L = e^{6} \). The general rule is that \( (1+a/x)^{bx} \to e^{ab} \), and here \( ab = 3 \times 2 \)