FE Electrical Mathematics Practice Problems

Answer: C) \(15x^2 - 2\cos x + e^x\)

Differentiate term by term with the Handbook's table. The power rule, entry 8, brings the exponent down and reduces it by one: \(d(5x^3)/dx = 5 \cdot 3x^2 = 15x^2\). Entry 16 gives \(d(\sin x)/dx = \cos x\), so \(d(-2\sin x)/dx = -2\cos x\), the constant \(-2\) riding along by entry 3. Entry 14 gives \(d(e^x)/dx = e^x\), the one function that is its own derivative.

$$f'(x) = 15x^2 - 2\cos x + e^x$$

Each term is handled independently because the derivative of a sum is the sum of the derivatives, entry 4. A check at \(x = 0\): \(f'(0) = 0 - 2 + 1 = -1\), and the function does start downhill, since \(f(0) = 1\) and \(f(0.1) = 0.005 - 0.1997 + 1.1052 = 0.9105\).

15x² + 2cos x + e^x has the sign of the sine term wrong: the derivative of \(\sin\) is \(+\cos\), and the \(-2\) in front stays.

15x³ - 2cos x + e^x multiplies by the exponent but never reduces it, leaving \(x^3\) where \(x^2\) belongs.

5x² - 2cos x + xe^{x-1} reduces the exponent without multiplying by it, and applies the power rule to \(e^x\) as if \(x\) were the base rather than the exponent; \(e^x\) is not a power of \(x\).

Answer: B) \(y^2 = 8x\)

Handbook: FE Reference Handbook 10.6, Mathematics → Conic Sections, Case 1. Parabola e = 1, p. 45 (PDF viewer: page 51):

$$(y-k)^2 = 2p(x-h);\quad \text{when } h = k = 0:\quad \text{Focus } \left(\tfrac{p}{2},\, 0\right),\quad \text{Directrix } x = -\tfrac{p}{2}$$

Step 1 – Which variable is squared? The directrix drawn is a vertical line, \(x = -2\). A parabola is symmetric about the line through its vertex perpendicular to the directrix, so the axis here is horizontal and the equation is the \(y^2\) form above. An \(x^2\) form belongs to a horizontal directrix such as \(y = -2\).

Step 2 – Find \(p\) from the directrix, not from the focus. The Handbook's directrix is \(x = -p/2\), and the drawing puts it at \(x = -2\):

$$-\frac{p}{2} = -2 \quad\Longrightarrow\quad p = 4$$

Step 3 – Write the equation.

$$y^2 = 2px = 2(4)x = 8x$$

Step 4 – Answer: (B).

Watch the Handbook's \(p\). It is not the vertex-to-focus distance. That page puts the focus at \((p/2, 0) = (2, 0)\), so the focus is 2 units from the vertex while \(p = 4\). Feeding the focal distance 2 straight into \(2px\) is the single most common way to get this question wrong, and it produces option (A).

Check it against the definition (\(e = 1\): every point of a parabola is the same distance from the focus as from the directrix). Take the point \((2, 4)\), which satisfies \(y^2 = 8x\) because \(4^2 = 8(2) = 16\):

$$\text{to the focus } (2,0):\ \sqrt{(2-2)^2 + 4^2} = 4 \qquad \text{to the line } x=-2:\ 2-(-2) = 4 \quad\checkmark$$

Step 5 – Eliminate the distractors:

(A) \(y^2 = 4x\) has \(2p = 4\), so \(p = 2\), putting its focus at \((1, 0)\) and its directrix at \(x = -1\) — half as far out as the line drawn. This is the answer you get by substituting the focal distance for \(p\).

(C) \(x^2 = 4y\) and (D) \(x^2 = 8y\) open upward, not sideways: their directrices are the horizontal lines \(y = -1\) and \(y = -2\). (D) even uses the right value of \(p\), but on the wrong axis — reading the given line as \(y = -2\) instead of \(x = -2\).

Answer: D) \(8\angle 90^\circ\)

Handbook: FE Reference Handbook 10.6, Mathematics → Roots, p. 39 (PDF viewer: page 45), read backwards: if \(w\) is a cube root of \(z\), then \(w^3 = z\). Cubing in polar form uses the multiplication rule from p. 38 — moduli multiply and angles add — so cubing cubes the modulus and triples the angle:

$$\left(r\angle\theta\right)^3 = r^3\angle 3\theta$$

Step 1 – Read one root off the plot. The marked root sits at a distance 2 from the origin, at \(30^\circ\) from the positive real axis:

$$w = 2\angle 30^\circ$$

Step 2 – Cube it.

$$z = w^3 = 2^3\angle\left(3 \times 30^\circ\right) = 8\angle 90^\circ$$

Step 3 – Answer: (D). In rectangular form that is \(j8\), straight up the imaginary axis.

Any of the three roots gives the same answer — that is what makes them roots of one number. The other two sit at \(150^\circ\) and \(270^\circ\), and

$$\left(2\angle 150^\circ\right)^3 = 8\angle 450^\circ = 8\angle 90^\circ \qquad \left(2\angle 270^\circ\right)^3 = 8\angle 810^\circ = 8\angle 90^\circ$$

since \(450^\circ\) and \(810^\circ\) differ from \(90^\circ\) by whole turns. Use whichever root is easiest to read.

What the picture tells you before any arithmetic. The \(k\) roots of a complex number are always equally spaced around a circle, \(360^\circ/k\) apart — here \(120^\circ\), which is why the three dots form an equilateral pattern at one common radius. If a plot ever shows roots at different radii, something is wrong.

Step 4 – Eliminate the distractors:

(A) \(6\angle 30^\circ\) makes both mistakes: the modulus multiplied by 3 instead of cubed, and the angle copied straight from the plot instead of tripled.

(B) \(6\angle 90^\circ\) triples the angle correctly but takes \(3 \times 2 = 6\) for the modulus. Cubing is \(2^3 = 8\); tripling and cubing agree only at 0 and 1.

(C) \(8\angle 30^\circ\) cubes the modulus correctly but leaves the angle as it was read off the figure. The exponent acts on both parts of the polar form, not just the modulus.

Answer: C) \(3\)

The Handbook's Initial Value Theorem is \(f(0^+) = \lim_{s\to\infty} sF(s)\). Multiply by \(s\) and compare the leading powers:

$$\lim_{s\to\infty} \dfrac{3s^2 + 7s}{s^2 + 4s + 3} = \lim_{s\to\infty} \dfrac{3 + 7/s}{1 + 4/s + 3/s^2} = 3$$

Only the highest powers survive as \(s \to \infty\), so the initial value is the ratio of the leading coefficients of \(sF(s)\). The full inverse confirms it: the denominator factors as \((s + 1)(s + 3)\), partial fractions give \(F(s) = 2/(s + 1) + 1/(s + 3)\), so \(f(t) = 2e^{-t} + e^{-3t}\) and \(f(0) = 2 + 1 = 3\). The theorem needs \(sF(s)\) to be proper for the limit to be finite, which it is here since the numerator and denominator are both of degree 2.

0 is the Final Value Theorem, \(\lim_{s\to 0} sF(s) = 0\): the value at \(t \to \infty\), which is indeed zero since both exponentials decay, but the question asks about \(t = 0\).

7/3 is \(F(0)\), the transform evaluated at \(s = 0\) without the factor \(s\) and at the wrong end of the \(s\) axis.

7 is the constant in the numerator, taken as if it were the initial value; the constant term is irrelevant as \(s \to \infty\).

Answer: B) \(4\)

First the Handbook's gradient, the vector of partial derivatives:

$$\dfrac{\partial\phi}{\partial x} = 2xyz + z^3, \qquad \dfrac{\partial\phi}{\partial y} = x^2z, \qquad \dfrac{\partial\phi}{\partial z} = x^2y + 3xz^2$$

At \((1, 1, 1)\) these are \(3\), \(1\) and \(4\), so \(\nabla\phi = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k}\). The direction vector is already a unit vector, \(|\mathbf{u}| = \tfrac{1}{3}\sqrt{4 + 4 + 1} = 1\), so the directional derivative is the Handbook's dot product:

$$\nabla\phi \cdot \mathbf{u} = \tfrac{1}{3}(3 \cdot 2 + 1 \cdot 2 + 4 \cdot 1) = \dfrac{12}{3} = 4$$

The directional derivative is the component of the gradient along \(\mathbf{u}\), \(|\nabla\phi|\cos\theta\), so it can never exceed \(|\nabla\phi| = \sqrt{26} = 5.10\), the rate of change in the gradient's own direction; here \(\cos\theta = 4/5.10 = 0.784\). Had \(\mathbf{u}\) not been a unit vector, dividing by its length would have been the missing step.

12 is the dot product with \(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), the factor \(\tfrac{1}{3}\) dropped; it exceeds \(|\nabla\phi|\), which no directional derivative can.

8/3 omits the \(\mathbf{k}\) term, \((6 + 2)/3\).

√26 is \(|\nabla\phi|\), the maximum rate of change over all directions, not the rate in the direction asked.