FE Electrical Mathematics Practice Problems

Answer: A) \(3\)

Both arguments are powers of 2, so the Handbook's definition of the logarithm, \(\log_b x = c\) where \(b^c = x\), evaluates each term directly:

$$32 = 2^5 \Rightarrow \log_2 32 = 5, \qquad 4 = 2^2 \Rightarrow \log_2 4 = 2$$

$$\log_2 32 - \log_2 4 = 5 - 2 = 3$$

The quotient identity gives the same thing in one step: \(\log_2 32 - \log_2 4 = \log_2(32/4) = \log_2 8 = 3\), since \(2^3 = 8\). When the arguments are not powers of the base, change the base with \(\log_2 x = \ln x/\ln 2\); here nothing but counting powers of 2 is needed.

5 is \(\log_2 32\) alone, the first term with the subtraction never carried out.

8 is the quotient \(32/4\) itself. The identity turns the difference of logarithms into the logarithm of the quotient; the logarithm still has to be taken.

28 is \(32 - 4\), the arguments subtracted. There is no identity for \(\log(x - y)\), and \(\log_2 28 = 4.81\) is not even a whole number, while the difference of two whole logarithms must be.

Answer: A) \(5\)

The Handbook's polar form writes \(z = c\angle\theta\) with \(c = \sqrt{a^2 + b^2}\): the magnitude is the hypotenuse of the right triangle with legs equal to the real and imaginary parts.

$$|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

The angle completes the polar form, \(\theta = \tan^{-1}(4/3) = 53.13^\circ\), so \(z = 5\angle 53.13^\circ\). The same number appears as \(z z^* = a^2 + b^2 = 25 = |z|^2\), the product of the number with its conjugate, which is the usual way the magnitude squared turns up in circuit work: the power delivered by a phasor is proportional to \(|z|^2\), not to \(|z|\).

25 is \(a^2 + b^2\) with the square root never taken, the magnitude squared.

7 is \(3 + 4\), the parts added without squaring; the magnitude of a sum of perpendicular components is never the plain sum.

12 is \(3 \times 4\), the product of the parts, which is not a quantity the polar form uses at all.

Answer: C) \(\{1, 2, 3, 4, 5, 6\}\)

The Handbook's symbol table defines \(A \cup B\) as the union of \(A\) and \(B\): every element that belongs to \(A\), or to \(B\), or to both. Listing each element once,

$$A \cup B = \{1, 2, 3, 4, 5, 6\}$$

The elements 3 and 4 belong to both sets but appear once, because a set has no repeated members. The union therefore has \(4 + 4 - 2 = 6\) elements: the two sizes added, minus the overlap that would otherwise be counted twice. That overlap is the intersection, \(A \cap B = \{3, 4\}\), the other symbol on the same Handbook line, and the two operations are the set versions of OR and AND.

{3, 4} is the intersection \(A \cap B\), the elements in both sets, the AND where the question asks for the OR.

{1, 2, 5, 6} is the symmetric difference, the elements in exactly one of the two sets; the union keeps the shared elements as well.

{1, 2} is \(A\) with the elements of \(B\) removed, the difference \(A - B\).

Answer: B) A hyperbola, because \(B^2 - 4AC > 0\)

Match the equation to the Handbook's general form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\): \(A = 1\), \(B = 0\), \(C = -4\). The type is decided by the discriminant:

$$B^2 - 4AC = 0 - 4(1)(-4) = +16 > 0$$

A positive discriminant is the Handbook's condition for a hyperbola. The shortcut is the signs of \(A\) and \(C\) when there is no \(xy\) term: opposite signs give a hyperbola, equal signs an ellipse, and one of them zero a parabola. Completing the squares shows the hyperbola explicitly, \((x + 1)^2 - 4(y - 1)^2 = 4\), with centre \((-1, 1)\), which is the Handbook's standard form of a hyperbola opening along the \(x\) axis. The linear terms \(2x\) and \(8y\) only shift the centre; they never change the type.

An ellipse comes from dropping the sign of \(C\) and computing \(-4(1)(4) = -16\); with \(C = -4\) the discriminant is positive.

A parabola reads \(B = 0\) as the parabola condition. The condition is \(B^2 - 4AC = 0\), and here it is 16; \(B = 0\) merely says the axes are not rotated.

A circle needs \(A = C\) as well as \(B = 0\); both squared terms are present, but their coefficients are 1 and \(-4\).

Answer: B) \(\dfrac{1}{2}\)

Direct substitution gives \((1 - 1 - 0)/0 = 0/0\), an indeterminate form, so the Handbook's L'Hospital's Rule applies: differentiate numerator and denominator separately and try the limit again.

$$\lim_{x \to 0} \dfrac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \dfrac{e^x - 1}{2x}$$

This is still \(0/0\), and the Handbook says to continue to the first quotient of derivatives that is not indeterminate:

$$= \lim_{x \to 0} \dfrac{e^x}{2} = \dfrac{1}{2}$$

The Maclaurin series on the Handbook's Taylor's Series entry gives the same answer without differentiating: \(e^x = 1 + x + x^2/2 + x^3/6 + \cdots\), so the numerator is \(x^2/2 + x^3/6 + \cdots\) and dividing by \(x^2\) leaves \(1/2 + x/6 + \cdots \to 1/2\). A numerical check at \(x = 0.01\): \((1.00501671 - 1.01)/0.0001 = 0.5017\).

0 comes from stopping after one application and substituting \(x = 0\) into the numerator \(e^x - 1\) alone, ignoring that the denominator \(2x\) is also zero there.

1 is \(e^0\) with the factor 2 from differentiating \(x^2\) twice lost along the way.

∞ is the guess that dividing by \(x^2\) must blow up; it would, if the numerator did not also vanish to second order.