Algebra Questions and Answers

Answer: A

The correct option is \(f(2,2) = g(2,2)\).

\(f(p, q)\) multiplies \(q\) copies of \(p\), so \(f(p, q) = p^{q}\). \(g(p, q)\) stacks \(q\) copies of \(p\) as a power tower. With two terms both give the same thing: \(f(2,2) = 2^{2} = 4\) and \(g(2,2) = 2^{2} = 4\).

The other options are false:

• \(g(2,2) = 4\), so \(f(g(2,2), 2) = f(4, 2) = 4^{2} = 16\) and \(f(2, g(2,2)) = f(2, 4) = 2^{4} = 16\). They are equal, so 'less than' fails.
• \(g(2,1) = 2\) and \(f(2,1) = 2\), so 'not equal' fails.
• \(f(3,2) = 3^{2} = 9\) while \(g(3,2) = 3^{3} = 27\), so \(f(3,2)\) is smaller, not greater.

Answer: B

The condition is \(\varphi(s) = 0\), \(d\varphi/ds = 0\) and \(d^{2}\varphi/ds^{2} = 0\) at \(s = -\sigma\).

A triple root at \(s = -\sigma\) means \(\varphi(s) = (s + \sigma)^{3}q(s)\), where \(q(-\sigma) \ne 0\). Differentiating, the factor \((s + \sigma)\) survives in \(\varphi\), \(\varphi'\) and \(\varphi''\), so all three vanish at \(-\sigma\); in \(\varphi'''\) it does not, so the third derivative is in general non-zero.

The option on \(s^{3} + \sigma^{3} = 0\) is wrong: that equation has \(-\sigma\) and two complex roots, and a repeated real root says nothing about those complex points. The option with the second and fourth derivatives leaves out the first derivative, which must vanish, and adds a fourth, which need not. The option with the third derivative is wrong because a root of multiplicity exactly three is the case where the third derivative does not vanish.

Answer: A

The value of \(p + q\) is 2.

Both \(p\) and \(q\) are positive integers, so each is at least 1. The second inequality is the tighter one:

\[ \begin{aligned} 3p + 2q &\lt 6 \\ p = 1, q = 1:\ 3 + 2 &= 5 \lt 6 \\ p = 1, q = 2:\ 3 + 4 &= 7 \end{aligned} \]

Any larger \(p\) or \(q\) pushes \(3p + 2q\) to 6 or more, so \(p = q = 1\) is the only candidate. It satisfies the first inequality too: \(1 - 4 = -3 \lt 4\). Hence \(p + q = 2\).

A sum of 1 would need one of the numbers to be 0, which is not positive, and sums of 3 or 4 need \(p = 2\) or \(q = 2\), which break \(3p + 2q \lt 6\).

Answer: B

The function is \(|2 - |x||\).

Read the plotted points: \(y = 2\) at \(x = 0\), \(y = 0\) at \(x = \pm 2\), and the lines climb back to \(y = 2\) at \(x = \pm 4\), giving a W shape symmetric about the y-axis. Check \(|2 - |x||\):

• \(x = 0\): \(|2 - 0| = 2\)
• \(x = \pm 2\): \(|2 - 2| = 0\)
• \(x = \pm 4\): \(|2 - 4| = 2\)

\(|2 - x|\) is a single V with its vertex at \(x = 2\) and is not symmetric (it equals 6 at \(x = -4\)). \(|2 + |x||\) is never below 2, so it cannot touch zero. \(2 - |x|\) has the right peak but goes negative beyond \(x = \pm 2\), reaching \(-2\) at \(x = \pm 4\), while the plot never drops below the axis.

Answer: A

The value of px + qy + rz is 0.

Let each ratio equal k, so \(x = k(pq - r^{2})\), \(y = k(qr - p^{2})\), \(z = k(rp - q^{2})\). Then

\[ \begin{aligned} px + qy + rz &= k[p^{2}q - pr^{2} + q^{2}r - qp^{2} + r^{2}p - rq^{2}] \\ &= k \times 0 = 0 \end{aligned} \]

The six terms cancel in pairs: \(p^{2}q\) with \(-qp^{2}\), \(-pr^{2}\) with \(r^{2}p\), and \(q^{2}r\) with \(-rq^{2}\). The result does not depend on k, so 1, pqr and \(p^{2} + q^{2} + r^{2}\) cannot be right for all values.

Answer: A

The value of px + qy + rz is 0.

Call the common value of the three ratios k, so x = k(pq - r²), y = k(qr - p²) and z = k(rp - q²). Then:

\[ \begin{aligned} px + qy + rz &= k[p^{2}q - pr^{2} + q^{2}r \\ &\quad - qp^{2} + r^{2}p - rq^{2}] \\ &= 0 \end{aligned} \]

Every term cancels with another (p²q with -qp², -pr² with r²p, q²r with -rq²), whatever k is. So the answer cannot be 1, pqr or p² + q² + r², which would all depend on the constants.

Answer: C

The sum is \(e - 1\).

The exponential series is

\[ \begin{aligned} e &= \frac{1}{0!} + \frac{1}{1!} \\ &+ \frac{1}{2!} + \frac{1}{3!} + \cdots \end{aligned} \]

The given series is the same except that it starts at \(\frac{1}{1!}\), so it leaves out the first term \(\frac{1}{0!} = 1\):

\[ \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \approx 1.718 \]

\(e\) would include the missing 1, \(1 + e\) adds it instead of removing it, and \(\pi\) has nothing to do with this series.

Answer: B

The graph is \(y = ||x| - 1| - 1\).

Read values off the graph: y = 0 at x = 0, y = -1 at x = ±1, y = 0 at x = ±2 and y = 1 at x = ±3. For this function, \(x = 0\) gives \(|0 - 1| - 1 = 0\); \(x = \pm 1\) gives \(0 - 1 = -1\); \(x = \pm 2\) gives \(1 - 1 = 0\); \(x = \pm 3\) gives \(2 - 1 = 1\). Every point fits.

\(||x| + 1| - 2\) gives -1 at x = 0, and \(||x| + 1| - 1\) gives 0 at x = 0 but never goes negative, so neither has the two dips to -1. \(||x - 1| - 1|\) is never negative at all.

Answer: C

\(x = 3\).

After the change the two fractions are \(\frac{3 + x}{4 + x}\) and \(\frac{15 - x}{17 - x}\). Setting them equal and cross-multiplying:

\[ \begin{aligned} (3 + x)(17 - x) &= (15 - x)(4 + x) \\ 51 + 14x - x^{2} &= 60 + 11x - x^{2} \\ 3x &= 9,\quad x = 3 \end{aligned} \]

Check: \(6/7\) and \(12/14 = 6/7\). With 1, 2 or 4 the fractions differ, for example \(4/5\) against \(14/16 = 7/8\) for x = 1.

Answer: A

The graph is P.

Split the line at the points where the inner absolute values change sign, \(x = -2\) and \(x = 1\):

\[ \begin{aligned} x \le -2:&\ \ |(-x-2) - (1-x)| = |-3| = 3 \\ -2 \le x \le 1:&\ \ |(x+2) - (1-x)| = |2x+1| \\ x \ge 1:&\ \ |(x+2) - (x-1)| = 3 \end{aligned} \]

So \(f\) is 3 on both outer pieces, and in the middle it falls along \(|2x+1|\) from 3 at \(x = -2\) to 0 at \(x = -0.5\) and rises back to 3 at \(x = 1\). That is the V between two flat arms in P.

Q is \(|x+2| - |x-1|\) without the outer modulus, so it goes negative. R and S grow without bound on the left, which no part of \(f\) does.