Engineering Mathematics

The determinant of the coefficient matrix is \(1 \cdot 1 - k \cdot k = 1 - k^{2}\). For every \(k \ne \pm 1\) it is non-zero, so the system has exactly one solution.

For \(k = 1\): \(x + y = 1\) and \(x + y = -1\) contradict each other, so there is no solution. For \(k = -1\): \(x - y = 1\) and \(-x + y = -1\) are the same equation, so there are infinitely many solutions.

• (A) True: only \(k = 1\) gives no solution.
• (B) False: no solution happens for \(k = 1\) alone.
• (C) False: infinitely many values (every \(k \ne \pm 1\)) give exactly one solution.
• (D) True: only \(k = -1\) gives infinitely many solutions.

Answer: A, D

A function that is differentiable at \(x = 1\) must be continuous there, and its left and right derivatives must be equal.

Continuity at \(x = 1\): \(a + b = 1 + 1 + 1 = 3\). Equal slopes: the left piece has slope \(a\), and the right piece has slope \(3x^{2} + 2x = 5\) at \(x = 1\), so \(a = 5\).

Hence \(b = 3 - 5 = -2\), that is, \(-2.0\).

Answer: −2.0

Use Bayes’ theorem, with \(F\) = the fake coin is chosen and \(HH\) = two heads.

\(P(F) = \frac{1}{5}\) and \(P(HH \mid F) = 1\); a regular coin is chosen with probability \(\frac{4}{5}\) and gives two heads with probability \(0.5 \times 0.5 = \frac{1}{4}\).

\(P(HH) = \frac{1}{5} \cdot 1 + \frac{4}{5} \cdot \frac{1}{4} = \frac{2}{5}\), so \(P(F \mid HH) = \dfrac{1/5}{2/5} = \frac{1}{2} = 0.50\).

Answer: 0.50

Squaring \(A\) gives a shortcut:

\[ A^2 = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I, \qquad A^{13} = (A^2)^6 A = 2^6 A = 64A. \]

The eigenvalues of \(A\) satisfy \(\det(A - \lambda I) = (1 - \lambda)(-1 - \lambda) - 1 = \lambda^2 - 2 = 0\), so \(\lambda = \pm\sqrt{2}\). Hence \(A^{13} = 64A\) has eigenvalues \(\pm 64\sqrt{2}\), the same as \((\pm\sqrt{2})^{13} = \pm 2^6\sqrt{2}\).

Answer: D

The total probability must be 1, which fixes \(C\):

\[\int_1^4 Cx^2\,dx = C \cdot \frac{4^3 - 1^3}{3} = 21C = 1 \;\Rightarrow\; C = \frac{1}{21}\]

\[P(2 \le x \le 3) = \frac{1}{21}\int_2^3 x^2\,dx = \frac{1}{21} \cdot \frac{3^3 - 2^3}{3} = \frac{19}{63} \approx 0.3016\]

Rounded to three decimal places, this is 0.302.

Answer: 0.302

Square the matrix first. The diagonal entries are \(1 \cdot 1 + 2 \cdot 2 = 5\) and \(2 \cdot 2 + (-1)(-1) = 5\), and both off-diagonal entries are \(1 \cdot 2 + 2 \cdot (-1) = 0\), so

\[A^{2} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 5I.\]

Then \(A^{8} = (A^{2})^{4} = 5^{4}I = 625I\). The other options are \(5^{2}I\), \(5^{3}I\) and \(5^{5}I\), which are \(A^{4}\), \(A^{6}\) and \(A^{10}\).

Answer: C

Integrate by parts with \(u = \ln t\) and \(dv = t \, dt\):

\[\int t \ln t \, dt = \frac{t^{2}}{2}\ln t - \frac{t^{2}}{4}.\]

So \(\int_{1}^{x} t \ln t \, dt = \frac{x^{2}}{2}\ln x - \frac{x^{2}}{4} + \frac{1}{4}\). Setting this equal to \(\frac{1}{4}\) gives \(\frac{x^{2}}{2}\ln x = \frac{x^{2}}{4}\). Since \(x \neq 0\), \(\ln x = \frac{1}{2}\), so \(x = \sqrt{e}\).

The other options fail this last check: \(e\), \(e^{2}\) and \(e - 1\) give \(\ln x = 1\), \(2\) and about \(0.54\), not \(\frac{1}{2}\).

Answer: A

Multiply \(L^{2} = L^{-1}\) on the right by \(L\) to get \(L^{3} = I\). Powers of \(L\) therefore repeat every three steps:

\[M = L^{8} = (L^{3})^{2}L^{2} = L^{2} = N.\]

So \(M - N\) is the zero matrix, and its determinant is 0. Non-singularity only guarantees that \(L^{-1}\) exists; it says nothing about the difference \(M - N\), so the non-zero options cannot occur.

Answer: A

• A true. \(PX = Q\) is \(L(UX) = Q\). Put \(Y = UX\): solve \(LY = Q\) by forward substitution, then \(UX = Y\) by back substitution.
• B true. \(L\) has a unit diagonal, so \(\det L = 1\) and \(\det P = \det L \cdot \det U = u_{11}u_{22}u_{33}\). If \(P\) is invertible this is non-zero, so \(U\) is invertible, and \(L\) always is.
• C true. If \(P\) is singular, \(u_{11}u_{22}u_{33} = \det P = 0\), so some \(u_{ii} = 0\).
• D false. A triangular matrix is symmetric only if it is diagonal. The symmetric \(P = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) has \(l_{21} = 1\) and \(u_{12} = 1\) with \(l_{12} = u_{21} = 0\), so neither factor is symmetric.

Answer: A, B, C

Square invariance means the roots \(\{\alpha, \beta\}\) and \(\{\alpha^{2}, \beta^{2}\}\) are the same pair. There are two ways to match them:

• \(\alpha^{2} = \alpha\) and \(\beta^{2} = \beta\): each root is 0 or 1, giving \(x^{2}\), \(x(x - 1)\) and \((x - 1)^{2}\).
• \(\alpha^{2} = \beta\) and \(\beta^{2} = \alpha\): then \(\alpha^{4} = \alpha\), so \(\alpha \in \{0, 1, \omega, \omega^{2}\}\), where \(\omega\) is a primitive cube root of unity. The only new polynomial is \((x - \omega)(x - \omega^{2}) = x^{2} + x + 1\).

So there are exactly 4 square invariant quadratics, and 2 of them, \(x^{2}\) and \((x - 1)^{2}\), have equal roots. The probability is \(2/4 = 0.5\).

Answer: 0.5