Engineering Mathematics

Let the cut be at \(U\), uniform on \((0,1)\). If \(U > 0.4\), the piece containing 0.4 is \((0, U)\), of length \(U\); if \(U < 0.4\), it is \((U, 1)\), of length \(1 - U\).

\[E = \int_{0}^{0.4} (1 - u)\,du + \int_{0.4}^{1} u\,du = (0.4 - 0.08) + \frac{1 - 0.16}{2} = 0.32 + 0.42 = 0.74\]

The result is above 0.5 because a fixed point is more likely to land in the longer of the two pieces.

Answer: 0.74

The curves \(y = x\) and \(y = x^{3}\) cross where \(x^{3} = x\), that is at \(x = -1, 0, 1\). Between these points the larger one switches:

• for \(x \lt -1\) and \(0 \lt x \lt 1\): \(f(x) = x\), slope 1;
• for \(-1 \lt x \lt 0\) and \(x \gt 1\): \(f(x) = x^{3}\), slope \(3x^{2}\).

At each crossing the slopes on the two sides differ: at \(x = \pm 1\) they are 1 and 3, and at \(x = 0\) they are 0 and 1. So the graph has a corner at each of these points, and \(f\) is not differentiable exactly on \(\{-1, 0, 1\}\). Elsewhere \(f\) equals a single polynomial near each point, so it is differentiable.

Answer: D

The product of all eigenvalues of a square matrix equals its determinant.

Here the rows are linearly dependent: each row exceeds the previous one by \((3, 3, 3)\), so \(R_{1} + R_{3} = 2R_{2}\). A matrix with dependent rows has determinant 0:

\[\det\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = 0.\]

So 0 is one of the eigenvalues, and the product of all the eigenvalues is 0, whatever the other two are.

Answer: B

By symmetry, 1 comes before 2 in exactly half of all permutations, so \(P(X) = \frac{1}{2}\); likewise \(P(Y) = \frac{1}{2}\).

The order inside the pair \(\{1, 2\}\) and the order inside the pair \(\{3, 4\}\) are each equally likely either way and do not influence each other, so \(P(X \cap Y) = \frac{1}{4} = P(X)\,P(Y)\). Hence \(X\) and \(Y\) are independent.

• A fails: both events can happen together (probability \(\frac{1}{4}\)).
• C fails: both can fail together, for example in 2, 1, 4, 3, ….
• D fails: the two events are equally likely.

Answer: B

• A: false. Independence needs \(P(A \cap B) = P(A)P(B) = 0.3 \times 0.5 = 0.15\), but \(P(A \cap B) = 0.1\).
• B: true. \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.5 - 0.1 = 0.7\).
• C: true. \(A\) splits into the disjoint parts \(A \cap B\) and \(A \cap B^{c}\), so \(P(A \cap B^{c}) = 0.3 - 0.1 = 0.2\).
• D: false. By De Morgan's law \(A^{c} \cap B^{c} = (A \cup B)^{c}\), so \(P(A^{c} \cap B^{c}) = 1 - 0.7 = 0.3\), not 0.4.

Answer: B, C

Since \(A\) has only \(n\) rows, \(\operatorname{rank}(A) \le n\). By rank–nullity, the solutions of \(Ax = \mathbf{0}\) form a subspace of dimension \(m - \operatorname{rank}(A) \ge m - n \ge 1\).

• (A) True: a basis of this subspace contains at least \(m - n\) linearly independent solutions.
• (B) False: it needs the dimension to be exactly \(m - n\). For the \(1 \times 3\) zero matrix every vector in \(\mathbb{R}^{3}\) is a solution, and \(m - n = 2\) vectors cannot span a 3-dimensional space.
• (C) False: for \(A = \begin{pmatrix} 1 & 1 \end{pmatrix}\) the solutions are \((t, -t)\), so no non-zero solution has a zero entry.
• (D) False: for \(A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\) every solution is \((0, 0, t)\), with at most one non-zero entry rather than \(n = 2\).

Answer: A

Let \(R_{1}\) and \(R_{2}\) be the events that the first and the second ball are red. Then

\[P(R_{1} \cap R_{2} \mid R_{1}) = \frac{P(R_{1} \cap R_{2})}{P(R_{1})} = \frac{\frac{10}{25} \times \frac{9}{24}}{\frac{10}{25}} = \frac{9}{24} = 0.375\]

In words: once a red ball is out, the bag holds 9 red and 15 blue balls, so the second ball is red with probability \(9/24\). The unconditional probability of two reds, \(\frac{10}{25} \times \frac{9}{24} = 0.15\), answers a different question.

Answer: 0.375

Let \(I = \int_{0}^{2} f(x)\,dx\). The substitution \(x \to 2 - x\) maps \([0, 2]\) onto itself, so also \(I = \int_{0}^{2} f(2 - x)\,dx\). Adding the two forms and using \(f(x) + f(2 - x) = 1\):

\[2I = \int_{0}^{2} \big[f(x) + f(2 - x)\big]\,dx = \int_{0}^{2} 1\,dx = 2,\]

so \(I = 1\). Geometrically, the graph is symmetric about the point \((1, \frac{1}{2})\), so the parts above and below the line \(y = \frac{1}{2}\) cancel and the integral is \(2 \times \frac{1}{2} = 1\). For example, \(f(x) = \frac{x}{2}\) satisfies the condition, and its integral from 0 to 2 is 1.

Answer: B

Each die shows one of 6 faces independently, so there are \(6^{6} = 46656\) equally likely outcomes.

For all six numbers to be distinct, the dice must show 1 to 6 in some order. The first die can take any of 6 values, the second any of the remaining 5, and so on, giving \(6! = 720\) favourable outcomes.

\[P = \frac{6!}{6^{6}} = \frac{720}{46656} = \frac{5}{324}\]

(Dividing 720 and 46656 by their common factor 144 gives 5 and 324.)

Answer: B

Because \(0 \le y \le 1\) and \(x \ge 0\), every outcome has \(z = xy \le x\). Taking expectations keeps the inequality, so \(\overline{z} \le \overline{x}\) always holds.

Options (A) to (C) compare \(E[xy]\) with \(E[x]\,E[y]\); the difference is the covariance of \(x\) and \(y\), which can have either sign when they are dependent. Take \(x\) uniform on \([0,1]\):

• \(y = x\): \(\overline{z} = E[x^{2}] = \tfrac{1}{3} > \tfrac{1}{4} = \overline{x}\,\overline{y}\), so (A) and (B) fail.
• \(y = 1 - x\): \(\overline{z} = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6} < \tfrac{1}{4} = \overline{x}\,\overline{y}\), so (C) fails.

Answer: D