Correct Answer: C. 5

Explanation

\((x * y) = \frac{x+y}{2}\)

\((3 * b) = \frac{3+b}{2}\)

\(x \circ y = \frac{x^{2}}{y}\)

\((\frac{3+b}{2}) \circ 48 = \frac{(\frac{3+b}{2})^{2}}{48} = \frac{1}{3}\)

\(\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}\)

\((3 + b)^{2} = \frac{48 \times 4}{3} = 64\)

\(b^{2} + 6b + 9 = 64 \implies b^{2} + 6b + 9 - 64 = 0\)

\(b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0\)

\(b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (} b + 11) = 0\)

Since b > 0, b - 5 = 0

b = 5.

Explanation

(a)

Let A and B be the reactions at P and Q respectively.

A + B = 80 + 350 + 1000 = 1430N

We take moments about Q.

Clockwise moments = 2.2A

Anticlockwise moments = \(80 \times 3 + 350 \times 1 + 1000 \times 0.6\)

= \(1190N\)

Both moments are equal

\(\therefore 2.2A = 1190 \implies A = \frac{1190}{2.2} = 540.9N\)

\(B = 1190N - 540.9N = 649.1N\)

The reactions at P and Q are 540.9N and 649.1N respectively.

(b)

Reaction at P : \((80 + 350 + 1000)N = 540.9N\)

\(\therefore \text{The new reaction at P} = (80 + 350 + 1200)N = \frac{1630}{1430} \times 540.9\)

= \(616.6N\)

Take moments at Q :

\(2.2 \times 616.6 = 80 \times 3 + 350 \times 1 + 1200x\)

\(1200x = 1356.5 - 590\)

\(1200x = 766.5\)

\(x = \frac{766.5}{1200}\)

= \(0.63875 \approxeq 0.64m\)

The 1200N should be placed 0.64m from Q.

Correct Answer: B. kn + 5m = 0

Explanation

Two lines are parallel if and only if their slopes are equal.

\(kx - 5y + 6 = 0 \implies 5y = kx + 6\)

\(y = \frac{k}{5}x + \frac{6}{5}\)

\(Slope = \frac{k}{5}\)

\(mx + ny - 1 = 0 \implies ny = 1 - mx\)

\(y = \frac{1}{n} - \frac{m}{n}x\)

\(Slope = -\frac{m}{n}\)

\(Parallel \implies \frac{k}{5} = -\frac{m}{n}\)

\(-5m = kn \implies 5m + kn = 0\)

Explanation

(a) \(T_{n} = a + (n - 1)d\) (terms of an AP)

\(T_{21} = a + 20d = 5\frac{1}{2}.... (1)\)

\(S_{n} = \frac{n}{2} (2a + (n - 1)d) = \frac{n}{2} (a + l)\)

Where a and l are the first and last terms respectively.

\(S_{21} = \frac{21}{2} (a + 5\frac{1}{2})\)\)

\(94\frac{1}{2} = \frac{21}{2} (a + 5\frac{21}{2})\)

\(189 = 21 (a + 5\frac{1}{2})\)

\(9 = a + 5\frac{1}{2} \implies a = 9 - 5\frac{1}{2} = 3\frac{1}{2}\)

(b) Put a in the equation (1),

\(3\frac{1}{2} + 20d = 5\frac{1}{2}\)

\(20d = 5\frac{1}{2} - 3\frac{1}{2} = 2\)

\(d = \frac{2}{20} = \frac{1}{10}\).

(c) \(S_{30} = \frac{30}{2} (2(3\frac{1}{2}) + (30 - 1)(0.1)\)

= \(15(9.9)\)

= \(148.5\)

Correct Answer: D. \(\frac{1}{2}(\text{Upper quartile - Lower quartile})\)

Correct Answer: C. \(6x - \frac{2}{x^{3}}\)

Explanation

\(y = 3x^{2} + \frac{1}{x^{2}} \equiv y = 3x^{2} + x^{-2}\)

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 2x^{-3}\)

= \(6x - \frac{2}{x^{3}}\)

Correct Answer: A. 24

Explanation

\(^{n}P_{4} = \frac{n!}{(n - 4)!}\)

\(^{n}C_{4} = \frac{n!}{(n - 4)! 4!}\)

\(\frac{^{n}P_{4}}{^{n}C_{4}} = \frac{n!}{(n - 4)!} ÷ \frac{n!}{(n - 4)! 4!}\)

= \(4! = 24\)

Explanation

(a) \(A(x, y) \to (2x - y , -5x + 3y)\)

Matrix \(A = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)

(b) \(A = \begin{pmatrix} 2 & -1 \\ -5 & 2 \end{pmatrix} ; B = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)

(i) \(A^{2} - B^{2} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} 9 & -5 \\ -25 & 14 \end{pamtrix} - \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}\)

= \(\begin{pmatrix} -5 & -10 \\ -50 & 5 \end{pmatrix}\)

= \(-5 \begin{pmatrix} 1 & 2 \\ 10 & -1 \end{pmatrix}\)

(ii) \(C = B^{2} A\)

= \(\begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)

= \(\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)

(iii) \(\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ 18 \end{pmatrix}\)

\(3x + y = 10 ... (1)\)

\(5x + 2y = 18 ... (2)\)

Multiply (1) by 2 :

\(6x + 2y = 20 ... (3)\)

\((3) - (2) : (6x + 2y) - (5x + 2y) = (20 - 18)\)

\(x = 2\)

Put x = 2 in (1) :

\(3x + y = 10 \implies 3(2) + y = 10\)

\(6 + y = 10 \implies y = 10 - 6 = 4\)

\(M(2, 4)\).

(c) C is the inverse of matrix A and A is the inverse of matrix C.

Correct Answer: C. \(\frac{2}{3}\)

Explanation

\(^{3n}C_{2} > 0 \implies \frac{3n!}{(3n - 2)! 2!} > 0\)

\(\frac{3n(3n - 1)(3n - 2)!}{(3n - 2)! 2} > 0\)

\(\frac{3n(3n - 1)}{2} > 0\)

\(3n(3n - 1) > 0 \implies n > 0; n > \frac{1}{3}\)

The least number in the option that satisfies \(n > 0; n > \frac{1}{3} = \frac{2}{3}\)

Explanation

P(a family has a car) = p = 65% = 0.65

P(a family has no car) = q = 35% = 0.35

For seven families and using the binomial probability distribution, we have

\((p + q)^{7} = p^{7} + 7p^{6} q + 21p^{5} q^{2} + 35p^{4} q^{3} + 35p^{3} q^{4} + 21p^{2} q^{5} + 7p q^{6} + q^{7}\)

(a) P(exactly 5 have cars) = \(21p^{5} q^{2} = 21(0.65)^{5} (0.35)^{2}\)

= \(21 \times 0.116029 \times 0.1225 = 0.29848\)

\(\approxeq 0.298\) (to 3 d.p)

(b) P(3 or 4 of them have cars) = \(35p^{4} q^{3} + 35p^{3} q^{4}\)

= \(35(0.65)^{4} (0.35)^{3} + 35(0.65)^{3} (0.35)^{4}\)

= \(35(0.27463)(0.015006) + 35(0.178506)(0.042875)\)

= \(0.144238 + 0.267871\)

\(\approxeq 0.412\) (to 3 d.p)

(c) P(at most 2 of them have cars) = P(0 cars) + P(1 car) + P(2 cars)

= \(q^{7} + 7p q^{6} + 21p^{2} q^{5}\)

= \((0.35)^{7} + 7(0.65) (0.35)^{6} + 21 (0.65)^{2} (0.35)^{5}\)

= \(0.000643 + 0.00836 + 0.04658\)

= \(0.5558 \approxeq 0.556\) (3 d.p)