Correct Answer: D. -4 log 2

Explanation

\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)

= \(\log 2^{-3} + \log 2^{-1}\)

= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)

Explanation

(a) \(\int_{1} ^{2} \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x\)

Let \(u^{2} = 5 - x^{2}\).

\(2u \mathrm {d} u = - 2x \mathrm {d} x \implies u \mathrm {d} u = - x \mathrm {d} x\)

\(\int \frac{x}{\sqrt{5 - x^{2}}} \mathrm {d} x = \int \frac{- u}{\sqrt{u^{2}}} \mathrm {d} u\)

= \(- \int \mathrm {d} u \)

= \(-[5 - x^{2}]_{1} ^{2}\)

= \((- (5 - 2^{2}) - (- (5 - 1^{2})))\)

= \(- 1 + 4 = 3\)

(b)(i) \(\begin{vmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{vmatrix}\)

= \(2(-3 + 4) + 3(0 + 2) + 1(0 - 1) = 7\)

(ii) \(2x - 3y + z = 10\)

\(0x + y - 2z = -7\)

\(x + 2y - 3z = -9\)

Let A be the matrix of coefficients.

\(A = \begin{pmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{pmatrix}\)

We write the set of equations in matrix form, i.e Ax = B.

\(\begin{pmatrix} 2 & -3 & 1 \\ 0 & 1 & -2 \\ 1 & 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10 \\ -7 \\ -9 \end{pmatrix}\).

We find the inverse of A, A\(^{-1}\).

Determinant of A = 7.

Co-factors:

\(A_{11} = + \begin{vmatrix} 1 & -2 \\ 2 & -3 \end{vmatrix} = +(-3 + 4) = 1\)

\(A_{12} = - \begin{vmatrix} 0 & -2 \\ 1 & 2 \end{vmatrix} = -(0 + 2) = -2\)

\(A_{13} = + \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = +(0 - 1) = -1\)

\(A_{21} = - \begin{vmatrix} -3 & 1 \\ 2 & -3 \end{vmatrix} = -(9 - 2) = -7\)

\(A_{22} = + \begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix} = +(-6 - 1) = -7\)

\(A_{23} = - \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = -(4 - (-3)) = -7\)

\(A_{31} = + \begin{vmatrix} -3 & 1 \\ 1 & -2 \end{vmatrix} = +(6 - 1) = 5\)

\(A_{32} = - \begin{vmatrix} 2 & 1 \\ 0 & -2 \end{vmatrix} = -(-4 - 0) = 4\)

\(A_{33} = + \begin{vmatrix} 2 & -3 \\ 0 & 1 \end{vmatrix} = +(2 - 0) = 2\)

\(\therefore C = \begin{pmatrix} 1 & -2 & -1 \\ -7 & -7 & -7 \\ 5 & 4 & 2 \end{pmatrix}\)

\(\text{adj A} = C^{T} = \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix}\)

\(A^{-1} = \frac{\text{adj A}}{|A|} = \frac{1}{7} \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix}\)

\(x = A^{-1} b = \frac{1}{7} \begin{pmatrix} 1 & -7 & 5 \\ -2 & -7 & 4 \\ -1 & -7 & 2 \end{pmatrix} \begin{pmatrix} 10 \\ -7 \\ -9 \end{pmatrix}\)

= \(\frac{1}{7} \begin{pmatrix} 14 \\ -7 \\ 21 \end{pmatrix}\)

\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)

\(x = 2; y = -1 ; z = 3\)

Correct Answer: B. \((-3 + 5\sqrt{3})N\)

Explanation

\(F = F\cos \theta + F\sin \theta\) (Resolving into its components)

\(10N = 10 \cos 60, 10 \sin 60\)

\(6N = 6 \cos 330, 6 \sin 330\)

\(R = F_{1} + F_{2} = (10 \cos 60, 10 \sin 60) + (6 \cos 330, 6 \sin 330)\)

The y- component : \(10 \sin 60 + 6 \sin 330 = 10 \times \frac{\sqrt{3}}{2} + 6 \times \frac{-1}{2}\)

= \((5\sqrt{3} - 3)N\)

Explanation

(a) 7 - \(\frac{6}{x} = 3 - 2x\)

Simplifying; \(x^2 + 2x - 2 = 0\)

x = 1 or x = -3

Substituting for x; y = 3 - 2(1) = 3 - 2 = 1 or y = 3 - 2(-3) = 3 + 6 = 9

The coordinate of the two points are (x y) = (1, 1), (-3, 9)

(b) (\(\frac{1 - 3}{2}, \frac{1 + 9}{2}\)) = (-1, 5)

The gradient of the point of intersection ; \(\frac{9 - 1}{-3 -1} = \frac{8}{-4}\) = -2

The gradient of the perpendicular bisector; \(\frac{1}{2}\)

Thus, the equation of the perpendicular bisector; y - 5 = \(\frac{1}{2}\) (x + 1)

Therefore, 2y - x - 11 = 0

Correct Answer: D. \(-\frac{5}{8}\)

Explanation

\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{2(2-3x)}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{4-6x}\)

Since the bases are equal, powers can be equated

= 2x - 1 = 4 - 6x

= 2x + 6x = 4 + 1

= 8x = 5

\(\therefore x = \frac{5}{8}\)

Explanation

(a) \(u = 5 - x^{2}\)

= \(\int_{1}^{2} \frac{x}{\sqrt{u}} dx\)

\(\frac{\mathrm d u}{\mathrm d x} = -2x\)

\(\mathrm {d} x = \frac{\mathrm d u}{-2x}\)

\(\implies \int_{1}^{2} \frac{x}{\sqrt{u}} \frac{\mathrm d u}{-2x}\)

= \(-\frac{1}{2} \int_{1}^{2} \frac{1}{\sqrt{u}} \mathrm {d} u\)

= \(-\frac{1}{2} \int_{1}^{2} u^{-\frac{1}{2}} \mathrm {d} u\)

= \(-\frac{1}{2} [2u^{1}{2}]|_{1}^{2}\)

= \(-\frac{1}{2} [2\sqrt{5 - x^{2}}]|_{1}^{2}\)

= \(-\frac{1}{2} [2 - 4]\)

= 1.

(b)\(y = px^{2} + qx\)

\(\frac{\mathrm d y}{\mathrm d x} = 2px + q = 6x + 7\)

\(\implies 2p = 6; p = 3\)

\(q = 7\)

Explanation

(b) Mode = 38.4

(c) Assumed mean, A = 43 days

Mean \(\bar{x} = A + \frac{\sum fd}{\sum f}\)

= \(43 + \frac{-145}{40}\)

= \(43 - 3.625\)

= \(39.375\)

(d) Number of batteries that will live for less than 31 days or more than 45 days = 4 + 6 + 2 = 12

\(%age = \frac{12}{40} \times 100% = 30%\)

Correct Answer: A. 8

Explanation

\(^{n}C_{3} = \frac{n!}{(n - 3)! 3!}\)

\(^{n}P_{2} = \frac{n!}{(n - 2)!}\)

\(\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}\)

\(\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}\)

Note that \((n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!\)

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

\(\frac{n - 2}{3!} = 1 \implies n - 2 = 6\)

\(n = 2 + 6 = 8\)

Correct Answer: A. 5

Explanation

\(f(x) = 3x^{2} + 18x + 32\)

\(\frac{\mathrm d y}{\mathrm d x} = 6x + 18 = 0\)

\(6x = -18 \implies x = -3\)

\(f(-3) = 3(-3^{2}) + 18(-3) + 32 = 27 - 54 + 32 = 5\)

Correct Answer: B. \(x = \frac{-1}{4}\)

Explanation

The line of symmetry of the curve is at the minimum point of the curve (ie y' = 0)

\(\frac{ \mathrm d}{ \mathrm d x} \left ( 5-x-2x^{2} \right)\) = -1 - 4x

If y' = 0, we have \(-1 - 4x = 0 \implies 4x = -1\)

\(x = \frac{-1}{4}\)