Correct Answer: A. \(\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Explanation

Let \(\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T^{-1}\)

\(T . T^{-1} = I\)

\(\begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\implies -2a - 5c = 1\)

\(-2b - 5d = 0 \implies b = \frac{-5d}{2}\)

\(3a + 8c = 0 \implies a = \frac{-8c}{3}\)

\(3b + 8d = 1\)

\(-2(\frac{-8c}{3}) - 5c = \frac{16c}{3} - 5c = \frac{c}{3} = 1 \implies c = 3\)

\(3(\frac{-5d}{2}) + 8d = \frac{-15d}{2} + 8d = \frac{d}{2} = 1 \implies d = 2\)

\(b = \frac{-5 \times 2}{2} = -5\)

\(a = \frac{-8 \times 3}{3} = -8\)

\(\therefore T^{-1} = \begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}\)

Correct Answer: D. 4

Explanation

\(\log_{3}(x - y) = 1 \implies x - y = 3^{1} = 3 .... (1)\)

\(\log_{3}(2x + y) = 2 \implies 2x + y = 3^{2} = 9 ..... (2)\)

From (1), y = x - 3

From (2), y = 9 - 2x

\(\implies 9 - 2x = x - 3\)

\(9 + 3 = x + 2x = 3x\)

\(x = 4\)

Explanation

(a)

(b) Mode = 33.5

Correct Answer: B. \(2x^{2} - 9x + 13 = 0\)

Explanation

Note: Given the sum of the roots and its product, we can get the equation using the formula:

\(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\). This will be used later on in the course of our solution.

Given equation: \(2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6\).

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}\)

\(\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3\)

Given the roots of the new equation as \((\alpha + 1)\) and \((\beta + 1)\), their sum and product will be

\((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}\)

\((\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}\)

The new equation is given by: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)

= \(x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0\)

Correct Answer: D. \(\frac{n+1-r}{r}\)

Explanation

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

\(^{n}C_{r - 1} = \frac{n!}{(n - (r - 1))! (r - 1)!}\)

\(^{n}C_{r} ÷ ^{n}C_{r - 1} = \frac{n!}{(n - r)! r!} ÷ \frac{n!}{(n-(r-1))!(r-1)!}\)

= \(\frac{n!}{(n-r)! r!} \times \frac{(n-(r-1)! (r-1)!}{n!}\)

= \(\frac{(n + 1 - r)! (r - 1)!}{(n - r)! r!}\)

= \(\frac{(n+1-r)(n-r)! (r-1)!}{(n-r)! r (r - 1)!}\)

= \(\frac{n + 1 - r}{r}\)

Explanation

(a)

(b) Scale : 2 cm to 60° on the x- axis

2 cm to 2 units on y- axis.

(c)(i) Maximum value of y is 3.60.

(ii) Graph : \(\frac{2}{3} \cos x + \sin x = \frac{5}{6}\)

Equation : \(2 \cos x + 3 \sin x = y\)

i.e. \(\frac{2}{3} \cos x + \sin x = \frac{y}{3} .... (2)\)

(2) - (1) : \(\frac{y}{3} - \frac{5}{6} = 0\)

\(\frac{y}{3} = \frac{5}{6} \implies y = 2.50\)

We draw the line y = 2.5 and find the x- values at the points of intersection of graph with y = 2.5.

x = 12° and 102°.

Explanation

Assumed mean, A = 19.

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

= \(19 + \frac{-54}{22}\)

= \(19 - 2.455\)

= \(16.545 \approxeq 16.5\) years.

(b) Standard deviation , \(\sigma = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd}{\sum f})}\)

= \(\sqrt{\frac{360}{22} - (\frac{-54}{22})}\)

= \(\sqrt{16.364 - (2.45)^{2}}\)

= \(\sqrt{16.364 - 6.025}\)

= \(\sqrt{10.339}\)

= 3.215 \(\approxeq\) 3.22 years.

Explanation

If the system is in equilibrium, then the weight of the body = the force pulling it

and the force = the vertical component of force F

since Force F is inclined to vertical then the vertical component = Fcos35º

the weight of the body = mg

Fcos35º = 18(10)

0.8192F = 180

F = \(\frac{180}{0.8192}\)

F = 219.7N

tension in the string

T = mg

T = 18kg × 10

T = 180N

Correct Answer: A. 5

Correct Answer: A. \(\frac{125}{288}\)

Explanation

\((216)^{-\frac{2}{3}} = (\frac{1}{216})^{\frac{2}{3}} = (\sqrt[3]{\frac{1}{216}})^{2} = (\frac{1}{6})^{2} = \frac{1}{36}\)

\((0.16)^{-\frac{3}{2}} = (\frac{100}{16})^{\frac{3}{2}} = (\sqrt{100}{16})^{3} = \frac{1000}{64}\)

\(\frac{1}{36} \times \frac{1000}{64} = \frac{125}{288}\)