Explanation

Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\).

\(T : \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}\)

\(2a + 3b = 3 ... (1)\)

\(4a + 5b = 5 ... (2)\)

\(2c + 3d = 4 ... (3)\)

\(4c + 5d = 6 .... (4)\)

Solving (1) & (2) :

\((1) \times 2 : 4a + 6b = 6 ... (5)\)

\((5) - (2) : b = 1\)

\(2a + 3(1) = 3 \implies 2a = 0; a = 0\)

\((3) \times 2 : 4c + 6d = 8 ... (6)\)

\((6) - (5) : d = 2\)

\(2c + 3(2) = 4 \implies 2c = -2; c = -1\)

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

(b) Let the inverse of A be A\(^{-1}\).

\(AA^{-1} = 1\)

\(\begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} w & x \\ y & z \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\begin{pmatrix} y & z \\ -w + 2y & -x + 2z \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\implies y = 1; z = 0\)

\(-w + 2y = 0 \implies -w + 2 =0\)

\(-w = -2 \implies w = 2\)

\(-x + 2z = 1 \implies -x + 2(0) = 1\)

\(-x = 1 \implies x = -1\)

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

(c) Let the point be (x, y).

\(A : x \to \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\)

\(0 + y = -1 \implies y = -1\)

\(-x + 2y = 1 \implies -x + 2(-1) = 1\)

\(-x - 2 = 1 \implies -x = 3\)

\(x = -3\).

The point is (-3, -1).

Correct Answer: B. 4.961g

Correct Answer: D. 2?3

Explanation

d = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

x1 = x, x2 = -x, y1 = 4, y2 = 3, d = 7

7 = \(\sqrt{(-x -x)^2 + (3 - 4)^2}\)

7 = \(\sqrt{(-2x)^2 + (-1)^2}\)

7 = ( \(\sqrt{4x^2 + 1}\)

square both sides

7\(^2\) = 4x\(^2\) + 1

collect like terms

4x\(^2\) = 49 - 1

4x\(^2\) = 48

x\(^2\) = \(\frac{48}{4}\)

x\(^2\) = 12

x = √12

x = 2√3

Correct Answer: A. n²

Explanation

\(^9{∫}_1\) \(\frac{x(2x-3)}{√x}\) dx = \(\frac{2x^2 - 3x)}{x^{1/2}}\)

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

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

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

= \(\frac{2x^{3/2}}{5/2}\) - \(\frac{3x^{1/2}}{3/2}\)

= \(^9{∫}_1\) [\(\frac{4x^{3/2}}{5}\) - \(\frac{6x^{1/2}}{3}\)]

[ \(\frac{4(9)^{3/2}}{5}\) - \(\frac{6^(9){1/2}}{3}\) ] - [ \(\frac{4(9)^{1/2}}{5}\) - \(\frac{6(1)^{1/2}}{3}\) ]

= [ \(\frac{4(27)}{5}\) - \(\frac{6(3)}{3}\) - [ \(\frac{4}{5}\) - 2 ]

= [ \(\frac{108}{5}\) - 2 ] - [ \(\frac{6}{5}\) ]

= \(\frac{78}{5}\) + \(\frac{6}{5}\) = \(\frac{84}{5}\)

16\(^{4/5}\)

Correct Answer: B. 10

Correct Answer: A. \(\frac{9}{4}\)

Explanation

x² - 3x + k(perfect square)

k = (-\(\frac{3}{2}\))2 ; k = \(\frac{9}{4}\)

Correct Answer: C. 12\(\sqrt{2cm}\)

Explanation

(a) Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\)

= \(\frac{96}{360} \times \frac{22}{7} \times 5^{2}\)

= \(20.95 cm^{2}\)

Area of triangle formed from the sector = \(\frac{1}{2} r^{2} \sin \theta\)

= \(\frac{1}{2} \times 5^{2} \times \sin 96\)

= \(12.43 cm^{2}\)

\(\therefore \text{The area of minor segment} = 20.95 - 12.43\)

= \(8.52 cm^{2}\)

(b) Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\)

= \(\frac{80}{360} \times \frac{22}{7} \times 7^{2}\)

= \(34.22 cm^{2}\)

Area of the square = \((14)^{2} = 196 cm^{2}\)

Area of the circle = \(\pi r^{2} = \frac{22}{7} \times 7^{2} = 154 cm^{2}\)

Area of the shaded portion in the square = \(196 - 154 = 42 cm^{2}\)

Total area of the shaded portions = \(42 + 34.22\)

= \(76.22 cm^{2} \approxeq 76.2 cm^{2}\)

Correct Answer: (B) 13.5