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).

Explanation

P: (x, y) → (2x+3y, 3x - y)

\(\begin{pmatrix} x\\ y \end{pmatrix}\) → \(\begin{pmatrix} -3x & 6y \\ 4x & + y \end{pmatrix}\)

= \(\begin{pmatrix} -3 & 6 \\ 4 & + 1 \end{pmatrix}\) \(\begin{pmatrix} x\\ y \end{pmatrix}\)

P = \(\begin{pmatrix} -3 & 6 \\ 4 & + 1 \end{pmatrix}\)

Q: (x, y) → (2x-3y, -4 - 6y)

\(\begin{pmatrix} x\\ y \end{pmatrix}\) → \(\begin{pmatrix} 2x & -3y \\ -4x & -6y \end{pmatrix}\)

= \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\) \(\begin{pmatrix} x\\ y \end{pmatrix}\)

Q = \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\)

(b) \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\) \(\begin{pmatrix} -2 \\ -3 \end{pmatrix}\)

= \(\begin{pmatrix} 2[-2] & -3[-3] \\ -4[-2] & -6[-3] \end{pmatrix}\)

= \(\begin{pmatrix} -4 & +9 \\ 8 & 18 \end{pmatrix}\)

= \(\begin{pmatrix} 5 \\ 26 \end{pmatrix}\)

: The image of (-2,-3) is (5,26)

(c) PQ = \(\begin{pmatrix} -3 & 6 \\ 4 & 1 \end{pmatrix}\) \(\begin{pmatrix} 2 & -3 \\ -4 & -6 \end{pmatrix}\)

\(\begin{pmatrix} -3[2] + 6[-4] & -3[-3] + 6[-6] \\ 4[2] + 1[-4] & 4[-3] + 1[-6] \end{pmatrix}\)

= \(\begin{pmatrix} -6 - 24 & 9 - 36 \\ 8 - 4 & -12 - 6 \end{pmatrix}\) = \(\begin{pmatrix} -30 & -27 \\ 4 & -18 \end{pmatrix}\)

= PQ: (x,y) → (-30x - 27y, 4x - 18y)

(d) \(\begin{pmatrix} -30 & -27 \\ 4 & -18 \end{pmatrix}\) \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\)

= \(\begin{pmatrix} -30[1] & + [-27][4] \\ 4[1] & + [-18][4] \end{pmatrix}\)

= \(\begin{pmatrix} -30 & -108 \\ 4 & -72 \end{pmatrix}\) = \(\begin{pmatrix} -138 \\ -68 \end{pmatrix}\)

: Image of the point (1,4) is (-138,-68)

(e) T = \(\begin{pmatrix} cos 225° & -sin 225 \\ sin 225° & cos 225° \end{pmatrix}\)

= \(\begin{pmatrix} -√2/2 & √2/2 \\ -√2/2 & -√2/2 \end{pmatrix}\) \(\begin{pmatrix} -√2 \\ 2√2 \end{pmatrix}\)

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

: The image of P\(^1\) is (3,-1)

Correct Answer: A. -77

Explanation

\(s = 3i - j; t = 2i + 3j\)

\( t - 3s = (2i + 3j) - 3(3i - j) = 2i + 3j - 9i + 3j = -7i + 6j\)

\(t + 3s = (2i + 3j) + 3(3i - j) = 2i + 3j + 9i - 3j = 11i\)

\((t - 3s).(t + 3s) = (-7i + 6j).(11i) = -77\)

Correct Answer: C. \(\frac{\sqrt{2} - \sqrt{6}}{4}\)

Explanation

\(\cos (x + y) = \cos x \cos y - \sin x \sin y \)

\(\cos (60 + 45) = \cos 60 \cos 45 - \sin 60 \sin 45\)

= \(\frac{1}{2} \times \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\)

= \(\frac{\sqrt{2} - \sqrt{6}}{4}\)

Correct Answer: C. 16.5

Explanation

The modal class = 14 - 16

The upper class boundary = \(\frac{16 + 17}{2} = \frac{33}{2} = 16.5\)

Correct Answer: C. \(2\sqrt{2}, (2, 1)\)

Explanation

Equation of a circle with radius r and centre (a, b).

= \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have

\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Comparing, with \(x^{2} + y^{2} - 4x - 2y - 3 = 0\)

\(2a = 4 \implies a = 2\)

\(2b = 2 \implies b = 1\)

\(r^{2} - a^{2} - b^{2} = 3 \implies r^{2} = 3 + 2^{2} + 1^{2} = 8\)

\(r = 2\sqrt{2}\)

Correct Answer: C. p=10, q=- 7

Explanation

If the body is in equilibrium then

F + R = N

3i - 12j + 7i + 5j = pi + qj

10i - 7j = pi + qj

p = 10, q = -7

Correct Answer: C. 3y + 2x + 5 = 0

Explanation

Given line: \(3x - 2y + 4 = 0 \implies 2y = 3x + 4\)

\(y = \frac{3}{2}x + 2\)

\(Gradient (\frac{\mathrm d y}{\mathrm d x}) = \frac{3}{2}\)

Gradient of perpendicular line = \(\frac{-1}{\frac{3}{2}} = \frac{-2}{3}\)

\(\implies \frac{y - (-3)}{x - 2} = \frac{-2}{3}\)

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

\(3(y + 3) = -2(x - 2) \implies 3y + 2x + 9 - 4 = 0\)

= \(3y + 2x + 5 = 0\)

Correct Answer: B. 6, 6

Explanation

\(Perimeter = 2(l + b) = 24\)

\(l + b = 12 \implies l = 12 - b\)

\(Area = (12 - b) \times b = 12b - b^{2}\)

\(\frac{\mathrm d A}{\mathrm d b} = 12 - 2b = 0\) (at maximum)

\(2b = 12 \implies b = 6\)

\(l = 12 - 6 = 6m\)

Correct Answer: A. Q`(5, -8)

Explanation

Q(3 -3, 2) = [-(-3) + 2, -4(2)]

= (3 + 2, -8)

= (5, -8)