Correct Answer: B. 4cm

Correct Answer: A. 2

Explanation

\(\frac{3x + 4}{(x - 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x - 2}\)

\(\frac{3x + 4}{(x - 2)(x + 3)}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}\)

=\(3x+4=P(x-2)+Q(x+3)\)

=\(3x+4=Px-2P+Qx+3Q\)

=\(3x+4=Px+Qx-2P+3Q\)

=\(3x+4=(P+Q)x-2P+3Q\)

Equating x and the constants

P+Q=3------(i)

2P+3Q=4-------(ii)

From equation (i)

P=3-Q------(iii)

Substitute (3-Q) for P inequation (ii)

=-2(3-Q)+3Q=4

=-6+2Q+3Q=4

=-6+5Q=4

=5Q=4+6

=5Q=10

∴Q=\(\frac{10}{5}=2\)

Correct Answer: D. (6,5)

Correct Answer: B. 56.3°

Explanation

\(a . b = |a||b|\cos \theta\)

\((5i + 12j).(2i + 3j) = -10 + 36 = 26\)

\(26 = |5i + 12j||-2i + 3j|\cos \theta\)

\(|5i + 12j| = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = 13\)

\(|2i + 3j| = \sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}\)

\(26 = 13(\sqrt{13})\cos \theta\)

\(\cos \theta = \frac{26}{13\sqrt{13}} = 0.5547\)

\(\theta = 56.3°\)

Explanation

(a)(i) The matrix of rotation of point anticlockwise through angle 60° about the origin is given by:

\(M = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} = \begin{pmatrix} \cos 60° & -\sin 60° \\ \sin 60° & \cos 60° \end{pmatrix}\)

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

(ii) Using the transformation,

\(T :P \to MP\) or \(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

\(T : \begin{pmatrix} 3 \\ 5 \end{pmatrix} \to \begin{pmatrix} \frac{1}{2} & \frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 3 \\ 5 \end{pmatrix}\)

Image of P, P' = \(\begin{pmatrix} \frac{3 + 5\sqrt{3}}{2} \\ \frac{3\sqrt{3} - 5}{2} \end{pmatrix}\)

(b) \(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 2x + 3y \\ 3x - y \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

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

(ii) \(N^{2} + aN + bI = 0\)

\(\begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} ^{2} + a \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + b \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + a \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + b \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

\(\begin{pmatrix} 4 + 9 & 6 - 3 \\ 6 - 3 & 9 + 1 \end{pmatrix} + \begin{pmatrix} 2a & 3a \\ 3a & -a \end{pmatrix} + \begin{pmatrix} b & 0 \\ 0 & b \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

Adding corresponding elements of each matrix,

\(13 + 2a + b = 0 \implies 2a + b = -13 ... (1)\)

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

\(2(-1) + b = -13 \implies -2 + b = -13\)

\(b = -13 + 2 = -11\)

\(\therefore a = -1 ; b = -11\)

Correct Answer: D. 133

Correct Answer: C. \(\sim q \implies \sim p\)

Correct Answer: D. 2\(\frac{1}{3}cm\)

Explanation

Length of arc = circumference of the base of the

cone \(\frac{\theta}{360} \times 2\pi R = 2 \pi r\)

\(\frac{\theta R}{360}\) = r

r = \(\frac{60 \times 14}{360}\)

= \(\frac{7}{3} = 2\frac{1}{3}\)cm

Correct Answer: C. 3< m< 4

Correct Answer: B. zero