Explanation

\(\overline{PQ}\) = \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\)

\(\overline{MR}\) = \(\begin{pmatrix} -5 \\ 2 \end{pmatrix}\) - \(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix} -5 & -x \\ 2 & - y \end{pmatrix}\)

\(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\) = \(\begin{pmatrix} -5 & -x \\ 2 & - y \end{pmatrix}\)

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6

The coordinate of M (-2,-6)

(b) \(\overline{PM}\) = \(\overline{OM}\) - \(\overline{OP}\)

= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -6 \\ -1 \end{pmatrix}\)

\(\overline{PQ}\) = \(\overline{OQ}\) - \(\overline{OP}\)

= \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\)

|\(\overline{PM}\)| = √(-6\(^2\) + [-1\(^2\)]) = √37

|\(\overline{PQ}\)| = √(-3\(^2\) + 8\(^2\)) = √73

cosØ = \(\frac{PM.PQ}{|\overline{PM}||\overline{PQ}|}\)

cosØ = \(\frac{[-6i - j].[-3i+8j]}{√37 . √73}\) = \(\frac{10}{√2710}\)

⇒ Ø = cos\(^{-1}\) \(\frac{10}{√2710}\)

: Ø = 78.91° ≈ 79°

Correct Answer: B. r and q are perpendicular

Explanation

The dot product of two perpendicular forces = 0

\((9i + 6j).(4i - 6j) = 36 - 36 = 0\)

Hence, r and q are perpendicular.

Correct Answer: B. \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)

Explanation

\(a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\); \(b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\)

\(\implies 2 \times a = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) and \(\frac{1}{4} \times b = \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\)

\(\therefore 2a - \frac{1}{4}b = \begin{pmatrix} 4 - \frac{-1}{4} \\ 6 - 1 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\)

Correct Answer: B. 4m

Explanation

\(v(t) = (3t^{2} - 2t) ms^{-1}\)

\(s(t) = \int v(t) \mathrm {d} t\)

= \(\int (3t^{2} - 2t) \mathrm {d} t = t^{3} - t^{2}\)

\(s(2) = 2^{3} - 2^{2} = 8 - 4 = 4m\)

Explanation

\(x^{3} + y^{3} - 2xy = 11\)

Differentiating implicitly,

\(3x^{2} + 3y^{2} \frac{\mathrm d y}{\mathrm d x} - 2y - 2x \frac{\mathrm d y}{\mathrm d x} = 0\)

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

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

At (2, -1) , Gradient = \(\frac{2(-1) - 3(2^{2})}{3(-1)^{2} - 2(2)}\)

= \(\frac{-2 - 12}{3 - 4}\)

= \(\frac{-14}{-1} = 14\)

Correct Answer: A. 10x + 1

Explanation

\(\frac{5x^ 3+x^2}{x}\) → \(\frac{5x^ 3}{x} + \frac{x^2}{x}\)

5x\(^2\) + x

Then dy/dx = 10x + 1

Explanation

(a) \(x^{2} + mx + 11 = 0\)

Roots are \(\alpha\) and \(\beta\).

\(\therefore \alpha + \beta = \frac{-b}{a} = - m & \alpha + \beta = \frac{c}{a} = 11\)

\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha \beta\)

\(27 = (-m)^{2} - 2(11)\)

\(m^{2} = 22 + 27 = 49\)

\(m = \pm 7\)

(b) Equation of line : \(2x + 3y = 1 i.e. y = \frac{1 - 2x}{3}\)

Substitute for y in the equation of the circle:

\(2x^{2} + 2(\frac{1 - 2x}{3})^{2} + 4x + 9(\frac{1 - 2x}{3}) - 9 = 0\)

\(2x^{2} + 2(\frac{1 - 4x + 4x^{2}}{9} + 4x + 3(1 - 2x) - 9 = 0\)

\(18x^{2} + 2 - 8x + 8x^{2} + 36x + 27 - 54x - 81 = 0\)

\(26x^{2} - 26x - 52 = 0\)

\(x^{2} - x - 2 = 0\)

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

\(x = \text{2 or -1}\)

Substitute for x in \(y = \frac{1 - 2x}{3}\); when x = 2,

\(y = \frac{1 - 2(2)}{3} = \frac{-3}{3} = -1\)

Coordinate = \((2, -1)\).

When x = -1,

\(y = \frac{1 - 2(-1)}{3} = \frac{3}{3} = 1\)

Coordinate = \((-1, 1)\)

Since Q lies in the fourth quadrant, we have P(-1, 1) and Q(2, -1).

Explanation

Spearmann's rank correlation coefficient = \(1 - \frac{6\sum D^{2}}{N(N^{2} - 1)}\)

= \(1 - \frac{6(36)}{8(64 - 1)}\)

= \(1 - \frac{216}{504}\)

= \(1 - 0.4286\)

= \(0.5714\)

\(\approxeq 0.57\).

Correct Answer: C. x\(^2\) + y\(^2\) - 10x + 8y + 25 =0

Explanation

(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)

x\(^2\) - 2hx + y\(^2\) - 2ky + h\(^2\) + k\(^2\) = r\(^2\)

x\(^2\) - 2(3)x + y\(^2\) - 2(-4) y + 5\(^2\) + (-4)\(^2\) = r\(^2\)

x\(^2\) - 10x + y\(^2\) + 8y + 25 + 16 = r\(^2\)

x\(^2\) - 10x + y\(^2\) + 8y + 41 = r\(^2\)

at point (5,0)

5\(^2\) - 10(5) + 0\(^2\) + 8(0) + 41 = r\(^2\)

25 - 50 + 41 = r\(^2\)

16 = r\(^2\)

r = \(\sqrt{16}\)

= 4

x\(^2\) + y\(^2\) - 10x + 8y + 25 = 0

Explanation

(a) \(y = 4x^{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2}\)

\(12x^{2} = 108 \implies x^{2} = 9\)

\(\therefore x = \pm 3\)

When x = -3, \(y = 4(-3^{3}) = -108\)

When x = 3, \(y = 4(3^{3}) = 108\)

\(\therefore P = (-3, -108) ; Q = (3, 108)\)

(b)(i) \(\sin (A - B) = \sin A \cos B - \sin B \cos A\)

\(\sin 15 = \sin (45 - 30) = \sin 45 \cos 30 - \sin 30 \cos 45\)

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

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

\(\cos (A - B) = \cos A \cos B + \sin A \sin B\)

\(\cos 15 = \cos (45 - 30) \)

= \(\cos 45 \cos 30 + \sin 45 \sin 30\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)

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

(b)(ii) \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

\(\tan 15 = \frac{\sqrt{6} - \sqrt{2}}{4} \div \frac{\sqrt{6} + \sqrt{2}}{4}\)

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

Ratinalizing, we have

\((\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}})(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}) = \frac{6 - 2\sqrt{3} - 2\sqrt{3} + 2}{6 - 2}\)

= \(\frac{8 - 4\sqrt{3}}{4}\)

= \(2 - \sqrt{3}\)