Correct Answer: B. 210°

Explanation

\(\pi = 180°\)

\(\frac{7\pi}{6} = \frac{7 \times 180}{6} \)

= \(210°\)

Explanation

\((x + 1)(x + 2) = k(3x + 7)\)

\(x^{2} + 3x + 2 = 3kx + 7k \)

\(x^{2} + 3x - 3kx + 2 - 7k = 0\)

\(x^{2} + (3 - 3k)x + 2 - 7k = 0\)

For equal roots, \(b^{2} - 4ac = 0, a \neq 0\)

\(\therefore (3 - 3k)^{2} - 4(1)(2 - 7k) = 0\)

\(9 - 18k + 9k^{2} - 8 + 28k = 0\)

\(9k^{2} + 10k + 1 = 0 \)

\(9k(k + 1) + 1(k + 1) = 0 \implies (9k + 1)(k + 1) = 0\)

\(\implies k = -\frac{1}{9} ; k = -1\)

Correct Answer: D. 6N

Explanation

F = m * a

d = t\(^2\) + 3t.

a = \(\frac{d^2d}{dt^2}\)

\(\frac{d[d]}{dt}\) = 2t + 3

\(\frac{d^2d}{dt^2}\) = 2m/s\(^2\)

a = 2m/s\(^2\)

F = m * a

F = 3 × 2 = 6N

Correct Answer: D. 23

Explanation

\(T_{n} = a + (n - 1)d\) (for a linear or arithmetic progression)

Given: \(T_{n} = 71, a = -6, d = -2\frac{1}{2} - (-6) = 3\frac{1}{2}\)

\(\implies 71 = -6 + (n - 1)\times 3\frac{1}{2}\)

\(71 = -6 + 3\frac{1}{2}n - 3\frac{1}{2} = -9\frac{1}{2} + 3\frac{1}{2}n\)

\(71 + 9\frac{1}{2} = 3\frac{1}{2}n \implies n = \frac{80\frac{1}{2}}{3\frac{1}{2}}\)

\(= 23\)

Explanation

(a) \(\begin{vmatrix} 5 & 2 & -3 \\ -1 & k & 6 \\ 3 & 9 & (k + 2) \end{vmatrix} = - 207\)

\(5[(k)(k + 2) - 54] - 2[(-1)(k + 2) - 18] - 3[-9 - 3k] = - 207\)

\(5k^{2} + 10k - 270 + 2k + 4 + 36 + 27 + 9k = - 207\)

\(5k^{2} + 21k - 203 + 207 = 0\)

\(5k^{2} + 21k + 4 = 0\)

\(5k^{2} + 20k + k + 4 = 0\)

\(5k(k + 4) + 1(k + 4) = 0\)

\((5k + 1)(k + 4) = 0\)

\(k = -\frac{1}{5}; k = -4\)

(b) Equation of curve:

\(x(y^{2} + 1) - y(x^{2} + 1) + 4 = 0\)

(i) Differentiating, we have

\(x(2y\frac{\mathrm d y}{\mathrm d x}) + y^{2} + 1 - y(2x) - (x^{2} + 1)\frac{\mathrm d y}{\mathrm d x} = 0\)

\(\frac{\mathrm d y}{\mathrm d x}(2xy) - \frac{\mathrm d y}{\mathrm d x}(x^{2} + 1) + y^{2} + 1 - 2xy = 0\)

\(\frac{\mathrm d y}{\mathrm d x}(2xy - x^{2} - 1) = 2xy - y^{2} - 1\)

The gradient of the curve at (x, y)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{2xy - y^{2} - 1}{2xy - x^{2} - 1}\)

(ii) Gradient of the tangent at (-1, -3) is

\(\frac{\mathrm d y}{\mathrm d x} = \frac{2(-1)(-3) - (-3)^{2} - 1}{2(-1)(-3) - (-1)^{2} - 1}\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{-4}{4} = -1\)

Equation : \(\frac{y - (-3)}{x - (-1)} = -1\)

\(y + 3 = - x - 1\)

\(y = - x -4\)

Correct Answer: C. \(3 ms^{-2}\)

Explanation

The two time differentiation of distance with respect to time will give the acceleration.

\(s = \frac{3}{2}t^{2} - 3t\)

\(\frac{\mathrm d s}{\mathrm d t} = v = 3t - 3\)

\(\frac{\mathrm d v}{\mathrm d t} = a = 3\)

Explanation

(a)(i) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)

= \(\frac{151.23}{100}\)

\(\approxeq 1.51cm\)

(ii) Mean deviation = \(\frac{\sum fd}{\sum f}\)

= \(\frac{3.31}{100}\)

= \(0.033\)

(b) p(height is gretaer than mean height) = \(\frac{24 + 14 + 6 + 1}{100}\)

= \(0.45\)

Correct Answer: B. \((\frac{2}{5}, 5\frac{2}{5})\)

Explanation

\((x = \frac{nx_{1} + mx_{2}}{n + m}, y = \frac{ny_{1} + my_{2}}{n + m})\)

Given P(-2, 3) and Q(4, 9),

\((\frac{2(4) + 3(-2)}{2 + 3}, \frac{2(9) + 3(3)}{2 + 3})\)

= \((\frac{2}{5}, \frac{27}{5})\)

= \((\frac{2}{5}, 5\frac{2}{5})\)

Correct Answer: B. \(x< -5\) or \(x > \frac{3}{2}\)

Explanation

\(2x^{2} + 7x - 15 > 0 \implies 2x^{2} - 3x + 10x - 15 > 0\)

\(x(2x - 3) + 5(2x - 3) > 0\)

\((x + 5)(2x - 3) > 0\)

For their product to be positive, they are either both +ve or -ve.

\(x + 5 > 0 \implies x > -5\)

\(2x - 3 > 0 \implies 2x > 3\)

\(x > \frac{3}{2}\)

Check:

\(x > -5: x = -3\)

\(2(-3)^{2} + 7(-3) - 15 = 18 - 21 - 15 = -18 < 0\) (Not satisfied)

\(\therefore x < -5\)

\(x > \frac{3}{2}: x = 2\)

\(2(2^{2}) + 7(2) - 15 = 8 + 14 - 15 = 7 > 0\) (Satisfied)

Explanation

(a) \(r = (10N, 090°) = \begin{pmatrix} 10 \cos 90° \\ 10 \sin 90° \end{pmatrix}\)

= \(\begin{pmatrix} 0 \\ 10 \end{pmatrix}\)

\(q = (8N, 135°) = \begin{pmatrix} 8 \cos 135° \\ 8 \sin 135° \end{pmatrix}\)

= \(\begin{pmatrix} -8 \cos 45° \\ 8 \sin 45° \end{pmatrix}\)

= \(\begin{pmatrix} -4\sqrt{2} \\ 4\sqrt{2} \end{pmatrix}\)

(b) \((r + q) = \begin{pmatrix} 0 \\ 10 \end{pmatrix} + \begin{pmatrix} -4\sqrt{2} \\ 4\sqrt{2} \end{pmatrix}\)

= \(\begin{pmatrix} -4\sqrt{2} \\ 10 + 4\sqrt{2} \end{pmatrix}\)

= \(\begin{pmatrix} -5.66 \\ 15.66 \end{pmatrix}\).