Correct Answer: A. 12

Explanation

\(Mean : \frac{x + 4 + z + t}{4} = 5 \implies x + 4 + z + t = 20\)

\(\implies x + z + t = 16 ... (1)\)

\(Median : \frac{4 + z}{2} = z \implies 4 + z = 2z\)

\(4 = z\)

From 1,

\(\implies x + 4 + t = 16 \)

\(x + t = 12\)

Correct Answer: C. 78

Correct Answer: A. -30

Explanation

\(\int (3x^{2} - 2x - 12) \mathrm {d} x = \frac{3x^{2 + 1}}{2 + 1} - \frac{2x^{1 + 1}}{2} - 12x\)

= \(x^{3} - x^{2} - 12x\)

\((x^{3} - x^{2} - 12x)|_{-2}^{3} = ((3^{3}) - (3^{2}) - 12(3)) - ((-2^{3}) - (-2^{2}) - 12(-2))\)

= \((27 - 9 - 36) - (-8 - 4 + 24) = -18 - 12 = -30\)

Explanation

\(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\)

Velocity, \(\frac{\mathrm d h}{\mathrm d t} = 20 - 3t - 2t^{2}\)

At point of rest, v = 0

\(2t^{2} + 3t - 20 = 0\)

\(\implies 2t^{2} - 5t + 8t - 20 = 0\)

\(t(2t - 5) + 4(2t - 5) = 0 \implies t = \frac{5}{2}; -4\)

The value of time cannot be negative hence \(t = \frac{5}{2}secs\)

(b) Maximum height, \(h_{m}\) is at time \(t = \frac{5}{2}s\).

\(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\)

= \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)

Correct Answer: A. 20

Correct Answer: D. 54°

Explanation

Ondo = \(\frac{126}{840} \times 360°\)

= \(54°\)

Correct Answer: D. -3

Explanation

\(g(x) = \frac{x + 1}{x + 2}, x \neq 2\)

Let y = x, then \(g(y) = \frac{y + 1}{y + 2}\)

Let x = g(y), so that \(x = \frac{y + 1}{y + 2}\)

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

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

\(y(x - 1) = 1 - 2x \implies y = \frac{1 - 2x}{x - 1}\)

\(y = g^{-1}(x) = \frac{1 - 2x}{x - 1}\)

\(g^{-1}(2) = \frac{1 - 2(2)}{2 - 1} = -3\)

Correct Answer: B. (4, 8, 16, 20)

Explanation

E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

P = {3, 6, 9, 12, 15, 18}

Q = {4, 8, 12, 16, 20}

P' = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}

P' \(\cap\) Q = {4, 8, 16, 20}

Correct Answer: D. 102°

Correct Answer: D. 7√3