Correct Answer: D. 13

Explanation

\(f(x) = 3x^{3} - 2x^{2} + 7x + 5\).

\(x - 1 = 0, x = 1\)

\(f(1) = 3(1)^{3} - 2(1)^{2} + 7(1) + 5 = 13\)

Correct Answer: B. \(\frac{3}{8}\)

Explanation

Let \(p(head) = p = \frac{1}{2}\) and \(p(tail) = q = \frac{1}{2}\)

\((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}\)

The probability of equal heads and tails = \(6p^{2}q^{2} = 6(\frac{1}{2}^{2})(\frac{1}{2}^{2})\)

= \(\frac{6}{16} = \frac{3}{8}\).

Correct Answer: C. 3

Explanation

\(y = 1 - 3x + 2x^{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = -3 + 6x^{2}\)

At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = -3 + 6(1^{2}) = -3 + 6 = 3\)

\(y = mx - 3 \implies \frac{\mathrm d y}{\mathrm d x} = m = 3\) (Tangent with equal gradient)

Explanation

(a) \(s = t^{2} - 6t + 5\)

Scale : On time- axis, 2 cm rep 2 s

On displacement, 2 cm rep 2m.

(b)(i) Velocity = \(\frac{\mathrm d s}{\mathrm d t} = Gradient\)

Velocity = 0 m/s when t = 3s

(ii) Average velocity = \(\frac{\text{total costume}}{\text{total time}}\)

\(\frac{5 + 4 + (4 - 3)}{4}\)

= \(\frac{10}{4} = 2.5 m/s\).

(iii) Total distance covered in the interval \(0 \leq x \leq 5\)

= 5 + 4 + 4 = 13m.

Explanation

P(y) = 2y + 3, q(y) = y\(^2\) - 2

q \(\cap\) p = qp(y)

= q(2y + 3)

= (2y + 3)\(^3\) - 2

= 4xy\(^2\) + 12y + 9 - 2

= 4y\(^2\) + 12y + 7

(b)

No. of ways = 2 x 5 x 5 x 5

= 250 ways

Explanation

(a) (i) \(F_{1} = \begin{pmatrix} -5 & 4 \end{pmatrix} N; F_{2} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} N; F_{3} = \begin{pmatrix} 2 & -1 \end{pmatrix} N ;F_{4} = \begin{pmatrix} 3 & -5 \end{pmatrix} N\)

\(R = \begin{pmatrix} -5 & 4 \end{pmatrix} N + \begin{pmatrix} 2 \\ 5 \end{pmatrix} N + \begin{pmatrix} 2 & -1 \end{pmatrix} N + \begin{pmatrix} 3 & -5 \end{pmatrix} N = \begin{pmatrix} 2 \\ 3 \end{pmatrix} N\)

(ii) Let the fifth force be F.

\(F + \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(F = \begin{pmatrix} -2 \\ -3 \end{pmatrix}\)

(b) (i)

Acceleration = a = 2.5 ms\(^{-2}\),

Let retardation be r ms\(^{-2}\).

\(a : r = 3 : 4\)

\(\therefore r = \frac{4}{3} \times 2.5 = 3\frac{1}{3} ms^{-2}\)

(ii) We use the formula \(v = u + at\) for the retardation period.

\(v = 0 m/s ; u = 30 m/s ; a = -3\frac{1}{3} m/s^{2} ; t = ?\)

\(0 = 30 - \frac{10}{3} t \)

\(t = \frac{30 \times 3}{10} = 9 secs\)

(iii) Total distance = area under graph.

\(A_{1} = \frac{1}{2} (4)(20 + 30) = 100m\)

\(A_{2} = 30 \times 8 = 240m\)

\(A_{3} = \frac{1}{2} (9) (30) = 135m\)

Total distance = 100m + 240m + 135m = 475m.

Correct Answer: C. \(\begin{pmatrix} 13 \\ 11 \end{pmatrix}\)

Explanation

\(\begin{pmatrix} 2 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix}\)

= \(\begin{pmatrix} 2 \times 2 + 3 \times 3 \\ 4 \times 2 + 1 \times 3 \end{pmatrix}\)

= \(\begin{pmatrix} 13 \\ 11 \end{pmatrix}\)

Correct Answer: D. -3

Explanation

The sum of deviations from the mean of a set of numbers equals 0.

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

\((k^2 + 6k + 9) + (k+7) - 2 + k + (k^2 + 4k + 4) = 0\)

\(2k^{2} + 12k + 18 = 0\)

\(2k^{2} + 6k + 6k + 18 = 2k(k + 3) + 6(k + 3) = 0\)

\(k = -3 (twice)\)

Explanation

(a) 5b, 4g, 3y = 12 balls.

(i) p(all 3 balls have the same colour) = p(all 3 blue or all 3 green or all 3 yellow)

= \(\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} + \frac{3}{12} \times \frac{2}{11} \times \frac{1}{10}\)

= \(\frac{60}{1320} + \frac{24}{1320} + \frac{6}{1320}\)

= \(\frac{90}{1320} = \frac{3}{44}\)

(ii) Two balls of the same colour :

Sample space for selecting with restriction (2 blue 1 green or 2 blue 1 yellow or 2 green 1 blue or 2 green 1 yellow or 2 yellow 1 blue or 2 yellow 1 green)

Ways : \(^{5}C_{2} \times ^{4}C_{1} + ^{5}C_{2} \times ^{3}C_{1} + ^{4}C_{2} \times ^{5}C_{1} + ^{4}C_{2} \times ^{3}C_{1} + ^{3}C_{2} \times ^{5}C_{1} + ^{3}C_2} \times ^{4}C_{1}\)

= \(40 + 30 + 30 + 18 + 15 + 12 = 145\)

Selecting without restriction 3 balls out of 12 balls : \(^{12}C_{3}\)

= \(\frac{12!}{3! 9!}\)

= \(220\)

Probability = \(\frac{\text{selection with restriction}}{\text{selection without restriction}} = \frac{145}{220}\)

= \(\frac{29}{44}\)

(b)

Spearman's rank correlation cofficient:

= \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)

= \(1 - \frac{6(16)}{7(7^{2} - 1)}\)

= \(1 - \frac{2}{7}\)

= \(0.714\)

Correct Answer: D. \(\frac{160}{729}\)

Explanation

\([\frac{1}{3}(2 + x)]^{6} = (\frac{2}{3} + \frac{x}{3})^{6}\)

The coefficient of \(x^{3}\) is

\(^{6}C_{3}(\frac{2}{3})^{3}(\frac{1}{3})^{3}x^{3} = (\frac{6!}{3!3!})(\frac{8}{27})(\frac{1}{27})x^{3}\)

= \(\frac{160}{729}\)