Correct Answer: A. 3.2

Explanation

Momentum of bodies with common velocity = \(m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v\)

\((10 \times 5) + (15 \times 2) = (10 + 15)v\)

\(50 + 30 = 25v \implies v = \frac{80}{25} = 3.2ms^{-1}\)

Explanation

To get f(x), multiply the divisor with the quotient and add the remainder.

\(2x^{3} + mx^{2} + nx + 11 = (x^{2} + 5x + 1)(2x - 5) + (30x + 16)\)

= \(2x^{3} - 5x^{2} + 10x^{2} - 25x + 2x - 5 + 30x + 16\)

= \(2x^{3} + 5x^{2} + 7x + 11\)

Therefore, m = 5 and n = 7.

Explanation

(a) \(f(x) = \int (4x - x^{2}) \mathrm {d} x\)

\(f(x) = 2x^{2} - \frac{x^{3}}{3} + c\)

\(f(3) = 2(3^{2}) - \frac{3^{3}}{3} + c\)

\(21 = 18 - 9 + c\)

\(c = 21 - 9 = 12\)

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

(b) \(T_{n} = a + (n - 1) d\) (terms of an A.P)

\(T_{2} = a + d ; T_{4} = a + 3d ; T_{8} = a + 7d\)

These are consecutive terms of a G.P

\(i.e. \frac{T_{4}}{T_{2}} = \frac{T_{8}}{T_{4}}\)

\((T_{4})^{2} = T_{2} \times T_{8}\)

\((a + 3d)^{2} = (a + d)(a + 7d)\)

\(a^{2} + 6ad + 9d^{2} = a^{2} + 8ad + 7d^{2}\)

\(6ad + 9d^{2} = 8ad + 7d^{2} \implies 2ad = 2d^{2}\)

\(a = d\)

\(T_{3} + T_{5} = 20\)

\(a + 2d + a + 4d = 20\)

\(a + 2a + a + 4a = 20 \implies 8a = 20\)

\(a = d = \frac{5}{2}\)

(i) First four terms of A.P = \(\frac{5}{2}, 5, \frac{15}{2}, 10\)

(ii) \(S_{n} = \frac{n}{2} (2a + (n - 1)d)\)

\(S_{10} = \frac{10}{2} (2(\frac{5}{2}) + (10 - 1)(\frac{5}{2}))\)

= \(5(5 + \frac{45}{2})\)

= \(\frac{275}{2} = 137.5\)

Correct Answer: B. 56.3°

Explanation

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

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

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

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

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

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

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

\(\theta = 56.3°\)

Correct Answer: A. A. (r ^ p) \(\to\) q

Correct Answer: C. \(\frac{23}{7}\)

Explanation

\(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) + \(\frac{9}{7(x-3)}\)

15-2x = R(x - 3) +9(x +4)/7

Put x =-4, we have 15 -2(-4) = -7R

23 -7R;

R = 23/7

Correct Answer: B. 12

Explanation

The total number of candidates = 40, therefore, the median occurs as the average between the 20th and 21st candidates. = \(\frac{3+3}{2} = 3\)

Number of candidates that scored above 3 : 9 + 3 = 12.

Explanation

(a)

Then \(\sum d_1^2\) = 50

The spearman's coefficient = 1 - \(\frac{6 \times 50}{10(100 - 1)}\) = 1 - 0.30303 = 0.697

(b) We conclude that there exist positive correlation between performance in entrance examination and school performance

Correct Answer: D. \(\frac{1}{5}(4i + 3j)\)

Explanation

\(\hat {n} = \frac{\overrightarrow{p}}{|p|}\)

where \(\hat {n}\) is the unit vector in the direction of p.

\(|p| = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\)

\(\hat {n} = \frac{1}{5} (4i + 3j)\)

Correct Answer: A. \(5\sqrt{3} - 3\)

Explanation

\(F = F \cos \theta i + F \sin \theta j\)

\(10N = 10 \cos 60 i + 10 \sin 60 j\)

\(6N = -6 \cos 330 i - 6 \sin 330 j\)

\(R_{x} = 10 \cos 60 - 6 \cos 330 \)

= \(10 \times \frac{1}{2} - 6 \times \frac{\sqrt{3}}{2}\)

= \(5 - 3\sqrt{3}\)