Correct Answer: C. 75°

Explanation

E\(G\)R = 50\(^o\) (Alternative angles)

n + 55\(^o\) + 50\(^o\) = 180\(^o\) (Sum of angles on a straight line)

n = 180\(^o\) - 105\(^o\)

n = 75\(^o\)

Correct Answer: A. \(-540x^{3}y^{3}\)

Explanation

Listing out in order, the terms of this expansion have coefficients: \(^{6}C_{6}, ^{6}C_{5}, ^{6}C_{4}, ^{6}C_{3}, ^{6}C_{2}, ^{6}C_{1}, ^{6}C_{0}\)

The 4th term = \(^{6}C_{3}(3x)^{3}(-y)^{3} = 20 \times 27x^{3} \times -y^{3}\)

= \(-540x^{3}y^{3}\).

Explanation

\(P = (4N, 030°) ; R = (10N, 300°)\)

Let \(F = (a N, \theta)\) be the force that will keep the particle in equilibrium.

Then P + R + F = 0.

\(\begin{pmatrix} 4 \cos 30 \\ 4 \sin 30 \end{pmatrix} + \begin{pmatrix} 10 \cos 300 \\ 10 \sin 300 \end{pmatrix} + \begin{a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).

\(\begin{pmatrix} 4(\frac{\sqrt{3}}{2}) \\ 4(\frac{1}{2}) \end{pmatrix} + \begin{pmatrix} 10(\frac{1}{2}) \\ 10(-\frac{\sqrt{3}}{2}) \end{pmatrix} + \begin{pmatrix} a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(\begin{pmatrix} 2\sqrt{3} + 5 \\ 2 - 5\sqrt{3} \end{pmatrix} + \begin{pmatrix} a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \begin{pmatrix}\)

\(\implies 2\sqrt{3} + 5 + a \cos \theta = 0 \)

\(2\sqrt{3} + 5 = -a \cos \theta\)

\(2 - 5\sqrt{3} = -a \sin \theta\)

\(\tan \theta = \frac{2 - 5\sqrt{3}}{2\sqrt{3} + 5} = \frac{-6.66}{8.486}\)

\(\tan \theta = -0.7848 \implies \theta = -38.13°\)

= \(141.87°\)

\(-a \cos \theta = 8.486\)

\(-a \cos (141.87) = -a (-0.7866) = 8.486\)

\(0.7866a = 8.486 \implies a = \frac{8.486}{0.7866} = 10.79N\)

\(\therefore F = (10.79N, 141.87°)\)

Correct Answer: A. 3

Explanation

\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2)

\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2)

x + 4y = 6......(iii)

3x - 4y = 2.....(iv) add (iii) and (iv)

4x = 8, x = \(\frac{8}{4}\) = 2

substitute x = 2 into equation (iii)

x + 4y = 6

2 + 4y = 6

4y = 6 - 2

4y = 4

y = \(\frac{4}{4}\)

= 1(x + y)

2 + 1 = 3

Explanation

(a) \(\log 3.6 = \log (\frac{18}{5})\)

\(\log 18 - \log 5 = \log (2 \times 3^{2}) - \log 5\)

\(\log 2 + \log 3^{2} - \log 5 = \log 2 + 2\log 3 - \log 5\)

= \(0.3010 + 2(0.4771) - 0.6990\)

= \(0.3010 + 0.9542 - 0.6990\)

= \(0.5562\).

(b) \(\frac{y}{y + 101} = \frac{11}{10010}\) (all in base 2)

Cross multiplying,

\(11(y + 101) = 10010y\)

\(11y + 1111 = 10010y\)

\(10010y - 11y = 1111\)

\(1111y = 1111\)

\(y = 1\).

Correct Answer: A. 42°

Correct Answer: B. 85

Explanation

Pass mark = 50%

No. of students that passed = f₅₀ + f₆₅ + f₈₀

= 45 + 25 + 15

= 85

Correct Answer: D. 13

Explanation

\(\lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3}\)

\(2x^{2} + x - 21 = 2x^{2} - 6x + 7x - 21 \) (by factorizing)

= \((2x + 7)(x - 3)\)

\(\therefore \lim\limits_{x \to 3} \frac{2x^{2} + x - 21}{x - 3} \equiv \lim\limits_{x \to 3} \frac{(2x+7)(x-3)}{x-3}\)

\(\lim\limits_{x \to 3} (2x + 7) = 2(3) + 7 = 13\)

Explanation

(a) \(f(x + 2) = 6x^{2} + 5x - 8\)

\(f(5) \implies x + 2 = 5\)

\(\therefore x = 3\)

\(f(5) = f(3 + 2) = 6(3^{2}) + 5(3) - 8\)

= \(54 + 15 - 8\)

= \(61\)

(b) \(\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}}\)

Rationalizing, we multiply the numerator and denominator by \(4\sqrt{2} + 2\sqrt{3}\)

\((\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 3\sqrt{3}})(\frac{4\sqrt{2} + 2\sqrt{3}}{4\sqrt{2} + 2\sqrt{3}})\)

= \(\frac{28(2) + 14\sqrt{6} + 12\sqrt{6} + 6(3)}{16(2) - 4(3)}\)

= \(\frac{74 + 26\sqrt{6}}{20}\)

= \(3.7 + 1.3\sqrt{6}\)

Correct Answer: B. 070°