Correct Answer: D. 32 m

Explanation

The height of the second building H = h + 24

tan θ = \(\frac {opp}{adj}\)

tan 30\(^o = \frac {h}{x}\)

\(\implies\frac{\sqrt 3}{3} = \frac {h}{x}\)

\(\implies x = \sqrt 3 = 3h\)

\(\implies x = \frac {3h}{\sqrt 3}\) ....(i)

tan 60\(^o = \frac {24}{x}\)

\(\implies\sqrt 3 = \frac {24}{x}\)

\(\implies x\sqrt 3 = 24\)

\(\implies x = \frac {24}{\sqrt 3}\) ....(ii)

Equate equation (i) and (ii)

\(\implies \frac {3h}{\sqrt 3} = \frac {24}{\sqrt 3}\)

\(\implies\) 3h = 24

\(\implies h = \frac {24}{3}\) = 8m

∴The height of the second building = 8 + 24 = 32m

Correct Answer: B. 27

Explanation

x = 3, y = 2, z = 4

3x² - 2y + z

= 3(3)² - 2(2) + 4

3(9) - 4 + 4 = 27

Correct Answer: C. 2

Correct Answer: A. 5

Explanation

\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\)

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

\(\frac{5}{3x} = \frac{1}{3}\)

3x = 15

x = \(\frac{15}{3}\)

= 5

Explanation

(a) \(px + qy = 5 \)

When x = 1 and y = -2, we have

\(p - 2q = 5 .... (1)\)

\(qx + py = -10 \)

When x = 1 and y = -2, we have

\(q - 2p = -10 .... (2)\)

Multiplying (1) by -2, we have

\(-2p + 4q = -10 ... (1a)\)

\(\equiv 4q - 2p = -10\)

(2) - (1a) : \(q - 4q = -10 + 10\)

\(-3q = 0\)

\(q = 0\)

Putting q = 0 in (1a), we have

\(-2p = -10 \implies p = 5\)

Hence, p = 5 and q = 0.

(b) \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\)

\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\)

\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\)

\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\)

\(11r = 11\)

\(r = 1\).

Correct Answer: A. 4

Explanation

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

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

\(\frac{(x - 3)(x + 1)}{x - 3} = x + 1\)

\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1)\) (L'Hopital rule)

\(\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4\)

Explanation

\(\cos (A + B) = \cos A \cos B - \sin A \sin B\)

\(\sin A = \frac{3}{5} ; \cos B = \frac{15}{17}\)

\(\cos A = -\frac{4}{5}\) (A is obtuse)

\(\sin B = \frac{8}{17}\)

\(\cos (A + B) = \cos A \cos B - \sin A \sin B\)

= \((\frac{-4}{5} \times \frac{15}{17}) - (\frac{3}{5} \times \frac{8}{17})\)

= \(\frac{-60}{85} - \frac{24}{85}\)

= \(\frac{-84}{85}\)

Correct Answer: B. an arc of a circle with PM as a chord

Explanation

(a)(i) \(T_{n} = ar^{n - 1}\) (terms of a G.P)

\(T_{1} + T_{2} + T_{3} = a + ar + ar^{2} = 7 ... (1)\)

\(a(ar)(ar^{2}) = 8 .... (2)\)

\((a^{3} r^{3}) = 8 \implies (ar)^{3} = 2^{3}\)

\(\therefore T_{2} = ar = 2\)

\(a + ar^{2} = 5 \implies a(1 + r^{2}) = 5\)

\(a = \frac{5}{1 + r^{2}}\)

\((\frac{5}{1 + r^{2}})(\frac{5r^{2}}{1 + r^{2}}) = 4 = 2^{2}\)

\(\frac{25r^{2}}{(1 + r^{2})^{2}} = 2^{2}\)

Taking square root of both sides, we have

\(\frac{5r}{1 + r^{2}} = 2 \implies 5r = 2 + 2r^{2}\)

\(2r^{2} - 5r + 2 = 0 \implies 2r^{2} - 4r - r + 2 = 0\)

\(2r(r - 2) - 1(r - 2) = 0 \implies r = \frac{1}{2} ; r = 2\)

Since it is a decreasing sequence, the common ratio \(r = \frac{1}{2}\).

(ii) \(T_{2} = 2 = ar\)

\(\frac{a}{2} = 2 \implies a = 4\)

\(T_{3} = ar^{2} = 4(\frac{1}{2})^{2}\)

= \(4(\frac{1}{4}) = 1\)

\(\therefore \text{The first 3 terms of the sequence are } 4, 2, 1.\)

(b) \(\int_{1} ^{5} (x + \frac{2}{x^{2}}) \mathrm {d} x\)

\(Height (h) = 1 ; y_{1} = 3.0 ; y_{5} = 5.08\)

\(y_{2} = 2.5 ; y_{3} = 3.222 ; y_{4} = 4.125\)

\(y_{1} + y_{5} = 3.0 + 5.08 = 8.08\)

\(y_{2} + y_{3} + y_{4} = 2.5 + 3.222 + 4.125 = 9.847\)

\(2(y_{2} + y_{3} + y_{4}) = 2(9.847) = 19.694\)

\(Area = \frac{1}{2} \times 1[8.08 + 19.694]\)

= \(\frac{1}{2} [27.774]\)

= \(13.887 \approxeq 13.89\) (to 2 d.p.)

Correct Answer: A. y = 4 ± k