Correct Answer: A. 20

Explanation

Since 2 students must be included, we have to arrange the remaining 3 students from the 6 students left

= \(^{6}C_{3} = \frac{6!}{(6-3)!3!}\)

= 20 ways

Explanation

No. of farmers = 6

No. of traders = 8

Total = 14

Pr(at least two farmers)

= \(\frac{^6C_2 \times ^8C_2}{^{14}C_4} + \frac{^6C_3 \times ^8C_1}{^{14}}+ \frac{^6C_4 \times ^8C_0}{^{14}C_4}\)

= \(\frac{15 \times 28}{1001} + \frac{20 \times 8}{1001} + \frac{15 \times 1}{1001}\)

= \(\frac{420}{1001} + \frac{160}{1001} + \frac{15}{1001}\)

= \(\frac{420 + 160 + 15}{1001}\)

= \(\frac{595}{1001}\)

(b) P(x) = \(\frac{1}{5}\)

p(x\(^1\)) = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)

p(y) = \(\frac{1}{2}\), p(y\(^1\))

= 1 - \(\frac{1}{12}\)

= \(\frac{11}{12}\)

p(2) = \(\frac{1}{6}\)

p(z\(^1\)) = 1 - \(\frac{1}{6}\)

= \(\frac{5}{6}\)

(i) Pr(None)

= \(\frac{4}{5} \times \frac{11}{12} \times \frac{5}{6} = \frac{11}{18}\)

(ii) Pr(all the factories)

= \(\frac{1}{5} \times \frac{1}{12} \times \frac{1}{6} = \frac{1}{360^o}\)

(iii) Pr (at least one factory)

= 1 - Pr (none) = \(\frac{1}{1} - \frac{11}{18}\)

= \(\frac{18 - 11}{18}\)

= \(\frac{7}{18}\)

Explanation

(a)(i) Using the Paschal triangle,

\((1 + x)^{7} = 1 + 7x + 21x^{2} + 35x^{3} + 35x^{4} + 21x^{5} + 7x^{6} + x^{7}\)

(ii) \(T_{5} = 35 ; T_{6} = 21 ; T_{7} = 7\)

\(d = T_{6} - T_{5} = 21 - 35 = -14\)

(b) \(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x\)

h = 1

\(y_{1} = 3.162, y_{2} = 6.0 , y_{3} = 8.832 , y_{4} = 11.662 , y_{5} = 14.491\)

\(y_{1} + y_{5} = 3.162 + 14.491 = 17.653\)

\(y_{2} + y_{3} + y_{4} = 6.0 + 8.832 + 11.662 = 26.494\)

\(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x = \frac{1}{2}[y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4})]\)

= \(\frac{1}{2} [17.653 + 2(26.494)]\)

= \(\frac{1}{2} [70.641]\)

= \(35.3205 \approxeq 35.32\)

Explanation

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

\(\frac{x}{y} = 2^{2} = 4 \implies x = 4y ... (1)\)

\(\log_{2} (x - 2y) = 3 \implies x - 2y = 2^{3} = 8 ... (2)\)

Putting (1) into (2),

\(4y - 2y = 8 \implies 2y = 8\)

\(y = 4\)

\(x = 4y \implies x = 4(4) = 16\)

\(x = 16 ; y = 4\)

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

(a) Let \(u = (x - 3)\) and \(v = (x^{2} + 5)\).

\(y = uv\)

\(\frac{\mathrm d u}{\mathrm d x} = 1 ; \frac{\mathrm d v}{\mathrm d x} = 2x\)

From the product rule,

\(\frac{\mathrm d y}{\mathrm d x} = v \frac{\mathrm d u}{\mathrm d x} + u \frac{\mathrm d v}{\mathrm d x}\)

\(\frac{\mathrm d y}{\mathrm d x} = (x^{2} + 5)(1) + (x - 3)(2x)\)

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

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

(b)(i) \(f(x) = x^{3} + ax^{2} + bx + 3\)

Since \((x + 1)^{2}\) is a factor, then \(f(-1) = 0\).

\(\therefore f(-1) = (-1)^{3} + a(-1)^{2} + b(-1) + 3\)

\(0 = -1 + a - b + 3 \implies a - b = -2 ... (1)\)

When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Since we already have 2 out of the three zeros of the polynomial, the third one will be (x + 3). [Check for yourself using the long division method].

\(\therefore f(-3) = 0\)

\(f(-3) = (-3)^{3} + a(-3)^{2} + b(-3) + 3\)

\(-27 + 9a - 3b + 3 = 0 \implies 9a - 3b = 24 ... (2)\)

From (1), \(a = -2 + b\)

\(\therefore 9(-2 + b) - 3b = 24\)

\(-18 + 9b - 3b = 24 \implies 6b = 42\)

\(b = 7 \implies a = -2 + 7 = 5\)

(a, b) = (5, 7).

(ii) We already have the three factors of the polynomial as \((x + 1)^{2}; (x + 3)\).

Hence the zeros of f(x) are -1, -1 and -3.

Correct Answer: C. 60

Explanation

\((2 + x)^{6} \)

\(x^{4} = ^{6}C_{2}(2^{2})(x^{4}) = 15 \times 4 = 60\)

Correct Answer: D. 56.25 m

Explanation

\(h = 45t - 9t^{2}\)

\(\frac{\mathrm d h}{\mathrm d t} = 45 - 18t = 0\)

\(45 = 18t \implies t = 2.5 s\)

\(h(2.5) = 45(2.5) - 9(2.5)^{2} = 112.5 - 56.25\)

= \(56.25 m.\)

Correct Answer: C. \((\frac{5}{8}, -\frac{3}{8})\)

Explanation

Equation : \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding : \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Given, \(4x^{2} + 4y^{2} - 5x + 3y - 2 = 0\)

Divide through by 4 to make the coefficient of \(x^{2}\) and \(y^{2}\) to be 1.

\(x^{2} + y^{2} - \frac{5}{4}x + \frac{3}{4}y - \frac{1}{2} = 0\)

Comparing, \(2a = \frac{5}{4} \implies a = \frac{5}{8}\)

\(2b = -\frac{3}{4} \implies b = -\frac{3}{8}\)

\((a, b) = (\frac{5}{8}, -\frac{3}{8})\)

Correct Answer: C. 56

Explanation

Number of matches possible = \(2 \times ^{8}C_{2}\) (Home and away repitition of the matches)

= \(2 \times \frac{8!}{(8 - 2)! 2!}\)

= \(2 \times 28\)

= 56