Correct Answer: C. 35.26°

Explanation

6 sin 2θ tan θ = 4, where 0º < θ < 90º

sin 2θ = 2sin θ cos θ and tanθ = \(\frac{sinθ}{cosθ}\)

= 6 x 2sin θ cos θ x \(\frac{sin θ}{cos θ} = 4\)

= \(sin^2 θ = 4\)

= \(sin^2 θ = \frac{4}{12}=\frac{1}{3}\)

=\(sin θ = \frac{\sqrt1}{3}=\frac{1}{\sqrt3}\)

= \(θ = sin^{-1}(\frac{1}{\sqrt3})\)

∴ θ = 35.26º

Correct Answer: A. \(\frac{7}{2}\)

Explanation

\(\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9}) = \lim\limits_{x \to 3} (\frac{x^{3} - 3x^{2} + 4x^{2} - 12x}{(x - 3)(x + 3)}\)

\(\lim\limits_{x \to 3} (\frac{(x^{2} + 4x)(x - 3)}{(x - 3)(x + 3)} = \lim\limits_{x \to 3} (\frac{x^{2} + 4x}{x + 3})\)

=\(\frac{3^{2} + 4(3)}{3 + 3} = \frac{21}{6} = \frac{7}{2}\)

Correct Answer: D. \(1\frac{1}{2}\)

Explanation

\(\int \frac{4}{x^{3}} \mathrm {d} x = \int 4x^{-3} \mathrm {d} x\)

\(\frac{4x^{-3 + 1}}{-2} = -2x^{-2}|_{1}^{2} = \frac{-2}{x^{2}}|_{1}^{2}\)

= \(\frac{-2}{2^{2}} - \frac{-2}{1^{2}} = -\frac{1}{2} + 2 = 1\frac{1}{2}\)

Correct Answer: C. \(9\pi\)

Explanation

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

Given that \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)

Expanding the equation of a circle, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Comparing this expansion with the given equation, we have

\(2a = 4 \implies a = 2\)

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

\(r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9\)

\(r = 3\)

\(Area = \pi r^{2} = \pi \times 3^{2}\)

= \(9\pi\)

Correct Answer: B. 6y + x - 31 = 0

Explanation

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

\(y^1 = \frac{dy}{dx} = 6x\)

Evaluating this derivative at x = 1 (since the point of interest is (1, 5)) gives us the slope of the tangent line at that point:

\(^mtangent = y^1(1) = 6 (1) = 6\)

Slope of the Normal Line \( ^mnorma l= - \frac{1}{^mtangent}\)

\(^mnormal = - \frac{1}{6}\)

\(y−y_1= ^mnormal⋅(x−x_1)\)

=y-5=-\(\frac{1}{6}(x-1)\)

=y-5=-\(\frac{1}{6}x+\frac{1}{6}\)

=y=-\(\frac{1}{6}x+\frac{1}{6}+5\)

Multiply through by 6

=6y=-x+1+30

∴6y+ x - 31=0

Correct Answer: D. 175

Explanation

From the five(5) men viable for the position of the chairman, one of them must be selected.

This implies; 4 men and 3 women are open for committee membership.

7 C\(_3\) = \(\frac{7!}{3!(7 - 3)!}\) =

\(\frac{7⋅6⋅5⋅4⋅3⋅2⋅1}{3.2.1 \times 4.3.2.1}\)

= 35 \(\times\) 5 = 175

Explanation

(a) \(^{18}C_{r} = ^{18}C_{r + 2}\)

\(\frac{18!}{r! (18 - r)!} = \frac{18!}{(r + 2)! (18 - r - 2)!}\)

\(r! (18 - r)! = (r + 2)(r + 1)r! (18 - r - 2)!\)

\((18 - r)! = (r + 2)(r + 1)(18 - r - 2)!\)

\((18 - r)(18 - r - 1)(18 - r - 2)! = (r + 2)(r + 1)(18 - r - 2)!\)

\((18 - r)(17 - r ) = r^{2} + 3r + 2\)

\(306 - 35r + r^{2} = r^{2} + 3r + 2\)

\(306 -2 = r^{2} - r^{2} + 3r + 35r\)

\(304 = 38r \implies r = 8\)

\(\therefore ^{8}C_{5} = \frac{8!}{5! (8 - 5)!}\)

\(\frac{8 \times 7 \times 6}{3 \times 2} = 56\)

(b) p(H) = 10% = 0.1 = a; p(H') = 0.9 = b

6 persons tested. We use the binomial probability distribution.

\((a + b)^{6} = a^{6} + 6a^{5}b + 15a^{4}b^{2} +20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6}\)

(i) p(exactly 5 tested positive to the virus) = \(6a^{5}b\)

= \(6 \times (0.1)^{5} \times (0.9) = 0.000054 \approxeq 0.0001\) (to 4 d.p)

(ii) p(none tested positive) = \(b^{6}\)

= \((0.9)^{6} = 0.531441 \approxeq 0.5314\) (to 4 d.p)

(iii) p(at most 2) = p(none) + p(one) + p(two)

= \(b^{6} + 6ab^{5} + 15a^{2}b^{4}\)

= \((0.9)^{6} + 6(0.1)(0.9)^{5} + 15(0.1^{2})(0.9^{4})\)

= \(0.531441 + 0.354294 + 0.098415 = 0.98415\)

\(\approxeq 0.9842\) (to 4 d.p)

Correct Answer: C. 144°

Explanation

\(\frac{4π}{2}\) = \(\frac{4}{5} * 180\)

= 144º

Correct Answer: A. 32

Explanation

Reacting force - Weight = Net force

R = W + ma = mg + ma

480 = 10p + 5p

480 = 15p

p = 32 kg

Correct Answer: B. \(\frac{2}{3}x^{\frac{3}{2}} + c\)

Explanation

\(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}\)

\(y = \int x^{\frac{1}{2}} \mathrm {d} x\)

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

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