Explanation

(a)

Let O(x, y) be the centre of the circle. O is the midpoint of the two points with given coordinates.

\(x = \frac{2 + 6}{2} = \frac{8}{2} = 4\)

\(y = \frac{1 + 7}{2} = \frac{8}{2} = 4\)

The centre is (4, 4).

(b) Radius of circle = \(\sqrt{(4 - 1)^{2} + (4 - 2)^{2}}\)

= \(\sqrt{3^{2} + 2^{2}}\)

= \(\sqrt{13} units\).

Equation of the circle : \((x - 4)^{2} + (y - 4)^{2} = (\sqrt{13})^{2}\)

\((x^{2} - 8x + 16) + (y^{2} - 8y + 16) = 13\)

\(x^{2} + y^{2} - 8x - 8y + 32 - 13 = 0\)

\(x^{2} + y^{2} - 8x - 8y + 19 = 0\)

Correct Answer: B. 1 or 2

Explanation

\(3^{2x} - 3^{x+2} = 3^{x+1} - 27\)

= \((3^{x})^{2} - (3^{x}).(3^{2}) = (3^{x}).(3^{1}) - 27\)

Let \(3^{x}\) be B; we have

= \(B^{2} - 9B - 3B + 27 = B^{2} - 12B + 27 = 0\).

Solving the equation, we have B = 3 or 9.

\(3^{x} = 3\) or \(3^{x} = 9\)

\(3^{x} = 3^{1}\) or \(3^{x} = 3^{2}\)

Equating, we have x = 1 or 2.

Explanation

(a)(i) Mean, \(\bar{x}\) = \(A + \frac{\sum fd}{\sum f}\)

= \(39 + \frac{-755}{190}\)

= \(39 - 3.974 = 35.026\)

(ii) Standard deviation \(SD = \sqrt{\frac{\sum fd^{2}}{N} - (\frac{\sum fd}{N})^{2}}\)

= \(\sqrt{\frac{17325}{190} - (\frac{-755}{190})^{2}}\)

= \(\sqrt{91.184 - (-3.974)^{2}}\)

\(\sqrt{91. 184 - 15.794} = \sqrt{75.39}\)

= \(8.68\)

(b) Total number of workers = 190

No of workers who are at most 36 years old = 12 + 24 + 30 + 37 = 103

p(award to a person at most 36 years old) = \(\frac{103}{190}\)

= 0.542

Explanation

Total number of board members = 14 members.

Number of editorial board members arrangement possible without restriction = \(^{14}C_{5}\) ways

= \(\frac{14!}{5! (14 - 5)!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2}\)

= \(\text{2002 ways}\).

(a) With 3 girls and 2 boys

= \(^{8}C_{3} \times ^{6}C_{2} \implies \frac{8!}{3! (8 - 3)!} \times \frac{6!}{2! (6 - 2)!}\)

= \(56 \times 15 = 840\)

p(3 girls and 2 boys) = \(\frac{840}{2002}\)

= \(\frac{60}{143}\)

(b) Editorial board with all girls = \(^{8}C_{5}\) ways

= \(\frac{8!}{5! (8 - 5)!} = 56\)

p(all girls ) = \(\frac{56}{2002}\)

With all boys = \(^{6}C_{5}\) ways

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

p(all boys) = \(\frac{6}{2002}\)

p(either all boys or all girls) = \(\frac{56}{2002} + \frac{6}{2002} = \frac{31}{1001}\).

Correct Answer: A. \(3\sqrt{2}\)

Correct Answer: C. \(\frac{1 - y}{x + 2y}\)

Explanation

Given \(y^{2} + xy - x = 0\)

Using the method of implicit differentiation, we have

\(2y\frac{\mathrm d y}{\mathrm d x} + x\frac{\mathrm d y}{\mathrm d x} + y - 1 = 0\)

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

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

Explanation

(b) (i) Lower quartile = 40 hours

(ii) Median = 49 hours

(iii) 40th percentile = 45.5 hours

(iv) Employees that spent at least 50 hours 30 mins = 80 - 45 = 35

Correct Answer: A. {21, 91, 221}

Explanation

multiples of 5 less than 20 = 5, 10 and 15

= [ 5, 10 and 15 ]x\(^2\) - x + 1

x\(^2\) - x + 1

when x = 5

5\(^2\) - 5 + 1; 25 - 5 + 1 --> 21

when x = 10

10\(^2\) - 10 + 1

100 - 9 = 91

when x = 15

15\(^2\) - 15 + 1

225 - 15 + 1 = 211

Correct Answer: C. 37\(^?\)

Explanation

y = 2x + 5

m1 = 2

2y = x - 6

tanθ = \(\frac{3}{2}\) ÷ (1+1)

tanθ = \(\frac{3}{2}\) ÷ 2

θ = \(tan^{-1} (\frac{3}{4})\)

θ = 36.87º

θ = 37º

Explanation

Position vector of A is (8i - 10j)

Position vector of B is (2i + 6j)

Position vector of C is (-10i + 4j).

Let position vector of N be (ai + bj).

ABCN is a parallelogram.

\(\therefore \overrightarrow{AB} = \overrightarrow{NC}\).

\(\begin{bmatrix} (2 - 8)i \\ (6 + 10)j \end{bmatrix} = \begin{bmatrix} (-10 - a)i \\ (4 - b)j \end{bmatrix}\)

Equating components,

\(2 - 8 = -10 - a \implies a = -4\)

\(6 + 10 = 4 - b \implies b = -12\)

The position vector of N is (-4i - 12j).

(b) \(\overrightarrow{AN} \cdot \overrightarrow{AB}\)

\(\overrightarrow{AN} = \bar{A} - \bar{N} \)

= \((8i - 10j) - (-4i - 12j)\)

= \(12i + 2j\)

\(\overrightarrow{AB} = \bar{A} - \bar{B}\)

= \((8i - 10j) - (2i + 6j)\)

= \(6i - 16j\)

\(\therefore \overrightarrow{AN} \cdot \overrightarrow{AB} = (12i + 2j) \cdot (6i - 16j)\)

= \(72 - 32 = 40\)

(c) Let \(\theta\) be the acute angle between \(\overrightarrow{AN}\) and \(\overrightarrow{AB}\).

\(\overrightarrow{AN} \cdot \overrightarrow{AB} = |AN||AB| \cos \theta\)

\(|AN| = |12i + 2j| = \sqrt{12^{2} + 2^{2}} = \sqrt{144 + 4} = \sqrt{148}\)

\(|AB| = |6i - 16j| = \sqrt{6^{2} + (-16)^{2}} = \sqrt{36 + 256} = \sqrt{292}\)

\(\therefore 40 = (\sqrt{148})(\sqrt{292}) \cos \theta\)

\(\cos \theta = \frac{40}{(\sqrt{148})(\sqrt{292})}\)

\(\cos \theta = \frac{40}{207.88} = 0.1924\)

\(\theta = \cos^{-1} (0.1924) = 78.907°\)

\(\approxeq 78.9°\)