Correct Answer: D. 1, -1/5

Explanation

5x² - 4x - 1 = 0

5x² 5x + x -1 = 0

5x(x - 1) + (x - 1) = 0

(5x + 1)(x -1) = 0

x = -1/5 or 1 = 0

Explanation

(a) < ZXY = 225° - 135°

= 90°.

\(\therefore /ZY/ = 80^{2} + 60^{2}\)

= \(6400 + 3600 = 10000\)

\(/ZY/ = \sqrt{10000}\)

= 100 km.

(b) From \(\Delta\) XYZ,

\(\tan \theta = \frac{opp}{adj}\)

\(\tan \theta = \frac{80}{60} = 1.333\)

\(\theta = \tan^{-1} (1.333)\)

\(\theta = 53.13°\)

Bearing of Z from Y = \(360° - (45° + 53.13°)\)

= \(360° - 98.13°\)

= \(261.87° \approxeq 262°\) (3 significant figures)

Explanation

Since PQRM is a parallelogram,

\(\overrightarrow{PQ} = \overrightarrow{MR}\)

\((4 - 1)i + (-5 - 3)j = (x + 5)i + (y - 2)j\)

Equating components, we have

\(3 = x + 5 \implies x = 3 - 5 = -2\)

\(-8 = y - 2 \implies y = -8 + 2 = -6\)

\(\therefore M = (-2i - 6j)\)

(b) \(|\overrightarrow{PM}| = \sqrt{(-2 - 4)^{2} + (- 6 + 5)^{2}} = \sqrt{37}\)

\(|\overrightarrow{PQ}| = \sqrt{(1 - 4)^{2} + (3 + 5)^{2}} = \sqrt{73}\)

(c) Let the angle be \(\theta\).

\(\overrightarrow{PM} . \overrightarrow{PQ} = |PM| |PQ| \cos \theta\)

\((-6i - j) . (-3i + 8j) = (\sqrt{37})(\sqrt{73}) \cos \theta\)

\(18 - 8 = \sqrt{2701} \cos \theta\)

\(\cos \theta = \frac{10}{\sqrt{2701}} = 0.1924\)

\(\theta = 78.907° \approxeq 78.9°\) (to 1 d.p)

(d) Area of PQRM = \((\sqrt{37})(\sqrt{73}) \sin 78.907°\)

= \(\sqrt{2701} \sin 78.907°\)

= \(51 \text{sq. units}\)

Correct Answer: B. N45,000

Correct Answer: C. x = 3 or -5

Explanation

f:x → 2x\(^2\) + 3x -7 and

g:x →5x\(^2\) + 7x - 6

If 3f(x) = g(x)

3(2x\(^2\) + 3x -7) = 5x\(^2\) + 7x - 6

6x\(^2\) + 9x - 21 = 5x\(^2\) + 7x - 6

6x\(^2\) - 5x\(^2\) + 9x - 7x - 21 + 6

x\(^2\) + 2x - 15 = 0

x = 3 or -5

Correct Answer: B. 3/24

Correct Answer: A. \(\frac{63}{65}\)

Explanation

\(sin x = \frac{4}{5}\) and \(cos y = \frac{12}{13}\)

x is obtuse i.e sin x = + ve while cos x = + ve

\(cos x=\frac{3}{5}==>cos x=-\frac {3}{5}(obtuse)\)

\(sin y= \frac{5}{13}\)

\(sin (x-y) = sin x\) \(cos y - cos x\) \(sin y\)

\(sin(x-y) = \frac{4}{5}\times\frac{12}{13}-(-\frac{3}{5})\times\frac{5}{13}\)

\(sin(x-y) = \frac{48}{65}-(-\frac{3}{13})\)

\(\therefore sin (x-y) = \frac{48}{65} + \frac{3}{13} = \frac{63}{65}\)

Correct Answer: B. x=1 and y=1

Explanation

x + y = 2 --- eqn I

3x - 2y = 1 --- eqn II

Multiply eqn by 2

2x + 2y = 2 --- eqn III

Subtract eqn II from eqnI

5x = 5 => x = 1

Substitute 1 for x in eqn 1

1 + y = 2 => y = 1

Explanation

(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {1, 2, 5, 7}

A' = U - A = {3, 4, 6, 8, 9, 10}

(ii) \(A \cap B = {1, 2, 5, 7} \cap {1, 3, 6, 7} = {1, 7}\)

\((A \cap B)' = {2, 3, 4, 5, 6, 8, 9, 10}\)

(iii) \(A \cup B = {1, 2, 5, 7} \cup {1, 3, 6, 7}\)

= \({1, 2, 3, 5, 6, 7}\)

\((A \cup B)' = {4, 8, 9,10}\)

(iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}.

(b) The nth term of the sequence is given as

\(U_{n} = \frac{n + 1}{n(n + 2)}, n = 1, 2,3\)

For the 15th term,

\(U_{15} = \frac{16}{15 \times 17}\)

(c)(i) \(T_{n} = a + (n - 1)d\)

\(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\)

= \(4n - 1\)

(ii) \(4n - 1 > 100\)

\(4n > 100 + 1 \implies 4n > 101\)

\(n > 25.25\)

The least term greater than 100 = 26th term.

(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {1, 2, 5, 7}

A' = U - A = {3, 4, 6, 8, 9, 10}

(ii) \(A \cap B = {1, 2, 5, 7} \cap {1, 3, 6, 7} = {1, 7}\)

\((A \cap B)' = {2, 3, 4, 5, 6, 8, 9, 10}\)

(iii) \(A \cup B = {1, 2, 5, 7} \cup {1, 3, 6, 7}\)

= \({1, 2, 3, 5, 6, 7}\)

\((A \cup B)' = {4, 8, 9,10}\)

(iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}.

(b) The nth term of the sequence is given as

\(U_{n} = \frac{n + 1}{n(n + 2)}, n = 1, 2,3\)

For the 15th term,

\(U_{15} = \frac{16}{15 \times 17}\)

(c)(i) \(T_{n} = a + (n - 1)d\)

\(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\)

= \(4n - 1\)

(ii) \(4n - 1 > 100\)

\(4n > 100 + 1 \implies 4n > 101\)

\(n > 25.25\)

The least term greater than 100 = 26th term.

Explanation

(d) /PM/ = 9.7 cm.