Correct Answer: B. a = -5, b = 14

Explanation

The terms of the sequence can be written as : \(u_{r} = ar + b\) in this case, being that they have a regular common difference for each of the r terms.

We can rewrite the sequence as \(a + b, 2a + b, 3a + b,...\) where a is the common difference of the sequence and b is a given constant gotten by solving

\(a + b = 9\) or \(2a + b = 4\) or any other one.

The common difference here is 4 - 9 = -1 - 4 = -5.

\(-5 + b = 9 \implies b = 9 + 5 = 14\)

\(\therefore\) The equation can be written as \(u_{r} = -5r + 14\).

Correct Answer: A. 2y - x -5 = 0

Explanation

We are given the equation \(x^{2} + y^{2} - 4x - 2y = 0\)

\(y = x^{2} + y^{2} - 4x - 2y \)

Using the method of implicit differentiation,

\(\frac{\mathrm d y}{\mathrm d x} = 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x}\)

For the tangent, \(\frac{\mathrm d y}{\mathrm d x} = 0\),

\(\therefore 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x} = 0\)

\((2y - 2)\frac{\mathrm d y}{\mathrm d x} = 4 - 2x \implies \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2x}{2y - 2}\)

At (1, 3), \(\frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}\)

Equation: \(\frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1\)

= \(2y - x - 6 + 1 = 2y - x - 5 = 0\)

Correct Answer: C. 3

Explanation

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = 6\)

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)\)

\(\begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = (2k + 3)\)

\(\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6\)

\(k^{2} - 2k - 3 = 0\), factorising, we have \(k + 1 = 0\) or \(k - 3 = 0\)

Since k > 0, k = 3.

Correct Answer: C. 6720

Explanation

The word has 8 letters with one letter repeated 3 times, therefore we have:

\(\frac{8!}{3!} = 6720\) ways.

Explanation

(a) \(p(Kunle) = \frac{1}{3} ; p(\text{not Kunle}) = \frac{2}{3}\)

\(p(Tayo) = \frac{1}{5} ; p(\text{not Tayo}) = \frac{4}{5}\)

\(p(\text{only one solve the question}) = p(\text{Kunle and not Tayo}) + p(\text{Tayo and not Kunle})\)

= \((\frac{1}{3} \times \frac{4}{5}) + (\frac{1}{5} \times \frac{2}{3})\)

= \(\frac{4}{15} + \frac{2}{15}\)

= \(\frac{6}{15} = \frac{2}{5}\)

(b) 10 persons to choose 8

Number of ways = \(^{10}C_{8}\)

= \(\frac{10!}{(10 - 8)! 8!}\)

= \(\frac{10 \times 9}{2}\)

= 45 ways.

Correct Answer: C. 291

Explanation

Note that every 4- digit and 5- digit numbers formed is greater than 150. Therefore, \(^{5}P_{5} + ^{5}P_{4} = \frac{5!}{(5 - 5)!} + \frac{5!}{(5 - 4)!} = 120 + 120 = 240\) numbers are greater than 150 already.

Among the 3- digit numbers, the numbers 123, 124, 125, 132, 134, 135, 142, 143 and 145 are removed from the ones that meet the criterion so we have:

\(^{5}P_{3} - 9 = \frac{5!}{(5 - 3)!} = 60 - 9 = 51 \implies 240 + 51 = 291\) numbers are greater than 150.

Explanation

\(\log_{3} x - 3\log_{x} 3 + 2 = 0\)

\(\log_{3} x - 3\log_{x} 3 + 2 = 0\)

Let \(\log_{3} x = a\), then \(\log_{x} 3 = \frac{1}{a}\).

\(a - \frac{3}{a} + 2 = 0\)

\(a^{2} + 2a - 3 = 0\)

\(a^{2} - a + 3a - 3 = 0\)

\(a(a - 1) + 3(a - 1) = 0\)

\((a - 1)(a + 3) = 0\)

\(\text{a = 1 or -3}\)

\(\log_{3} x = 1 \implies x = 3^{1} = 3\)

\(\log_{3} x = -3 \implies x = 3^{-3} = \frac{1}{27}\)

Correct Answer: D. \(\frac{-4}{5}\)

Explanation

Given \(\sin \theta = \frac{3}{5} \implies opp = 3, hyp = 5\)

Using Pythagoras' Theorem, we have \( adj = \sqrt{5^{2} - 3^{2}} = \sqrt{16} = 4\)

\(\therefore \cos \theta = \frac{4}{5}, 0° < \theta < 90°\)

In the quadrant where \(180° - \theta\) lies is the 2nd quadrant and here, only \(\sin \theta = +ve\).

\(\therefore \cos (180 - \theta) = -ve = \frac{-4}{5}\)

Correct Answer: D. 672

Explanation

Let the power of \(2x^{2}\) be t and the power of \(\frac{1}{x} \equiv x^{-1}\) = 9 - t.

\((2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}\)

Dealing with x alone, we have

\((x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0\)

\(3t - 9 = 0 \therefore t = 3\)

The binomial expansion is then,

\(^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}\)

= 84 x 8

= 672

Correct Answer: B. \(2x^{2} - 9x + 13 = 0\)

Explanation

Note: Given the sum of the roots and its product, we can get the equation using the formula:

\(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\). This will be used later on in the course of our solution.

Given equation: \(2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6\).

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}\)

\(\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3\)

Given the roots of the new equation as \((\alpha + 1)\) and \((\beta + 1)\), their sum and product will be

\((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}\)

\((\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}\)

The new equation is given by: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)

= \(x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0\)