2019 - JAMB Mathematics Past Questions and Answers - page 7

μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

A = {3, 6, 9, 12, 15, 18}

B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

A ∩ B = {3, 9, 15}

For the committee to include 2 females, we must have 3 males, so that there should be a total of 5 members.

thus, \(^4C_3 \times ^3C_2\)

= \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\)

= 4 × 3

= 12 ways

\(\sqrt{28} + \sqrt{112} - \sqrt{k} = \sqrt{175}\)

\(\sqrt{4 \times 7} + \sqrt{16 \times 7} - \sqrt{k} = \sqrt{25 \times 7}\)

\(2\sqrt{7} + 4\sqrt{7} - \sqrt{k} = 5\sqrt{7}\)

\(6\sqrt{7} - 5\sqrt{7} = \sqrt{k}\)

\(\sqrt{k} = \sqrt{7}\)

\(\implies k = 7\)

\(\frac{2\log_{3} 9 \times \log_{3} 81^{-2}}{\log_{5} 625}\)

= \(\frac{2\log_3 3^2 \times \log_3 (3^4)^{-2}}{\log_5 (5^4)}\)

= \(\frac{4\log_3 3 \times -8\log_3 3}{4\log_5 5}\)

= -8.

\(\frac{2 + 2x}{3} - 2 \geq \frac{4x - 6}{5}\)

\(\frac{2 + 2x - 6}{3} \geq \frac{4x - 6}{5}\)

\(\frac{2x - 4}{3} \geq \frac{4x - 6}{5}\)

\(5(2x - 4) \geq 3(4x - 6)\)

\(10x - 20 \geq 12x - 18\)

\(10x - 12x \geq -18 + 20\)

\(-2x \geq 2\)

\(x \leq -1\)

\(x^2 - 7x + 10 \leq 0\)

Solving for \(x^2 - 7x + 10 = 0\), we have, (x - 5)(x - 2) \(\leq\) 0.

Given conditions:

case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0.

\(\implies\) x \(\leq\) 5; x \(\geq\) 2.

2 \(\leq\) x \(\leq\) 5.

Take x = 3,

3\(^2\) - 7(3) + 10 = 9 - 21 + 10

= -2 \(\leq\) 0.

\(\therefore\) 2 \(\leq\) x \(\leq\) 5.

Given:

q(x) [quotient],

d(x) [divisor]

r(x) [remainder],

the polynomial is determined by multiplying the quotient and the divisor, then adding the remainder.

The polynomial = (x\(^2\) - x - 5)(3x - 1) + 7.

= (3x\(^3\) - x\(^2\) - 3x\(^2\) + x - 15x + 5) + 7

= (3x\(^3\) - 4x\(^2\) - 14x + 5) + 7

= 3x\(^3\) - 4x\(^2\) - 14x + 12

By dividing a polynomial p(x) by (x - a), you get a remainder = p(a)

Then for 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2).

= 2(2)\(^2\) + 2 - 3

= 8 + 2 - 3

= 7.

\(\begin{vmatix} 2 & -5 & 3 \ x & 1 & 4 \ 0 & 3 & 2 \end{vmatrix} = 132\)

\(\implies 2 \begin{vmatrix} 1 & 4 \ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \ 0 & 3 \end{vmatrix} = 132\)

\(2(2 - 12) + 5(2x) + 3(3x) = 132\)

\(-20 + 10x + 9x = 132\)

\(19x = 152\)

\(x = 8\)

\(A = \begin{vmatrix} 3 & -2 \ 1 & 6 \end{vmatrix}\)

|A| = (3 x 6) - (-2 x 1)

= 18 + 2 = 20.

A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \ -1 & 3 \end{vmatrix}\)

= \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)

= \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)