Mathematics 2019 Jamb Examination

First, list the elements of each set within the universal range: Set A (multiples of 3): {3, 6, 9, 12, 15, 18} Set B (odd numbers): {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} The intersection \( A \cap B \) consists of elements common to both: {3, 9, 15}.

To include exactly 2 females in a committee of 5, we must also select 3 males (since 5 - 2 = 3). Ways to choose 2 females from 3: \( \displaystyle ^3C_2 = 3 \) Ways to choose 3 males from 4: \( \displaystyle ^4C_3 = 4 \) Total ways = \( 3 \times 4 = 12 \).

Simplify each surd into its basic form: \( \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7} \) \( \sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7} \) \( \sqrt{175} = \sqrt{25 \times 7} = 5\sqrt{7} \) Substituting back: \( 2\sqrt{7} + 4\sqrt{7} - \sqrt{k} = 5\sqrt{7} \) \( 6\sqrt{7} - \sqrt{k} = 5\sqrt{7} \implies \sqrt{k} = \sqrt{7} \implies k = 7 \).

Evaluate each part: 1. \( \log_3 9 = 2 \) (since \( 3^2=9 \)) 2. \( \log_3 81^{-2} = -2 \log_3 81 = -2(4) = -8 \) 3. \( \log_5 625 = 4 \) (since \( 5^4=625 \)) Calculation: \( \frac{2(2) \times (-8)}{4} = \frac{4 \times (-8)}{4} = -8 \).

Multiply every term by the LCM of 3 and 5, which is 15: \( 5(2 + 2x) - 15(2) \ge 3(4x - 6) \) \( 10 + 10x - 30 \ge 12x - 18 \) \( 10x - 20 \ge 12x - 18 \) \( -2 \ge 2x \implies -1 \ge x \text{ or } x \le -1 \).

Factor the quadratic: \( (x - 2)(x - 5) \le 0 \). The roots are 2 and 5. For the product of two factors to be less than or equal to zero (negative), the value of x must lie between the roots. Result: \( 2 \le x \le 5 \).

Using the Division Transformation rule: \( P(x) = d(x) \cdot q(x) + r(x) \) \( P(x) = (3x - 1)(x^2 - x - 5) + 7 \) \( P(x) = [3x^3 - 3x^2 - 15x - x^2 + x + 5] + 7 \) \( P(x) = 3x^3 - 4x^2 - 14x + 12 \).

Using the Remainder Theorem, to find the remainder when \( f(x) \) is divided by \( (x - a) \), calculate \( f(a) \). Here, \( a = 2 \): \( f(2) = 2(2)^2 + (2) - 3 \) \( f(2) = 2(4) + 2 - 3 = 8 + 2 - 3 = 7 \).

Expand along the first column: \( 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - x \begin{vmatrix} -5 & 3 \\ 3 & 2 \end{vmatrix} + 0 = 132 \) \( 2(2 - 12) - x(-10 - 9) = 132 \) \( 2(-10) - x(-19) = 132 \) \( -20 + 19x = 132 \implies 19x = 152 \implies x = 8 \).

First find the determinant \( |A| = (3 \times 6) - (-2 \times 1) = 18 + 2 = 20 \). The adjoint is found by swapping leading diagonal elements and negating the others: \( \begin{pmatrix} 6 & 2 \\ -1 & 3 \end{pmatrix} \). Inverse \( A^{-1} = \frac{1}{20} \begin{pmatrix} 6 & 2 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} \frac{3}{10} & \frac{1}{10} \\ -\frac{1}{20} & \frac{3}{20} \end{pmatrix} \).