2017 - JAMB Mathematics Past Questions and Answers - page 5

x * y is an operation on 3x + 2y − 1

Find 3A − 1

x = 3, y = −1

3 * − 1 on 3x + 2y − 1

3(3) + 2(−1) −1

= 9 − 2 − 1

= 6

Gradient of line joining points (3, 2), (1, 4)

Gradient = \(\frac{\text{Change in Y}}{\text{Change in X}}\)

= \(\frac{y_2 - Y_1}{x_2 - x_1}\)\)

(X1, Y1) = (3, 2)

(X2, Y2) = (1, 4)

Gradient = \(\frac{4 − 2}{1 + 3}\)

= \(\frac{2}{-2}\)

= −1

(3√64a³)\(^{-1}\)

\(\frac{1}{(3√64a^3)

= \(\frac{1}{4a}\)

\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= m + n√6

\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)


\(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\)

\(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\)

= \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\)

= \(\frac{0 - 4 \sqrt{6} - 6}{5}\)

= \(\frac{10 - 5 \sqrt{6}}{5}\)

= − 2 + √6

∴ m + n\(\sqrt{6}\) = − 2 + √6

m = − 2, n = 1

\(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) = \(\frac{\beta -\alpha}{\alpha \beta}\)

3x² + 5x + 5x − 2 = 0.

Sum of root = α + β

Product of root = αβ

x² + \(\frac{5x}{3}\) − \(\frac{2}{3}\) = 0

αβ = − \(\frac{-2}{3}\)

α + β = \(\frac{5}{3}\)

∴ \(\frac{\alpha + \beta}{\alpha \beta}\) = − \(\frac{\frac{5}{3}}{\frac{2}{3}}}\)


= − \(\frac{2}{3}\) × \(\frac{3}{3}\)

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

0.4, −0.4, 0.3, 0.47, −0.53, 0.2, −0.2

Range is the difference between the highest and lowest value

i.e Highest − Lowest

− 0.53, −0.4, −0.2, 0.2, 0.3, 0.4, 0.47

0.47 is the highest

− 0.53 is the lowest

∴ = 0.47 − (− 0.53)

∴0.47 + 0.53

= 1.0

1 − (\(\frac{1}{5}\) + 1\(\frac{2}{3}\)) + (5 + 1\(\frac{2}{3}\))

1 − (\(\frac{1}{5}\) × \(\frac{5}{3}\)) + (5 + \(\frac{5}{3}\))

1 − \(\frac{1}{3}\) + \(\frac{20}{3}\)

= \(\frac{22}{3}\)

(2x² - x + 1) × (3 - 2x);

3(2x² - x + 1) - 2x (2x² - x + 1)

6x² - 3x + 3 - 4x³ + 2x² - 2x

-4x³ + 8x² -5x + 3