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
Users' Answers & CommentsGradient 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
Users' Answers & Commentsfind the values of m and n respectively
(\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
Users' Answers & Comments(\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})
Users' Answers & Comments0.4, −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
Users' Answers & Comments1 − ((\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})
Users' Answers & Comments(2x2 - x + 1) × (3 - 2x);
3(2x2 - x + 1) - 2x (2x2 - x + 1)
6x2 - 3x + 3 - 4x3 + 2x2 - 2x
-4x3 + 8x2 -5x + 3
Users' Answers & Comments