Mathematics 2017 Past Questions | JAMB
Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.
41. The operation * on the set R of real number is defined by x * y = 3x + 2y − 1, find 3* − 1
- A. 9
- B. – 9
- C. 6
- D. – 6
Correct Option: Answer is C
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
42. Find the gradient of the line joining the points (3, 2) and (1, 4)
- A. 3/2
- B. 2/1
- C. -1
- D. 3/2
Correct Option: Answer is C
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
43. Simplify (3√64a3)\(^{−1}\)
- A. 4a
- B. \(\frac{1}{8a}\)
- C. 8a
- D. \(\frac{1}{4a}\)
Correct Option: Answer is D
44. If \(\frac{2 \sqrt{3} – \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) = m + n √ 6,
find the values of m and n respectively
- A. 1, − 2
- B.− 2, n = 1
- C. \(\frac{-2}{5}\), 1
- D. \(\frac{2}{3}\)
Correct Option: Answer is B
\(\frac{2 \sqrt{3} – \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= 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
45. If α and β are the roots of the equation 3x2 + bx − 2 = 0. Find the value of \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\)
- A. \(\frac{-5}{3}\)
- B. \(\frac{-2}{3}\)
- C. \(\frac{1}{2}\)
- D. \(\frac{5}{2}\)
Correct Option: Answer is D
\(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) = \(\frac{\beta -\alpha}{\alpha \beta}\)
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 = \(\frac{5}{3}\)
∴ \(\frac{\alpha + \beta}{\alpha \beta}\) = − \(\frac{\frac{5}{3}}{\frac{2}{3}}}\)
= − \(\frac{2}{3}\) × \(\frac{3}{3}\)
= \(\frac{5}{2}\)
46. Find the range of the following set of numbers 0.4, −0.4, 0.3, 0.47, −0.53, 0.2 and −0.2
- A. 1.03
- B. 0.07
- C. 0.03
- D. 1.0
Correct Option: Answer is D
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
47. Evaluate 1 − (\(\frac{1}{5}\) x 1\(\frac{2}{3}\)) + (5 + 1\(\frac{2}{3}\))
- A. 4
- B. 3
- C. \(\frac{22}{3}\)
- D. 3 \(\frac{2}{3}\)
Correct Option: Answer is C
1 − (\(\frac{1}{5}\) x 1\(\frac{2}{3}\)) + (5 + 1\(\frac{2}{3}\))
1 − (\(\frac{1}{5}\) x \(\frac{5}{3}\)) + (5 + \(\frac{5}{3}\))
1 − \(\frac{1}{3}\) + \(\frac{20}{3}\)
= \(\frac{22}{3}\)
48. What is the product of 2x2 − x + 1 and 3 − 2x
- A. 4x3 − 8x2 + 5x + 3
- B. −4x3 + 8x2 − 5x + 3
- C. −4x3 − 8x2 + 5x + 3
- D. 4x3 + 8x2 − 5x + 3
Correct Option: Answer is C
(2x2 – x + 1) × (3 – 2x);
3(2x2 – x + 1) – 2x (2x2 – x + 1)
6x2 – 3x + 3 – 4x3 + 2x2 – 2x
-4x3 + 8x2 -5x + 3