Year :
2014
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 47 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
If \(\begin{vmatrix}-x & 12 \\-1 & 4 \end{vmatrix} = - 12\), find x. A. -6 B. -2 C. 3 D. 6 Detailed Solution\(\begin{vmatrix}-x & 12 \\-1 & 4\end{vmatrix} = - 12\)-4x - (-1)12 = -12 -4x + 12 = -12 -4x = -12 - 12 -4x = - 24 x = 6 There is an explanation video available below. |
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| 22. |
Find the value of \(\begin{vmatrix}0 & 3 & 2 \\1 & 7 & 8 \\0 & 5 & 4\end{vmatrix}\). A. 12 B. 10 C. -1 D. -2 Detailed Solution\(0 \begin{vmatrix}7 & 8 \\5 & 4\end{vmatrix} -3 \begin{vmatrix}1 & 8 \\0 & 4\end{vmatrix} +2 \begin{vmatrix}1 & 7 \\0 & 5\end{vmatrix}\)= 0(28 - 40) - 3(4 - 0) + 2(5 - 0) = 0(-12) - 3(4) + 2(5) = 0 - 12 + 10 = -2 There is an explanation video available below. |
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| 23. |
How many sides has a regular polygon whose interior angle is 135°? A. 12 B. 10 C. 9 D. 8 Detailed SolutionIf each interior angle of the polygon is 135°, then each exterior angle is 180° - 135° = 45°.Hence, number of sides = \(\frac{360°}{\text{one exterior angle}}\) \(\frac{360°}{45°}\) = 8 There is an explanation video available below. |
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| 24. |
A cylindrical tank has a capacity of 6160m3. What is the depth of the tank if the radius of its base is 28cm? A. 8.0m B. 7.5m C. 5.0m D. 2.5m |
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| 25. |
Find the mid point of S(-5, 4) and T(-3, -2) A. -4, 2 B. 4, -2 C. -4, 1 D. 4, -1 |
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| 26. |
The gradient of a line joining (x,4) and (1,2) is \(\frac{1}{2}\). Find the value of x A. 5 B. 3 C. -3 D. -5 |
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| 27. |
Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis. A. \(\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}\) B. \(\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}\) C. \(\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}\) D. \(\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}\) Detailed Solutiony - 4x + 3 = 0When y = 0, 0 - 4x + 3 = 0 Then -4x = -3 x = 3/4 So the line cuts the x-axis at point (3/4, 0). When x = 0, y - 4(0) + 3 = 0 Then y + 3 = 0 y = -3 So the line cuts the y-axis at the point (0, -3) Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is; \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\) \([\frac{1}{2}(\frac{3}{4}), \frac |
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| 28. |
If \(\sin\theta = \frac{12}{13}\), find the value of \(1 + \cos\theta\) A. \(\frac{25}{13}\) B. \(\frac{18}{13}\) C. \(\frac{8}{13}\) D. \(\frac{5}{13}\) |
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| 29. |
If y = 4x3 - 2x2 + x, find \(\frac{\delta y}{\delta x}\) A. 8x2 - 2x + 1 B. 8x2 - 4x + 1 C. 12x2 - 2x + 1 D. 12x2 - 4x + 1 |
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| 30. |
If y = cos 3x, find \(\frac{\delta y}{\delta x}\) A. \(\frac{1}{3} \sin 3x\) B. \(-\frac{1}{3} \sin 3x\) C. 3 sin 3x D. -3 sin 3x |
| 21. |
If \(\begin{vmatrix}-x & 12 \\-1 & 4 \end{vmatrix} = - 12\), find x. A. -6 B. -2 C. 3 D. 6 Detailed Solution\(\begin{vmatrix}-x & 12 \\-1 & 4\end{vmatrix} = - 12\)-4x - (-1)12 = -12 -4x + 12 = -12 -4x = -12 - 12 -4x = - 24 x = 6 There is an explanation video available below. |
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| 22. |
Find the value of \(\begin{vmatrix}0 & 3 & 2 \\1 & 7 & 8 \\0 & 5 & 4\end{vmatrix}\). A. 12 B. 10 C. -1 D. -2 Detailed Solution\(0 \begin{vmatrix}7 & 8 \\5 & 4\end{vmatrix} -3 \begin{vmatrix}1 & 8 \\0 & 4\end{vmatrix} +2 \begin{vmatrix}1 & 7 \\0 & 5\end{vmatrix}\)= 0(28 - 40) - 3(4 - 0) + 2(5 - 0) = 0(-12) - 3(4) + 2(5) = 0 - 12 + 10 = -2 There is an explanation video available below. |
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| 23. |
How many sides has a regular polygon whose interior angle is 135°? A. 12 B. 10 C. 9 D. 8 Detailed SolutionIf each interior angle of the polygon is 135°, then each exterior angle is 180° - 135° = 45°.Hence, number of sides = \(\frac{360°}{\text{one exterior angle}}\) \(\frac{360°}{45°}\) = 8 There is an explanation video available below. |
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| 24. |
A cylindrical tank has a capacity of 6160m3. What is the depth of the tank if the radius of its base is 28cm? A. 8.0m B. 7.5m C. 5.0m D. 2.5m |
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| 25. |
Find the mid point of S(-5, 4) and T(-3, -2) A. -4, 2 B. 4, -2 C. -4, 1 D. 4, -1 |
| 26. |
The gradient of a line joining (x,4) and (1,2) is \(\frac{1}{2}\). Find the value of x A. 5 B. 3 C. -3 D. -5 |
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| 27. |
Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis. A. \(\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}\) B. \(\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}\) C. \(\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}\) D. \(\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}\) Detailed Solutiony - 4x + 3 = 0When y = 0, 0 - 4x + 3 = 0 Then -4x = -3 x = 3/4 So the line cuts the x-axis at point (3/4, 0). When x = 0, y - 4(0) + 3 = 0 Then y + 3 = 0 y = -3 So the line cuts the y-axis at the point (0, -3) Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is; \([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\) \([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\) \([\frac{1}{2}(\frac{3}{4}), \frac |
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| 28. |
If \(\sin\theta = \frac{12}{13}\), find the value of \(1 + \cos\theta\) A. \(\frac{25}{13}\) B. \(\frac{18}{13}\) C. \(\frac{8}{13}\) D. \(\frac{5}{13}\) |
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| 29. |
If y = 4x3 - 2x2 + x, find \(\frac{\delta y}{\delta x}\) A. 8x2 - 2x + 1 B. 8x2 - 4x + 1 C. 12x2 - 2x + 1 D. 12x2 - 4x + 1 |
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| 30. |
If y = cos 3x, find \(\frac{\delta y}{\delta x}\) A. \(\frac{1}{3} \sin 3x\) B. \(-\frac{1}{3} \sin 3x\) C. 3 sin 3x D. -3 sin 3x |