Year :
2000
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
3y = 4x - 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K. A. -4/3 B. -3/4 C. 3/4 D. 4/3 Detailed SolutionGrad of 3y = 4x - 1y = 4x/3 - 1/3 Grad = 4/3 Grad of Ky = x + 3 y = x/k + 3/4 Grad = 1/k Since two lines are perpendicular, 1/k = -3/4 -3k = 4 k = -4/3 |
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| 22. |
If P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is A. A straight line B. a circle C. the bisector of angle PXQ D. the perpendicular bisector of PQ |
D |
| 23. |
In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon. A. 8 B. 6 C. 4 D. 3 Detailed Solution2x + x = 180°, => 3x = 180°, and thus x = 60°Each exterior angle = 60° but size of ext. angle = 360°/n Therefore 60° = 360°/n n = 360°/60° = 6 sides |
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| 24. |
A predator moves in a circle of radius √2 centre (0,0), while a prey moves along the line y = x. If 0 \(\leq\) x \(\leq\) 2, at which point(s) will they meet? A. (1,1) only B. (1,1) and (1,2) C. (0,0) and (1,1) D. (√2,√2) only Detailed Solutionx2 + y2 = (√2)2x2 + y2 = 2 but y = x Thus; x2 + x2 = 2 2x2 = 2 x2 = + or - 1 But x2 + y2 = 2 12 + y2 = 2 1 + y2 = 2 y2 = 2 - 1 y2 = 1 y = + or - 1 Thus point (x,y) = (1,1) only. |
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| 25. |
Find the value of \(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta\) A. π B. π/2 C. -π/2 D. -π Detailed Solution\(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta = \int^{\pi}_{0}\frac{-sin^{2}\theta}{sin^{2}\theta}\\ = \int^{\pi}_{0}d\theta = -\pi\) |
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| 26. |
If y = 2x - sin2x, find dy/dx when x = π/4 A. π B. -π C. π/2 D. -π/2 Detailed Solutiony = 2x cos2x - sin2xdy/dx = 2 cos2x +(-2x sin2x) - 2 cos2x = 2 cos2x - 2x sin2x - cos2x = -2x sin2x = -2 x (π/4) sin2 x (π/4) = -(π/2) x 1 = -(π/2) |
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| 27. |
A bowl is designed by revolving completely the area enclosed by y = x2 - 1, y = 3 and x ≥ 0 around the axis. What is the volume of this bowl? A. 7π cubic units B. 15π/2 cubic units C. 8π cubic units D. 17π/2 cubic units Detailed Solution∫30 π(y+1) dy = π[y2 + y30= π(9/2 + 3) = 15π/2 |
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| 28. |
If the volume of a hemisphere is increasing at a steady rate of 18π m\(^{3}\) s\(^{-1}\), at what rate is its radius changing when its is 6m? A. 2.50m/s B. 2.00 m/s C. 0.25 m/s D. 0.20 m/s Detailed Solution\(V = \frac{2}{3} \pi r^{3}\)Given: \(\frac{\mathrm d V}{\mathrm d t} = 18\pi m^{3} s^{-1}\) \(\frac{\mathrm d V}{\mathrm d t} = \frac{\mathrm d V}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(18\pi = 2\pi r^{2} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d r}{\mathrm d t} = \frac{18\pi}{2\pi r^{2}} = \frac{9}{r^{2}}\) The rate of change of the radius when r = 6m, \(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\) = \(0.25 ms^{-1}\) |
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| 29. |
X and Y are two events. The probability of X or Y is 0.7 and that of X is 0.4. If X and Y are independent, find the probability of Y. A. 0.30 B. 0.50 C. 0.57 D. 1.80 Detailed SolutionP (X or Y) = P(X) + P(Y), when they are independent as given.0.7 = 0.4 + P(Y) P(Y) = 0.7 - 0.4 = 0.30 |
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| 30. |
If the mean of the numbers 0, (x+2), (3x+6), and (4x+8) is 4, find their mean deviation. A. 0 B. 2 C. 3 D. 4 Detailed SolutionMean = {0 + (x+2) + (3x+6) + (4x+8)}/4 = 4=> 0 + (x+2) + (3x+6) + (4x+8) = 16 8x + 16 = 16 x = 0 Now prepare a table showing the deviation of each of 0, (x+2), (3x+6) and (4x+8), adding the deviations will give 12. Thus M.D = 12/4 = 3 |
| 21. |
3y = 4x - 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K. A. -4/3 B. -3/4 C. 3/4 D. 4/3 Detailed SolutionGrad of 3y = 4x - 1y = 4x/3 - 1/3 Grad = 4/3 Grad of Ky = x + 3 y = x/k + 3/4 Grad = 1/k Since two lines are perpendicular, 1/k = -3/4 -3k = 4 k = -4/3 |
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| 22. |
If P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is A. A straight line B. a circle C. the bisector of angle PXQ D. the perpendicular bisector of PQ |
D |
| 23. |
In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon. A. 8 B. 6 C. 4 D. 3 Detailed Solution2x + x = 180°, => 3x = 180°, and thus x = 60°Each exterior angle = 60° but size of ext. angle = 360°/n Therefore 60° = 360°/n n = 360°/60° = 6 sides |
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| 24. |
A predator moves in a circle of radius √2 centre (0,0), while a prey moves along the line y = x. If 0 \(\leq\) x \(\leq\) 2, at which point(s) will they meet? A. (1,1) only B. (1,1) and (1,2) C. (0,0) and (1,1) D. (√2,√2) only Detailed Solutionx2 + y2 = (√2)2x2 + y2 = 2 but y = x Thus; x2 + x2 = 2 2x2 = 2 x2 = + or - 1 But x2 + y2 = 2 12 + y2 = 2 1 + y2 = 2 y2 = 2 - 1 y2 = 1 y = + or - 1 Thus point (x,y) = (1,1) only. |
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| 25. |
Find the value of \(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta\) A. π B. π/2 C. -π/2 D. -π Detailed Solution\(\int^{\pi}_{0}\frac{cos^{2}\theta-1}{sin^{2}\theta}d\theta = \int^{\pi}_{0}\frac{-sin^{2}\theta}{sin^{2}\theta}\\ = \int^{\pi}_{0}d\theta = -\pi\) |
| 26. |
If y = 2x - sin2x, find dy/dx when x = π/4 A. π B. -π C. π/2 D. -π/2 Detailed Solutiony = 2x cos2x - sin2xdy/dx = 2 cos2x +(-2x sin2x) - 2 cos2x = 2 cos2x - 2x sin2x - cos2x = -2x sin2x = -2 x (π/4) sin2 x (π/4) = -(π/2) x 1 = -(π/2) |
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| 27. |
A bowl is designed by revolving completely the area enclosed by y = x2 - 1, y = 3 and x ≥ 0 around the axis. What is the volume of this bowl? A. 7π cubic units B. 15π/2 cubic units C. 8π cubic units D. 17π/2 cubic units Detailed Solution∫30 π(y+1) dy = π[y2 + y30= π(9/2 + 3) = 15π/2 |
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| 28. |
If the volume of a hemisphere is increasing at a steady rate of 18π m\(^{3}\) s\(^{-1}\), at what rate is its radius changing when its is 6m? A. 2.50m/s B. 2.00 m/s C. 0.25 m/s D. 0.20 m/s Detailed Solution\(V = \frac{2}{3} \pi r^{3}\)Given: \(\frac{\mathrm d V}{\mathrm d t} = 18\pi m^{3} s^{-1}\) \(\frac{\mathrm d V}{\mathrm d t} = \frac{\mathrm d V}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(18\pi = 2\pi r^{2} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d r}{\mathrm d t} = \frac{18\pi}{2\pi r^{2}} = \frac{9}{r^{2}}\) The rate of change of the radius when r = 6m, \(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\) = \(0.25 ms^{-1}\) |
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| 29. |
X and Y are two events. The probability of X or Y is 0.7 and that of X is 0.4. If X and Y are independent, find the probability of Y. A. 0.30 B. 0.50 C. 0.57 D. 1.80 Detailed SolutionP (X or Y) = P(X) + P(Y), when they are independent as given.0.7 = 0.4 + P(Y) P(Y) = 0.7 - 0.4 = 0.30 |
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| 30. |
If the mean of the numbers 0, (x+2), (3x+6), and (4x+8) is 4, find their mean deviation. A. 0 B. 2 C. 3 D. 4 Detailed SolutionMean = {0 + (x+2) + (3x+6) + (4x+8)}/4 = 4=> 0 + (x+2) + (3x+6) + (4x+8) = 16 8x + 16 = 16 x = 0 Now prepare a table showing the deviation of each of 0, (x+2), (3x+6) and (4x+8), adding the deviations will give 12. Thus M.D = 12/4 = 3 |