Year :
2002
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
A circle with radius 5cm has its radius increasing at the rate of 0.2m/s. What will be the corresponding increase in the area? A. 2π B. 5π C. π D. 4π Detailed SolutionArea of the circle (A) = \(\pi r^{2}\)\(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\) \(r = 5cm\); \(\frac{\mathrm d r}{\mathrm d t} = 0.2\) \(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} . \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d A}{\mathrm d t} = 2\pi \times 5 \times 0.2 = 2\pi\) |
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| 12. |
If \(y = x^2 - \frac{1}{x}\). find dy/dx A. 2x - (1/x2) B. 2x + x2 C. 2x - x2 D. 2x + (1/x2) Detailed Solution\(y = x^{2} - \frac{1}{x} = x^{2} - x^{-1}\)\(\frac{\mathrm d y}{\mathrm d x} = 2x - (- x^{-2})\) = \(2x + \frac{1}{x^{2}}\) |
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| 13. |
Solve for x in the equation x\(^3\) - 5x\(^2\) - x + 5 = 0 A. 1, - 1, or 5 B. 1, 1, or -5 C. -1, 1, or -5 D. 1, 1, or 5 Detailed Solutionx\(^3\) - 5x\(^2\) - x + 5 = 0.\(x^{2}(x - 5) - 1(x - 5) = 0\) \((x^2 - 1)(x - 5) = 0 \implies (x - 1)(x + 1)(x - 5) = 0\) \(\therefore x = 1, -1, 5\) |
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| 14. |
The time taken to do a piece of work is inversely proportional to the number of men employed. if it takes 45 men to do a piece of work in 5 days, how long will it take 25 men? A. 15 days B. 12 days C. 5 days D. 9 days Detailed SolutionThe time (t) to do the work is inversely proportional to the number of workers (n).\(\implies t \propto \frac{1}{n}\) \(t = \frac{k}{n}\) \(5 = \frac{k}{45} \implies k = 45 \times 5 = 225\) \(\therefore t = \frac{225}{n}\) For 25 men, \(t = \frac{225}{25} = 9\) \(\therefore\) 25 men will do the work in 9 days. |
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| 15. |
Find the range of values of x for which \(\frac{(x+2)}{4}-\frac{2x-3}{3}<4\) A. x > -6 B. x > -3 C. x < 8 D. x < 4 Detailed Solution\(\frac{x + 2}{4} - \frac{2x - 3}{3} < 4\)\(\frac{3(x + 2) - 4(2x - 3)}{12} < 4\) \(3x + 6 - 8x + 12 < 48\) \(18 - 5x < 48 \implies -5x < 48 - 18\) \(-5x < 30 \implies x > -6\) |
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| 16. |
Find the maximum value of y in the equation y = 1 - 2x - 3x\(^2\) A. 5/4 B. 5/3 C. 3/4 D. 4/3 Detailed Solution\(y = 1 - 2x - 3x^2\)At maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = -2 - 6x\) \(-2 - 6x = 0 \implies -2 = 6x\) \(x = -\frac{1}{3}\) At x = \(-\frac{1}{3}\), y = \(1 - 2(-\frac{1}{3}) - 3(-\frac{1}{3})^{2}\) = \(1 + \frac{2}{3} - \frac{1}{3}\) = \(\frac{4}{3}\) |
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| 17. |
If the 9th term of an A.P is five times the 5th term, find the relationship between a and d. A. 2a + 2 = 0 B. 3a + 5d = 0 C. a + 3d = 0 D. a + 2d = 0 Detailed Solution\(T_{n} = a + (n - 1) d\) (nth term of an AP).\(T_{9} = 5T_{5}\) \(a + 8d = 5(a + 4d) \implies a + 8d = 5a + 20d\) \(5a - a + 20d - 8d = 0 \implies 4a + 12d = 0\) \(a + 3d = 0\) |
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| 18. |
Make r subject of the formula given that \(\frac{x}{r+a}=\frac{a}{r}\) A. \(\frac{a^{2}}{(x-a)}\) B. \(\frac{a^{2}}{(x+a)}\) C. \(\frac{a}{x-a}\) D. \(\frac{a}{x+a}\) Detailed Solution\(\frac{x}{r + a} = \frac{a}{r}\)\(\implies rx = ar + a^{2}\) \(rx - ra = a^{2} \implies r = \frac{a^2}{x - a}\) |
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| 19. |
The inverse function f(x) = 3x + 4 is A. (x-4)/3 B. (x-5)/5 C. (x+3)/4 D. (x+4)/3 Detailed SolutionLet y = f(x).\(y = 3x + 4 \implies x = \frac{y - 4}{3}\) Replace y with x. \(y = \frac{x - 4}{3} \) |
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| 20. |
If -2 is the solution of the equation 2x + 1 - 3c = 2c + 3x - 7, find the value of c. A. 4 B. 3 C. 2 D. 1 Detailed Solution-2 is the solution implies x = -2.2x + 1 - 3c = 2c + 3x - 7 2(-2) + 1 - 3c = 2c + 3(-2) - 7 -4 + 1 - 3c = 2c - 6 - 7 -3 + 13 = 2c + 3c \(\implies\) c = 2. |
| 11. |
A circle with radius 5cm has its radius increasing at the rate of 0.2m/s. What will be the corresponding increase in the area? A. 2π B. 5π C. π D. 4π Detailed SolutionArea of the circle (A) = \(\pi r^{2}\)\(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\) \(r = 5cm\); \(\frac{\mathrm d r}{\mathrm d t} = 0.2\) \(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} . \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d A}{\mathrm d t} = 2\pi \times 5 \times 0.2 = 2\pi\) |
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| 12. |
If \(y = x^2 - \frac{1}{x}\). find dy/dx A. 2x - (1/x2) B. 2x + x2 C. 2x - x2 D. 2x + (1/x2) Detailed Solution\(y = x^{2} - \frac{1}{x} = x^{2} - x^{-1}\)\(\frac{\mathrm d y}{\mathrm d x} = 2x - (- x^{-2})\) = \(2x + \frac{1}{x^{2}}\) |
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| 13. |
Solve for x in the equation x\(^3\) - 5x\(^2\) - x + 5 = 0 A. 1, - 1, or 5 B. 1, 1, or -5 C. -1, 1, or -5 D. 1, 1, or 5 Detailed Solutionx\(^3\) - 5x\(^2\) - x + 5 = 0.\(x^{2}(x - 5) - 1(x - 5) = 0\) \((x^2 - 1)(x - 5) = 0 \implies (x - 1)(x + 1)(x - 5) = 0\) \(\therefore x = 1, -1, 5\) |
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| 14. |
The time taken to do a piece of work is inversely proportional to the number of men employed. if it takes 45 men to do a piece of work in 5 days, how long will it take 25 men? A. 15 days B. 12 days C. 5 days D. 9 days Detailed SolutionThe time (t) to do the work is inversely proportional to the number of workers (n).\(\implies t \propto \frac{1}{n}\) \(t = \frac{k}{n}\) \(5 = \frac{k}{45} \implies k = 45 \times 5 = 225\) \(\therefore t = \frac{225}{n}\) For 25 men, \(t = \frac{225}{25} = 9\) \(\therefore\) 25 men will do the work in 9 days. |
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| 15. |
Find the range of values of x for which \(\frac{(x+2)}{4}-\frac{2x-3}{3}<4\) A. x > -6 B. x > -3 C. x < 8 D. x < 4 Detailed Solution\(\frac{x + 2}{4} - \frac{2x - 3}{3} < 4\)\(\frac{3(x + 2) - 4(2x - 3)}{12} < 4\) \(3x + 6 - 8x + 12 < 48\) \(18 - 5x < 48 \implies -5x < 48 - 18\) \(-5x < 30 \implies x > -6\) |
| 16. |
Find the maximum value of y in the equation y = 1 - 2x - 3x\(^2\) A. 5/4 B. 5/3 C. 3/4 D. 4/3 Detailed Solution\(y = 1 - 2x - 3x^2\)At maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\) \(\frac{\mathrm d y}{\mathrm d x} = -2 - 6x\) \(-2 - 6x = 0 \implies -2 = 6x\) \(x = -\frac{1}{3}\) At x = \(-\frac{1}{3}\), y = \(1 - 2(-\frac{1}{3}) - 3(-\frac{1}{3})^{2}\) = \(1 + \frac{2}{3} - \frac{1}{3}\) = \(\frac{4}{3}\) |
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| 17. |
If the 9th term of an A.P is five times the 5th term, find the relationship between a and d. A. 2a + 2 = 0 B. 3a + 5d = 0 C. a + 3d = 0 D. a + 2d = 0 Detailed Solution\(T_{n} = a + (n - 1) d\) (nth term of an AP).\(T_{9} = 5T_{5}\) \(a + 8d = 5(a + 4d) \implies a + 8d = 5a + 20d\) \(5a - a + 20d - 8d = 0 \implies 4a + 12d = 0\) \(a + 3d = 0\) |
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| 18. |
Make r subject of the formula given that \(\frac{x}{r+a}=\frac{a}{r}\) A. \(\frac{a^{2}}{(x-a)}\) B. \(\frac{a^{2}}{(x+a)}\) C. \(\frac{a}{x-a}\) D. \(\frac{a}{x+a}\) Detailed Solution\(\frac{x}{r + a} = \frac{a}{r}\)\(\implies rx = ar + a^{2}\) \(rx - ra = a^{2} \implies r = \frac{a^2}{x - a}\) |
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| 19. |
The inverse function f(x) = 3x + 4 is A. (x-4)/3 B. (x-5)/5 C. (x+3)/4 D. (x+4)/3 Detailed SolutionLet y = f(x).\(y = 3x + 4 \implies x = \frac{y - 4}{3}\) Replace y with x. \(y = \frac{x - 4}{3} \) |
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| 20. |
If -2 is the solution of the equation 2x + 1 - 3c = 2c + 3x - 7, find the value of c. A. 4 B. 3 C. 2 D. 1 Detailed Solution-2 is the solution implies x = -2.2x + 1 - 3c = 2c + 3x - 7 2(-2) + 1 - 3c = 2c + 3(-2) - 7 -4 + 1 - 3c = 2c - 6 - 7 -3 + 13 = 2c + 3c \(\implies\) c = 2. |