Year :
1992
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
31 - 40 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 31. |
Evaluate \(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\) A. 7 B. 2 C. 3 D. 4 Detailed Solution\(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)\(\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}\) = \(\frac{(x^{2} + 3x - 2)}{x + 2}\) \(\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}\) = \(\frac{2^{2} + 3(2) - 2}{2 + 2}\) = \(\frac{4 + 6 - 2}{4} = 2\) |
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| 32. |
If y = x sin x, Find \(\frac{d^2 y}{d^2 x}\) A. 2 cosx - x sinx B. sinx + x cosx C. sinx - x cosx D. x sinx - 2 cosx Detailed Solution\(y = x \sin x\)\(\frac{\mathrm d y}{\mathrm d x} = x \cos x + \sin x\) \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = x (- \sin x) + \cos x + \cos x\) = \(2 \cos x - x \sin x\) |
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| 33. |
Ice forms on a refrigerator ice-box at the rate of (4 - 06t)g per minute after t minutes. If initially there are 2g of ice in the box, find the mass of ice formed in 5 minutes A. 19.5 B. 17.0 C. 14.5 D. 12.5 Detailed Solution\(\frac{dm}{dt}\) = 4 - 0.6t\(\int\)dm = \(\int\)(4 - 0.6t)dt m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g ∴ c = 2 m = \(4t - 0.3t^2 + 2\), when t = 5 minutes m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\) = 14.5 |
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| 34. |
Obtain a maximum value of the function f(x) x3 - 12x + 11 A. -5 B. -2 C. 2 D. 27 Detailed Solutionf(x) = x3 - 12x + 11\(\frac{df(x)}{dx)}\) = 3x2 - 12 = 0 ∴ 3x2 - 12 = 0 \(\to\) x2m = 4 x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15 = f(-2) = (-8) + 24 + 11 = 35 - 8 = 27 ∴ maximum value = 27 |
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| 35. |
A student blows a balloon and its volume increases at a rate of \(\pi\)(20 - t2)cm3S-1 after t seconds. If the initial volume is 0 cm3, find the volume of the balloon after 2 seconds A. 37.00\(\pi\) B. 37.33\(\pi\) C. 40.00\(\pi\) D. 42.67\(\pi\) Detailed Solution\(\frac{dv}{dt}\) = \(\pi\)(20 - t2)cm2S-1\(\int\)dv = \(\pi\)(20 - t2)dt V = \(\pi\) \(\int\)(20 - t2)dt V = \(\pi\)(20 \(\frac{t}{3}\) - t3) + c when c = 0, V = (20t - \(\frac{t^3}{3}\)) after t = 2 seconds V = \(\pi\)(40 - \(\frac{8}{3}\) = \(\pi\)\(\frac{120 - 8}{3}\) = \(\frac{112}{3}\) = 37.33\(\pi\) |
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| 36. |
Evaluate the integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}} 2 \cos 2x \mathrm {d} x\) A. -\(\frac{1}{2}\) B. -1 C. \(\frac{1}{2}\) D. 1 Detailed Solution\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)= \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\) = \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\) = \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\) = \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\) = \(1 - \frac{1}{2} = \frac{1}{2}\) |
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| 37. |
A store keeper checked his stock of five commodities and arrived at the following statics A. 216o B. 108o C. 68o D. 62o Detailed SolutionH will represent \(\frac{108}{630}\) x \(\frac{360^o}{1}\) ≈ 62° |
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| 38. |
\(\begin{array}{c|c} x & 2 & 4 & 6 & 8\\ \hline f & 4 & y & 6 & 5 \end{array}\) A. 2, 1 B. 1, 2 C. 1, 5 D. 5, 2 Detailed SolutionMean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)= \(\frac{5.2}{1}\) = \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\) = \(\frac{5.2}{1}\) = \(\frac{84 + 4y}{15 + y}\) = 5.2(15 + y) = 84 + 4y = 5.2 x 15 + 5.2y = 84 + 4y = 78 + 5.2y = 84 = 4y = 5.2y - 4y = 84 - 78 1.2y = 6 y = \(\frac{6}{1.2}\) = \(\frac{60}{12}\) = 5 |
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| 39. |
\(\begin{array}{c|c} \text{No. of children} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of families} & 7 & 11 & 6 & 7 & 7 & 5 & 3 \end{array}\) A. 2, 1 B. 1, 2 C. 1, 5 D. 5, 2 Detailed SolutionFrom the table, the mode = 1.The median = 2. |
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| 40. |
If the scores of 3 students in a test are 5, 6 and 7, find the standard deviation of their scores A. \(\frac{2}{3}\) B. \(\frac{2}{3}\sqrt{3}\) C. \(\sqrt{\frac{2}{3}}\) D. \(\sqrt{\frac{3}{2}}\) Detailed Solution(x) = \(\frac{5 + 6 + 7}{3}\)= \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\\hline 5 & 5 - 6 & 1\\ 6 & 6 - 6 & 0 \\ 7 & 7 - 6 & 1\\ \hline & & 2\end{array}\) S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x) = \(\sqrt{\frac{2}{3}}\) |
| 31. |
Evaluate \(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\) A. 7 B. 2 C. 3 D. 4 Detailed Solution\(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)\(\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}\) = \(\frac{(x^{2} + 3x - 2)}{x + 2}\) \(\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}\) = \(\frac{2^{2} + 3(2) - 2}{2 + 2}\) = \(\frac{4 + 6 - 2}{4} = 2\) |
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| 32. |
If y = x sin x, Find \(\frac{d^2 y}{d^2 x}\) A. 2 cosx - x sinx B. sinx + x cosx C. sinx - x cosx D. x sinx - 2 cosx Detailed Solution\(y = x \sin x\)\(\frac{\mathrm d y}{\mathrm d x} = x \cos x + \sin x\) \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = x (- \sin x) + \cos x + \cos x\) = \(2 \cos x - x \sin x\) |
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| 33. |
Ice forms on a refrigerator ice-box at the rate of (4 - 06t)g per minute after t minutes. If initially there are 2g of ice in the box, find the mass of ice formed in 5 minutes A. 19.5 B. 17.0 C. 14.5 D. 12.5 Detailed Solution\(\frac{dm}{dt}\) = 4 - 0.6t\(\int\)dm = \(\int\)(4 - 0.6t)dt m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g ∴ c = 2 m = \(4t - 0.3t^2 + 2\), when t = 5 minutes m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\) = 14.5 |
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| 34. |
Obtain a maximum value of the function f(x) x3 - 12x + 11 A. -5 B. -2 C. 2 D. 27 Detailed Solutionf(x) = x3 - 12x + 11\(\frac{df(x)}{dx)}\) = 3x2 - 12 = 0 ∴ 3x2 - 12 = 0 \(\to\) x2m = 4 x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15 = f(-2) = (-8) + 24 + 11 = 35 - 8 = 27 ∴ maximum value = 27 |
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| 35. |
A student blows a balloon and its volume increases at a rate of \(\pi\)(20 - t2)cm3S-1 after t seconds. If the initial volume is 0 cm3, find the volume of the balloon after 2 seconds A. 37.00\(\pi\) B. 37.33\(\pi\) C. 40.00\(\pi\) D. 42.67\(\pi\) Detailed Solution\(\frac{dv}{dt}\) = \(\pi\)(20 - t2)cm2S-1\(\int\)dv = \(\pi\)(20 - t2)dt V = \(\pi\) \(\int\)(20 - t2)dt V = \(\pi\)(20 \(\frac{t}{3}\) - t3) + c when c = 0, V = (20t - \(\frac{t^3}{3}\)) after t = 2 seconds V = \(\pi\)(40 - \(\frac{8}{3}\) = \(\pi\)\(\frac{120 - 8}{3}\) = \(\frac{112}{3}\) = 37.33\(\pi\) |
| 36. |
Evaluate the integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{12}} 2 \cos 2x \mathrm {d} x\) A. -\(\frac{1}{2}\) B. -1 C. \(\frac{1}{2}\) D. 1 Detailed Solution\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\)= \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\) = \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\) = \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\) = \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\) = \(1 - \frac{1}{2} = \frac{1}{2}\) |
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| 37. |
A store keeper checked his stock of five commodities and arrived at the following statics A. 216o B. 108o C. 68o D. 62o Detailed SolutionH will represent \(\frac{108}{630}\) x \(\frac{360^o}{1}\) ≈ 62° |
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| 38. |
\(\begin{array}{c|c} x & 2 & 4 & 6 & 8\\ \hline f & 4 & y & 6 & 5 \end{array}\) A. 2, 1 B. 1, 2 C. 1, 5 D. 5, 2 Detailed SolutionMean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)= \(\frac{5.2}{1}\) = \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\) = \(\frac{5.2}{1}\) = \(\frac{84 + 4y}{15 + y}\) = 5.2(15 + y) = 84 + 4y = 5.2 x 15 + 5.2y = 84 + 4y = 78 + 5.2y = 84 = 4y = 5.2y - 4y = 84 - 78 1.2y = 6 y = \(\frac{6}{1.2}\) = \(\frac{60}{12}\) = 5 |
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| 39. |
\(\begin{array}{c|c} \text{No. of children} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of families} & 7 & 11 & 6 & 7 & 7 & 5 & 3 \end{array}\) A. 2, 1 B. 1, 2 C. 1, 5 D. 5, 2 Detailed SolutionFrom the table, the mode = 1.The median = 2. |
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| 40. |
If the scores of 3 students in a test are 5, 6 and 7, find the standard deviation of their scores A. \(\frac{2}{3}\) B. \(\frac{2}{3}\sqrt{3}\) C. \(\sqrt{\frac{2}{3}}\) D. \(\sqrt{\frac{3}{2}}\) Detailed Solution(x) = \(\frac{5 + 6 + 7}{3}\)= \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\\hline 5 & 5 - 6 & 1\\ 6 & 6 - 6 & 0 \\ 7 & 7 - 6 & 1\\ \hline & & 2\end{array}\) S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x) = \(\sqrt{\frac{2}{3}}\) |