Year :
1991
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Find the range of values of x which satisfy the inequality \(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\) < 1 A. x < \(\frac{12}{13}\) B. x < 13 C. x < 9 D. \(\frac{13}{12}\) Detailed Solution\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1= \(\frac{6x + 4x + 3x < 12}{12}\) i.e. 13 x < 12 = x < \(\frac{12}{13}\) |
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| 22. |
Find the positive number n, such that thrice its square is equal to twelve times the number A. 1 B. 2 C. 3 D. 4 Detailed Solution3n2 = 12n= 3n2 - 12n = 0 = 3n(n - 4) = 0 ∴ n = 4 |
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| 23. |
What is the nth term of the progression 27, 9, 3,......? A. 27\(\frac{1}{3}\) n - 1 B. 3n + 2 C. 27 + 18(n - 1) D. 27 + 6(n - 1) Detailed SolutionGiven 27, 9, 3,......this is a G.Pr = \(\frac{9}{27}\) = \(\frac{1}{3}\) T = arn - 1 = 27\(\frac{1}{3}\) n - 1 |
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| 24. |
Solve the equation (x - 2) (x - 3) = 12 A. 2, 3 B. 3, 6 C. -1, 6 D. 1, -6 Detailed Solution(x - 2) (x - 3) = 12x2 - 3x - 2x + 6 = 12 x2 - 5x - 6 = 0 (x +1)(x - 6) = 0 x = -1 or 6 |
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| 25. |
Simplify \(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\) A. -2x - 2\(\sqrt{x (1 + x)}\) B. 1 + 2x + 2\(\sqrt{x (1 + x)}\) C. \(\sqrt{x (1 + x)}\) D. 1 + 2x - 2\(\sqrt{x (1 + x)}\) Detailed Solution\(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)= \((\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}) (\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}})\) = \(\frac{(1 + x) + \sqrt{x(1 + x)} + \sqrt{x(1 + x)} + x}{(1 + x) - x}\) = \(\frac{1 + 2x + 2\sqrt{x(1 + x)}}{1}\) = \(1 + 2x + 2\sqrt{x(1 + x)}\) |
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| 26. |
Evaluate x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) A. (x2 - 1)-\(\frac{1}{2}\) B. (x2 - 1)1 C. (x2 - 1) D. (x2 - 1)-1 Detailed Solutionx2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\)= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\) = \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\) = (x2 - 1)-\(\frac{1}{2}\) |
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| 27. |
Find the gradient of the line passing through the points (-2, 0) and (0, -4) A. 2 B. -4 C. -2 D. 4 Detailed SolutionGiven (-2, 0) and (0, -4)Gradient = \(\frac{-4 - 0}{0 - (-2)}\) = \(\frac{-4}{2}\) = -2 |
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| 28. |
At what value of x is the function y = x2 - 2x - 3 minimum? A. 1 B. -1 C. -4 D. 4 Detailed SolutionFor y = ax2 - bx + c for minimum y\(\frac{dy}{dx}\) = 2x - 2 = \(\frac{dy}{dx}\) = 0 ∴ 2x - 2 = 0 x = 1 |
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| 29. |
Find the sum of the first 20 terms in an arithmetic progression whose first term is 7 and last term is 117. A. 2480 B. 1240 C. 620 D. 124 Detailed SolutionGiven the first and last term of an A.P, the sum of the terms is given by\(S_{n} = \frac{n}{2} [a + l]\) where a = first term; l = last term and n = number of terms. \(\therefore S_{20} = \frac{20}{2} [7 + 117]\) = \(10 (124)\) = 1240 |
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| 30. |
The area of a square is 144 sq cm. Find the length of its diagonal A. 11\(\sqrt{3cm}\) B. 12cm C. 12\(\sqrt{2cm}\) D. 13cm Detailed SolutionBD = \(\sqrt{x^2 + x^2}\)= \(\sqrt{12^2 + 12^2}\) = \(\sqrt{144 + 144}\) = 2(144) = 12\(\sqrt{2cm}\) |
| 21. |
Find the range of values of x which satisfy the inequality \(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\) < 1 A. x < \(\frac{12}{13}\) B. x < 13 C. x < 9 D. \(\frac{13}{12}\) Detailed Solution\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1= \(\frac{6x + 4x + 3x < 12}{12}\) i.e. 13 x < 12 = x < \(\frac{12}{13}\) |
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| 22. |
Find the positive number n, such that thrice its square is equal to twelve times the number A. 1 B. 2 C. 3 D. 4 Detailed Solution3n2 = 12n= 3n2 - 12n = 0 = 3n(n - 4) = 0 ∴ n = 4 |
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| 23. |
What is the nth term of the progression 27, 9, 3,......? A. 27\(\frac{1}{3}\) n - 1 B. 3n + 2 C. 27 + 18(n - 1) D. 27 + 6(n - 1) Detailed SolutionGiven 27, 9, 3,......this is a G.Pr = \(\frac{9}{27}\) = \(\frac{1}{3}\) T = arn - 1 = 27\(\frac{1}{3}\) n - 1 |
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| 24. |
Solve the equation (x - 2) (x - 3) = 12 A. 2, 3 B. 3, 6 C. -1, 6 D. 1, -6 Detailed Solution(x - 2) (x - 3) = 12x2 - 3x - 2x + 6 = 12 x2 - 5x - 6 = 0 (x +1)(x - 6) = 0 x = -1 or 6 |
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| 25. |
Simplify \(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\) A. -2x - 2\(\sqrt{x (1 + x)}\) B. 1 + 2x + 2\(\sqrt{x (1 + x)}\) C. \(\sqrt{x (1 + x)}\) D. 1 + 2x - 2\(\sqrt{x (1 + x)}\) Detailed Solution\(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)= \((\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}) (\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}})\) = \(\frac{(1 + x) + \sqrt{x(1 + x)} + \sqrt{x(1 + x)} + x}{(1 + x) - x}\) = \(\frac{1 + 2x + 2\sqrt{x(1 + x)}}{1}\) = \(1 + 2x + 2\sqrt{x(1 + x)}\) |
| 26. |
Evaluate x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) A. (x2 - 1)-\(\frac{1}{2}\) B. (x2 - 1)1 C. (x2 - 1) D. (x2 - 1)-1 Detailed Solutionx2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\)= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\) = \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\) = (x2 - 1)-\(\frac{1}{2}\) |
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| 27. |
Find the gradient of the line passing through the points (-2, 0) and (0, -4) A. 2 B. -4 C. -2 D. 4 Detailed SolutionGiven (-2, 0) and (0, -4)Gradient = \(\frac{-4 - 0}{0 - (-2)}\) = \(\frac{-4}{2}\) = -2 |
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| 28. |
At what value of x is the function y = x2 - 2x - 3 minimum? A. 1 B. -1 C. -4 D. 4 Detailed SolutionFor y = ax2 - bx + c for minimum y\(\frac{dy}{dx}\) = 2x - 2 = \(\frac{dy}{dx}\) = 0 ∴ 2x - 2 = 0 x = 1 |
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| 29. |
Find the sum of the first 20 terms in an arithmetic progression whose first term is 7 and last term is 117. A. 2480 B. 1240 C. 620 D. 124 Detailed SolutionGiven the first and last term of an A.P, the sum of the terms is given by\(S_{n} = \frac{n}{2} [a + l]\) where a = first term; l = last term and n = number of terms. \(\therefore S_{20} = \frac{20}{2} [7 + 117]\) = \(10 (124)\) = 1240 |
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| 30. |
The area of a square is 144 sq cm. Find the length of its diagonal A. 11\(\sqrt{3cm}\) B. 12cm C. 12\(\sqrt{2cm}\) D. 13cm Detailed SolutionBD = \(\sqrt{x^2 + x^2}\)= \(\sqrt{12^2 + 12^2}\) = \(\sqrt{144 + 144}\) = 2(144) = 12\(\sqrt{2cm}\) |