Year :
1986
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Solve the equation 3x2 + 6x - 2 = 0 A. x = -1 \(\pm\) \(\frac{\sqrt{3}}{3}\) B. x = -1 \(\pm\) \(\frac{\sqrt{15}}{3}\) C. x = -2 \(\pm\) 2 D. x = 3 \(\pm\) \(\frac{\sqrt{3}}{15}\) Detailed Solution3x2 + 6x - 2 = 0Using almighty formula i.e. x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\) a = 3, b = 6, c = -2 x = \(\frac{-6 \pm \sqrt{6^2 - 4(3)(-2)}}{2(3)}\) x = \(\frac{-6 \pm \sqrt{36 + 24}}{6}\) x = \(\frac{-6 \pm \sqrt{60}}{6}\) x = \(\frac{-6 \pm \sqrt{4 \times 15}}{6}\) x = \(-1 \pm \frac{\sqrt{15}}{3}\) |
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| 22. |
Simplify \(\frac{1}{5x + 5}\) + \(\frac{1}{7x+ 7}\) A. \(\frac{12}{35x + 1}\) B. \(\frac{1}{35(x + 1)}\) C. \(\frac{12x}{35(x + 7)}\) D. \(\frac{12}{35x + 35}\) Detailed Solution\(\frac{1}{5x + 5}\) + \(\frac{1}{7x+ 7}\) = \(\frac{1}{5(x + 1)}\) + \(\frac{1}{7(x + 1)}\)= \(\frac{7 + 5}{35(x + 1)}\) = \(\frac{12}{35(x + 1)}\) |
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| 23. |
Factorize (4a + 3)2 - (3a - 2)2 A. (a + 1)(a + 5) B. (a - 5)(7a - 1) C. (a + 5)(7a + 1) D. a(7a + 1) Detailed Solution(4a + 3)2 - (3a - 2)2 = a2 - b2= (a + b)(a - b) = [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)] = [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)] = (7a + 1)(a + 5) ∴ (a + 5)(7a + 1) |
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| 24. |
If \(5^{(x + 2y)} = 5\) and \(4^{(x + 3y)} = 16\), find \(3^{(x + y)}\). A. 7 B. 1 C. 3 D. 27 Detailed Solution\(5^{(x + 2y)} = 5\)∴ x + 2y = 1.....(i) \(4^{(x + 3y)} = 16 = 4^2\) x + 3y = 2 .....(ii) x + 2y = 1.....(i) x + 3y = 2......(ii) y = 1 Substitute y = 1 into equation (i) \(x + 2y = 1 \implies x + 2(1) = 1\) \(x + 2 = 1 \implies x = -1\) \(\therefore 3^{(x + y)} = 3^{(-1 + 1)}\) \(3^{0} = 1\) |
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| 25. |
Simplify \(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\) A. \(\frac{2x}{(x - 2)(x + 2)(x^2 - 4)}\) B. \(\frac{2x}{x^2 - 4}\) C. \(\frac{x}{x^2 - 4}\) D. \(\frac{4x}{x^2 - 4}\) Detailed Solution\(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\)= \(\frac{(x + 2) + (x - 2) + 2x}{(x + 2)(x - 2)}\) = \(\frac{4x}{x^2 - 4}\) |
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| 26. |
Make U the subject of the formula S = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\) A. u = \(\frac{12}{2s^2}\) B. u = \(\frac{12}{2s+ w}\) C. u = \(\frac{12}{2s^2 + w}\) D. u = \(\frac{12}{2s^2}\) + w Detailed SolutionS = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\)S = \(\frac{12 - uw}{2u}\) 2us2 = 12 - uw u(2s2 + w) = 12 u = \(\frac{12}{2s^2 + w}\) |
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| 27. |
Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0 A. 1 and 4 B. -2 and 2 C. 0 and 1 D. -1 and 0 Detailed Solution16x - 5 x 4x + 4 = 0(4x)2 - 5(4x) + 4 = 0 let 4x = y y2 - 5y + 4 = 0 (y - 4)(y - 1) = 0 y = 4 or 1 4x = 4 x = 1 4x = 1 i.e. 4x = 4o, x = 0 ∴ x = 1 or 0 |
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| 28. |
If \(\frac{a}{c}\) = \(\frac{c}{d}\) = k, find the value of \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\) A. 3k2 B. 2k - k2 C. \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\) D. k2 Detailed Solution\(\frac{a}{c}\) = \(\frac{c}{d}\) = k∴ \(\frac{a}{b}\) = bk \(\frac{c}{d}\) = k ∴ c = dk = \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\) = \(\frac{3(bk)^2 - (bk)(dk) + dk^2}{3b^2 - bd + a^2}\) = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\) k = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\) |
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| 29. |
At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1? A. (-2, -3) and(\(\frac{1}{2}\), 0) B. (1, 0), (1, 3) C. (4, 0) and (0,1) D. 2, 0) and (0,1) |
A |
| 30. |
If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\) A. \(\frac{b^2}{a^2}\) B. \(\frac{a^2}{b^2}\) C. \(\frac{a^2 + b^2}{b^2 - a^2}\) D. \(\frac{2a^2 + b^2}{a^2 + b^2}\) Detailed Solutioncos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\)Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\) 1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\) = \(\frac{a^2 + b^2 - a^2}{a^2}\) = \(\frac{b^2}{a^2}\) |
| 21. |
Solve the equation 3x2 + 6x - 2 = 0 A. x = -1 \(\pm\) \(\frac{\sqrt{3}}{3}\) B. x = -1 \(\pm\) \(\frac{\sqrt{15}}{3}\) C. x = -2 \(\pm\) 2 D. x = 3 \(\pm\) \(\frac{\sqrt{3}}{15}\) Detailed Solution3x2 + 6x - 2 = 0Using almighty formula i.e. x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\) a = 3, b = 6, c = -2 x = \(\frac{-6 \pm \sqrt{6^2 - 4(3)(-2)}}{2(3)}\) x = \(\frac{-6 \pm \sqrt{36 + 24}}{6}\) x = \(\frac{-6 \pm \sqrt{60}}{6}\) x = \(\frac{-6 \pm \sqrt{4 \times 15}}{6}\) x = \(-1 \pm \frac{\sqrt{15}}{3}\) |
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| 22. |
Simplify \(\frac{1}{5x + 5}\) + \(\frac{1}{7x+ 7}\) A. \(\frac{12}{35x + 1}\) B. \(\frac{1}{35(x + 1)}\) C. \(\frac{12x}{35(x + 7)}\) D. \(\frac{12}{35x + 35}\) Detailed Solution\(\frac{1}{5x + 5}\) + \(\frac{1}{7x+ 7}\) = \(\frac{1}{5(x + 1)}\) + \(\frac{1}{7(x + 1)}\)= \(\frac{7 + 5}{35(x + 1)}\) = \(\frac{12}{35(x + 1)}\) |
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| 23. |
Factorize (4a + 3)2 - (3a - 2)2 A. (a + 1)(a + 5) B. (a - 5)(7a - 1) C. (a + 5)(7a + 1) D. a(7a + 1) Detailed Solution(4a + 3)2 - (3a - 2)2 = a2 - b2= (a + b)(a - b) = [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)] = [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)] = (7a + 1)(a + 5) ∴ (a + 5)(7a + 1) |
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| 24. |
If \(5^{(x + 2y)} = 5\) and \(4^{(x + 3y)} = 16\), find \(3^{(x + y)}\). A. 7 B. 1 C. 3 D. 27 Detailed Solution\(5^{(x + 2y)} = 5\)∴ x + 2y = 1.....(i) \(4^{(x + 3y)} = 16 = 4^2\) x + 3y = 2 .....(ii) x + 2y = 1.....(i) x + 3y = 2......(ii) y = 1 Substitute y = 1 into equation (i) \(x + 2y = 1 \implies x + 2(1) = 1\) \(x + 2 = 1 \implies x = -1\) \(\therefore 3^{(x + y)} = 3^{(-1 + 1)}\) \(3^{0} = 1\) |
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| 25. |
Simplify \(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\) A. \(\frac{2x}{(x - 2)(x + 2)(x^2 - 4)}\) B. \(\frac{2x}{x^2 - 4}\) C. \(\frac{x}{x^2 - 4}\) D. \(\frac{4x}{x^2 - 4}\) Detailed Solution\(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\)= \(\frac{(x + 2) + (x - 2) + 2x}{(x + 2)(x - 2)}\) = \(\frac{4x}{x^2 - 4}\) |
| 26. |
Make U the subject of the formula S = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\) A. u = \(\frac{12}{2s^2}\) B. u = \(\frac{12}{2s+ w}\) C. u = \(\frac{12}{2s^2 + w}\) D. u = \(\frac{12}{2s^2}\) + w Detailed SolutionS = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\)S = \(\frac{12 - uw}{2u}\) 2us2 = 12 - uw u(2s2 + w) = 12 u = \(\frac{12}{2s^2 + w}\) |
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| 27. |
Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0 A. 1 and 4 B. -2 and 2 C. 0 and 1 D. -1 and 0 Detailed Solution16x - 5 x 4x + 4 = 0(4x)2 - 5(4x) + 4 = 0 let 4x = y y2 - 5y + 4 = 0 (y - 4)(y - 1) = 0 y = 4 or 1 4x = 4 x = 1 4x = 1 i.e. 4x = 4o, x = 0 ∴ x = 1 or 0 |
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| 28. |
If \(\frac{a}{c}\) = \(\frac{c}{d}\) = k, find the value of \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\) A. 3k2 B. 2k - k2 C. \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\) D. k2 Detailed Solution\(\frac{a}{c}\) = \(\frac{c}{d}\) = k∴ \(\frac{a}{b}\) = bk \(\frac{c}{d}\) = k ∴ c = dk = \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\) = \(\frac{3(bk)^2 - (bk)(dk) + dk^2}{3b^2 - bd + a^2}\) = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\) k = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\) |
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| 29. |
At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1? A. (-2, -3) and(\(\frac{1}{2}\), 0) B. (1, 0), (1, 3) C. (4, 0) and (0,1) D. 2, 0) and (0,1) |
A |
| 30. |
If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\) A. \(\frac{b^2}{a^2}\) B. \(\frac{a^2}{b^2}\) C. \(\frac{a^2 + b^2}{b^2 - a^2}\) D. \(\frac{2a^2 + b^2}{a^2 + b^2}\) Detailed Solutioncos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\)Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\) 1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\) = \(\frac{a^2 + b^2 - a^2}{a^2}\) = \(\frac{b^2}{a^2}\) |