Year :
1999
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 45 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Find the tangent to the acute angle between the lines 2x + y = 3 and 3x - 2y = 5. A. -7/4 B. 7/8 C. 7/4 D. 7/2 Detailed SolutionLet \(\phi\) be the angle between the two lines.tan \(\phi\) = \(\frac{m_1 - m_2}{1 + m_1 m_2}\) where m\(_1\) = slope of line 1; m\(_2\) = slope of line 2. Line 1: 2x + y = 3 \(\implies\) y = 3 - 2x. Line 2: 3x - 2y = 5 \(\implies\) -2y = 5 - 3x. y = \(\frac{3}{2}\)x - \(\frac{5}{2}\). m\(_1\) = -2, m\(_2\) = \(\frac{3}{2}\). tan \(\phi\) = \(\frac{-2 - \frac{3}{2}}{1 + (-2 \times \frac{3}{2})}\) = \(\frac{\frac{-7}{2}}{-2}\) \(\therefore\) Tan \(\phi\) = \(\frac{7}{4}\). |
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| 22. |
From a point P, the bearings of two points Q and R are N670W and N230E respectively. If the bearing of R from Q is N680E and PQ = 150m, calculate PR A. 120m B. 140m C. 150m D. 160m |
C |
| 23. |
Find the equation of the locus of a point P(x,y) such that PV = PW, where V = (1,1) and W = (3,5) A. 2x + 2y = 9 B. 2x + 3y = 8 C. 2x + y = 9 D. x + 2y = 8 Detailed SolutionThe locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance.PV = PW \(\sqrt{(x-1)^{2} + (y-1)^{2}} = \sqrt{(x-3)^{2} + (y-5)^{2}}\). Squaring both sides of the equation, (x-1)2 + (y-1)2 = (x-3)2 + (y-5)2. x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25 Collecting like terms and solving, x + 2y = 8. |
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| 24. |
Find the area bounded by the curve y = x(2-x). The x-axis, x = 0 and x = 2. A. 4 sq units B. 2 sq units C. \(\frac{4}{3}sq\hspace{1 mm}units\) D. \(\frac{1}{3}sq\hspace{1 mm}units\) Detailed Solution\(y = x(2-x) \Rightarrow y= 2x - x^{2};\int^{2}_{0}(2x-x^{2} = (x^{2}-\frac{x{3}}{3})^{2}\\ solving further gives (4 - \frac{1}{3} * 8) - (0) = \frac{4}{3} sq\hspace{1 mm}unit\) |
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| 25. |
Evaluate: \(\int^{z}_{0}(sin x - cos x) dx \hspace{1mm} A. \(\sqrt{2 +1}\) B. \(\sqrt{2 }-1\) C. \(-\sqrt{2 }-1\) D. \(1-\sqrt{2}\) |
B |
| 26. |
Find the volume of solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the x-axis. A. 81 π cubic units B. 36 π cubic units C. 18 π cubic units D. 9 π cubic units Detailed Solution\(y = 2x \\ V = \int\pi^{2}dy \\ but\hspace{1mm}y = 2x \\ V = \int\pi4x^{2}dx\\ V = \frac{4(3)^{3}\pi}{3}-\frac{4(3)^{3}\pi}{3}\\V=\frac{4*27\pi}{3} = 36\pi \hspace{1mm}cubic\hspace{1mm}units\) |
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| 27. |
What is the derivative of t2 sin (3t - 5) with respect to t? A. 6t cos (3t - 5) B. 2t sin (3t - 5) - 3t2 cos (3t - 5) C. 2t sin (3t - 5) + 3t2 cos (3t - 5) D. 2t sin (3t - 5) + t2 cos 3t Detailed Solutiont2 sin (3t - 5) = 2t sin ( 3t - 5) + t2 x 3 cos (3t - 5) = 2t sin (3t - 5) + 3t2 cos (3t - 5).Detailed answer below was provided by Ifechuks, a female prospective student of Okopoly. |
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| 28. |
Evaluate \(\int^{1}_{-2}(x-1)^{2}dx\) A. \(\frac{-10}{3}\) B. 7 C. 9 D. 11 |
C |
| 29. |
Find the value of x for which the function y = x3 - x has a minimum value. A. \(-\sqrt{3}\) B. \(-\sqrt{\frac{3}{3}}\) C. \(\sqrt{\frac{3}{3}}\) D. \(\sqrt{3}\) |
C |
| 30. |
If the minimum value of y = 1 + hx - 3x2 is 13, find h. A. 13 B. 12 C. 11 D. 10 |
B |
| 21. |
Find the tangent to the acute angle between the lines 2x + y = 3 and 3x - 2y = 5. A. -7/4 B. 7/8 C. 7/4 D. 7/2 Detailed SolutionLet \(\phi\) be the angle between the two lines.tan \(\phi\) = \(\frac{m_1 - m_2}{1 + m_1 m_2}\) where m\(_1\) = slope of line 1; m\(_2\) = slope of line 2. Line 1: 2x + y = 3 \(\implies\) y = 3 - 2x. Line 2: 3x - 2y = 5 \(\implies\) -2y = 5 - 3x. y = \(\frac{3}{2}\)x - \(\frac{5}{2}\). m\(_1\) = -2, m\(_2\) = \(\frac{3}{2}\). tan \(\phi\) = \(\frac{-2 - \frac{3}{2}}{1 + (-2 \times \frac{3}{2})}\) = \(\frac{\frac{-7}{2}}{-2}\) \(\therefore\) Tan \(\phi\) = \(\frac{7}{4}\). |
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| 22. |
From a point P, the bearings of two points Q and R are N670W and N230E respectively. If the bearing of R from Q is N680E and PQ = 150m, calculate PR A. 120m B. 140m C. 150m D. 160m |
C |
| 23. |
Find the equation of the locus of a point P(x,y) such that PV = PW, where V = (1,1) and W = (3,5) A. 2x + 2y = 9 B. 2x + 3y = 8 C. 2x + y = 9 D. x + 2y = 8 Detailed SolutionThe locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance.PV = PW \(\sqrt{(x-1)^{2} + (y-1)^{2}} = \sqrt{(x-3)^{2} + (y-5)^{2}}\). Squaring both sides of the equation, (x-1)2 + (y-1)2 = (x-3)2 + (y-5)2. x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25 Collecting like terms and solving, x + 2y = 8. |
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| 24. |
Find the area bounded by the curve y = x(2-x). The x-axis, x = 0 and x = 2. A. 4 sq units B. 2 sq units C. \(\frac{4}{3}sq\hspace{1 mm}units\) D. \(\frac{1}{3}sq\hspace{1 mm}units\) Detailed Solution\(y = x(2-x) \Rightarrow y= 2x - x^{2};\int^{2}_{0}(2x-x^{2} = (x^{2}-\frac{x{3}}{3})^{2}\\ solving further gives (4 - \frac{1}{3} * 8) - (0) = \frac{4}{3} sq\hspace{1 mm}unit\) |
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| 25. |
Evaluate: \(\int^{z}_{0}(sin x - cos x) dx \hspace{1mm} A. \(\sqrt{2 +1}\) B. \(\sqrt{2 }-1\) C. \(-\sqrt{2 }-1\) D. \(1-\sqrt{2}\) |
B |
| 26. |
Find the volume of solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the x-axis. A. 81 π cubic units B. 36 π cubic units C. 18 π cubic units D. 9 π cubic units Detailed Solution\(y = 2x \\ V = \int\pi^{2}dy \\ but\hspace{1mm}y = 2x \\ V = \int\pi4x^{2}dx\\ V = \frac{4(3)^{3}\pi}{3}-\frac{4(3)^{3}\pi}{3}\\V=\frac{4*27\pi}{3} = 36\pi \hspace{1mm}cubic\hspace{1mm}units\) |
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| 27. |
What is the derivative of t2 sin (3t - 5) with respect to t? A. 6t cos (3t - 5) B. 2t sin (3t - 5) - 3t2 cos (3t - 5) C. 2t sin (3t - 5) + 3t2 cos (3t - 5) D. 2t sin (3t - 5) + t2 cos 3t Detailed Solutiont2 sin (3t - 5) = 2t sin ( 3t - 5) + t2 x 3 cos (3t - 5) = 2t sin (3t - 5) + 3t2 cos (3t - 5).Detailed answer below was provided by Ifechuks, a female prospective student of Okopoly. |
|
| 28. |
Evaluate \(\int^{1}_{-2}(x-1)^{2}dx\) A. \(\frac{-10}{3}\) B. 7 C. 9 D. 11 |
C |
| 29. |
Find the value of x for which the function y = x3 - x has a minimum value. A. \(-\sqrt{3}\) B. \(-\sqrt{\frac{3}{3}}\) C. \(\sqrt{\frac{3}{3}}\) D. \(\sqrt{3}\) |
C |
| 30. |
If the minimum value of y = 1 + hx - 3x2 is 13, find h. A. 13 B. 12 C. 11 D. 10 |
B |