Year :
2001
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 50 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius. A. 6.5 units B. 13.0 units C. 3.5 units D. 7.0 units Detailed SolutionPQ\(^2\) = (x2 - x1)\(^2\) + (y2 - y1)\(^2\)= 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = √169 = 13 But PQ = diameter = 2r, r = PQ/2 = 6.5 units |
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| 22. |
The bearings of P and Q from a common point N are 020° and 300° respectively. If P and Q are also equidistant from N, find the bearing of P from Q. A. 040° B. 070° C. 280° D. 320° Detailed SolutionHint: Simply make a sketch of the question and find the bearing to be 070° |
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| 23. |
A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m? A. 23 m B. 25 m C. 20 m D. 22 m Detailed SolutionCapacity = 3080 m3Base diameter = 14 m, thus radius = 7 m. Volume of cylinder = capacity of cylinder = πr2h. = 3080 22/7 x 72 x h = 3080 h = 3080/(22x7) h = 20 m |
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| 24. |
Find the locus of a point which moves such that its distance from the line y = 4 is a constant, k. A. y = 4 \(\pm\) k B. y = k \(\pm\) 4 C. y = 4 + k D. y = k - 4 |
A |
| 25. |
The chord ST of a circle is equal to the radius, r, of the circle. Find the length of arc ST. A. πr/6 B. πr/2 C. πr/12 D. πr/3 Detailed SolutionSince ST = the radius r, => SOT is equilateral.Each angle = 60° Length of arc ST = θ/360 x 2πr = 60/360 x 2 x π x r = πr/3 |
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| 26. |
If the gradient of the curve y = 2kx2 + x + 1 at x = 1 is 9, find k. A. 4 B. 3 C. 2 D. 1 Detailed Solutiony = 2kx2 + x + 1Gradient function dy/dx = 4kx + 1 at dy/dx = 9, => 9 = 4kx + 1 => 8 = 4kx, and k = 8/4x. At x = 1, k = 8/4 = 2 |
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| 27. |
Evaluate \(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\) A. 3/5(2x-3)5/3 + k B. 6/5(2x-3)5/3 + k C. 2x-3+k D. 2(2x-3)+k Detailed Solution\(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\)Let \(u = 2x - 3\) \(\mathrm d u = 2 \mathrm d x\) = \(\int u^{\frac{2}{3}} \mathrm d u\) = \(\frac{u^{\frac{5}{3}}}{\frac{5}{3}} + k\) = \(\frac{3}{5} (2x - 3)^{\frac{5}{3}} + k\) |
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| 28. |
Differentiate \((2x+5)^{2} (x-4)\) with respect to x. A. 4(2x+5)(x-4) B. 4(2x+5)(4x-3) C. (2x+5)(2x-13) D. (2x+5)(6x-11) Detailed Solution\(y = (2x + 5)^{2} (x - 4)\)\(\frac{\mathrm d y}{\mathrm d x} = (2x + 5)^{2} (1) + (x - 4)(2)(2)(2x + 5)\) = \((2x + 5)(2x + 5 + 4x - 16)\) = \((2x + 5)(6x - 11)\) |
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| 29. |
Find the area bounded by the curves y = 4 - x2 and y = 2x + 1 A. units B. units C. units D. units Detailed SolutionHint:y = 4 - x\(^2\) and y = 2x + 1 => 4 - x2 = 2x + 1 => x2 + 2x - 3 = 0 (x+3)(x-1) = 0 thus x = 1 or x = -3. Integrating x\(^2\) + 2x - 3 = 3x - x\(^2\) - \(\frac{x^3}{3}\) from (1, to -3) : 3 (1) - 1\(^2\) - \(\frac{1^3}{3}\) - 3 (-3) - -3\(^2\) - \(\frac{-3^3}{3}\) = \(\frac{5}{3}\) + 9 will give 32/3 = 10 \(\frac{2}{3}\) |
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| 30. |
Find the rate of change of the volume, V of a sphere with respect to its radius, r when r = 1. A. 12π B. 4π C. 24π D. 8π Detailed SolutionVolume of sphere, V = 4/3 x πr3Rate of change of V = dv/dr Thus if V = 4/3 x πr3, => dv/dr = 4πr2 At r = 1, Rate = 4 x π x 1 = 4π |
| 21. |
P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius. A. 6.5 units B. 13.0 units C. 3.5 units D. 7.0 units Detailed SolutionPQ\(^2\) = (x2 - x1)\(^2\) + (y2 - y1)\(^2\)= 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = √169 = 13 But PQ = diameter = 2r, r = PQ/2 = 6.5 units |
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| 22. |
The bearings of P and Q from a common point N are 020° and 300° respectively. If P and Q are also equidistant from N, find the bearing of P from Q. A. 040° B. 070° C. 280° D. 320° Detailed SolutionHint: Simply make a sketch of the question and find the bearing to be 070° |
|
| 23. |
A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m? A. 23 m B. 25 m C. 20 m D. 22 m Detailed SolutionCapacity = 3080 m3Base diameter = 14 m, thus radius = 7 m. Volume of cylinder = capacity of cylinder = πr2h. = 3080 22/7 x 72 x h = 3080 h = 3080/(22x7) h = 20 m |
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| 24. |
Find the locus of a point which moves such that its distance from the line y = 4 is a constant, k. A. y = 4 \(\pm\) k B. y = k \(\pm\) 4 C. y = 4 + k D. y = k - 4 |
A |
| 25. |
The chord ST of a circle is equal to the radius, r, of the circle. Find the length of arc ST. A. πr/6 B. πr/2 C. πr/12 D. πr/3 Detailed SolutionSince ST = the radius r, => SOT is equilateral.Each angle = 60° Length of arc ST = θ/360 x 2πr = 60/360 x 2 x π x r = πr/3 |
| 26. |
If the gradient of the curve y = 2kx2 + x + 1 at x = 1 is 9, find k. A. 4 B. 3 C. 2 D. 1 Detailed Solutiony = 2kx2 + x + 1Gradient function dy/dx = 4kx + 1 at dy/dx = 9, => 9 = 4kx + 1 => 8 = 4kx, and k = 8/4x. At x = 1, k = 8/4 = 2 |
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| 27. |
Evaluate \(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\) A. 3/5(2x-3)5/3 + k B. 6/5(2x-3)5/3 + k C. 2x-3+k D. 2(2x-3)+k Detailed Solution\(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\)Let \(u = 2x - 3\) \(\mathrm d u = 2 \mathrm d x\) = \(\int u^{\frac{2}{3}} \mathrm d u\) = \(\frac{u^{\frac{5}{3}}}{\frac{5}{3}} + k\) = \(\frac{3}{5} (2x - 3)^{\frac{5}{3}} + k\) |
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| 28. |
Differentiate \((2x+5)^{2} (x-4)\) with respect to x. A. 4(2x+5)(x-4) B. 4(2x+5)(4x-3) C. (2x+5)(2x-13) D. (2x+5)(6x-11) Detailed Solution\(y = (2x + 5)^{2} (x - 4)\)\(\frac{\mathrm d y}{\mathrm d x} = (2x + 5)^{2} (1) + (x - 4)(2)(2)(2x + 5)\) = \((2x + 5)(2x + 5 + 4x - 16)\) = \((2x + 5)(6x - 11)\) |
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| 29. |
Find the area bounded by the curves y = 4 - x2 and y = 2x + 1 A. units B. units C. units D. units Detailed SolutionHint:y = 4 - x\(^2\) and y = 2x + 1 => 4 - x2 = 2x + 1 => x2 + 2x - 3 = 0 (x+3)(x-1) = 0 thus x = 1 or x = -3. Integrating x\(^2\) + 2x - 3 = 3x - x\(^2\) - \(\frac{x^3}{3}\) from (1, to -3) : 3 (1) - 1\(^2\) - \(\frac{1^3}{3}\) - 3 (-3) - -3\(^2\) - \(\frac{-3^3}{3}\) = \(\frac{5}{3}\) + 9 will give 32/3 = 10 \(\frac{2}{3}\) |
|
| 30. |
Find the rate of change of the volume, V of a sphere with respect to its radius, r when r = 1. A. 12π B. 4π C. 24π D. 8π Detailed SolutionVolume of sphere, V = 4/3 x πr3Rate of change of V = dv/dr Thus if V = 4/3 x πr3, => dv/dr = 4πr2 At r = 1, Rate = 4 x π x 1 = 4π |