Year :
1995
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 46 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Find the sum to infinity of the following sequence \(1, \frac{9}{10}, (\frac{9}{10})^{2}, (\frac{9}{10})^{3}\) A. \(\frac{1}{10}\) B. \(\frac{9}{10}\) C. \(\frac{10}{9}\) D. 10 Detailed Solution\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a G.P)\(a = 1; r = \frac{9}{10}\) \(S_{\infty} = \frac{1}{1 - \frac{9}{10}}\) = \(\frac{1}{\frac{1}{10}} = 10\) |
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| 22. |
\(\begin{vmatrix} -2 & 1 & 1 \\ 2 & 1 & k \\1 & 3 & -1 \end{vmatrix}\) = 23 A. 1 B. 2 C. 3 D. 4 Detailed Solution-2(-1 - 3k) - 1(-2 -k) + 1(6 - 1) = 237k = 14 k = 2 |
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| 23. |
If x = \(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\) and y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\). Find xy. A. \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\) B. \(\begin{pmatrix} 2 & 7 \\ 4 & 17 \end{pmatrix}\) C. \(\begin{pmatrix} 10 & 4 \\ 4 & 6 \end{pmatrix}\) D. \(\begin{pmatrix} 4 & 3 \\ 10 & 9 \end{pmatrix}\) Detailed Solution\(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\); y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\).= \(\begin{pmatrix} 2 + 8 & 1 + 6 \\ 0 + 12 & 0 + 9 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\) |
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| 24. |
In a triangle XYZ, < YXZ = 44° and < XYZ = 112°. Calculate the acute angle between the internal bisectors of < XYZ and < XZY. A. 12o B. 56o C. 68o D. 78o |
C |
| 25. |
Find the distance between two towns P(45°N, 30°W) and Q(15°S, 30°W) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)] A. \(\frac{1100}{3}\)km B. \(\frac{2200}{3}\)km C. \(\frac{22000}{3}\)km D. \(\frac{1100}{3}\)km Detailed SolutionAngular difference = 45° + 15° = 60°.\(D = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 7000\) = \(\frac{22000}{3} km\) |
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| 26. |
Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR A. x + 2y - 1= -0 B. 2x + y - 3 = 0 C. x - 2y - 3 = 0 D. 2x + y - 1 = 0 Detailed SolutionLine PQ : x - 2y + 4 = 02y = x + 4 \(\implies y = \frac{x}{2} + 2 \) Slope = \(\frac{1}{2}\) Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\) Line QR: \(y = mx + b\) \(y = -2x + b\) Point of intersection: (1, -1) \(-1 = -2(1) + b \implies b = -1 + 2 = 1\) \(y = -2x + 1 \implies y + 2x - 1 = 0\) \(QR: 2x + y - 1 = 0\) |
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| 27. |
P is on the locus of points equidiatant from two given points X and Y. UV is a straight line throuh Y parallel to the locus. If < PYU is 40°, find < XPY. A. 100o B. 80 o C. 50 o D. 40 o Detailed Solution< PYU = < YPQ = 40°(Alternative) = PQ\(\parallel\)UV< XPQ = < YPQ = 40°, Since QX = QY therefore < XPY = 80° |
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| 28. |
A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank. A. 60m B. 30 \(\sqrt{3}\)m C. 20 \(\sqrt{3}\)m D. 10 \(\sqrt{3}\)m Detailed Solutionh = 30 tan 60o= 30\(\sqrt{3}\) |
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| 29. |
The derivative of cosec x is A. tan x cosec x B. -cot x cosec x C. tan x sec x D. -cot x sec x Detailed Solution\(\csc x = \frac{1}{\sin x}\)Using the quotient rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\) = \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\) = \(\frac{- \cos x}{\sin^{2} x}\) = \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\) = \(- \cot x \csc x\) |
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| 30. |
For what value of x is the tangent to the curve y = x\(^2\) - 4x + 3 parallel to the x-axis? A. 3 B. 2 C. 1 D. 6 Detailed SolutionAt the point where the tangent is parallel to the x- axis, the slope of the curve = 0.Given: \(y = x^{2} - 4x + 3\) \(\frac{\mathrm d y}{\mathrm d x} = 2x - 4\) \(2x - 4 = 0 \implies 2x = 4 \) \(x = 2\) When x = 2, the tangent of the curve is parallel to the x- axis. |
| 21. |
Find the sum to infinity of the following sequence \(1, \frac{9}{10}, (\frac{9}{10})^{2}, (\frac{9}{10})^{3}\) A. \(\frac{1}{10}\) B. \(\frac{9}{10}\) C. \(\frac{10}{9}\) D. 10 Detailed Solution\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a G.P)\(a = 1; r = \frac{9}{10}\) \(S_{\infty} = \frac{1}{1 - \frac{9}{10}}\) = \(\frac{1}{\frac{1}{10}} = 10\) |
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| 22. |
\(\begin{vmatrix} -2 & 1 & 1 \\ 2 & 1 & k \\1 & 3 & -1 \end{vmatrix}\) = 23 A. 1 B. 2 C. 3 D. 4 Detailed Solution-2(-1 - 3k) - 1(-2 -k) + 1(6 - 1) = 237k = 14 k = 2 |
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| 23. |
If x = \(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\) and y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\). Find xy. A. \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\) B. \(\begin{pmatrix} 2 & 7 \\ 4 & 17 \end{pmatrix}\) C. \(\begin{pmatrix} 10 & 4 \\ 4 & 6 \end{pmatrix}\) D. \(\begin{pmatrix} 4 & 3 \\ 10 & 9 \end{pmatrix}\) Detailed Solution\(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\); y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\).= \(\begin{pmatrix} 2 + 8 & 1 + 6 \\ 0 + 12 & 0 + 9 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\) |
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| 24. |
In a triangle XYZ, < YXZ = 44° and < XYZ = 112°. Calculate the acute angle between the internal bisectors of < XYZ and < XZY. A. 12o B. 56o C. 68o D. 78o |
C |
| 25. |
Find the distance between two towns P(45°N, 30°W) and Q(15°S, 30°W) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)] A. \(\frac{1100}{3}\)km B. \(\frac{2200}{3}\)km C. \(\frac{22000}{3}\)km D. \(\frac{1100}{3}\)km Detailed SolutionAngular difference = 45° + 15° = 60°.\(D = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 7000\) = \(\frac{22000}{3} km\) |
| 26. |
Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR A. x + 2y - 1= -0 B. 2x + y - 3 = 0 C. x - 2y - 3 = 0 D. 2x + y - 1 = 0 Detailed SolutionLine PQ : x - 2y + 4 = 02y = x + 4 \(\implies y = \frac{x}{2} + 2 \) Slope = \(\frac{1}{2}\) Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\) Line QR: \(y = mx + b\) \(y = -2x + b\) Point of intersection: (1, -1) \(-1 = -2(1) + b \implies b = -1 + 2 = 1\) \(y = -2x + 1 \implies y + 2x - 1 = 0\) \(QR: 2x + y - 1 = 0\) |
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| 27. |
P is on the locus of points equidiatant from two given points X and Y. UV is a straight line throuh Y parallel to the locus. If < PYU is 40°, find < XPY. A. 100o B. 80 o C. 50 o D. 40 o Detailed Solution< PYU = < YPQ = 40°(Alternative) = PQ\(\parallel\)UV< XPQ = < YPQ = 40°, Since QX = QY therefore < XPY = 80° |
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| 28. |
A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank. A. 60m B. 30 \(\sqrt{3}\)m C. 20 \(\sqrt{3}\)m D. 10 \(\sqrt{3}\)m Detailed Solutionh = 30 tan 60o= 30\(\sqrt{3}\) |
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| 29. |
The derivative of cosec x is A. tan x cosec x B. -cot x cosec x C. tan x sec x D. -cot x sec x Detailed Solution\(\csc x = \frac{1}{\sin x}\)Using the quotient rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\) = \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\) = \(\frac{- \cos x}{\sin^{2} x}\) = \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\) = \(- \cot x \csc x\) |
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| 30. |
For what value of x is the tangent to the curve y = x\(^2\) - 4x + 3 parallel to the x-axis? A. 3 B. 2 C. 1 D. 6 Detailed SolutionAt the point where the tangent is parallel to the x- axis, the slope of the curve = 0.Given: \(y = x^{2} - 4x + 3\) \(\frac{\mathrm d y}{\mathrm d x} = 2x - 4\) \(2x - 4 = 0 \implies 2x = 4 \) \(x = 2\) When x = 2, the tangent of the curve is parallel to the x- axis. |