Year :
1988
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
31 - 40 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 31. |
List the integral values of x which satisfy the inequality -1 < 5 - 2x \(\geq\) 7 A. -1, 0, 1, 2 B. 0, 1, 2, 3 C. -0, 1, 2, 3 D. -1, 0, 2, 3 Detailed Solution-1 < 5 - 2x \(\geq\) 7 = -1 < 5 -2x and 5 - 2x \(\leq\) 7= 2x < 5 + 1 and 5 - 7 \(\leq\) 2x = x < 3 and -1 \(\leq\) x Integral value of x are -1, 0, 1, 2 |
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| 32. |
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is A. (\(\frac{-1}{8}, \frac{19}{24}\)) B. 8, \(\frac{24}{19}\) C. -8, \(\frac{24}{19}\) D. (\(\frac{19}{24}, \frac{-1}{8}\)) Detailed Solution3x - 5y = 3, 2y - 6x = -5-5y + 3x = 3........{i} x 2 2y - 6x = -5.........{ii} x 5 Substituting for x in equation (i) -5y + 3(\(\frac{19}{24}\)) = 3 -5y + 3 x \(\frac{19}{24}\) = 3 -5y = \(\frac{3 - 19}{8}\) -5 = \(\frac{24 - 19}{8}\) = \(\frac{5}{8}\) y = \(\frac{5}{8 \times 5}\) y = \(\frac{-1}{8}\) (x, y) = (\(\frac{19}{24}, \frac{-1}{8}\) |
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| 33. |
The solution of the quadratic equation px2 + qx + b = 0 is A. \(\sqrt{\frac{-b \pm b^2 - 4ac}{2a}}\) B. \(\frac{-b \pm \sqrt{ p^2 - 4pb}}{2a}\) C. \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\) D. \(\frac{-q \pm \sqrt{ p^2 - 4bp}}{2p}\) Detailed Solutionpx2 + qx + b = 0Using almighty formula \(\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}\).........(i) Where a = p, b = q and c = b substitute for this value in equation (i) = \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\) |
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| 34. |
Simplify \(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) A. \(\frac{x}{(x +2)(x - 3)}\) B. \(\frac{2}{(x + 5)(x - 3)}\) C. \(\frac{2}{(x + 1)(x + 3)}\) D. \(\frac{2}{(x - 1)(x - 3)}\) Detailed Solution\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 1)(x + 2)}\) + \(\frac{1}{(x + 1)(x + 1)}\)\(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\) \(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2}{(x + 1)(x + 3)}\) |
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| 35. |
Simplify \(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) A. \(\frac{a - b}{2a + b}\) B. \(\frac{2a + 7b}{a - b}\) C. \(\frac{2a - 7b}{a + b}\) D. \(\frac{2a - 7b}{a - b}\) Detailed Solution\(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) = \(\frac{(2a)^2 - (7b)^2}{(a - b)(2a + 7b)}\)= \(\frac{(2a + 7b)(2a - 7b)}{(a - b)(2a + 7b)}\) = \(\frac{2a - 7b}{a - b}\) |
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| 36. |
If cot \(\theta\) = \(\frac{x}{y}\), find cosec\(\theta\) A. \(\frac{1}{y}\)(x2 + y2) B. \(\frac{x}{y}\) C. \(\frac{1}{y}\)\(\sqrt{x^2 + y^2}\) D. \(\frac{x - y}{y}\) Detailed Solution\(\cot \theta = \frac{x}{y}\)\(\implies \tan \theta = \frac{y}{x}\) \(opp = y; adj = x\) Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\) \(Hyp^{2} = y^{2} + x^{2}\) \(Hyp = \sqrt{y^{2} + x^{2}}\) \(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\) \(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\) = \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\) |
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| 37. |
In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle A. \(\sqrt{3}\)cm B. \(\sqrt{7}\)cm C. 3cm D. 7cm Detailed SolutionPR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120oPR2 = 12 + 22 - 2(1)(2) x - cos 60o = 5 - 2(1)(2) x -\(\frac{1}{2}\) = 5 + 2 = 7 PR = \(\sqrt{7}\)cm |
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| 38. |
If \(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\), find \(\tan \theta\). A. 3 B. \(\frac{3\sqrt{7}}{7}\) C. 3\(\sqrt{7}\) D. \(\sqrt{7}\) Detailed Solution\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i)from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii) Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \) = \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\) \(2\cos^2 \theta = \frac{7}{8}\) \(\cos^2 \theta = \frac{7}{2 \times 8}\) \(\frac{7}{16} = \cos \theta\) \(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\) but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\) \(opp^2 = hyp^2 - adj^2\) \(opp^2 = 4^2 - (\sqrt{7})^{2}\) = 16 - 7 opp = \(\sqrt{9}\) = 3 \(\tan \theta = \frac{\te |
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| 39. |
If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm find the volume of the metal used in making the pipe A. 120\(\pi\)cm3 B. 110\(\pi\)cm3 C. 60\(\pi\)cm3 D. 50\(\pi\)cm3 Detailed SolutionThe volume of the pipe is equal to the area of the cross section and length.let outer and inner radii be R and r respectively. Area of the cross section = (R2 - r2) where R = 6 and r = 6 - 1 = 5cm Area of the cross section = (62 - 52)\(\pi\) = (36 - 25)\(\pi\)cm sq vol. of the pipe = \(\pi\) (R2 - r2)L where length (L) = 10 volume = 11\(\pi\) x 10 = 110\(\pi\)cm3 |
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| 40. |
PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place A. 2.7 B. 4.33 C. 3.1 D. 3.3 Detailed Solution\(\bigtriangleup\)PUS is right angled\(\frac{US}{5}\) = sin60o US = 5 x \(\frac{\sqrt{3}}{2}\) = 2.5\(\sqrt{3}\) = 4.33cm |
| 31. |
List the integral values of x which satisfy the inequality -1 < 5 - 2x \(\geq\) 7 A. -1, 0, 1, 2 B. 0, 1, 2, 3 C. -0, 1, 2, 3 D. -1, 0, 2, 3 Detailed Solution-1 < 5 - 2x \(\geq\) 7 = -1 < 5 -2x and 5 - 2x \(\leq\) 7= 2x < 5 + 1 and 5 - 7 \(\leq\) 2x = x < 3 and -1 \(\leq\) x Integral value of x are -1, 0, 1, 2 |
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| 32. |
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is A. (\(\frac{-1}{8}, \frac{19}{24}\)) B. 8, \(\frac{24}{19}\) C. -8, \(\frac{24}{19}\) D. (\(\frac{19}{24}, \frac{-1}{8}\)) Detailed Solution3x - 5y = 3, 2y - 6x = -5-5y + 3x = 3........{i} x 2 2y - 6x = -5.........{ii} x 5 Substituting for x in equation (i) -5y + 3(\(\frac{19}{24}\)) = 3 -5y + 3 x \(\frac{19}{24}\) = 3 -5y = \(\frac{3 - 19}{8}\) -5 = \(\frac{24 - 19}{8}\) = \(\frac{5}{8}\) y = \(\frac{5}{8 \times 5}\) y = \(\frac{-1}{8}\) (x, y) = (\(\frac{19}{24}, \frac{-1}{8}\) |
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| 33. |
The solution of the quadratic equation px2 + qx + b = 0 is A. \(\sqrt{\frac{-b \pm b^2 - 4ac}{2a}}\) B. \(\frac{-b \pm \sqrt{ p^2 - 4pb}}{2a}\) C. \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\) D. \(\frac{-q \pm \sqrt{ p^2 - 4bp}}{2p}\) Detailed Solutionpx2 + qx + b = 0Using almighty formula \(\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}\).........(i) Where a = p, b = q and c = b substitute for this value in equation (i) = \(\frac{-q \pm \sqrt{ q^2 - 4bp}}{2p}\) |
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| 34. |
Simplify \(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) A. \(\frac{x}{(x +2)(x - 3)}\) B. \(\frac{2}{(x + 5)(x - 3)}\) C. \(\frac{2}{(x + 1)(x + 3)}\) D. \(\frac{2}{(x - 1)(x - 3)}\) Detailed Solution\(\frac{1}{x^2 + 5x + 6}\) + \(\frac{1}{x^2 + 3x + 2}\) = \(\frac{1}{(x + 1)(x + 2)}\) + \(\frac{1}{(x + 1)(x + 1)}\)\(\frac{(x + 1)+ (x + 3)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{x + 1 + x + 3}{(x + 1)(x + 2)(x + 3)}\) \(\frac{2x + 4}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2(x + 2)}{(x + 1)(x + 2)(x + 3)}\) = \(\frac{2}{(x + 1)(x + 3)}\) |
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| 35. |
Simplify \(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) A. \(\frac{a - b}{2a + b}\) B. \(\frac{2a + 7b}{a - b}\) C. \(\frac{2a - 7b}{a + b}\) D. \(\frac{2a - 7b}{a - b}\) Detailed Solution\(\frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2}\) = \(\frac{(2a)^2 - (7b)^2}{(a - b)(2a + 7b)}\)= \(\frac{(2a + 7b)(2a - 7b)}{(a - b)(2a + 7b)}\) = \(\frac{2a - 7b}{a - b}\) |
| 36. |
If cot \(\theta\) = \(\frac{x}{y}\), find cosec\(\theta\) A. \(\frac{1}{y}\)(x2 + y2) B. \(\frac{x}{y}\) C. \(\frac{1}{y}\)\(\sqrt{x^2 + y^2}\) D. \(\frac{x - y}{y}\) Detailed Solution\(\cot \theta = \frac{x}{y}\)\(\implies \tan \theta = \frac{y}{x}\) \(opp = y; adj = x\) Using Pythagoras theorem, \(Hyp^{2} = Opp^{2} + Adj^{2}\) \(Hyp^{2} = y^{2} + x^{2}\) \(Hyp = \sqrt{y^{2} + x^{2}}\) \(\sin \theta = \frac{y}{\sqrt{y^{2} + x^{2}}}\) \(\therefore \csc \theta = \frac{\sqrt{y^{2} + x^{2}}}{y}\) = \(\frac{1}{y}(\sqrt{y^{2} + x^{2}})\) |
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| 37. |
In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle A. \(\sqrt{3}\)cm B. \(\sqrt{7}\)cm C. 3cm D. 7cm Detailed SolutionPR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120oPR2 = 12 + 22 - 2(1)(2) x - cos 60o = 5 - 2(1)(2) x -\(\frac{1}{2}\) = 5 + 2 = 7 PR = \(\sqrt{7}\)cm |
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| 38. |
If \(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\), find \(\tan \theta\). A. 3 B. \(\frac{3\sqrt{7}}{7}\) C. 3\(\sqrt{7}\) D. \(\sqrt{7}\) Detailed Solution\(\cos^2 \theta + \frac{1}{8} = \sin^2 \theta\)..........(i)from trigometric ratios for an acute angle, where \(cos^{2} \theta + \sin^2 \theta = 1\) ........(ii) Substitute for equation (i) in (i) = \(\cos^2 \theta + \frac{1}{8} = 1 - \cos^2 \theta \) = \(\cos^2 \theta + \cos^2 \theta = 1 - \frac{1}{8}\) \(2\cos^2 \theta = \frac{7}{8}\) \(\cos^2 \theta = \frac{7}{2 \times 8}\) \(\frac{7}{16} = \cos \theta\) \(\sqrt{\frac{7}{16}}\) = \(\frac{\sqrt{7}}{4}\) but cos \(\theta\) = \(\frac{\text{adj}}{\text{hyp}}\) \(opp^2 = hyp^2 - adj^2\) \(opp^2 = 4^2 - (\sqrt{7})^{2}\) = 16 - 7 opp = \(\sqrt{9}\) = 3 \(\tan \theta = \frac{\te |
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| 39. |
If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm find the volume of the metal used in making the pipe A. 120\(\pi\)cm3 B. 110\(\pi\)cm3 C. 60\(\pi\)cm3 D. 50\(\pi\)cm3 Detailed SolutionThe volume of the pipe is equal to the area of the cross section and length.let outer and inner radii be R and r respectively. Area of the cross section = (R2 - r2) where R = 6 and r = 6 - 1 = 5cm Area of the cross section = (62 - 52)\(\pi\) = (36 - 25)\(\pi\)cm sq vol. of the pipe = \(\pi\) (R2 - r2)L where length (L) = 10 volume = 11\(\pi\) x 10 = 110\(\pi\)cm3 |
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| 40. |
PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place A. 2.7 B. 4.33 C. 3.1 D. 3.3 Detailed Solution\(\bigtriangleup\)PUS is right angled\(\frac{US}{5}\) = sin60o US = 5 x \(\frac{\sqrt{3}}{2}\) = 2.5\(\sqrt{3}\) = 4.33cm |