Mathematics (Core), 1997, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1997

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

11 - 20 of 45 Questions

#	Question	Ans
11.	Find the minimum value of X² - 3x + 2 for all real values of x A. -\(\frac{1}{4}\) B. -\(\frac{1}{2}\) C. \(\frac{1}{4}\) D. \(\frac{1}{2}\) Show Content Detailed Solution y = X² - 3x + 2, \(\frac{dy}{dx}\) = 2x - 3 at turning pt, \(\frac{dy}{dx}\) = 0 ∴ 2x - 3 = 0 ∴ x = \(\frac{3}{2}\) \(\frac{d^2y}{dx^2}\) = \(\frac{d}{dx}\)(\(\frac{d}{dx}\)) = 270 ∴ y_min = 2\(\frac{3}{2}\) - 3\(\frac{3}{2}\) + 2 = \(\frac{9}{4}\) - \(\frac{9}{2}\) + 2 = -\(\frac{1}{4}\)	A
12.	Make F the subject of the formula t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\) A. \(\frac{gv-t^2}{gt^2}\) B. \(\frac{gt^2}{gv-t^2}\) C. \(\frac{v}{\frac{1}{t^2} - \frac{1}{g}}\) D. \(\frac{gv}{t^2 - g}\) Show Content Detailed Solution t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\) t² = \(\frac{v}{\frac{1}{f} + \frac{1}{g}}\) = \(\frac{vfg}{ftg}\) \(\frac{1}{f} + \frac{1}{g}\) = \(\frac{v}{t^2}\) = (g + f)t² = vfg gt² = vfg - ft² gt² = f(vg - t²) f = \(\frac{gt^2}{gv-t^2}\)	B
13.	What value of g will make the expression 4x² - 18xy + g a perfect square? A. 9 B. \(\frac{9y^2}{4}\) C. 81y^2 D. \(\frac{18y^2}{4}\) Show Content Detailed Solution 4x² - 18xy + g = g \(\to\) (\(\frac{18y}{4}\))² = \(\frac{18y^2}{4}\)	D
14.	Find the value of k if \(\frac{5 + 2r}{(r + 1)(r - 2)}\) expressed in partial fraction is \(\frac{k}{r - 2}\) + \(\frac{L}{r + 1}\) where K and L are constants A. 3 B. 2 C. 1 D. -1 Show Content Detailed Solution 5 + 2r = k(r + 1) + L(r - 2) but r - 2 = 0 and r = 2 9 = 3k k = 3	A
15.	Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1 A. x < - \(\frac{3}{4}\) B. x > - \(\frac{4}{3}\) C. x > - \(\frac{3}{4}\) D. x > - 12 Show Content Detailed Solution \(\frac{f(x)}{g(x)}\) < 1 ∴ \(\frac{2x +4}{6x + 7}\) < 1 = 2x + 4 < 6x + 7 = 6x + 7 > 2x + 4 = 6x - 2x > 4 - 7 = 4x > -3 ∴ x > -\(\frac{3}{4}\)	C
16.	Find the range of values of x which satisfies the inequality 12x² < x + 1 A. -\(\frac{1}{4}\) < x < \(\frac{1}{3}\) B. \(\frac{1}{4}\) < x < -\(\frac{1}{3}\) C. -\(\frac{1}{3}\) < x < \(\frac{1}{4}\) D. -\(\frac{1}{4}\) < x < - \(\frac{1}{3}\) Show Content Detailed Solution 12x² < x + 1 12 - x - 1 < 0 12x² - 4x + 3x - 1 < 0 4x(3x - 1) + (3x - 1) < 0 Case 1 (+, -) 4x + 1 > 0, 3x - 1 < 0 x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\) Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0 -\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)	A
17.	Sn is the sum of the first n terms of a series given by Sn = n\(^2\) - 1. Find the nth term A. 4n + 1 B. 4n - 1 C. 2n + 1 D. 2n - 1 Show Content Detailed Solution \(S_{n} = n^{2} - 1\) \(T_{n} = S_{n} - S_{n - 1}\) \(S_{n - 1} = (n - 1)^{2} - 1\) = \(n^{2} - 2n + 1 - 1\) = \(n^{2} - 2n\) \(S_{n} - S_{n - 1} = (n^{2} - 1) - (n^{2} - 2n)\) = \(2n - 1\)	D
18.	The nth term of a sequence is given 3^{1 - n} , find the sum of the first terms of the sequence. A. \(\frac{13}{9}\) B. 1 C. \(\frac{1}{3}\) D. \(\frac{1}{9}\) Show Content Detailed Solution Tn = 3^{1 - n} S₃ = 3^{1 - 1} + 3^{1 - 2} + 3^{1 - 3} = 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\) = \(\frac{13}{9}\)	A
19.	Two binary operations \(\ast\) and \(\oplus\) are defines as m \(\ast\) n = mn - n - 1 and m \(\oplus\) n = mn + n - 2 for all real numbers m, n. Find the value of 3 \(\oplus\) (4 \(\ast\) 50) A. 60 B. 57 C. 54 D. 42 Show Content Detailed Solution m \(\ast\) n = mn - n - 1, m \(\oplus\) n = mn + n - 2 3 \(\oplus\) (4 \(\ast\) 5) = 3 \(\oplus\) (4 x 5 - 5 - 1) = 3 \(\oplus\) 14 3 \(\oplus\) 14 = 3 x 14 + 14 - 2 = 54	C
20.	If X \(\ast\) Y = X + Y - XY, find x when (x \(\ast\) 2) + (x \(\ast\) 3) = 68 A. 24 B. 22 C. -12 D. -21 Show Content Detailed Solution x \(\ast\) y = x + y - xy (x \(\ast\) 2) + (x \(\ast\) 3) = 68 = x + 2 - 2x + x + 3 - 3x = 86 3x = 63 x = -21	D

11.	Find the minimum value of X² - 3x + 2 for all real values of x A. -\(\frac{1}{4}\) B. -\(\frac{1}{2}\) C. \(\frac{1}{4}\) D. \(\frac{1}{2}\) Show Content Detailed Solution y = X² - 3x + 2, \(\frac{dy}{dx}\) = 2x - 3 at turning pt, \(\frac{dy}{dx}\) = 0 ∴ 2x - 3 = 0 ∴ x = \(\frac{3}{2}\) \(\frac{d^2y}{dx^2}\) = \(\frac{d}{dx}\)(\(\frac{d}{dx}\)) = 270 ∴ y_min = 2\(\frac{3}{2}\) - 3\(\frac{3}{2}\) + 2 = \(\frac{9}{4}\) - \(\frac{9}{2}\) + 2 = -\(\frac{1}{4}\)	A
12.	Make F the subject of the formula t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\) A. \(\frac{gv-t^2}{gt^2}\) B. \(\frac{gt^2}{gv-t^2}\) C. \(\frac{v}{\frac{1}{t^2} - \frac{1}{g}}\) D. \(\frac{gv}{t^2 - g}\) Show Content Detailed Solution t = \(\sqrt{\frac{v}{\frac{1}{f} + \frac{1}{g}}}\) t² = \(\frac{v}{\frac{1}{f} + \frac{1}{g}}\) = \(\frac{vfg}{ftg}\) \(\frac{1}{f} + \frac{1}{g}\) = \(\frac{v}{t^2}\) = (g + f)t² = vfg gt² = vfg - ft² gt² = f(vg - t²) f = \(\frac{gt^2}{gv-t^2}\)	B
13.	What value of g will make the expression 4x² - 18xy + g a perfect square? A. 9 B. \(\frac{9y^2}{4}\) C. 81y^2 D. \(\frac{18y^2}{4}\) Show Content Detailed Solution 4x² - 18xy + g = g \(\to\) (\(\frac{18y}{4}\))² = \(\frac{18y^2}{4}\)	D
14.	Find the value of k if \(\frac{5 + 2r}{(r + 1)(r - 2)}\) expressed in partial fraction is \(\frac{k}{r - 2}\) + \(\frac{L}{r + 1}\) where K and L are constants A. 3 B. 2 C. 1 D. -1 Show Content Detailed Solution 5 + 2r = k(r + 1) + L(r - 2) but r - 2 = 0 and r = 2 9 = 3k k = 3	A
15.	Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1 A. x < - \(\frac{3}{4}\) B. x > - \(\frac{4}{3}\) C. x > - \(\frac{3}{4}\) D. x > - 12 Show Content Detailed Solution \(\frac{f(x)}{g(x)}\) < 1 ∴ \(\frac{2x +4}{6x + 7}\) < 1 = 2x + 4 < 6x + 7 = 6x + 7 > 2x + 4 = 6x - 2x > 4 - 7 = 4x > -3 ∴ x > -\(\frac{3}{4}\)	C

16.	Find the range of values of x which satisfies the inequality 12x² < x + 1 A. -\(\frac{1}{4}\) < x < \(\frac{1}{3}\) B. \(\frac{1}{4}\) < x < -\(\frac{1}{3}\) C. -\(\frac{1}{3}\) < x < \(\frac{1}{4}\) D. -\(\frac{1}{4}\) < x < - \(\frac{1}{3}\) Show Content Detailed Solution 12x² < x + 1 12 - x - 1 < 0 12x² - 4x + 3x - 1 < 0 4x(3x - 1) + (3x - 1) < 0 Case 1 (+, -) 4x + 1 > 0, 3x - 1 < 0 x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\) Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0 -\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)	A
17.	Sn is the sum of the first n terms of a series given by Sn = n\(^2\) - 1. Find the nth term A. 4n + 1 B. 4n - 1 C. 2n + 1 D. 2n - 1 Show Content Detailed Solution \(S_{n} = n^{2} - 1\) \(T_{n} = S_{n} - S_{n - 1}\) \(S_{n - 1} = (n - 1)^{2} - 1\) = \(n^{2} - 2n + 1 - 1\) = \(n^{2} - 2n\) \(S_{n} - S_{n - 1} = (n^{2} - 1) - (n^{2} - 2n)\) = \(2n - 1\)	D
18.	The nth term of a sequence is given 3^{1 - n} , find the sum of the first terms of the sequence. A. \(\frac{13}{9}\) B. 1 C. \(\frac{1}{3}\) D. \(\frac{1}{9}\) Show Content Detailed Solution Tn = 3^{1 - n} S₃ = 3^{1 - 1} + 3^{1 - 2} + 3^{1 - 3} = 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\) = \(\frac{13}{9}\)	A
19.	Two binary operations \(\ast\) and \(\oplus\) are defines as m \(\ast\) n = mn - n - 1 and m \(\oplus\) n = mn + n - 2 for all real numbers m, n. Find the value of 3 \(\oplus\) (4 \(\ast\) 50) A. 60 B. 57 C. 54 D. 42 Show Content Detailed Solution m \(\ast\) n = mn - n - 1, m \(\oplus\) n = mn + n - 2 3 \(\oplus\) (4 \(\ast\) 5) = 3 \(\oplus\) (4 x 5 - 5 - 1) = 3 \(\oplus\) 14 3 \(\oplus\) 14 = 3 x 14 + 14 - 2 = 54	C
20.	If X \(\ast\) Y = X + Y - XY, find x when (x \(\ast\) 2) + (x \(\ast\) 3) = 68 A. 24 B. 22 C. -12 D. -21 Show Content Detailed Solution x \(\ast\) y = x + y - xy (x \(\ast\) 2) + (x \(\ast\) 3) = 68 = x + 2 - 2x + x + 3 - 3x = 86 3x = 63 x = -21	D