Mathematics (Core), 1993, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1993

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

11 - 20 of 51 Questions

#	Question	Ans
11.	Make x the subject of the relation \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) A. \(\frac{p + q}{a(p - q)}\) B. \(\frac{p - q}{a(p + q)}\) C. \(\frac{p - q}{apq}\) D. \(\frac{pq}{a(p - q)}\) Show Content Detailed Solution \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication, q(1 + ax) = p(1 - ax) q + qax = p - pax qax + pax = p - q ∴ x = \(\frac{p - q}{a(p + q)}\)	B
12.	Which of the following is a factor of 15 + 7x - 2x² A. x + 3 B. x - 3 C. x - 5 D. x + 5 Show Content Detailed Solution Factorize 15 + 7x - 2x² (5 - x)(3 + 2x); suppose 15 + 7x = 2x² = 0 ∴ (5 - x)(3 + 2x) = 0 x = 5 or x = -\(\frac{3}{2}\) Since 5 is a root, then (x - 5) is a factor	C
13.	Evaluate (x + \(\frac{1}{x}\) + 1)² - (x + \(\frac{1}{x}\) + 1)² A. 4x² B. (\(\frac{2}{x}\) + 2)² C. 4 D. 4(1 + x) Show Content Detailed Solution (x + \(\frac{1}{x}\) + 1)² - (x + \(\frac{1}{x}\) + 1)² = (x + \(\frac{1}{x}\) + 1 + x + \(\frac{-1}{x}\) - 1)(x - \(\frac{1}{x}\) + 1 - x + \(\frac{1}{x}\) + 1) = (2x) (2 + \(\frac{2}{x}\)) = 2x x 2(1 + \(\frac{1}{x}\)) 4x (1 + \(\frac{1}{x}\)) = 4x + 4 = 4(1 + x)	D
14.	Solve the following simultaneous equation for x. x² + y - 5 = 0, y - 7x + 3 = 0 A. -2, 4 B. 2, 4 C. -1,8 D. 1, -8 Show Content Detailed Solution x² + y - 5 = 0.....(i) y - 7x + 3 = 0.........(ii) y = 7x - 3, substituting the value of y in equation (i) x² + (7x - 3) - 5 = 0 x² + 7x + 3 = 0 (x + 8)(x - 1) = 0 x = -8 or 1	D
15.	Solve the following equation (3x - 2)(5x - 4) = (3x - 2)² A. \(\frac{-2}{3}\), 1 B. 1 C. \(\frac{2}{3}\), 1 D. \(\frac{2}{3}\), \(\frac{4}{5}\) Show Content Detailed Solution (3x - 2)(5x - 4) = (3x - 2)² = 5x² - 22x + 6 = 9x² = 12x + 4 6x² - 10x + 4 = 0 6x² - 6x - 4x + 4 = 0 6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0 x = 1 or \(\frac{2}{3}\)	C
16.	If the function f is defined by f(x + 2) = 2x\(^2\) + 7x - 5, find f(-1) A. -10 B. -8 C. 4 D. 10 Show Content Detailed Solution f(x + 2) = f(-1) f(-1) is gotten when x = -3. \(2x^{2} + 7x - 5\) at x = -3. f(-1) = \(2(-3)^{2} + 7(-3) - 5\) = \(18 - 21 - 5 = -8\)	B
17.	Divide the expression x³ + 7x² - x - 7 by -1 + x² A. -x³ + 7x² - x - 7 B. -x³ = 7x + 7 C. x - 7 D. x + 7	D
18.	Simplify \(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) A. \(\frac{1}{p - q}\) B. \(\frac{-1}{p + q}\) C. \(\frac{1}{pq}\) D. \(\frac{1}{pq(p - q)}\) Show Content Detailed Solution \(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) ÷ \(\frac{p^2 - q^2}{pq}\) \(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 - q^2}\) \(\frac{-(p - q)}{(p + q)(p - q)}\) = \(\frac{-1}{p + q}\)	B
19.	Solve the inequality y² - 3y > 18 A. -3 < y < 6 B. y < -3 or y > 6 C. y > -3 or y > 6 D. y < 3 or y < 6 Show Content Detailed Solution y² - 3y > 18 = 3y - 18 > 0 y² - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0 = (y + 3) (y - 6) > 0 Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0 = y > -3 y > 6 Case 2 (-, -) \(\to\) (y + 3) < 0, (y - 6) < 0 = y < -3, y < 6 Combining solution in case 1 and Case 2 = x < -3y < 6 = -3 < y < 6	A
20.	If x is negative, what is the range of values of x within which \(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) A. 3 < x < 4 B. -4 < x < -3 C. -2 < x < -1 D. -3 < x < 0 Show Content Detailed Solution \(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\) = (x + 1)(x + 3)² > 3(x + 3) = (x + 1)[x² + 6x + 9] > 3(x + 3) x³ + 7x² + 15x + 9 > 3x + 9 = x³ + 7x² + 12x > 0 = x(x + 3)9x + 4) > 0 Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0 = x > -4 (solution only) Case 2 (+, -, -) = x > 0, x + 4 < 0 = x > 0, x < -3, x < -4 = x < -3(solution only) Case 3 (	B

11.	Make x the subject of the relation \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) A. \(\frac{p + q}{a(p - q)}\) B. \(\frac{p - q}{a(p + q)}\) C. \(\frac{p - q}{apq}\) D. \(\frac{pq}{a(p - q)}\) Show Content Detailed Solution \(\frac{1 + ax}{1 - ax}\) = \(\frac{p}{q}\) by cross multiplication, q(1 + ax) = p(1 - ax) q + qax = p - pax qax + pax = p - q ∴ x = \(\frac{p - q}{a(p + q)}\)	B
12.	Which of the following is a factor of 15 + 7x - 2x² A. x + 3 B. x - 3 C. x - 5 D. x + 5 Show Content Detailed Solution Factorize 15 + 7x - 2x² (5 - x)(3 + 2x); suppose 15 + 7x = 2x² = 0 ∴ (5 - x)(3 + 2x) = 0 x = 5 or x = -\(\frac{3}{2}\) Since 5 is a root, then (x - 5) is a factor	C
13.	Evaluate (x + \(\frac{1}{x}\) + 1)² - (x + \(\frac{1}{x}\) + 1)² A. 4x² B. (\(\frac{2}{x}\) + 2)² C. 4 D. 4(1 + x) Show Content Detailed Solution (x + \(\frac{1}{x}\) + 1)² - (x + \(\frac{1}{x}\) + 1)² = (x + \(\frac{1}{x}\) + 1 + x + \(\frac{-1}{x}\) - 1)(x - \(\frac{1}{x}\) + 1 - x + \(\frac{1}{x}\) + 1) = (2x) (2 + \(\frac{2}{x}\)) = 2x x 2(1 + \(\frac{1}{x}\)) 4x (1 + \(\frac{1}{x}\)) = 4x + 4 = 4(1 + x)	D
14.	Solve the following simultaneous equation for x. x² + y - 5 = 0, y - 7x + 3 = 0 A. -2, 4 B. 2, 4 C. -1,8 D. 1, -8 Show Content Detailed Solution x² + y - 5 = 0.....(i) y - 7x + 3 = 0.........(ii) y = 7x - 3, substituting the value of y in equation (i) x² + (7x - 3) - 5 = 0 x² + 7x + 3 = 0 (x + 8)(x - 1) = 0 x = -8 or 1	D
15.	Solve the following equation (3x - 2)(5x - 4) = (3x - 2)² A. \(\frac{-2}{3}\), 1 B. 1 C. \(\frac{2}{3}\), 1 D. \(\frac{2}{3}\), \(\frac{4}{5}\) Show Content Detailed Solution (3x - 2)(5x - 4) = (3x - 2)² = 5x² - 22x + 6 = 9x² = 12x + 4 6x² - 10x + 4 = 0 6x² - 6x - 4x + 4 = 0 6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0 x = 1 or \(\frac{2}{3}\)	C

16.	If the function f is defined by f(x + 2) = 2x\(^2\) + 7x - 5, find f(-1) A. -10 B. -8 C. 4 D. 10 Show Content Detailed Solution f(x + 2) = f(-1) f(-1) is gotten when x = -3. \(2x^{2} + 7x - 5\) at x = -3. f(-1) = \(2(-3)^{2} + 7(-3) - 5\) = \(18 - 21 - 5 = -8\)	B
17.	Divide the expression x³ + 7x² - x - 7 by -1 + x² A. -x³ + 7x² - x - 7 B. -x³ = 7x + 7 C. x - 7 D. x + 7	D
18.	Simplify \(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) A. \(\frac{1}{p - q}\) B. \(\frac{-1}{p + q}\) C. \(\frac{1}{pq}\) D. \(\frac{1}{pq(p - q)}\) Show Content Detailed Solution \(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) ÷ \(\frac{p^2 - q^2}{pq}\) \(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 - q^2}\) \(\frac{-(p - q)}{(p + q)(p - q)}\) = \(\frac{-1}{p + q}\)	B
19.	Solve the inequality y² - 3y > 18 A. -3 < y < 6 B. y < -3 or y > 6 C. y > -3 or y > 6 D. y < 3 or y < 6 Show Content Detailed Solution y² - 3y > 18 = 3y - 18 > 0 y² - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0 = (y + 3) (y - 6) > 0 Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0 = y > -3 y > 6 Case 2 (-, -) \(\to\) (y + 3) < 0, (y - 6) < 0 = y < -3, y < 6 Combining solution in case 1 and Case 2 = x < -3y < 6 = -3 < y < 6	A
20.	If x is negative, what is the range of values of x within which \(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) A. 3 < x < 4 B. -4 < x < -3 C. -2 < x < -1 D. -3 < x < 0 Show Content Detailed Solution \(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\) = (x + 1)(x + 3)² > 3(x + 3) = (x + 1)[x² + 6x + 9] > 3(x + 3) x³ + 7x² + 15x + 9 > 3x + 9 = x³ + 7x² + 12x > 0 = x(x + 3)9x + 4) > 0 Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0 = x > -4 (solution only) Case 2 (+, -, -) = x > 0, x + 4 < 0 = x > 0, x < -3, x < -4 = x < -3(solution only) Case 3 (	B