Mathematics (Core), 1994, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1994

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

11 - 20 of 48 Questions

#	Question	Ans
11.	Factorize a²x - b²y - b²x + a²y A. (a - b)(x + y) B. (y - x)(a - b)(a + b) C. (x - y)(a - b)(a + b) D. (x + y)(a - b)(a + b) Show Content Detailed Solution a²x - b²y - b²x + a²y = a²x - b²x - b²y + a²y Rearrange = x(a² - b²) + y(a² - b²) = (x + y)(a² - b²) = (x + y)(a + b)(a + b)	D
12.	Find the values of p and q such that (x - 1)and (x - 3) are factors of px3 + qx2 + 11x - 6 A. -1, -6 B. 1, -6 C. 1, 6 D. 6, -1 Show Content Detailed Solution Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero \(\therefore (x - 1) = 0\) x = 1 Substitute in the polynomial the value x = 1 = \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\) p + q + 5 = 0 .....(i) Also since x - 3 is a factor, \(\therefore\) x - 3 = 0 x = 3 Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\) 27p + 9q = -27 ......(2) Combine eqns. (i) and (ii) Multiply equation (i) by 9 to eliminate q 9p + 9q = -45 Subtract (ii) from (i), \(18p = 18\) \(\therefore\) p = 1 Put p = 1 in (i), \(1 + q = -5 \implies q = -6\) \((p, q) = (1, -6)\)	B
13.	If a = 1, b = 3, solve for x in the equation \(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) A. \(\frac{4}{3}\) B. \(\frac{2}{3}\) C. \(\frac{3}{2}\) D. \(\frac{3}{4}\) Show Content Detailed Solution \(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) \(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\) ∴ 3(1 - x) = x - 3 3 - 3x = x - 3 Rearrange 6 = 4x; x = \(\frac {6}{4}\) = \(\frac{3}{2}\)	C
14.	Solve for r in the following equation \(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) A. 3 B. 4 C. 5 D. 6 Show Content Detailed Solution \(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) Multiply through by r(r -1) which is the LCM = (r)(r + 1) + 2(r)(r - 1) = 3(r - 1)(r + 1) = r² + r + 2r² - 2r 3r² - 3 = 3r² r = 3r² - 3 -r = -3 ∴ r = 3	A
15.	Find P if \(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\) A. \(\frac{-2}{3}\) B. \(\frac{-5}{3}\) C. \(\frac{5}{3}\) D. \(\frac{2}{3}\) Show Content Detailed Solution \(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\) Multiply both sides by LCM i.e. (1 - x(x + 2)) ∴ x - 3 = p(x + 2) + Q(1 - x) When x = +1 (+1) - 3 = p(+1 + 2) + Q(1 - 1) -2 = 3p + 0(Q) 3p = -2 ∴ p = \(\frac{-2}{3}\)	A
16.	Find the range of values of x for which \(\frac{1}{x}\) > 2 is true A. x < \(\frac{1}{2}\) B. x < 0 or x < \(\frac{1}{2}\) C. 0 < x < \(\frac{1}{2}\) D. 1 < x < 2 Show Content Detailed Solution \(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2 x > 2x² = 2x² < x = 2x² - x < 0 = x(2x - 10 < 0 Case 1(+, -) = x > 0, 2x - 1 < 0 x > 0, x < \(\frac{1}{2}\) (solution) Case 2(-, 4) = x < 0, 2x - 1 > 0 x < 0, x , \(\frac{1}{2}\) = 0	C
17.	If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference. A. \(\frac{12}{5}\) B. \(\frac{5}{3}\) C. -2 D. 2 Show Content Detailed Solution In an AP, Tn = a + (n - 1)d T₆ = a + 5d = 11 The first term = a = 1 ∴ T₆ = 1 + 5d = 11 5d = 11 - 1 5d = 10 ∴ d = 2	D
18.	Find the value of log₁₀ r + log₁₀ r² + log₁₀ r⁴ + log₁₀ r⁸ + log₁₀ r¹⁶ + log₁₀ r³² = 63 A. 10^-1 B. 10^o C. 10 D. 10² Show Content Detailed Solution log₁₀ r + log₁₀ r² + log₁₀ r⁴ + log₁₀ r⁸ + log₁₀ r¹⁶ + log₁₀ r³² = 63 log₁₀r⁶³ = 63 ⁶³ = 10⁶³ ∴ r = 10	C
19.	Find the nth term of the sequence 3, 6, 10, 15, 21..... A. \(\frac{n(n - 1)}{2}\) B. \(\frac{n(n + 1)}{2}\) C. \(\frac{(n + 1)(n + 2)}{2}\) D. n(2n + 1) Show Content Detailed Solution \(\frac{(n + 1)(n + 2)}{2}\) If n = 1, the expression becomes 3 n = 2, the expression becomes 6 n = 4, the expression becomes 15 n = 5, the expression becomes 21	C
20.	A binary operation \(\oplus\) is defines on the set of all positive integers by a \(\oplus\) b = ab for all positive integers a, b. Which of the following properties does NOT hold? A. closure B. identity C. positive D. inverse Show Content Detailed Solution a \(\oplus\) b = ab The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab The number 1 is the identity element under multiplication	D

11.	Factorize a²x - b²y - b²x + a²y A. (a - b)(x + y) B. (y - x)(a - b)(a + b) C. (x - y)(a - b)(a + b) D. (x + y)(a - b)(a + b) Show Content Detailed Solution a²x - b²y - b²x + a²y = a²x - b²x - b²y + a²y Rearrange = x(a² - b²) + y(a² - b²) = (x + y)(a² - b²) = (x + y)(a + b)(a + b)	D
12.	Find the values of p and q such that (x - 1)and (x - 3) are factors of px3 + qx2 + 11x - 6 A. -1, -6 B. 1, -6 C. 1, 6 D. 6, -1 Show Content Detailed Solution Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero \(\therefore (x - 1) = 0\) x = 1 Substitute in the polynomial the value x = 1 = \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\) p + q + 5 = 0 .....(i) Also since x - 3 is a factor, \(\therefore\) x - 3 = 0 x = 3 Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\) 27p + 9q = -27 ......(2) Combine eqns. (i) and (ii) Multiply equation (i) by 9 to eliminate q 9p + 9q = -45 Subtract (ii) from (i), \(18p = 18\) \(\therefore\) p = 1 Put p = 1 in (i), \(1 + q = -5 \implies q = -6\) \((p, q) = (1, -6)\)	B
13.	If a = 1, b = 3, solve for x in the equation \(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) A. \(\frac{4}{3}\) B. \(\frac{2}{3}\) C. \(\frac{3}{2}\) D. \(\frac{3}{4}\) Show Content Detailed Solution \(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) \(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\) ∴ 3(1 - x) = x - 3 3 - 3x = x - 3 Rearrange 6 = 4x; x = \(\frac {6}{4}\) = \(\frac{3}{2}\)	C
14.	Solve for r in the following equation \(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) A. 3 B. 4 C. 5 D. 6 Show Content Detailed Solution \(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) Multiply through by r(r -1) which is the LCM = (r)(r + 1) + 2(r)(r - 1) = 3(r - 1)(r + 1) = r² + r + 2r² - 2r 3r² - 3 = 3r² r = 3r² - 3 -r = -3 ∴ r = 3	A
15.	Find P if \(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\) A. \(\frac{-2}{3}\) B. \(\frac{-5}{3}\) C. \(\frac{5}{3}\) D. \(\frac{2}{3}\) Show Content Detailed Solution \(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\) Multiply both sides by LCM i.e. (1 - x(x + 2)) ∴ x - 3 = p(x + 2) + Q(1 - x) When x = +1 (+1) - 3 = p(+1 + 2) + Q(1 - 1) -2 = 3p + 0(Q) 3p = -2 ∴ p = \(\frac{-2}{3}\)	A

16.	Find the range of values of x for which \(\frac{1}{x}\) > 2 is true A. x < \(\frac{1}{2}\) B. x < 0 or x < \(\frac{1}{2}\) C. 0 < x < \(\frac{1}{2}\) D. 1 < x < 2 Show Content Detailed Solution \(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2 x > 2x² = 2x² < x = 2x² - x < 0 = x(2x - 10 < 0 Case 1(+, -) = x > 0, 2x - 1 < 0 x > 0, x < \(\frac{1}{2}\) (solution) Case 2(-, 4) = x < 0, 2x - 1 > 0 x < 0, x , \(\frac{1}{2}\) = 0	C
17.	If the 6th term of an arithmetic progression is 11 and the first term is 1, find the common difference. A. \(\frac{12}{5}\) B. \(\frac{5}{3}\) C. -2 D. 2 Show Content Detailed Solution In an AP, Tn = a + (n - 1)d T₆ = a + 5d = 11 The first term = a = 1 ∴ T₆ = 1 + 5d = 11 5d = 11 - 1 5d = 10 ∴ d = 2	D
18.	Find the value of log₁₀ r + log₁₀ r² + log₁₀ r⁴ + log₁₀ r⁸ + log₁₀ r¹⁶ + log₁₀ r³² = 63 A. 10^-1 B. 10^o C. 10 D. 10² Show Content Detailed Solution log₁₀ r + log₁₀ r² + log₁₀ r⁴ + log₁₀ r⁸ + log₁₀ r¹⁶ + log₁₀ r³² = 63 log₁₀r⁶³ = 63 ⁶³ = 10⁶³ ∴ r = 10	C
19.	Find the nth term of the sequence 3, 6, 10, 15, 21..... A. \(\frac{n(n - 1)}{2}\) B. \(\frac{n(n + 1)}{2}\) C. \(\frac{(n + 1)(n + 2)}{2}\) D. n(2n + 1) Show Content Detailed Solution \(\frac{(n + 1)(n + 2)}{2}\) If n = 1, the expression becomes 3 n = 2, the expression becomes 6 n = 4, the expression becomes 15 n = 5, the expression becomes 21	C
20.	A binary operation \(\oplus\) is defines on the set of all positive integers by a \(\oplus\) b = ab for all positive integers a, b. Which of the following properties does NOT hold? A. closure B. identity C. positive D. inverse Show Content Detailed Solution a \(\oplus\) b = ab The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a \(\oplus\) b = ab; b \(\oplus\) a = ba = ab The number 1 is the identity element under multiplication	D