Mathematics (Core), 1992, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1992

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

11 - 20 of 49 Questions

#	Question	Ans
11.	Solve the equation: \(y - 11\sqrt{y} + 24 = 0\) A. 8, 3 B. 64, 9 C. 6, 4 D. 9, -8 Show Content Detailed Solution \(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\) Squaring both sides, \(y^{2} + 48y + 576 = 121y\) \(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\) \(y^{2} - 64y - 9y + 576 = 0\) \(y(y - 64) - 9(y - 64) = 0\) \((y - 9)(y - 64) = 0\) \(\therefore \text{y = 64 or y = 9}\)	B
12.	Make t the subject of formula S = ut + \(\frac{1}{2} at^2\) A. \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\)) B. \(\frac{1}{a}\) {u \(\pm\) (U² - 2as)} C. \(\frac{1}{a}\) {u \(\pm\) \(\sqrt{2as}\)} D. \(\frac{1}{a}\) {-u + \(\sqrt{( 2as)}\)} Show Content Detailed Solution Given S = ut + \(\frac{1}{2} at^2\) S = ut + \(\frac{1}{2} at^2\) ∴ 2S = 2ut + at² = at² + 2ut - 2s = 0 t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\) = -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\) = \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))	A
13.	A man invested a sum of N280.00 partly at 5% and partly at 4%. if the total interest is N12.80 per annum, find the amount invested at 5% A. 14.00 B. 120.00 C. 140.00 D. 160.00 Show Content Detailed Solution Let the amounts invested at 4% and 5% respectively be x and y. \(\therefore x + y = 280 ... (i)\) Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\) Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\) \(\therefore 0.04x + 0.05y = 12.80\) \(\implies 4x + 5y = 1280 ... (ii)\) From (i), \(x = 280 - y\). Put into (ii), \(4(280 - y) + 5y = 1280\) \(1120 - 4y + 5y = 1280\) \(1120 + y = 1280 \implies y = 1280 - 1120 = N160\) \(\therefore\) N160 was invested at the rate of 5% per annum.	D
14.	If x + 1 is a factor of x³ + 3x² + kx + 4, find the value of k A. 6 B. -6 C. 8 D. -8 Show Content Detailed Solution x + 1 is a factor of x³ + 3x² + kx + 4 Let f(x) = x³ + 3x² + kx + 4 ∴ f(-1) = (-1)³ + 3(-1)² + k(-1) + 4 = 0 -1 + 3 - k + 4 = 0 ∴ k = 6	A
15.	Resolve \(\frac{3}{x^2 + x - 2}\) into partial fractions A. \(\frac{1}{x - 1} - \frac{1}{x + 2}\) B. \(\frac{1}{x + 1} + \frac{1}{x - 2}\) C. \(\frac{1}{x + 1} - \frac{1}{x - 2}\) D. \(\frac{1}{x - 2} + \frac{1}{x + 2}\) Show Content Detailed Solution \(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\) \(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\) A(x + 2) + B(x - 1) = 3 when x = 1, 3A = 3 \(\to\) a = 1 when x = -2, -3B = 3 \(\to\) B = -1 = \(\frac{1}{x - 1} - \frac{1}{x + 2}\)	A
16.	Find all values of x satisfying the inequality -11 \(\leq\) 4 - 3x \(\leq\) 28 A. -5 \(\leq\) x v 8 B. 5 \(\leq\) x \(\leq\) 8 C. -8 \(\leq\) x \(\leq\) 5 D. -5 < x \(\leq\) 8 Show Content Detailed Solution To solve -11 \(\leq\) 4 - 3x \(\leq\) 28 -11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28 15 \(\leq\) -3x \(\leq\) 24 = 15 \(\geq\) 3x - 3x \(\geq\) -24 -5 \(\geq\) x, x \(\geq\) -8 i.e. x \(\leq\) 5 ∴ -8 \(\leq\) x \(\leq\) 5	C
17.	Find the sum to infinity to the following series 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + ..... A. 1270 B. 190 C. 18 D. 9 Show Content Detailed Solution 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + ..... a = 3 r = \(\frac{2}{3}\) s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\) = \(\frac{3}{\frac{1}{3}}\) = 3 x 3 = 9	D
18.	What is the n-th term of the sequence 2, 6, 12, 20...? A. 4n - 2 B. 2(3^{n - 1}) C. n² + n D. n² + 3n + 2 Show Content Detailed Solution Given that 2, 6, 12, 20...? the nth term = n\(^2\) + n check: n = 1, u1 = 2 n = 2, u2 = 4 + 2 = 6 n = 3, u3 = 9 + 3 = 12 ∴ n = 4, u4 = 16 + 4 = 20	C
19.	For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms A. 157 B. 187 C. 197 D. 200 Show Content Detailed Solution a = 2, d = 3 and n = 11 To find S_{n/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)] = \(\frac{11}{2}\) [2(2) + (11 - 1) 3] = \(\frac{11}{2}\)n [4 + 10(3)] = \(\frac{11}{2}\)(34) = 11 x 17 = 187}	B
20.	If the binary operation \(\ast\) is defined by m \(\ast\) n = mn + m + n for any real number m and n, find the identity of the elements under this operation A. e = 1 B. e = -1 C. e = -2 D. e = 0 Show Content Detailed Solution Identity(e) : a \(\ast\) e = a m \(\ast\) e = m...(i) m \(\ast\) e = me + m + e Because m \(\ast\) e = m : m = me + m + e m - m = e(m + 1) e = \(\frac{0}{m + 1}\) e = 0	B

11.	Solve the equation: \(y - 11\sqrt{y} + 24 = 0\) A. 8, 3 B. 64, 9 C. 6, 4 D. 9, -8 Show Content Detailed Solution \(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\) Squaring both sides, \(y^{2} + 48y + 576 = 121y\) \(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\) \(y^{2} - 64y - 9y + 576 = 0\) \(y(y - 64) - 9(y - 64) = 0\) \((y - 9)(y - 64) = 0\) \(\therefore \text{y = 64 or y = 9}\)	B
12.	Make t the subject of formula S = ut + \(\frac{1}{2} at^2\) A. \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\)) B. \(\frac{1}{a}\) {u \(\pm\) (U² - 2as)} C. \(\frac{1}{a}\) {u \(\pm\) \(\sqrt{2as}\)} D. \(\frac{1}{a}\) {-u + \(\sqrt{( 2as)}\)} Show Content Detailed Solution Given S = ut + \(\frac{1}{2} at^2\) S = ut + \(\frac{1}{2} at^2\) ∴ 2S = 2ut + at² = at² + 2ut - 2s = 0 t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\) = -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\) = \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))	A
13.	A man invested a sum of N280.00 partly at 5% and partly at 4%. if the total interest is N12.80 per annum, find the amount invested at 5% A. 14.00 B. 120.00 C. 140.00 D. 160.00 Show Content Detailed Solution Let the amounts invested at 4% and 5% respectively be x and y. \(\therefore x + y = 280 ... (i)\) Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\) Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\) \(\therefore 0.04x + 0.05y = 12.80\) \(\implies 4x + 5y = 1280 ... (ii)\) From (i), \(x = 280 - y\). Put into (ii), \(4(280 - y) + 5y = 1280\) \(1120 - 4y + 5y = 1280\) \(1120 + y = 1280 \implies y = 1280 - 1120 = N160\) \(\therefore\) N160 was invested at the rate of 5% per annum.	D
14.	If x + 1 is a factor of x³ + 3x² + kx + 4, find the value of k A. 6 B. -6 C. 8 D. -8 Show Content Detailed Solution x + 1 is a factor of x³ + 3x² + kx + 4 Let f(x) = x³ + 3x² + kx + 4 ∴ f(-1) = (-1)³ + 3(-1)² + k(-1) + 4 = 0 -1 + 3 - k + 4 = 0 ∴ k = 6	A
15.	Resolve \(\frac{3}{x^2 + x - 2}\) into partial fractions A. \(\frac{1}{x - 1} - \frac{1}{x + 2}\) B. \(\frac{1}{x + 1} + \frac{1}{x - 2}\) C. \(\frac{1}{x + 1} - \frac{1}{x - 2}\) D. \(\frac{1}{x - 2} + \frac{1}{x + 2}\) Show Content Detailed Solution \(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\) \(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\) A(x + 2) + B(x - 1) = 3 when x = 1, 3A = 3 \(\to\) a = 1 when x = -2, -3B = 3 \(\to\) B = -1 = \(\frac{1}{x - 1} - \frac{1}{x + 2}\)	A

16.	Find all values of x satisfying the inequality -11 \(\leq\) 4 - 3x \(\leq\) 28 A. -5 \(\leq\) x v 8 B. 5 \(\leq\) x \(\leq\) 8 C. -8 \(\leq\) x \(\leq\) 5 D. -5 < x \(\leq\) 8 Show Content Detailed Solution To solve -11 \(\leq\) 4 - 3x \(\leq\) 28 -11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28 15 \(\leq\) -3x \(\leq\) 24 = 15 \(\geq\) 3x - 3x \(\geq\) -24 -5 \(\geq\) x, x \(\geq\) -8 i.e. x \(\leq\) 5 ∴ -8 \(\leq\) x \(\leq\) 5	C
17.	Find the sum to infinity to the following series 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + ..... A. 1270 B. 190 C. 18 D. 9 Show Content Detailed Solution 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + ..... a = 3 r = \(\frac{2}{3}\) s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\) = \(\frac{3}{\frac{1}{3}}\) = 3 x 3 = 9	D
18.	What is the n-th term of the sequence 2, 6, 12, 20...? A. 4n - 2 B. 2(3^{n - 1}) C. n² + n D. n² + 3n + 2 Show Content Detailed Solution Given that 2, 6, 12, 20...? the nth term = n\(^2\) + n check: n = 1, u1 = 2 n = 2, u2 = 4 + 2 = 6 n = 3, u3 = 9 + 3 = 12 ∴ n = 4, u4 = 16 + 4 = 20	C
19.	For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms A. 157 B. 187 C. 197 D. 200 Show Content Detailed Solution a = 2, d = 3 and n = 11 To find S_{n/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)] = \(\frac{11}{2}\) [2(2) + (11 - 1) 3] = \(\frac{11}{2}\)n [4 + 10(3)] = \(\frac{11}{2}\)(34) = 11 x 17 = 187}	B
20.	If the binary operation \(\ast\) is defined by m \(\ast\) n = mn + m + n for any real number m and n, find the identity of the elements under this operation A. e = 1 B. e = -1 C. e = -2 D. e = 0 Show Content Detailed Solution Identity(e) : a \(\ast\) e = a m \(\ast\) e = m...(i) m \(\ast\) e = me + m + e Because m \(\ast\) e = m : m = me + m + e m - m = e(m + 1) e = \(\frac{0}{m + 1}\) e = 0	B