Mathematics (Core), 1980, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1980

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

31 - 40 of 49 Questions

#	Question	Ans
31.	Solve the equation for the positive values of \(\theta\) less than 360^o. 3 tan \(\theta\) + 2 = -1 A. 135^o or 315^o B. 45^o or 135^o C. 315^o or 180^o D. 315v + 45^o E. 360^o or 315^o Show Content Detailed Solution 3 tan \(\theta\) + 2 = -1 3 tan \(\theta\) \(\frac{-3}{3}\) = -1 \(\theta\) = tan -1(-1) \(\theta\) = 360^o - 45^o = 315^o \(\theta\) = 180 - 45^o = 135^o	A
32.	Given log 2 = 0.69, log3 = 1, 10 and log7 = 1.90, all to a fixed base, find log 10.5 to the same base without using tables. A. 1.03 B. 2.31 C. 3.69 D. 10.5 E. 25 Show Content Detailed Solution log 10.5 = log \(\frac{21}{2}\) = log 21 - log 2 = log(3 x 7) - log 2 = log 3 + log 7 - log 2 = 1.10 + 1.90 - 0.69 = 3 - 0.69 = 2.31	B
33.	Simplify 10² + log₁₀5 A. 500 B. 2 log₁₀ 5 C. 10 D. 25 E. log₁₀5 x 10¹⁰⁰ Show Content Detailed Solution 10² + log₁₀5 = log₁₀ 10¹⁰⁰ + log₁₀5 = log₁₀5 x 10¹⁰⁰	E
34.	Find the missing numerator \(\frac{5}{x + 1}\) - \(\frac{3}{1 - x}\) - \(\frac{7x - 1}{x^2 - 1}\) = \(\frac{?}{x + 1}\). A. -1 B. x - 1 C. \(\frac{3(1 - 5x)}{x + 1}\) D. 1 E. 3(1 - 5x) Show Content Detailed Solution \(\frac{5}{x + 1} - \frac{3}{1 - x} - \frac{7x - 1}{x^{2} - 1} = \frac{?}{x + 1}\) \(\frac{5(x - 1) - 3(- (x + 1)) - 7x - 1}{x^{2} - 1}\) = \(\frac{5x - 5 + 3x + 3 - 7x - 1}{x^{2} - 1}\) = \(\frac{x - 1}{x^{2} - 1}\) = \(\frac{x - 1}{(x - 1)(x + 1)}\) = \(\frac{1}{x + 1}\). The numerator = 1.	D
35.	The expression x3 - 4x2 + cx + d is such that x + 1 is its factor, and its value is 1 when x is -2. Find c and d. A. c = 4 and d = -9 B. c = -4 and d = 9 C. c = -20 and d = -15 D. c = 20 and d = -15 E. c = -20 and d = 15 Show Content Detailed Solution F(X) = x3 - 4x2 + cx + d = (X + 1) Q(X) + R x = -1, R = 0,f(-1) = -13 - 4(-1)2 + c(-1) + d = 0 -1 - 4 - c + d = 0 d - c = 5................(i) f(-2) = -23 - 4(-2)2 + c(-2) + d = 1 = -8 - 16 - 2c + d = 1 -8 - 16 - 2c + d = 1 -24 - 2c + d = 1 d - 2c = 1 + 24 d - 2c = 25.................(ii) \(\frac{d - c = 5}{-c = 20}\) d - c = 5 c = -20 d - (-20) = 5 d + 20 = 5	C
36.	If a function is defined by f(x + 1) = 3x2 - x + 4, Find f(0). A. 4 B. 6 C. 9 D. 8 E. 2 Show Content Detailed Solution f(x + 1) = 3x2 - x + 4 f(0) = f(x + 1) x + 1 = 0 ===> x = -1 f(0) = 3(-1)2 - (-1) + 4 f(0) = 3 + 1 + 4 = 8	D
37.	A cylindrical motor of height 12cm has uniform thickness of 2cm. If the diameter of its outer cross section is 10cm, Find the volume of the constituent material. (take \(\pi\) = \(\frac{22}{7}\) A. \(\frac{6600}{7}\)cm³ B. 270cm³ C. 660cm³ D. \(\frac{4224}{7}\)cm³ E. \(\frac{1980}{7}\)cm³ Show Content Detailed Solution V = \(\pi\)r²h = \(\frac{22}{7}\) x 5² x 12 - \(\frac{22}{7}\) x 3² x 12 = \(\frac{22}{7}\) x 12(5² - 3²) = \(\frac{4224}{7}\)	D
38.	If x varies inversely as y, and y varies directly as the square root of z, and z varies directly \(\frac{1}{w^2}\) write down in words how x varies with w A. x varies inversely as w² B. x varies directly as w² C. x varies directly as w D. x varies inversely as w E. x varies directly as square root of w	B
39.	Write the decimal number 39 to base 2 A. 10011 base 2 B. 110111 base 2 C. 111001 base 2 D. 100101 base 2 E. 19.5 base 2 Show Content Detailed Solution \(\begin{array}{c\|c} 2 & 39\\ 2 & 19 R 1\\ 2 & 9 R 1\\2 & 4 R 0\\2 & 1 R 0\\ 2 & 1 R 1\end{array}\) = 10011 base 2	A
40.	If PN is perpendicular to QR, find the value of tan x. A. \(\frac{5}{9}\) B. \(\frac{3}{5}\) C. \(\frac{3}{4}\) D. \(\frac{4}{3}\) E. \(\frac{5}{3}\) Show Content Detailed Solution By Pythagoras ON = 4 NR = 5 Tan x = \(\frac{3}{5}\)	B

31.	Solve the equation for the positive values of \(\theta\) less than 360^o. 3 tan \(\theta\) + 2 = -1 A. 135^o or 315^o B. 45^o or 135^o C. 315^o or 180^o D. 315v + 45^o E. 360^o or 315^o Show Content Detailed Solution 3 tan \(\theta\) + 2 = -1 3 tan \(\theta\) \(\frac{-3}{3}\) = -1 \(\theta\) = tan -1(-1) \(\theta\) = 360^o - 45^o = 315^o \(\theta\) = 180 - 45^o = 135^o	A
32.	Given log 2 = 0.69, log3 = 1, 10 and log7 = 1.90, all to a fixed base, find log 10.5 to the same base without using tables. A. 1.03 B. 2.31 C. 3.69 D. 10.5 E. 25 Show Content Detailed Solution log 10.5 = log \(\frac{21}{2}\) = log 21 - log 2 = log(3 x 7) - log 2 = log 3 + log 7 - log 2 = 1.10 + 1.90 - 0.69 = 3 - 0.69 = 2.31	B
33.	Simplify 10² + log₁₀5 A. 500 B. 2 log₁₀ 5 C. 10 D. 25 E. log₁₀5 x 10¹⁰⁰ Show Content Detailed Solution 10² + log₁₀5 = log₁₀ 10¹⁰⁰ + log₁₀5 = log₁₀5 x 10¹⁰⁰	E
34.	Find the missing numerator \(\frac{5}{x + 1}\) - \(\frac{3}{1 - x}\) - \(\frac{7x - 1}{x^2 - 1}\) = \(\frac{?}{x + 1}\). A. -1 B. x - 1 C. \(\frac{3(1 - 5x)}{x + 1}\) D. 1 E. 3(1 - 5x) Show Content Detailed Solution \(\frac{5}{x + 1} - \frac{3}{1 - x} - \frac{7x - 1}{x^{2} - 1} = \frac{?}{x + 1}\) \(\frac{5(x - 1) - 3(- (x + 1)) - 7x - 1}{x^{2} - 1}\) = \(\frac{5x - 5 + 3x + 3 - 7x - 1}{x^{2} - 1}\) = \(\frac{x - 1}{x^{2} - 1}\) = \(\frac{x - 1}{(x - 1)(x + 1)}\) = \(\frac{1}{x + 1}\). The numerator = 1.	D
35.	The expression x3 - 4x2 + cx + d is such that x + 1 is its factor, and its value is 1 when x is -2. Find c and d. A. c = 4 and d = -9 B. c = -4 and d = 9 C. c = -20 and d = -15 D. c = 20 and d = -15 E. c = -20 and d = 15 Show Content Detailed Solution F(X) = x3 - 4x2 + cx + d = (X + 1) Q(X) + R x = -1, R = 0,f(-1) = -13 - 4(-1)2 + c(-1) + d = 0 -1 - 4 - c + d = 0 d - c = 5................(i) f(-2) = -23 - 4(-2)2 + c(-2) + d = 1 = -8 - 16 - 2c + d = 1 -8 - 16 - 2c + d = 1 -24 - 2c + d = 1 d - 2c = 1 + 24 d - 2c = 25.................(ii) \(\frac{d - c = 5}{-c = 20}\) d - c = 5 c = -20 d - (-20) = 5 d + 20 = 5	C

36.	If a function is defined by f(x + 1) = 3x2 - x + 4, Find f(0). A. 4 B. 6 C. 9 D. 8 E. 2 Show Content Detailed Solution f(x + 1) = 3x2 - x + 4 f(0) = f(x + 1) x + 1 = 0 ===> x = -1 f(0) = 3(-1)2 - (-1) + 4 f(0) = 3 + 1 + 4 = 8	D
37.	A cylindrical motor of height 12cm has uniform thickness of 2cm. If the diameter of its outer cross section is 10cm, Find the volume of the constituent material. (take \(\pi\) = \(\frac{22}{7}\) A. \(\frac{6600}{7}\)cm³ B. 270cm³ C. 660cm³ D. \(\frac{4224}{7}\)cm³ E. \(\frac{1980}{7}\)cm³ Show Content Detailed Solution V = \(\pi\)r²h = \(\frac{22}{7}\) x 5² x 12 - \(\frac{22}{7}\) x 3² x 12 = \(\frac{22}{7}\) x 12(5² - 3²) = \(\frac{4224}{7}\)	D
38.	If x varies inversely as y, and y varies directly as the square root of z, and z varies directly \(\frac{1}{w^2}\) write down in words how x varies with w A. x varies inversely as w² B. x varies directly as w² C. x varies directly as w D. x varies inversely as w E. x varies directly as square root of w	B
39.	Write the decimal number 39 to base 2 A. 10011 base 2 B. 110111 base 2 C. 111001 base 2 D. 100101 base 2 E. 19.5 base 2 Show Content Detailed Solution \(\begin{array}{c\|c} 2 & 39\\ 2 & 19 R 1\\ 2 & 9 R 1\\2 & 4 R 0\\2 & 1 R 0\\ 2 & 1 R 1\end{array}\) = 10011 base 2	A
40.	If PN is perpendicular to QR, find the value of tan x. A. \(\frac{5}{9}\) B. \(\frac{3}{5}\) C. \(\frac{3}{4}\) D. \(\frac{4}{3}\) E. \(\frac{5}{3}\) Show Content Detailed Solution By Pythagoras ON = 4 NR = 5 Tan x = \(\frac{3}{5}\)	B