Mathematics (Core), 1984, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

1984

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

21 - 30 of 45 Questions

#	Question	Ans
21.	If f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 and g(y) = \(\sqrt{5 + y}\), find g [f(3)] and f[g(4)]. A. 3 and 4 B. -3 and 4 C. -3 and -4 D. 3 and -4 E. 0 and 5 Show Content Detailed Solution f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 = (2 + 3) (x - 3) + 4 = 5(x - 3) + 4 = 5x - 15 + 4 = 5x - 11 f(3) = 5 x 3 - 11 = 4 g(f(3)) = g(4) = \(\sqrt{5 + 4}\) = \(\sqrt{9}\) = 3 g(4) = 3 f(g(4)) = f(3) = 4 g[f(3)] and f[g(4)] = 3 and 4 respectively.	A
22.	The quadratic equation whose roots are 1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\) is? A. x² + (1 - \(\sqrt{13}\)x + 1 + \(\sqrt{13}\) = 0 B. x² - 2x - 12 = 0 C. x² - 2x + 12 = 0 D. x² + 12 + 2x² = 0 Show Content Detailed Solution 1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\) sum of roots - \(1 + \sqrt{13} + 1 - \sqrt{13} = 2\) Product of roots = (1 - \(\sqrt{13}\)) (1 + \(\sqrt{13}\)) = -12 x2 - (sum of roots) x + (product of roots) = 0 x2 - 2x - 12 = 0	B
23.	Find a factor which is common to all three binomial expressions 4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 A. 4a + 6b B. 4a - 6b C. 2a + 3b D. 2a - 3b E. none Show Content Detailed Solution 4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b) 8a3 + 27b3 = (2a)3 + (3b)3 = (2a + 3b)(4a - 6ab = 9a2) (4a + 6b)2 = 2(2a + 3b)2	C
24.	If (x - 2) and (x + 1) are factors of the expression x³ + px² + qx + 1, what is the sum of p and q A. 9 B. -3 C. 3 D. \(\frac{17}{3}\) E. \(\frac{2}{3}\) Show Content Detailed Solution x³ + px² + qx + 1 = (x - 1) Q(x) + R x - 2 = 0, x = 2, R = 0, 4p + 2p = -9........(i) x³ + px² + qx + 1 = (x - 1)Q(x) + R -1 + p - q + 1 = 0 p - q = 0.......(ii) Solve the equation simultaneously p = \(\frac{-3}{2}\) q = \(\frac{-3}{2}\) p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\) = \(\frac{-6}{2}\) = -3	B
25.	A cone is formed by bending a sector of a circle having an angle of 210^o. Find the radius of the base of the cone if the diameter of the circle is 12cm. A. 12cm B. 7.00cm C. 1.75cm D. 21cm E. 3.50cm Show Content Detailed Solution If diameter of the circle = 12cm; radius of the circle(L) = \(\frac{12}{2}\) = 6cm \(\frac{\theta}{360}\) = \(\frac{r}{L}\) where \(\theta\) = 210\(\theta\), L = 6cm \(\frac{210}{360}\) = \(\frac{r}{6}\) where r = radius of the base of the cone V = \(\frac{1260}{360}\) = 3.50cm	E
26.	The sides of a triangle are(x + 4)cm, xcm and (x - 4)cm, respectively If the cosine of the largest angle is \(\frac{1}{5}\), find the value of x A. 24cm B. 20cm C. 28cm D. 7cm E. \(\frac{88}{7}\) Show Content Detailed Solution < B is the largest since the side facing it is the largest, i.e. (x + 4)cm Cosine B = \(\frac{1}{5}\) = 0.2 given b² - a² + c² - 2a Cos B Cos B = \(\frac{a^2 + c^2 - b^2}{2ac}\) \(\frac{1}{5}\) = \(\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}\) \(\frac{1}{5}\)= \(\frac{x(x - 16)}{2x(x - 4)}\) \(\frac{1}{5}\) = \(\frac{x - 16}{2x - 8}\) = 5(x - 16) = 2x - 8 3x = 72 x = \(\frac{72}{3}\) = 24	A
27.	If a = \(\frac{2x}{1 - x}\) and b = \(\frac{1 + x}{1 - x}\), then a2 - b2 in the simplest form is A. \(\frac{3x + 1}{x - 1}\) B. \(\frac{3x^2 - 1}{(x - 1)}\)² C. x\(\frac{3x - 2}{1 - x}\) D. \(\frac{3x^2 - 1}{(x - 1)}\) Show Content Detailed Solution a2 - b2 = (\(\frac{2x}{1 - x}\))2 - (\(\frac{1 + x}{1 - x}\))2 = (\(\frac{2x}{1 - x} + \frac{1 + x}{1 - x}\))(\(\frac{2x}{1 - x} - \frac{1 + x}{1 - X}\)) = (\(\frac{3x + 1}{1 - x}\))(\(\frac{x - 1}{1 - x}\)) = \(\frac{3x + 1}{x - 1}\)	A
28.	Simplify (1 + \(\frac{\frac{x - 1}{1}}{\frac{1}{x + 1}}\))(x + 2) A. (x² - 1)(x + 2) B. x²(x + 2) C. ² + 34 D. \(\frac{3x^2 - 1}{(x - 1)}\) Show Content Detailed Solution \((1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)\) \((x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1\) \((1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)\)	B
29.	If pq + 1 = q² and t = \(\frac{1}{p}\) - \(\frac{1}{pq}\) express t in terms of q A. \(\frac{1}{p - q}\) B. \(\frac{1}{q - 1}\) C. \(\frac{1}{q + 1}\) D. 1 + 0 E. \(\frac{1}{1 - q}\) Show Content Detailed Solution Pq + 1 = q²......(i) t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii) p = \(\frac{q^2 - 1}{q}\) Sub for p in equation (ii) t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\) t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\) t = \(\frac{q - 1}{q^2 - 1}\) = \(\frac{q - 1}{(q + 1)(q - 1)}\) = \(\frac{1}{q + 1}\)	C
30.	The cumulative frequency function of the data below is given below by the equation y = cf(x). What is cf(5)? \(\begin{array}{c\|c} Score(n) & Frequency(f)\\\hline 3 & 30\\ 4 & 32\\ 5 & 30\\ 6 & 35\\8 & 20 \end{array}\) A. 30 B. 32 C. 55 D. 62 E. 92 Show Content Detailed Solution If y = cf(x) cf(5) = 30 + 32 + 30 = 92	E

21.	If f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 and g(y) = \(\sqrt{5 + y}\), find g [f(3)] and f[g(4)]. A. 3 and 4 B. -3 and 4 C. -3 and -4 D. 3 and -4 E. 0 and 5 Show Content Detailed Solution f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 = (2 + 3) (x - 3) + 4 = 5(x - 3) + 4 = 5x - 15 + 4 = 5x - 11 f(3) = 5 x 3 - 11 = 4 g(f(3)) = g(4) = \(\sqrt{5 + 4}\) = \(\sqrt{9}\) = 3 g(4) = 3 f(g(4)) = f(3) = 4 g[f(3)] and f[g(4)] = 3 and 4 respectively.	A
22.	The quadratic equation whose roots are 1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\) is? A. x² + (1 - \(\sqrt{13}\)x + 1 + \(\sqrt{13}\) = 0 B. x² - 2x - 12 = 0 C. x² - 2x + 12 = 0 D. x² + 12 + 2x² = 0 Show Content Detailed Solution 1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\) sum of roots - \(1 + \sqrt{13} + 1 - \sqrt{13} = 2\) Product of roots = (1 - \(\sqrt{13}\)) (1 + \(\sqrt{13}\)) = -12 x2 - (sum of roots) x + (product of roots) = 0 x2 - 2x - 12 = 0	B
23.	Find a factor which is common to all three binomial expressions 4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 A. 4a + 6b B. 4a - 6b C. 2a + 3b D. 2a - 3b E. none Show Content Detailed Solution 4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b) 8a3 + 27b3 = (2a)3 + (3b)3 = (2a + 3b)(4a - 6ab = 9a2) (4a + 6b)2 = 2(2a + 3b)2	C
24.	If (x - 2) and (x + 1) are factors of the expression x³ + px² + qx + 1, what is the sum of p and q A. 9 B. -3 C. 3 D. \(\frac{17}{3}\) E. \(\frac{2}{3}\) Show Content Detailed Solution x³ + px² + qx + 1 = (x - 1) Q(x) + R x - 2 = 0, x = 2, R = 0, 4p + 2p = -9........(i) x³ + px² + qx + 1 = (x - 1)Q(x) + R -1 + p - q + 1 = 0 p - q = 0.......(ii) Solve the equation simultaneously p = \(\frac{-3}{2}\) q = \(\frac{-3}{2}\) p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\) = \(\frac{-6}{2}\) = -3	B
25.	A cone is formed by bending a sector of a circle having an angle of 210^o. Find the radius of the base of the cone if the diameter of the circle is 12cm. A. 12cm B. 7.00cm C. 1.75cm D. 21cm E. 3.50cm Show Content Detailed Solution If diameter of the circle = 12cm; radius of the circle(L) = \(\frac{12}{2}\) = 6cm \(\frac{\theta}{360}\) = \(\frac{r}{L}\) where \(\theta\) = 210\(\theta\), L = 6cm \(\frac{210}{360}\) = \(\frac{r}{6}\) where r = radius of the base of the cone V = \(\frac{1260}{360}\) = 3.50cm	E

26.	The sides of a triangle are(x + 4)cm, xcm and (x - 4)cm, respectively If the cosine of the largest angle is \(\frac{1}{5}\), find the value of x A. 24cm B. 20cm C. 28cm D. 7cm E. \(\frac{88}{7}\) Show Content Detailed Solution < B is the largest since the side facing it is the largest, i.e. (x + 4)cm Cosine B = \(\frac{1}{5}\) = 0.2 given b² - a² + c² - 2a Cos B Cos B = \(\frac{a^2 + c^2 - b^2}{2ac}\) \(\frac{1}{5}\) = \(\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}\) \(\frac{1}{5}\)= \(\frac{x(x - 16)}{2x(x - 4)}\) \(\frac{1}{5}\) = \(\frac{x - 16}{2x - 8}\) = 5(x - 16) = 2x - 8 3x = 72 x = \(\frac{72}{3}\) = 24	A
27.	If a = \(\frac{2x}{1 - x}\) and b = \(\frac{1 + x}{1 - x}\), then a2 - b2 in the simplest form is A. \(\frac{3x + 1}{x - 1}\) B. \(\frac{3x^2 - 1}{(x - 1)}\)² C. x\(\frac{3x - 2}{1 - x}\) D. \(\frac{3x^2 - 1}{(x - 1)}\) Show Content Detailed Solution a2 - b2 = (\(\frac{2x}{1 - x}\))2 - (\(\frac{1 + x}{1 - x}\))2 = (\(\frac{2x}{1 - x} + \frac{1 + x}{1 - x}\))(\(\frac{2x}{1 - x} - \frac{1 + x}{1 - X}\)) = (\(\frac{3x + 1}{1 - x}\))(\(\frac{x - 1}{1 - x}\)) = \(\frac{3x + 1}{x - 1}\)	A
28.	Simplify (1 + \(\frac{\frac{x - 1}{1}}{\frac{1}{x + 1}}\))(x + 2) A. (x² - 1)(x + 2) B. x²(x + 2) C. ² + 34 D. \(\frac{3x^2 - 1}{(x - 1)}\) Show Content Detailed Solution \((1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)\) \((x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1\) \((1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)\)	B
29.	If pq + 1 = q² and t = \(\frac{1}{p}\) - \(\frac{1}{pq}\) express t in terms of q A. \(\frac{1}{p - q}\) B. \(\frac{1}{q - 1}\) C. \(\frac{1}{q + 1}\) D. 1 + 0 E. \(\frac{1}{1 - q}\) Show Content Detailed Solution Pq + 1 = q²......(i) t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii) p = \(\frac{q^2 - 1}{q}\) Sub for p in equation (ii) t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\) t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\) t = \(\frac{q - 1}{q^2 - 1}\) = \(\frac{q - 1}{(q + 1)(q - 1)}\) = \(\frac{1}{q + 1}\)	C
30.	The cumulative frequency function of the data below is given below by the equation y = cf(x). What is cf(5)? \(\begin{array}{c\|c} Score(n) & Frequency(f)\\\hline 3 & 30\\ 4 & 32\\ 5 & 30\\ 6 & 35\\8 & 20 \end{array}\) A. 30 B. 32 C. 55 D. 62 E. 92 Show Content Detailed Solution If y = cf(x) cf(5) = 30 + 32 + 30 = 92	E