Paper 1  Objectives  51 Questions
JAMB Exam
Year: 1993
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
Integrate \(\frac{1  x}{x^3}\) with respect to x A. \(\frac{x  x^2}{x^4}\) + k B. \(\frac{4}{x^4}  \frac{3 + k}{x^3}\) C. \(\frac{1}{x}  \frac{1}{2x^2}\) + k D. \(\frac{1}{3x^2}  \frac{1}{2x}\) + k
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Detailed Solution\(\int \frac{1  x}{x^3}\)= \(\int^{1}_{x^3}  \int^{x}_{x^3}\) = x^{3} dx  x^{2}dx = \(\frac{1}{2x^2} + \frac{1}{x}\) 

2. 
Change 71_{10} to base 8 A. 107_{8} B. 106_{8} C. 71_{8} D. 17_{8}
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Detailed Solution\(\begin{array}{cc} 8 & 71 \\ 8 & 8 \text{rem} 7\\ 8 & 1 \text{rem} 0\end{array}\)= 107_{8} 

3. 
Evaluate \(\frac{3524}{0.05}\) correct to 3 significant figures A. 705 B. 70,000 C. 70, 480 D. 70, 500
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Detailed Solution\(\frac{3524}{0.05}\) = 70480\(\approx\) 70500(3 s.g) 

4. 
If 9^{(x  \(\frac{1}{2}\))} 3x^{2} A. \(\frac{1}{2}\) B. 1 C. 2 D. 3
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Detailed Solution9^{(x  \(\frac{1}{2}\))} 3x^{2} = 3^{2(x  \(\frac{1}{2}\))} = 3x^{2}∴ 2(x  \(\frac{1}{2}\)) = x^{2} 2x  1 = x^{2} hence x^{2}  2x + 1 = 0 (x  1)(x  1) = 0 x = 1 

5. 
Solve for y in the equation 10^1 x 5(2y  2) x 4(y  1) = 1 A. \(\frac{3}{4}\) B. \(\frac{5}{4}\) C. \(\frac{2}{3}\) D. 5
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Detailed Solution10y x 5(2y  2) x 4(y  1) = 1but 10y  (5 x 2)y = 5y x 2y = (Law of indices) 5y x 2y x 5(2y  2) x 4(y  1) = 1 but 4(y  1) = 22(y  1) = 2y  2 (Law of indices) 5y x 5(2y 2) x 2( 2) = 1 5(3y 2) x 2y x 2(2y 2) = 1 = 5(3y 2) x 2(3y 2) = 1 But ao = 1 10(3y 2) = 10o 3y  2 = 0 ∴ y = \(\frac{2}{3}\) 

6. 
Simplify \(\frac{1}{√3  2}\)  \(\frac{1}{√3 + 2}\) A. 3 B. \(\frac{2}{3}\) C. 7 D. 4
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Detailed Solution\(\frac{1}{√3  2}\)  \(\frac{1}{√3 + 2}\)L.C.M = (3 2) (3 + 2) ∴ \(\frac{1}{\sqrt{3  2}}\)  \(\frac{1}{\sqrt{3  2}}\) = \(\frac{\sqrt{3 + 2}  \sqrt{3  2}}{\sqrt{3  2} + \sqrt{3  2}}\) \(\frac{√3 + 2  √3 + 2}{3  2√3 + 2√3  4}\) = \(\frac{4}{3  2}\) = \(\frac{4}{1}\) = 4 

7. 
If 2log_{3} y + log_{3} x^{2} = 4, then y is A. 4  log_{3} B. \(\frac{4}{log_3 x}\) C. \(\frac{4}{x}\) D. \(\pm\) \(\frac{9}{x}\)
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Detailed Solution2log_{3}y + log_{3}x^{2} = 4log_{3}y^{2} + log_{3}x^{2} = 4 ∴ log_{3} (x^{2}y^{2}) = log_{3}81(correct all to base 4) x^{2}y^{2} = 81 ∴ xy = \(\pm\)9 ∴ y = \(\pm\)\(\frac{9}{x}\) 

8. 
Solve without using tables log_{5}(62.5)  log_{5}(\(\frac{1}{2}\)) A. 3 B. 4 C. 5 D. 8
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Detailed Solutionlog_{5}(62.5)  log_{5}(\(\frac{1}{2}\))= log_{5}\(\frac{(62.5)}{\frac{1}{2}}\)  log_{5}(2 x 62.5) = log_{5}(125) = log_{5}5^{3}  3log_{5}5 = 3 

9. 
If N225.00 yields N27.00 in x years simple interest at the rate of 4% per annum, find x A. 3 B. 4 C. 12 D. 17
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Detailed SolutionPrincipal = N255.00, Interest = 27.00year = x Rate: 4% ∴ 1 = \(\frac{PRT}{100}\) 27 = \(\frac{225 \times 4 \times T}{100}\) 2700 = 900T T = 3 years 

10. 
If \(\sqrt{x^2 + 9}\) = x + 1, solve for x A. 5 B. 4 C. 3 D. 2 E. 1
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Detailed Solution\(\sqrt{x^2 + 9}\) = x + 1x^{2} + 9 = (x + 1)^{2} + 1 0 = x^{2} + 2x + 1  x^{2}  9 = 2x  8 = 0 2(x  4) = 0 x = 4 

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. \(\frac{pq}{a(p  q)}\)
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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 = \(\frac{p  q}{a(p + q)}\) 
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