Mathematics (Core), 2010, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2010

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 49 of 49 Questions

#	Question	Ans
41.	Find the gradient of the line passing through the points P(1, 1) and Q(2, 5). A. 3 B. 2 C. 5 D. 4 Show Content Detailed Solution Let (x1, y1) = (1, 1) and (x2, y2)= (2, 5) then gradient m of \(\bar{PQ}\) is m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{5 - 1}{2 - 1}\) = \(\frac{4}{1}\) = 4 There is an explanation video available below.	D
42.	If cot\(\theta\) = \(\frac{8}{15}\), where \(\theta\) is acute, find sin\(\theta\) A. \(\frac{8}{17}\) B. \(\frac{15}{17}\) C. \(\frac{16}{17}\) D. \(\frac{13}{17}\) Show Content Detailed Solution cot\(\theta\) = \(\frac{1}{\cos \theta}\) = \(\frac{8}{15}\)(given) tan\(\theta\) = \(\frac{15}{18}\) By Pythagoras theorem, x2 = 152 + 82 x2 = 225 + 64 = 289 x = \(\sqrt{289}\) = 17 Hence sin\(\theta\) = \(\frac{15}{x}\) = \(\frac{15}{17}\) There is an explanation video available below.	B
43.	If y = x sinx, find \(\frac{dy}{dx}\) A. sin x - x cosx B. sinx + x cosx C. sinx - cosx D. sinx + cosx Show Content Detailed Solution If y = x sinx, then Let u = x and v = sinx \(\frac{du}{dx}\) = 1 and \(\frac{dv}{dx}\) = cosx Hence by the product rule, \(\frac{dy}{dx}\) = v \(\frac{du}{dx}\) + u\(\frac{dv}{dx}\) = (sin x) x 1 + x cosx = sinx + x cosx There is an explanation video available below.	B
44.	Evaluate \(\int^{2}_{0}(x^3 + x^2)\)dx. A. 4\(\frac{5}{6}\) B. 6\(\frac{2}{3}\) C. 1\(\frac{5}{6}\) D. 2\(\frac{5}{6}\) Show Content Detailed Solution \(\int^{2}_{0}(x^3 + x^2)\)dx = \(\int^{2}_{0}\)(\(\frac{x^4}{4} + {\frac {x^3}{3}}\)) = (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\)) = (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0 = \(\frac{80}{12} = {\frac {20}{3}}\) or 6\(\frac{2}{3}\) There is an explanation video available below.	B
45.	In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women? A. 100 B. 200 C. 30 D. 50 Show Content Detailed Solution A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in \(^{5}C_{2}\) x \(^{6}C_{3}\) ways = \(\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}\) = \(\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}\) = \(\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}\) = \(\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}}\) = 10 x \(\frac{6 \times 20}{6}\) = 200 There is an explanation video available below.	B
46.	If three unbiased coins are tossed, find the probability that they are all heads A. \(\frac{1}{2}\) B. \(\frac{1}{3}\) C. \(\frac{1}{9}\) D. \(\frac{1}{8}\) Show Content Detailed Solution P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\) Using the binomial prob. distribution, (H + T)3 = H3 + 3H2T1 + 3HT2 + T3 Hence the probability that three heads show in a toss of the three coins is H3 = (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) There is an explanation video available below.	D
47.	Find the standard deviation of 2, 3, 5 and 6 A. √6 B. √10 C. √\(\frac{2}{5}\) D. √\(\frac{5}{2}\) Show Content Detailed Solution \(\begin{array}& x & x - \bar{x} & (x - \bar{x})^2 \\2 & -2 & 4 \\ 3 & -1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4\\ \hline \sum x = 16 & & \sum (x - \bar{x}^2) = 10 \end{array}\) ___________________________________ \(\bar{x}\) = \(\frac{\sum x }{N}\) = \(\frac{16}{4}\) = 4 S = \(\sqrt{\frac {(x - \bar{x})^2}{N}}\) = \(\sqrt{\frac {(10)}{4}}\) = \(\sqrt{\frac {(5)}{2}}\) There is an explanation video available below.	D
48.	W is directly proportional to U. If W = 5 when U = 3, find U when W = \(\frac{2}{7}\) A. \(\frac{6}{35}\) B. \(\frac{10}{21}\) C. \(\frac{21}{10}\) D. \(\frac{35}{6}\) Show Content Detailed Solution W \(\alpha\) U W = ku u = \(\frac{w}{k}\); \(\frac{2}{7}\) x \(\frac{3}{5}\) = \(\frac{6}{35}\) There is an explanation video available below.	A
49.	Determine the value of x for which (x2 - 1)>0 A. x < -1 or x > 1 B. -1 < x < 1 C. x > 0 D. x < -1 Show Content Detailed Solution x(x - 1) > 0 x < -1 or x > 1 There is an explanation video available below.	A

41.	Find the gradient of the line passing through the points P(1, 1) and Q(2, 5). A. 3 B. 2 C. 5 D. 4 Show Content Detailed Solution Let (x1, y1) = (1, 1) and (x2, y2)= (2, 5) then gradient m of \(\bar{PQ}\) is m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{5 - 1}{2 - 1}\) = \(\frac{4}{1}\) = 4 There is an explanation video available below.	D
42.	If cot\(\theta\) = \(\frac{8}{15}\), where \(\theta\) is acute, find sin\(\theta\) A. \(\frac{8}{17}\) B. \(\frac{15}{17}\) C. \(\frac{16}{17}\) D. \(\frac{13}{17}\) Show Content Detailed Solution cot\(\theta\) = \(\frac{1}{\cos \theta}\) = \(\frac{8}{15}\)(given) tan\(\theta\) = \(\frac{15}{18}\) By Pythagoras theorem, x2 = 152 + 82 x2 = 225 + 64 = 289 x = \(\sqrt{289}\) = 17 Hence sin\(\theta\) = \(\frac{15}{x}\) = \(\frac{15}{17}\) There is an explanation video available below.	B
43.	If y = x sinx, find \(\frac{dy}{dx}\) A. sin x - x cosx B. sinx + x cosx C. sinx - cosx D. sinx + cosx Show Content Detailed Solution If y = x sinx, then Let u = x and v = sinx \(\frac{du}{dx}\) = 1 and \(\frac{dv}{dx}\) = cosx Hence by the product rule, \(\frac{dy}{dx}\) = v \(\frac{du}{dx}\) + u\(\frac{dv}{dx}\) = (sin x) x 1 + x cosx = sinx + x cosx There is an explanation video available below.	B
44.	Evaluate \(\int^{2}_{0}(x^3 + x^2)\)dx. A. 4\(\frac{5}{6}\) B. 6\(\frac{2}{3}\) C. 1\(\frac{5}{6}\) D. 2\(\frac{5}{6}\) Show Content Detailed Solution \(\int^{2}_{0}(x^3 + x^2)\)dx = \(\int^{2}_{0}\)(\(\frac{x^4}{4} + {\frac {x^3}{3}}\)) = (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\)) = (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0 = \(\frac{80}{12} = {\frac {20}{3}}\) or 6\(\frac{2}{3}\) There is an explanation video available below.	B
45.	In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women? A. 100 B. 200 C. 30 D. 50 Show Content Detailed Solution A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in \(^{5}C_{2}\) x \(^{6}C_{3}\) ways = \(\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}}\) = \(\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}}\) = \(\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}}\) = \(\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}}\) = 10 x \(\frac{6 \times 20}{6}\) = 200 There is an explanation video available below.	B

46.	If three unbiased coins are tossed, find the probability that they are all heads A. \(\frac{1}{2}\) B. \(\frac{1}{3}\) C. \(\frac{1}{9}\) D. \(\frac{1}{8}\) Show Content Detailed Solution P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\) Using the binomial prob. distribution, (H + T)3 = H3 + 3H2T1 + 3HT2 + T3 Hence the probability that three heads show in a toss of the three coins is H3 = (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) There is an explanation video available below.	D
47.	Find the standard deviation of 2, 3, 5 and 6 A. √6 B. √10 C. √\(\frac{2}{5}\) D. √\(\frac{5}{2}\) Show Content Detailed Solution \(\begin{array}& x & x - \bar{x} & (x - \bar{x})^2 \\2 & -2 & 4 \\ 3 & -1 & 1 \\ 5 & 1 & 1 \\ 6 & 2 & 4\\ \hline \sum x = 16 & & \sum (x - \bar{x}^2) = 10 \end{array}\) ___________________________________ \(\bar{x}\) = \(\frac{\sum x }{N}\) = \(\frac{16}{4}\) = 4 S = \(\sqrt{\frac {(x - \bar{x})^2}{N}}\) = \(\sqrt{\frac {(10)}{4}}\) = \(\sqrt{\frac {(5)}{2}}\) There is an explanation video available below.	D
48.	W is directly proportional to U. If W = 5 when U = 3, find U when W = \(\frac{2}{7}\) A. \(\frac{6}{35}\) B. \(\frac{10}{21}\) C. \(\frac{21}{10}\) D. \(\frac{35}{6}\) Show Content Detailed Solution W \(\alpha\) U W = ku u = \(\frac{w}{k}\); \(\frac{2}{7}\) x \(\frac{3}{5}\) = \(\frac{6}{35}\) There is an explanation video available below.	A
49.	Determine the value of x for which (x2 - 1)>0 A. x < -1 or x > 1 B. -1 < x < 1 C. x > 0 D. x < -1 Show Content Detailed Solution x(x - 1) > 0 x < -1 or x > 1 There is an explanation video available below.	A