2007 - JAMB Mathematics Past Questions and Answers - page 2

(3 x 5) + (4(y - 1)) + (5y) + (6 x 9) + (7 x 4) + 8 = 200

15 + 4y - 4 + 5y + 54 + 28 + 8 = 200

9y + 105 - 4 = 200

9y + 101 = 200

9y = 200 - 101

9y = 99

y = 11

mean((\bar{x})) = (\frac{\sum x}{n})

= (\frac{7 + 5 + 2 + 1 + 2 + 11}{6}) = 10

M.D = (\frac{7 + 5 + 2 + 1 + 2 + 11}{6})

= (\frac{28}{6})

= 4.7

a = 2, r = (\frac{{\frac{3}{2}}}{3})

= (\frac{3}{4})

S(\infty) = (\frac{a}{1 - r})

(\frac{2}{1 - \frac{3}{4}})

= (\frac{2}{\frac{1}{4}})

= 8

W (\alpha) L², (\frac{W}{L^2}) = k(constant)

(\frac{6}{42}) = k

(\frac{W}{(\sqrt{17})^2})

W = (\frac{17 \times 6}{16})

= (\frac{51}{8})

= 6(\frac{3}{8})

a (\Delta) b = a + b + 1

a (\Delta) a^-1 = e

7 (\Delta) 7^-1 = 7 + 7^-1

therefore 7^-1 = -1 - 8

= -9

the inverse of 7 under the operation (\Delta) is -9

-3 (x - 2) < -2(x + 3), -3x + 6 < -2x - 6

6 + 6 < 3x - 2x, 12 < x or x > 12

f(x) = 3x - 2, f(P) = 3p - 2I

= 3(\begin{pmatrix} 2 & 1 \ -1 & 0 \end{pmatrix}) - 2(\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}) = (\begin{pmatrix} 6 & 3 \ -3 & 0 \end{pmatrix}) - (\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix})

= (\begin{pmatrix} 6 & -2 \ -3 & 0 \end{pmatrix})(\begin{pmatrix} 3 & -0 \ 0 & -2 \end{pmatrix}) = (\begin{pmatrix} 4 & 3 \ -3 & -2 \end{pmatrix})

2t² + t - 15 = 2t² - 5t + 6t - 15

= t(2t - 5) + 3(2t - 5) = (t + 3)(2t - 5)

x (\oplus) y = xy + x + y

(-(\frac{3}{4})) (\oplus) 6 = -(\frac{3}{4.6}) - (\frac{3}{4}) + 6

= -(\frac{9}{2}) - (\frac{3}{4}) + 6

= -(\frac{3}{4})

(x² + x - 12) (\geq) 0 , (x - 3)(x + 4) (\geq) 0

For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign

.i.e. x - 3 (\geq) 0 and x + 4 (\geq) 0

or x - 3(\leq) 0 and x + 4 (\leq) 0

when x (\geq) 3, the condition is satisfied

when x (\geq) -4, the condition is not satisfied.

when x (\leq) 3, the condition is not satisfied

when x (\leq) -4 , the condition is not satisfied. Thus, the solution of the inequality is x (\geq) 3 or x (\leq) -4 ,