2012 - JAMB Mathematics Past Questions and Answers - page 3

\(\begin{vmatrix} 5 & 3 \ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \ 4 & 5 \end{vmatrix}\)

10 - 3x = 15 - 20

-3x = 15 - 20 - 10

-3x = -15

x = 5

Recall that a unit matrice is a diagonal matrix in which the elements in the leading diagonal is unity. Therefore,

I₃ = \(\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\)

I₃ = \(+1\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \ 0 & 0 \end{vmatrix} \)

I₃ = +1(1 - 0) - 0(0 - 0) + 0(0 - 0)

= 1(1)

= 1

Volume of cuboid = L x b x h

= 0.76cm x 2.6cm x 0.82cm

= 1.62cm³

PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)

P₁ (4, 3), P₂ (x, y)

y = 2x + 4 .....(1)

y = 7 - x .....(2)

Substitute (2) in (1)

7 - x = 2x + 4

7 - 4 = 2x + x

3 = 3x

x = 1

Substitute in eqn (2)

y = 7 - x

y = 7 - 1

y = 6

P₂ (1, 6)

Distance between 2 points is given as

D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)

D = \(\sqrt{(-3)^2 + (3)^2}\)

D = \(\sqrt{9 + 9}\)

D = \(\sqrt{18}\)

D = \(\sqrt{9 \times 2}\)

D = \(3\sqrt{2}\)

\(\frac{y - y_1}{x + x_1}\) = \(\frac{y_2 - y_1}{x - x_1}\)

\(\frac{y - 1}{x + 2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)

= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)

y = 2x + 5

\(\theta\) = 135^o

Cos 135^o = Cos(90 + 45)^o

= cos90^ocos45^o - sin90^osin45^o

= 0cos45^o - (1 x \(\frac{\sqrt{2}}{2}\))

= \(-\frac{\sqrt{2}}{2}\)

y = x² - \(\frac{1}{x}\)

y = x² - x^-1

\(\frac{\delta y}{\delta x}\) = 2x + x^-2


\(\frac{\delta y}{\delta x}\) = 2x + \(\frac{1}{x^2}\)