2014 - JAMB Mathematics Past Questions and Answers - page 4

If y = 4x³ - 2x² + x, then;

(\frac{\delta y}{\delta x}) = 3(4x²) - 2(2x) + 1

= 12x² - 4x + 1

y = cos 3x

Let u = 3x so that y = cos u

Now, (\frac{\delta y}{\delta x} = 3),

(\frac{\delta y}{\delta x} = -sin u)

By the chain rule,

(\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x})

(\frac{\delta y}{\delta x} = (-\sin u) (3))

(\frac{\delta y}{\delta x} = -3 \sin u)

(\frac{\delta y}{\delta x} = -3 \sin 3x)

y = x² - 2x - 3,

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

But at minimum point,(\frac{\delta y}{\delta x} = 0),

Which means 2x - 2 = 0

2x = 2

x = 1.

Hence the minimum value of y = x² - 2x - 3 is;

y_min = (1)² - 2(1) - 3

y_min = 1 - 2 - 3

y_min = -4

(\int \sin 2x dx = \frac{1}{2} (-\cos 2x) + k)

(- \frac{1}{2} \cos 2x + k)

(\int (2x + 3)^{\frac{1}{2}} \delta x)

let u = 2x + 3, (\frac{\delta y}{\delta x} = 2)

(\delta x = \frac{\delta u}{2})

Now (\int (2x + 3)^{\frac{1}{2}} \delta x = \int u^{\frac{1}{2}}.{\frac{\delta x}{2}})

( = \frac{1}{2} \int u^{\frac{1}{2}} \delta u)

( = \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k)

( = \frac{1}{3} u^{\frac{3}{2}} + k)

( = \frac{1}{3} (2x + 3)^{\frac{3}{2}} + k)

Mean x = (\frac{\sum x}{n})

= [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1) (\div)] 5

= [11 - 1 + 3t - 3t] (\div) 5

= 10 (\div) 5

= 2

First arrange the numbers in order of magnitude;

1,2,3,3,4,5,5,5,5,6,7,8,9,9,10

Hence the median = 5

Mean x = (\frac{\sum x}{n})

( = \frac{5 + 4 + 3 + 2 + 1}{5})

( = \frac{15}{5})

= 3

(\begin{array}{c|c}

x & d = x - 3 & d^2 \

\hline

5 & 2 & 4 \

4 & 1 & 1 \

3 & 0 & 0 \

2 & -1 & 1 \

1 & -2 & 4 \

\hline

& & \sum d^2 + 10

\end{array})

Hence, standard deviation;

( = \sqrt{\frac{\sum d^2}{n}} = \sqrt{\frac{10}{5}})

( = \sqrt{2})

A team of 2 girls can be selected from 7 girls in (^7C_3)

( = \frac{7!}{(7 - 3)! 3!})

( = \frac{7!}{4! 3!} ways)