JAMB Mathematics Past Questions & Answers - Page 334

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

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

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

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

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

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

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

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

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

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

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

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

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,

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

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

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

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