JAMB Mathematics Past Questions & Answers - Page 340

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,

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

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

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

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

If y = 4x³ - 2x² + x, then;

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

= 12x² - 4x + 1

No explanation has been provided for this answer.

y - 4x + 3 = 0

When y = 0, 0 - 4x + 3 = 0

Then -4x = -3

x = 3/4

So the line cuts the x-axis at point (3/4, 0).

When x = 0, y - 4(0) + 3 = 0

Then y + 3 = 0

y = -3

So the line cuts the y-axis at the point (0, -3)

Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

\([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]\)

\([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]\)

\([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]\)

\([\frac{3}{8}, \frac{-3}{2}]\)