2019 - JAMB Mathematics Past Questions and Answers - page 8
If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).
y = 8x\(^3\) - 3x\(^2\) + 7x - 1
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)
= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)
\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)
Differentiate \(\frac{2x}{\sin x}\) with respect to x.
Find the distance between the points C(2, 2) and D(5, 6).
= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)
= \(\sqrt{3^2 + 4^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= 5 units
Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)
7y + 4x + 5 = 0
7y - 4x - 5 = 0
3y - 5x + 2 = 0
3y + 5x - 2 = 0
the equation is 4y = 7x + 3
\(\implies y = \frac{7}{4} x + \frac{3}{4}\)
the slope = coefficient of x = \(\frac{7}{4}\)
the slope of a perpendicular line = \(\frac{-1}{\frac{7}{4}}\)
= \(\frac{-4}{7}\)
for perpendicular line passes (-3, 1)
\(\therefore\) using the equation of line \(y = mx + b\)
m = slope
b = intercept.
\(y = \frac{-4}{7} x + b\)
we substitute y = 1 and x = -3 in the equation to find the intercept:
\(1 = \frac{-4}{7} (-3) + b\)
\(1 = \frac{12}{7} + b\)
\(b = \frac{-5}{7}\)
\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)
\(7y + 4x + 5 = 0\)
| Marks | 1 | 2 | 3 | 4 | 5 |
| Frequency | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y |
The table above is the distribution of data with mean equals to 3. Find the value of y.
| Marks (x) | 1 | 2 | 3 | 4 | 5 | |
| Frequency (f) | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y | 3y + 2 |
| fx | 2y - 2 | 2y - 2 | 9y - 12 | 12 - 4y | 30 - 10y | 26 - y |
Mean = \(\frac{\sum fx}{\sum f}\)
\(3 = \frac{26 - y}{3y + 2}\)
\(3(3y + 2) = 26 - y\)
\(9y + 6 = 26 - y\)
\(9y + y = 26 - 6\)
\(10y = 20 \implies y = 2\)