JAMB Mathematics Past Questions & Answers - Page 12

log (y + 8) + log (y - 8) = 2log 3 + 2log 5

⇒ log (y + 8)(y - 8) = 2log 3 + 2log 5 (log ab = log a + log b)

⇒ log (y$^2$ -8y + 8y - 64) = log 3$^2$ + log 5$^2$

⇒ log (y$^2$ - 64) = log 3$^2$ + log 5$^2$

⇒ log (y$^2$ - 64) = log 9 + log 25

⇒ log (y$^2$ - 64) = log (9 × 25)

⇒ log(y$^2$ - 64) = log 225

⇒ y$^2$ - 64 = 225

⇒ y$^2$ = 225 + 64

⇒ y$^2$ = 289

⇒ y = ±√289

∴ y = ±17

$402_y = 102_ten$

⇒ 4 x y$^2$ + 0 x y$^1$ + 2 x y$^0$ = 102

⇒ 4y$^2$ + 0 + 2 x 1 = 102

⇒ 4y$^2$ + 2 = 102

⇒ 4y$^2$ = 102 - 2

⇒ 4y$^2$ = 100

⇒ y$^2 = \frac {100}{4}$

⇒ y$^2$ = 25

∴ y = √25 = 5

To find the point of intersection, equate the two equations

⇒ 2x$^2$ + 10 = 4x + 16

⇒ 2x$^2$ + 10 - 4x - 16 = 0

⇒ 2x$^2$ - 4x - 6 = 0

Factorize

⇒ 2(x + 1)(x - 3) = 0

∴ x = -1 or 3

The curves will intersect at -1 and 3

Area = $\int^b_a$ [upper function] - [lower function] dx

= $\int^3_1 4x + 16 - (2x^2 + 10) dx$

= $\int^3_1 -2x^2 + 4x + 6dx$

= $(\frac {-2x^3}{3} + 2x^2 + 6x)^3_1$

= $(\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))$

= 18 - $(\frac {10}{3})$

= 18 + $\frac {10}{3}$

= $\frac {64}{3}$

The first five prime numbers are 2, 3, 5, 7 and 11

M.D = $\frac {\sum (x - x̄)}{n}$

n = 5

$x̄ = \frac {\sum x}{n} = \frac {2 + 3 + 5 + 7 + 11}{5} = \frac {28}{5} = 5.6$

M.D = $\frac {(2 - 5.6) + (3 - 5.6) + (5 - 5.6) + (7 - 5.6) + (11 - 5.6)}{5}$

= $\frac {(-3.6) + (-2.6) + (-0.6) + (1.4) + (5.4)}{5} = \frac {13.6}{5}$

Take the absolute value of all numbers in bracket

= $\frac {3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac {13.6}{5}$

$\therefore M.D = 2.72$

y = $\sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)$

= $2x^{5/3} - x^{8/3}$

Now, we can differentiate the function

$\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}$