JAMB Mathematics Past Questions & Answers - Page 312

y = x3 + x2 - x + 1

$\frac{dy}{dx}$ = $\frac{d(x^3)}{dx}$ + $\frac{d(x^2)}{dx}$ - $\frac{d(x)}{dx}$ + $\frac{d(1)}{dx}$

$\frac{dy}{dx}$ = 3x2 + 2x - 1 = 0

$\frac{dy}{dx}$ = 3x2 + 2x - 1

At the maximum point $\frac{dy}{dx}$ = 0

3x2 + 2x - 1 = 0

(3x2 + 3x) - (x - 1) = 0

3x(x + 1) -1(x + 1) = 0

(3x - 1)(x + 1) = 0

therefore x = $\frac{1}{3}$ or -1

For the maximum point

$\frac{d^2y}{dx^2}$ < 0

$\frac{d^2y}{dx^2}$ 6x + 2

when x = $\frac{1}{3}$

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

= 2 + 2 = 4

$\frac{d^2y}{dx^2}$ > o which is the minimum point

when x = -1

$\frac{d^2y}{dx^2}$ = 6(-1) + 2

= -6 + 2 = -4

-4 < 0

therefore, $\frac{d^2y}{dx^2}$ < 0

the minimum point is 1/3

Mode = L₁ + ($\frac{D_1}{D_1 + D_2}$)C

D₁ = frequency of modal class - frequency of the class before it

D₁ = 5 - 2 = 3

D₂ = frequency of modal class - frequency of the class that offers it

D₂ = 5 - 3 = 2

L₁ = lower class boundary of the modal class

L₁ = 5 - 5

C is the class width = 8 - 5.5 = 3

Mode = L₁ + ($\frac{D_1}{D_1 + D_2}$)C

= 5.5 + $\frac{3}{2 + 3}$C

= 5.5 + $\frac{3}{5}$ x 3

= 5.5 + $\frac{9}{5}$

= 5.5 + 1.8

= 7.3 $\approx$ = 7

2x + 1 $\frac{d(3x + 1)}{\mathrm d x}$ + (3x + 1) $\frac{d(2x + 1)}{\mathrm d x}$

2x + 1 (3) + (3x + 1) (2)

6x + 3 + 6x + 2 = 12x + 5

tan$\theta$ = $\frac{100}{100}$ = 1

$\theta$ = tan^-1(1) = 45^o

The bearing of x from z is N45^oE or 135^o

tan$\theta$ = $\frac{3}{4}$

from Pythagoras tippet, the hypotenus is T

i.e. 3, 4, 5.

then sin $\theta$ = $\frac{3}{5}$ and cos$\theta$ = $\frac{4}{5}$

cos$\theta$ - sin$\theta$

$\frac{4}{5}$ - $\frac{3}{5}$ = $\frac{1}{5}$