WAEC Further Mathematics Past Questions & Answers - Page 3

$m=4.56kg;t=16s;v_1=(10ms^{-1},060^o);v_2=(50ms^{-1},060^o)$

Notice that there is no change in the direction of the velocity

So,we don't need to write it in a vector form

$a=\frac{∆v}{t}=\frac{50 - 10}{16}$

$a=\frac{40}{16}=2.5ms^{-2}$

F = ma = 4.56 x 2.5

∴ F = 11.4 N

$\frac{3x + 4}{(x - 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x - 2}$

$\frac{3x + 4}{(x - 2)(x + 3)}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}$

=$3x+4=P(x-2)+Q(x+3)$

=$3x+4=Px-2P+Qx+3Q$

=$3x+4=Px+Qx-2P+3Q$

=$3x+4=(P+Q)x-2P+3Q$

Equating x and the constants

P+Q=3------(i)

2P+3Q=4-------(ii)

From equation (i)

P=3-Q------(iii)

Substitute (3-Q) for P inequation (ii)

=-2(3-Q)+3Q=4

=-6+2Q+3Q=4

=-6+5Q=4

=5Q=4+6

=5Q=10

∴Q=$\frac{10}{5}=2$

The general form of a quadratic equation is:

$x^2$-(sum of roots) = 0

For equal roots it's:$x^2 - 2(α)+(α)^2=0$

$3x^2+px+12=0$

Divide through by 3

= $x^2+\frac{p}{3}x+4=0$

=$x^2-(-\frac{p}{3})x+4=0$

$So,α^2=4$

= α = √4 = ±2

Also,-$\frac{p}{3}= 2α$

When α = 2

-$\frac{p}{3} = 2(2)=4$

=p=-12

When α =-2

-$\frac{p}{3}=2(-2)=-4$

= p = 12

∴ values of p = ±12

The general form of a quadratic equation is:

$x^2$ -(sum of roots)$x$ +(product of roots) = 0

$7x^2+12x-4=0$

Divide through by 7

=$x^2+\frac{12}{7}x-\frac{4}{7}=0$

=$x^2-(-\frac{12}{7})x+(-\frac{4}{7})=0$

$\therefore$ sum of roots = $-\frac{12}{7}$, and products of roots =$-\frac{4}{7}$

α + β = $-\frac{12}{7}, αβ = -\frac{4}{7}$

$\frac{αβ}{(α + β)^2} = \frac{\frac{-4}{7}}{(\frac{-12}{7})^2}$

=$\frac{\frac{-4}{7}}{\frac{144}{49}}=-\frac{4}{7}\times\frac{49}{144}$

$\therefore - \frac{7}{36}$

Using substitution method, Let $u = 2x + 1$

$\frac{du}{dx}=2==>du=2dx==>dx=\frac{du}{2}$

=$\int\frac{u^3}{2} du = \frac{1}{2}\int u^3 du$

=$\frac{1}{2}(\frac{u^4}{4})=\frac{u^4}{8}$

$\therefore\frac{1}{8} (2x + 1)^4 + k$