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 \(u = 2x + 1\)

\(\frac{du}{dx}=2==>du=2dx==>dx=\frac{du}{2}\)

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

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

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