WAEC Further Mathematics Past Questions & Answers - Page 5

8√2, 16√2, 32√2, ..

\(a = 8\sqrt2; r =\frac{T_2}{T_1}=\frac{16\sqrt2}{8\sqrt2}=2\)

\(T_n=ar^{n-1}\)

\(T_n=8\sqrt2 \times 2^{n-1}\)

\(T_n=2^3\times2^{n-1}\times\sqrt2\)

\(T_n=2^{3+n-1}\times\sqrt2\)

\(\therefore T_n= 2^{(n+2)}\sqrt2\)

6 sin 2θ tan θ = 4, where 0º < θ < 90º

sin 2θ = 2sin θ cos θ and tanθ = \(\frac{sinθ}{cosθ}\)

= 6 x 2sin θ cos θ x \(\frac{sin θ}{cos θ} = 4\)

= \(sin^2 θ = 4\)

= \(sin^2 θ = \frac{4}{12}=\frac{1}{3}\)

=\(sin θ = \frac{\sqrt1}{3}=\frac{1}{\sqrt3}\)

= \(θ = sin^{-1}(\frac{1}{\sqrt3})\)

∴ θ = 35.26º

r = (10 N, 200º) and n = (16 N, 020º)

In rectangular form:

r = 10cos 200ºi + 10sin 200ºj = -9.397i - 3.420j

n = 16cos 20ºi + 16sin 20ºj = 15.035i + 5.472j

3r = -28.191i - 10.260j

2n = 30.070i + 10.945j

3r - 2n = (-28.191i - 10.260j) - (30.070i + 10.945j)

3r - 2n = -58.261i - 21.205j

|3r - 2n| = √((-58.261)\(^2\) + (-21.205)\(^2\)) = 62 N

\(tan θ =\frac{-21.205}{-58.261} = 0.3640\)

\(θ = tan^{-1} (0.3640) = 20^o\)

∴ (62 N , 020º)

\(\frac{log √27 - log √8}{log 3 - log 2}\)

= \(\frac{log √3^3 - log √2^3}{log 3 - log 2}\)

= \(\frac{log3^{3/2} - log2^{3/2}}{log3}\)

=\(\frac{^3/_2(log 3 - log 2)}{log 3 - log 2}\)

\(\therefore\frac{3}{2}\)

 x̄ \(=\frac{∑fx}{∑f}= 3\)

\(=\frac{(1 \times 2)+(2 \times k)+(3 \times 1)+(4 \times 1)+(5 \times 2)}{2 + k + 1 + 1 + 2}= 3\)

\(=\frac{2 + 2k + 3 + 4 + 10}{6 + k} = 3\)

\(=\frac{19 + 2k}{6 + k} = 3\)

\(=\frac{19 + 2k}{6 + k} = \frac{3}{1}\)

=19+2k=3(6+k)

=19+2k=18+3k

=2k-3k=18-19

=-k=-1

∴k=1