2013 - JAMB Mathematics Past Questions and Answers - page 2

Let f(p) = 6p³ - p² - 47p + 30

Then by the remainder theorem,

(p - 3): f(3) = remainder R,

i.e. f(3) = 6(3)³ - (3)² - 47(3) + 30 = R

162 - 9 - 141 + 30 = R

192 - 150 = R

R = 42

P \(\propto\) mu, p \(\propto \frac{1}{q}\)

p = muk ................ (1)

p = \(\frac{1}{q}k\).... (2)

Combining (1) and (2), we get

P = \(\frac{mu}{q}k\)

4 = \(\frac{m \times u}{1}k\)

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

So, P = \(\frac{mu}{q} \times \frac{2}{3} = \frac{2mu}{3q}\)

Hence, P = \(\frac{2 \times 6 \times 4}{3 \times \frac{8}{5}}\)

P = \(\frac{2 \times 6 \times 4 \times 5}{3 \times 8}\)

p = 10

\(r \propto \frac{1}{\sqrt{s}}, r \propto \frac{1}{\sqrt{t}}\)

\(r \propto \frac{1}{\sqrt{s}}\) ..... (1)

\(r \propto \frac{1}{\sqrt{t}}\) ..... (2)

Combining (1) and (2), we get

\(r = \frac{k}{\sqrt{s} \times \sqrt{t}} = \frac{k}{\sqrt{st}}\)

This gives \(\sqrt{st} = \frac{k}{r}\)

By taking the square of both sides, we get

st = \(\frac{k^2}{r^2}\)

s = \(\frac{k^2}{r^{2}t}\)

3(x + 2) > 6(x + 3)

3x + 6 > 6x + 18

3x - 6x > 18 - 6

-3x > 12

x < -4

|x - 2| < 3 implies

-(x - 2) < 3 .... or .... +(x - 2) < 3

-x + 2 < 3 .... or .... x - 2 < 3

-x < 3 - 2 .... or .... x < 3 + 2

x > 1 .... or .... x < 1

combining the two inequalities results, we get;

-1 < x < 5

Using S_n = \(a\frac{r^2 - 1}{r - 1}\)

we get S₂ = 3 = \(a\frac{r^2 - 1}{r - 1}\)

giving 3(r - 1) = a(r² - 1)

3r - 3 = ar² - a

ar² - 3r - a = -3 ..... (1)

ar + ar² = -6 ..... (2)

From (2), a = \(\frac{-6}{(r + r^2)}\)

Substitute \(\frac{-6}{(r + r^2)}\) for a in (1)

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

Multiply through by (r + r²) to get

-6r² - 3r(r + r²) + 6 = -3(r + r²)

-6r² - 3r² - 3r³ + 6 = -3r - 3r²

Equating to zero, we have

3r³ - 3r² + 3r² + 6r² - 3r - 6 = 0

This reduces to;

3r³ + 6r² - 3r - 6 = 0

3(r³ + 2r² - r - 2) = 0

By the factor theorem,

(r + 2): f(-2) = (-2)³ + 2(-2)² - (-2) - 2

-8 + 8 + 2 - 2 = 0

giving r = -2 as the only valid value of r for the G.P.

From (3), = \(\frac{-6}{-2 + (-2)^2} = \frac{-6}{-2 + 4}\)

a = -6/2 = -3

Hence (a + r) = (-3 + -2) = -5

Using T_n = \(\frac{3n + 1}{n + 1}\),

T₁ = \(\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}\)

T₂ = \(\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}\)

T₃ = \(\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}\)

x * y = x + 2y (given)

3 * 4 = 3 + 2(4) = 11

Hence, 2 * (3 * 4) = 2 * 11

= 2 + 2(11)

= 2 + 22

= 24

2P + Q = 2\(\begin{pmatrix} 5 & 3 \ 2 & 1 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \ 3 & 5 \end{pmatrix}\)

= \(\begin{pmatrix} 10 & 6 \ 4 & 2 \end{pmatrix}\) + \(\begin{pmatrix} 4 & 2 \ 3 & 5 \end{pmatrix}\)

= \(\begin{pmatrix} 14 & 8 \ 7 & 7 \end{pmatrix}\)

Let A = \(\begin{pmatrix} 5 & 3 \ 6 & 4 \end{pmatrix}\)

Then |A| = \(\begin{pmatrix} 5 & 3 \ 6 & 4 \end{pmatrix}\) = 20 - 18 = 2

Hence A^-1 = \(\frac{1}{|A|}\begin{pmatrix} 4 & -3 \ -6 & 5 \end{pmatrix}\)

= \(\frac{1}{2}\begin{pmatrix} 4 & -3 \ -6 & 5 \end{pmatrix}\)

= \(\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}\)

= \(\begin{vmatrix} 2 & -\frac{3}{2} \ -3 & \frac{5}{2} \end{vmatrix}\)