Further Mathematics questions and answers

Further Mathematics Questions and Answers

Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.

116.

Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)

A.

63.616

B.

63.167

C.

62.628

D.

62.629

Correct answer is B

Put 1 + (0.998) = 1 + x;

x = (0.998)

Hence; 1 + 6(0.998) + 15(0.998)(0.998)\(^2\) + 20(0.998)\(^3\) + 15(0.998)\(^4\) + 6(0.998)\(^5\) + (0.998)\(^6\)

= 1 + 5.988 + 14.790 + 19.880 + 14.880 + 5.940 + 0.990

≈ 63.167

117.

g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4

find f {g (-3)}.

A.

37

B.

1

C.

-3

D.

-179

Correct answer is A

g(-3) = 2(-3) + 3 = -6 + 3 = -3

F(-3) = 3 (-3)\(^2\) - 2(-3) + 4

= 27 + 6 + 4 = 37

118.

Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\) 

A.

\(\frac{7}{5}\)

B.

\(\frac{6}{7}\)

C.

\(\frac{-7}{6}\)

D.

\(\frac{-6}{7}\)

Correct answer is A

(\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)

3\(^{-2(x + 2)}\) =  3\(^{5(x - 2)}\);

-2(x+2) = 5(x - 2)

-2x -4 = 5x - 10

-7x = -6

x =  \(\frac{6}{7}\)

119.

Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

A.

log \(\frac{2}{7}\)

B.

log 2

C.

log \(\frac{2}{9}\)

D.

log 9

Correct answer is C

\(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

= log 2 + log 4 - log 36

= log \(\frac{8}{36}\)

= log \(\frac{2}{9}\)

120.

Given that M = \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) and N = \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\), calculate (3M - 2N)

A.

\(\begin{pmatrix} 1 & 6 \\ 1 & 18 \end{pmatrix}\)

B.

\(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\)

C.

\(\begin{pmatrix} 1 & 6 \\ -1 & -18 \end{pmatrix}\)

D.

\(\begin{pmatrix} -1 & -6 \\ -1 & -18 \end{pmatrix}\)

Correct answer is B

3M = 3 \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\)

 = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\)

2N = 2 \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) 

= \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)

(3M - 2N) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)