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.

156.

Given that X  : R \(\to\) R is defined by x = \(\frac{y + 1}{5 - y}\) , y \(\in\) R, find the domain of x

A.

{y : y \(\in\) R, y \(\neq\) 0}

B.

{y : y \(\in\) R, y \(\neq\) 1}

C.

{y : y \(\in\) R, y \(\neq\) 5}

D.

{y : y \(\in\) R, y \(\neq\) 7}

Correct answer is C

No explanation has been provided for this answer.

157.

If  \(\begin{pmatrix} p+q & 1\\ 0 & p-q \end {pmatrix}\) = \(\begin{pmatrix} 2 & 1 \\ 0 & 8 \end{pmatrix}\)

Find the values of p and q

A.

p = 5, q = 3

B.

p = 5, q = -3

C.

p = -5, q = -3

D.

p = -5, q = 3

Correct answer is B

p + q = 2

p - q = 8

\(\overline{\frac{2p}{2} = \frac{10}{2}}\)

p = 5

from p + q = 2

5 + q = 2

q = 2 - 5 

= -3

158.

If \(\int^3_0(px^2 + 16)dx\) = 129. Find the value of p

A.

9

B.

8

C.

7

D.

6

Correct answer is A

\(\int^3_0(px^2 + 16)\) = 129

\(\frac{px^2 + 1}{0 + 1} + 16x|^3_0 = 129\)

\(\frac{px^3}{3} + 16x|^3_0 = 129\)

(\(\frac{p(3)^3}{3} + 16(3)\)) - 0 = 129

9p + 48 = 129

9p = 129 - 48 

\(\frac{9p}{9} = \frac{81}{9}\) 

p = 9

159.

If cos x = -0.7133, find the values of x between 0\(^o\) and 360\(^o\) 

A.

44.5\(^o\) , 224.5\(^o\)

B.

123.5\(^o\) , 190.5\(^o\)

C.

135.5\(^o\) , 213.5\(^o\)

D.

135.5\(^o\) , 224.5\(^o\)

Correct answer is D

cos x = -0.7133

x = cos \(^{-1}\)(0.7133) 

= 44.496\(^o\) 

x = 180 - 44.495\(^o\) 

x = 135.5\(^o\)

and x = 180\(^o\)  + 44.495\(^o\) 

= 224.5\(^o\)

160.

Find the inverse of \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)

A.

\(\begin{pmatrix} 5 & 1 \\ -3 & 2 \end{pmatrix}\)

B.

\(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)

C.

\(\begin{pmatrix} -5 & 2 \\ -1 & 3 \end{pmatrix}\)

D.

\(\begin{pmatrix} 5 & 1 \\ 2 & 3 \end{pmatrix}\)

Correct answer is B

Let A = \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)

|A| = (3 x 2 - 5 x 1) 

= 6 - 5

= 1

A\(^{-1}\) = \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)