Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

1,761.

The vectors a and b are given in terms of two perpendicular units vectors i and j on a plane by a = 2i - 3j, b = -i + 2j. Find the magnitude of the vector a + 3b

A.

2

B.

4

C.

\(\sqrt{10}\)

D.

35

E.

2.2

Correct answer is C

a = 2i - 3j - 3i + 6j

= -i + 3j

= \(\sqrt{5 \times 2}\)

= \(\sqrt{10}\)

1,763.

Evaluate without using tables sin(-1290º)

A.

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

B.

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

C.

\(\frac{2}{2}\)

D.

1

E.

\(\frac{1}{2}\)

Correct answer is E

sin(-1290º) = -sin(1290º)

sin([3*360] + 210)

where sin 360 = 0 and sin 210 = 180 + 30 ⇔ -30º

-sin ([3* 0] + [-30]) 

-sin(-30)

sin30 = \(\frac{1}{2}\)

1,764.

If x4 - kx3 + 10x2 + 1x - 3 is divisible by (x - 1), and if when it is divided by (x + 2) the remainder is 27, find the constants k and 1

A.

k = -7, 1 = -15

B.

k = -15, 1 = -7

C.

k = \(\frac{15}{3}\) , 1 = -7

D.

k = \(\frac{7}{3}\) , 1 = -17

Correct answer is A

If k = -7 is put as -15, the equation x4 - kx3 + 10x2 + 1x - 3 becomes x4 - (7x3) + 10x2 + (15)-3 = x4 + 7x3 + 10x2 - 15x - 3

This equation is divisible by (x - 1) and (x + 2) with the remainder as 27

k = -7, 1 = -15

1,765.

Simplify \(\frac{(a^2 - \frac{1}{a}) (a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)

A.

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

B.

a-\(\frac{1}{3}\)

C.

\(a^{2}\) + 1

D.

a

E.

a\(\frac{1}{3}\)

Correct answer is E

\(\frac{(a^2 - \frac{1}{a})(a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)

= \(\frac{(\frac{a^2 - 1}{a})(\frac{a^2 + 1}{a^{\frac{2}{3}}})}{a^{4} - \frac{1}{a^2}}\)

= \(\frac{a^4 - 1}{a^{\frac{5}{3}}}\) x \(\frac{a^2}{a^4 - 1}\)

= a\(\frac{1}{3}\)