WAEC Further Mathematics Past Questions & Answers - Page 31

151.

If V = plog\(_x\), (M + N), express N in terms of X, P, M and V

A.

N = X\(^{\frac{v}{p}}\) - M

B.

N = X\(^{\frac{p}{v}}\) - M

C.

N = X\(^{\frac{v}{p}}\) + M

D.

N = X\(^{\frac{p}{v}}\) + M

Correct answer is A

\(\frac{v}{p} = \frac{p}{p} log _x(M + N)\)

\(\log_x(M + N) = \frac{v}{p}\)

\(x^{\frac{v}{p}} = M + N\)

N = X\(^{\frac{v}{p}}\) - M 

152.

Given that 2x + 3y - 10 and 3x = 2y - 11, calculate the value of (x - y). 

A.

5

B.

3

C.

-3

D.

-5

Correct answer is D

2x + 3y = 10 .......x2

3x - 2y = -11 ........x3

4x + 6y = 20 

9x - 6y = -33

\(\overline{\frac{13x}{13} = \frac{-13}{13}}\), x = 1

from 

2x + 3y = 10

2(-1) + 3y = 10

-2 + 3y = 10

3y = 10 + 2

\(\frac{3y}{3} = \frac{12}{3}\), y = 4

x - y = -1 - 4

= -5

153.

Differentiate \(\frac{x}{x + 1}\) with respect to x. 

A.

\(\frac{x}{x + 1}\)

B.

\(\frac{-1}{x + 1}\)

C.

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

D.

\(\frac{1}{(x + 1)^2}\)

Correct answer is D

\(\frac{d}{dx}(\frac{x }{x + 1}\)) = \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\) 

u = x, \(\frac{du}{dx}\) = 1, v = x + 1, \(\frac{dv}{dx}\) = 1

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

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

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

154.

If \(\frac{6x + k}{2x^2 + 7x - 15}\)  = \(\frac{4}{x + 5} - \frac{2}{2x - 3}\). Find the value of k. 

A.

- 21

B.

- 22

C.

- 24

D.

- 25

Correct answer is B

\(\frac{6x + k}{2x^2 + 7x - 15} = \frac{4}{x + 5} - \frac{2}{2x - 3}\)

6x + k = 4 (2x - 3) - 2(x + 5)

6x + k = 8x - 12 - 2x - 10 

6x + k = 6x - 22 

k = - 22 

155.

Simplify; \(\frac{\sqrt{5} + 3}{4 - \sqrt{10}}\) 

A.

\(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + 2

B.

\(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\)

C.

\(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2

D.

\(\frac{2}{3}\)\(\sqrt{5}\) - \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2

Correct answer is C

\(\frac{(\sqrt{5} + 3)(4 + \sqrt{10})}{(4 - \sqrt{10})(4 + \sqrt{10})}\)

= \(\frac{4\sqrt{5} + \sqrt{50} + 12 + 3\sqrt{10}}{4^2 - (\sqrt{10})^2}\)

= \(\frac{4\sqrt{5} + 5\sqrt{2} + 12 + 3\sqrt{10}}{16 - 10}\)

= \(\frac{4 \sqrt{5}}{6} + \frac{5 \sqrt{2}}{6} + \frac{12}{6} + \frac{3\sqrt{10}}{6}\)

= \(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2