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WAEC Further Mathematics 2014 Paper

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Question : 16

Total: 40

A function is defined by

h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}

h^{-1}\left(\frac{1}{2}\right)

.

$6$
$\frac{11}{6}$
$\frac{11}{4}$
$\frac{5}{3}$

Solution:

Concept:

The inverse value

h^{-1}\left(\frac12\right)

is the input

x

that makes

h(x)=\frac12

.

So solve the equation directly instead of deriving the full inverse.

Explanation:

Given:

h(x)=2-\frac{1}{2x-3}

, for

x\neq \frac32

.

Set

h(x)=\frac12

:

2-\frac{1}{2x-3}=\frac12

.

Subtract

2

from both sides:

-\frac{1}{2x-3}=-\frac32

.

Multiply both sides by

-1

:

\frac{1}{2x-3}=\frac32

.

Taking reciprocals gives:

2x-3=\frac23

.

Add

3

to both sides:

2x=\frac{11}{3}

.

Divide by

2

:

x=\frac{11}{6}

.

Verification:

h\left(\frac{11}{6}\right)=2-\frac{1}{\frac{11}{3}-3}=2-\frac{1}{\frac23}=2-\frac32=\frac12

.

Therefore,

h^{-1}\left(\frac12\right)=\frac{11}{6}

.

Answer:

\frac{11}{6}

(Option B).

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