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WAEC Further Mathematics 2015 Paper

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Question : 31

Total: 40

Given that

y = 4 - 9x

\Delta x = 0.1

\Delta y

.

9.0
0.9
-0.3
-0.9

Solution:

Concept:

For a small change

\Delta x

, the approximate change in

y

is given by

\Delta y \approx \frac{dy}{dx} \cdot \Delta x

.

Explanation:

The given function is

y = 4 - 9x

.

Differentiate

y

with respect to

x

term by term.

The derivative of the constant

4

is

0

.

The derivative of

-9x

is

-9

.

So,

\frac{dy}{dx} = -9

.

For small changes,

\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x}

.

Substitute

\frac{dy}{dx} = -9

and

\Delta x = 0.1

.

Thus,

\frac{\Delta y}{0.1} = -9

.

Multiply both sides by

0.1

:

\Delta y = -9 \times 0.1

.

Therefore,

\Delta y = -0.9

.

Answer:

\Delta y = -0.9

Correct option: D.

-0.9

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