Question

If $x$ is so small that the values of $x^n$, $n\geq2$ are negligible, then the approximate value of \[ \frac{\sqrt{2-3x}}{(3+2x)}(x+1) \] is

• A. $\dfrac{\sqrt{2}}{3}\left(1-\dfrac{5x}{12}\right)$
• B. $\dfrac{\sqrt{2}}{3}(1+5x)$
• C. $\dfrac{\sqrt{2}}{3}(1-7x)$
• D. $\dfrac{\sqrt{2}}{3}(1+6x)$

Hint:

For approximation problems, keep only first-degree terms in $x$ and ignore all higher powers.

Answer 1

The Correct Option is A

Step 1: Set up the approximations.
Since $x$ is very small, drop $x^2$ and higher. We use $(1+u)^n\approx1+nu$ and $\dfrac{1}{1+u}\approx1-u$.
Step 2: Approximate the square root.
$\sqrt{2-3x}=\sqrt2\sqrt{1-\dfrac{3x}{2}}\approx\sqrt2\left(1-\dfrac{3x}{4}\right)$, using power $\dfrac12$.
Step 3: Approximate the denominator.
$\dfrac{1}{3+2x}=\dfrac{1}{3}\cdot\dfrac{1}{1+\frac{2x}{3}}\approx\dfrac{1}{3}\left(1-\dfrac{2x}{3}\right)$.
Step 4: Multiply the three factors.
The expression is $\dfrac{\sqrt2}{3}\left(1-\dfrac{3x}{4}\right)\left(1-\dfrac{2x}{3}\right)(1+x)$.
Step 5: Keep only first-order terms.
Add the small parts: $-\dfrac{3x}{4}-\dfrac{2x}{3}+x$. Over a common base of $12$ this is $\dfrac{-9x-8x+12x}{12}=-\dfrac{5x}{12}$.
Step 6: Write the final form.
So the value is $\dfrac{\sqrt2}{3}\left(1-\dfrac{5x}{12}\right)$. \[ \boxed{\frac{\sqrt2}{3}\left(1-\frac{5x}{12}\right)} \]