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

Concept: For very small $x$, neglect: \[ x^2,x^3,\ldots \] Useful approximations: \[ (1+x)^n\approx1+nx \] \[ \frac{1}{1+x}\approx1-x \]

Step 1: Simplify the square root term.
\[ \sqrt{2-3x} = \sqrt{2} \sqrt{1-\frac{3x}{2}} \] Using: \[ \sqrt{1+t}\approx1+\frac{t}{2} \] we get: \[ \sqrt{1-\frac{3x}{2}} \approx 1-\frac{3x}{4} \] Hence: \[ \sqrt{2-3x} \approx \sqrt{2}\left(1-\frac{3x}{4}\right) \]

Step 2: Simplify denominator.
\[ \frac{1}{3+2x} = \frac{1}{3} \cdot \frac{1}{1+\frac{2x}{3}} \] Using: \[ \frac{1}{1+t}\approx1-t \] \[ \approx \frac{1}{3} \left( 1-\frac{2x}{3} \right) \]

Step 3: Multiply all factors.
Expression becomes: \[ \frac{\sqrt{2}}{3} \left( 1-\frac{3x}{4} \right) \left( 1-\frac{2x}{3} \right) (1+x) \] Neglecting higher powers: \[ = \frac{\sqrt{2}}{3} \left[ 1-\frac{3x}{4}-\frac{2x}{3}+x \right] \] \[ = \frac{\sqrt{2}}{3} \left[ 1-\frac{9x+8x-12x}{12} \right] \] \[ = \frac{\sqrt{2}}{3} \left( 1-\frac{5x}{12} \right) \] Hence, \[ \boxed{ \dfrac{\sqrt{2}}{3} \left( 1-\dfrac{5x}{12} \right) } \]