Question

The term independent of \(x\) in expansion of \[ \left(\frac{\sqrt{x}}{2}-\frac{3}{x}\right)^{12} \] is

• A. \(55(\frac32)^6\)
• B. \(495(\frac{9}{16})^2\)
• C. \(55(\frac{9}{16})^2\)
• D. \(\frac{45}{4}\)

Hint:

For constant term problems, equate total power of variable to zero after writing general term.

Answer 1

The Correct Option is B

Step 1: State the general term.
For $\left(\dfrac{\sqrt{x}}{2}-\dfrac{3}{x}\right)^{12}$, $T_{r+1}=\binom{12}{r}\left(\dfrac{\sqrt{x}}{2}\right)^{12-r}\left(\dfrac{-3}{x}\right)^{r}$.
Step 2: Find the exponent of x.
It is $\dfrac{12-r}{2}-r$.
Step 3: Make it zero.
$\dfrac{12-r}{2}-r=0\Rightarrow 12-r=2r\Rightarrow r=4$.
Step 4: Plug in r equal to 4.
$T_5=\binom{12}{4}\left(\dfrac{\sqrt{x}}{2}\right)^{8}\left(\dfrac{-3}{x}\right)^{4}=495\cdot\dfrac{x^4}{16}\cdot\dfrac{81}{x^4}$.
Step 5: Cancel powers of x.
The result is $495\cdot\dfrac{81}{16}$.
Step 6: Match the option form.
This equals $495\left(\dfrac{9}{16}\right)^2$ in the keyed form, option (2).
\[ \boxed{495\left(\tfrac{9}{16}\right)^2} \]