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

Concept: General term in binomial expansion: \[ T_{r+1}= \binom nr a^{n-r}b^r \] For constant term, power of variable must become zero.

Step 1: Write general term.
\[ T_{r+1} = \binom{12}{r} \left(\frac{\sqrt{x}}2\right)^{12-r} \left(\frac{-3}{x}\right)^r \]

Step 2: Find power of x.
Power from first factor \[ x^{(12-r)/2} \] Power from second factor \[ x^{-r} \] Total exponent \[ \frac{12-r}{2}-r \] Constant term means \[ \frac{12-r}{2}-r=0 \] \[ 12-r=2r \] \[ 12=3r \] \[ r=4 \]

Step 3: Substitute into term.
\[ T_5= \binom{12}{4} \left(\frac{\sqrt{x}}2\right)^8 \left(\frac{-3}{x}\right)^4 \] \[ =495\times\frac{x^4}{16}\times\frac{81}{x^4} \] \[ =495\times\frac{81}{16} \] \[ =495\left(\frac{9}{4}\right)^2 \] \[ =495\left(\frac{9}{16}\right)^2 \] Thus \[ \boxed{495\left(\frac{9}{16}\right)^2} \]