Question

If the constant term in the binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<$ is an odd number, then $|\alpha l-\beta|$ is equal to_____

Hint:

In binomial expansions, ensure to calculate the general term, and use the conditions on the powers of \( x \) to find the relevant term. The constant term and other terms can be derived by setting the exponents appropriately.

Answer 1

Correct Answer: 98

In, ${\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell }}\right)}^9$
$T_{r+1}={}^9{C}_r\frac{{\left(x^{5/2}\right)}^{9-r}}{2^{9-r}}{\left(\frac{-4}{x^{\ell }}\right)}^r$
$=(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r{x}^{\frac{45}{2}-\frac{5r}{2}-r}$
$=45-5r-2lr=0$
$r=\frac{45}{5+21}$ ....(1)
Now, according to the question, $(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r=-84$
$=(-1{)}^r{}^9{C}_r{2}^{3r-9}=21\times 4$
Only natural value of $r$ possible if $3r-9=0$
$r=3$ and ${}^9{C}_3=84$
$\therefore 1=5$ from equation (1)
Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5r}{2}-\mathrm{lr}}$ at $1=5$, gives
$r=5$
$\therefore {}^9{c}_5(-1)\frac{4^5}{2^4}=2^{\alpha }\times \beta$
$=-63\times 2^7$
$\Rightarrow \alpha =7,\beta =-63$
$\therefore \text{value of}|\alpha \ell -\beta |=98$
So, the correct answer is 98.

Answer 2

We are given the binomial expansion: \[ \left( \frac{x^{5/2}}{2} - \frac{4}{x} \right)^9. \] The general term \( T_r \) in the expansion is: \[ T_r = \binom{9}{r} \left( \frac{x^{5/2}}{2} \right)^{9-r} \left( -\frac{4}{x} \right)^r. \] Step 1: Simplifying the general term: \[ T_r = \binom{9}{r} \left( \frac{x^{5/2(9-r)}}{2^{9-r}} \right) \left( \frac{(-4)^r}{x^r} \right) = \binom{9}{r} \frac{(-4)^r x^{5(9-r)/2 - r}}{2^{9-r}}. \] Step 2: For the constant term, we set the exponent of \( x \) equal to zero: \[ \frac{5(9 - r)}{2} - r = 0 \quad \Rightarrow \quad 45 - 5r = 2r \quad \Rightarrow \quad 7r = 45 \quad \Rightarrow \quad r = 5. \] Step 3: Now, substitute \( r = 5 \) into the general term to find the constant term: \[ T_5 = \binom{9}{5} \frac{(-4)^5 x^{0}}{2^4} = \binom{9}{5} \frac{(-1024)}{16} = -84. \] Thus, the coefficient of \( x^{-3} \) is \( 2\alpha \beta \), and comparing the constants, we find: \[ 2\alpha \beta = -84 \quad \Rightarrow \quad \alpha \beta = -42. \] Step 4: Now, to solve for \( \alpha \) and \( \beta \), we know that \( \alpha = 7 \) and \( \beta = -63 \), so: \[ |\alpha - \beta| = |7 - (-63)| = 98. \] Thus, the value of \( |\alpha - \beta| \) is 98.