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Step-by-Step Solution
Step 1: Write down the given expression
The expression is:
\biggl(\frac{x^{\tfrac{5}{2}}}{2} \;-\; \frac{4}{x^{l}}\biggr)^{9}.
We need to find the constant term in this binomial expansion and also the coefficient of x^{-3l} .
Step 2: Identify the general term of the binomial expansion
For a binomial (A + B)^{n} , the general term (the (r+1) th term) is:
T_{r+1} = \binom{n}{r} \, A^{n-r} \, B^{r} \,.
Here, we have:
n = 9
A = \frac{x^{\tfrac{5}{2}}}{2}
B = -\,\frac{4}{x^{l}}
So, the (r+1) th term is:
T_{r+1}
= \binom{9}{r}\!\Bigl(\frac{x^{\tfrac{5}{2}}}{2}\Bigr)^{9-r}
\Bigl(-\frac{4}{x^{l}}\Bigr)^{r}.
Step 3: Simplify the general term
We have two parts to handle: the power of x and the powers of the constants.
T_{r+1}
= \binom{9}{r} \, \Bigl(\frac{1}{2^{\,9-r}}\Bigr)\, x^{\tfrac{5}{2}(9-r)}
\;\times\; (-1)^{r}\, 4^{\,r}\, x^{-\,l\,r}.
Notice that 4^{\,r} = (2^2)^{r} = 2^{2r}. Combine powers of 2:
T_{r+1}
= \binom{9}{r} \, (-1)^{r} \, 2^{2r} \, 2^{-(9-r)}
\; x^{\tfrac{5}{2}(9-r) - l\,r}.
So,
T_{r+1}
= \binom{9}{r} \, (-1)^{r} \, 2^{2r - (9-r)} \; x^{\tfrac{5}{2}(9-r) - l\,r}
= \binom{9}{r} \, (-1)^{r} \, 2^{3r - 9}
\; x^{\tfrac{45 - 5r}{2} \;-\; l\,r}.
Step 4: Determine the condition for the constant term
The constant term occurs when the overall power of x is zero:
\tfrac{45 - 5r}{2} \;-\; l\,r \;=\; 0.
Multiply both sides by 2:
45 - 5r = 2\,l\,r
\quad\Longrightarrow\quad
45 = 5r + 2\,l\,r
\quad\Longrightarrow\quad
45 = r\,(5 + 2l).
Step 5: Use the information about the constant term being -84
The constant term (for that specific r ) in the expansion (ignoring the x part, since x^0=1 ) is:
\binom{9}{r} \, (-1)^{r} \, 2^{3r - 9}.
According to the problem, this value is -84, and at the same time r\,(5 + 2l)=45. Testing integer values of r from 0 to 9 soon reveals a consistent solution when:
r = 3 and l = 5, satisfying 3\,(5 + 2\times 5) = 3\times 15 = 45.
Thus, we identify r=3 and l=5.
Step 6: Verify the constant term equals -84 with r=3, l=5
Substitute r=3 into the expression \binom{9}{r} \, (-1)^{r} \, 2^{3r - 9} :
\binom{9}{3} \, (-1)^3 \, 2^{3\times 3 - 9}
= \binom{9}{3} \, (-1) \, 2^{9 - 9}
= \binom{9}{3} \,(-1)\, 2^{0}
= \binom{9}{3} \times (-1) \times 1
= 84 \times (-1)
= -84.
This matches the constant term given in the problem, confirming l=5.
Step 7: Find the term where the exponent of x is -3l
We need to find the coefficient of x^{-3l}. Since l=5, that means we want x^{-15}. From the general exponent in the term:
\tfrac{45 - 5r}{2} \;-\; l\,r = -15.
Substitute l=5 :
\tfrac{45 - 5r}{2} - 5\,r = -15.
Multiply everything by 2 to clear the fraction:
45 - 5r - 10r = -30
\quad\Longrightarrow\quad
45 - 15r = -30
\quad\Longrightarrow\quad
-15r = -75
\quad\Longrightarrow\quad
r = 5.
Step 8: Compute the coefficient for r = 5
With r=5, the corresponding term is:
T_{6}
= \binom{9}{5}\, (-1)^5 \, 2^{3\cdot 5 - 9}
= \binom{9}{5}\, (-1)\, 2^{15 - 9}
= \binom{9}{5}\, (-1)\, 2^6.
We know \binom{9}{5} = \binom{9}{4} = 126. Thus,
T_{6}
= 126 \times (-1) \times 64
= -8064.
The problem states that this coefficient can be written as 2^{\alpha}\,\beta, where \beta < 0 is an odd integer. Factor out powers of 2 from -8064:
-8064
= -63 \times 128
= -63 \times 2^{7}.
Hence, \alpha = 7 and \beta = -63. (Note that -63 is indeed an odd integer, and it is negative.)
Step 9: Evaluate |\alpha\,l - \beta|
We have l=5, \alpha=7, and \beta=-63. Therefore:
\alpha l - \beta
= 7 \times 5 - (-63)
= 35 + 63
= 98.
Taking the absolute value yields the same result:
|\alpha l - \beta| \;=\;\bigl|\,98\,\bigr|
= 98.
Therefore, the final answer is 98.
