How did we deduce directly from ?

Since is divisible by an odd number times , factoring 84 gives . For , trying small values of : leaves , and , which directly matches .

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Binomial Theorem: JEE Main Mathematics Question with Solution

If the constant term in the binomial expansion of

\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}}\right)^9

-84

and the coefficient of

x^{-3 l}

2^\alpha \beta

, where

\beta<0

is an odd number, then

|\alpha l-\beta|

is equal to

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Step-by-step solutionView

Correct answer

The value of $|\alpha l - \beta|$ is 98.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given the binomial expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^\ell}\right)^9$ . The constant term is $-84$ , and the coefficient of $x^{-3\ell}$ is $2^\alpha \beta$ where $\beta < 0$ is an odd number.

02goal

Find the values of $\ell$ , $\alpha$ , and $\beta$ , and then calculate $|\alpha\ell - \beta|$ .

03approach

Write down the general term $T_{r+1} = {^9C_r} \left(\frac{x^{5/2}}{2}\right)^{9-r} \left(-\frac{4}{x^\ell}\right)^r$ . Determine $r$ and $\ell$ using the given constant term value $-84$ . Then find the index $r'$ corresponding to the power $x^{-3\ell}$ , factor out the powers of 2 to match $2^\alpha \beta$ , and compute $|\alpha\ell - \beta|$ .

04execute

The general term is: $T_{r+1} = {^9C_r} \left(\frac{1}{2}\right)^{9-r} (-4)^r x^{\frac{5(9-r)}{2} - \ell r} = {^9C_r} (-1)^r 2^{3r-9} x^{\frac{45-5r}{2} - \ell r}$ For the constant term, the coefficient is ${^9C_r} (-1)^r 2^{3r-9} = -84$ . Since $-84 = -1 \times 84 = -1 \times {^9C_3} = {^9C_3} (-1)^3 2^0$ , we equate powers of 2: $3r - 9 = 0 \implies r = 3$ . Then ${^9C_3} (-1)^3 2^0 = 84 \times (-1) = -84$ , which matches perfectly. Since the power of $x$ must be 0 for the constant term: $\frac{45 - 5(3)}{2} - 3\ell = 0 \implies 15 - 3\ell = 0 \implies \ell = 5$

05execute

Now, we need the coefficient of $x^{-3\ell} = x^{-15}$ . Set the power of $x$ to $-15$ : $\frac{45 - 5r}{2} - 5r = -15 \implies 45 - 15r = -30 \implies 15r = 75 \implies r = 5$ Substitute $r = 5$ into the coefficient: $\text{coeff} = {^9C_5} \left(\frac{1}{2}\right)^{9-5} (-4)^5 = 126 \times \frac{1}{16} \times (-1024) = 126 \times (-64) = -8064$ Express this in the form $2^\alpha \beta$ where $\beta < 0$ is an odd number: $-8064 = -(2^7 \times 63) = 2^7 \times (-63)$ Since $\beta = -63$ is an odd negative integer, we have: $\alpha = 7, \quad \beta = -63$

06execute

Compute the final expression $|\alpha \ell - \beta|$ : $|\alpha \ell - \beta| = |7(5) - (-63)| = |35 + 63| = 98$

✓verify

Check: ${^9C_5} = 126 = 2 \times 63$ . Factor of powers of 2: $2^1 \times 2^{-4} \times 2^{10} = 2^7$ . So $2^7 \times (-63)$ , which gives $\alpha = 7$ , $\beta = -63$ . Odd condition on $\beta$ holds. $\alpha\ell - \beta = 35 - (-63) = 98$ .

Hints that build this answer step by step

What is the general term $T_{r+1}$ in the expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^\ell}\right)^9$ ?
$T_{r+1} = \binom{9}{r} (-1)^r 2^{3r-9} x^{\frac{45 - (5 + 2\ell)r}{2}}$
Given that the constant term is $-84$ , what are the values of $r$ and $\ell$ ?
$r = 3$ and $\ell = 5$
With $\ell = 5$ , which value of $r$ produces the term containing $x^{-3\ell} = x^{-15}$ , and what are $\alpha$ and $\beta$ ?
$r = 5$ , giving $\alpha = 7$ and $\beta = -63$
What is the final value of $|\alpha \ell - \beta|$ ?
$98$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

How did we deduce

r = 3

directly from

{^9C_r} (-1)^r 2^{3r-9} = -84

Since ${^9C_3} = 84$ is divisible by an odd number times $2^2$ , factoring 84 gives $2^2 \times 21$ . For ${^9C_r} 2^{3r-9} = 84$ , trying small values of $r$ : ${^9C_3} = 84$ leaves $2^{3(3)-9} = 2^0 = 1$ , and $(-1)^3 = -1$ , which directly matches $-84$ .

Answer

The value of $|\alpha l - \beta|$ is 98.

Why each option works or fails

Step-by-step solution

given: Given the binomial expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^\ell}\right)^9$ . The constant term is $-84$ , and the coefficient of $x^{-3\ell}$ is $2^\alpha \beta$ where $\beta < 0$ is an odd number.
goal: Find the values of $\ell$ , $\alpha$ , and $\beta$ , and then calculate $|\alpha\ell - \beta|$ .
approach: Write down the general term $T_{r+1} = {^9C_r} \left(\frac{x^{5/2}}{2}\right)^{9-r} \left(-\frac{4}{x^\ell}\right)^r$ . Determine $r$ and $\ell$ using the given constant term value $-84$ . Then find the index $r'$ corresponding to the power $x^{-3\ell}$ , factor out the powers of 2 to match $2^\alpha \beta$ , and compute $|\alpha\ell - \beta|$ .
execute: The general term is: $T_{r+1} = {^9C_r} \left(\frac{1}{2}\right)^{9-r} (-4)^r x^{\frac{5(9-r)}{2} - \ell r} = {^9C_r} (-1)^r 2^{3r-9} x^{\frac{45-5r}{2} - \ell r}$ For the constant term, the coefficient is ${^9C_r} (-1)^r 2^{3r-9} = -84$ . Since $-84 = -1 \times 84 = -1 \times {^9C_3} = {^9C_3} (-1)^3 2^0$ , we equate powers of 2: $3r - 9 = 0 \implies r = 3$ . Then ${^9C_3} (-1)^3 2^0 = 84 \times (-1) = -84$ , which matches perfectly. Since the power of $x$ must be 0 for the constant term: $\frac{45 - 5(3)}{2} - 3\ell = 0 \implies 15 - 3\ell = 0 \implies \ell = 5$
execute: Now, we need the coefficient of $x^{-3\ell} = x^{-15}$ . Set the power of $x$ to $-15$ : $\frac{45 - 5r}{2} - 5r = -15 \implies 45 - 15r = -30 \implies 15r = 75 \implies r = 5$ Substitute $r = 5$ into the coefficient: $\text{coeff} = {^9C_5} \left(\frac{1}{2}\right)^{9-5} (-4)^5 = 126 \times \frac{1}{16} \times (-1024) = 126 \times (-64) = -8064$ Express this in the form $2^\alpha \beta$ where $\beta < 0$ is an odd number: $-8064 = -(2^7 \times 63) = 2^7 \times (-63)$ Since $\beta = -63$ is an odd negative integer, we have: $\alpha = 7, \quad \beta = -63$
execute: Compute the final expression $|\alpha \ell - \beta|$ : $|\alpha \ell - \beta| = |7(5) - (-63)| = |35 + 63| = 98$
verify: Check: ${^9C_5} = 126 = 2 \times 63$ . Factor of powers of 2: $2^1 \times 2^{-4} \times 2^{10} = 2^7$ . So $2^7 \times (-63)$ , which gives $\alpha = 7$ , $\beta = -63$ . Odd condition on $\beta$ holds. $\alpha\ell - \beta = 35 - (-63) = 98$ .

Binomial Theorem: JEE Main Mathematics Question with Solution

What feels right?

The value of $|\alpha l - \beta|$ is 98.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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