JEE MainMathematics

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Finding Coefficients and Unknown Exponents in Binomial Expansions

Q: 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 × 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.

If the constant term in the binomial expansion of ((x^(5/2))/2-4/(x^ell))^9 is -84 and the coefficient of x^(-3 l) is 2^α β, where β<0 is an odd number, then |α l-β| is equal to

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Correct answer

The value of $∣ α l - β ∣$ is 98.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

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

02Goal

Find the values of $ℓ$ , $α$ , and $β$ , and then calculate $∣ α ℓ - β ∣$ .

03Approach

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

04Execute

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

05Execute

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

06Execute

Compute the final expression $∣ α ℓ - β ∣$ : $∣ α ℓ - β ∣ = ∣ 7 (5) - (- 63) ∣ = ∣ 35 + 63 ∣ = 98$

✓Verify

Check: $^{9} C_{5} = 126 = 2 \times 63$ . Factor of powers of 2: $2^{1} \times 2^{- 4} \times 2^{10} = 2^{7}$ . So $2^{7} \times (- 63)$ , which gives $α = 7$ , $β = - 63$ . Odd condition on $β$ holds. $α ℓ - β = 35 - (- 63) = 98$ .

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

How did we deduce

r = 3

directly from

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

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

Answer

The value of $∣ α l - β ∣$ is 98.

Why each option works or fails

Step-by-step solution

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

Finding Coefficients and Unknown Exponents in Binomial Expansions

What feels right?

The value of $∣ α l - β ∣$ is 98.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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