If the coefficient of x^7 in (ax - 1 / bx^2)^13 and the coefficie… — Solution | JEE Mathematics

MCQMediumJEE 2023General Term

If the coefficient of $x^7$ in $(ax - 1 / bx^2)^{13}$ and the coefficient of $x^{-5}$ in $(ax + 1 / bx^2)^{13}$ are equal, then $a^4b^4$ is equal to:

Answer

Correct answer:A

Standard Method

Given: The coefficients of $x^7$ in $(ax - 1 / bx^2)^{13}$ and of $x^{-5}$ in $(ax + 1 / bx^2)^{13}$ are equal.

Find: The value of $a^4b^4$ .

The general term used in the solution is

T_{r+1} = \binom{13}{r} a^{13-r} \left(-\frac{1}{b}\right)^r x^{13-3r}

For the coefficient of $x^7$ ,

13 - 3r = 7 \implies r = 2

So the coefficient is

\binom{13}{2} \frac{a^{11}}{b^2}

For the coefficient of $x^{-5}$ ,

13 - 3r = -5 \implies r = 6

So the coefficient is

\binom{13}{6} \frac{a^7}{b^6}

Since these coefficients are equal,

\binom{13}{2} \frac{a^{11}}{b^2} = \binom{13}{6} \frac{a^7}{b^6}

Therefore,

a^4 \cdot b^4 = \frac{\binom{13}{6}}{\binom{13}{2}}

Now simplify the binomial coefficients:

\binom{13}{2} = 78

and the solution gives

\frac{\binom{13}{6}}{\binom{13}{2}} = 22

Hence,

a^4b^4 = 22

Therefore, the correct option is A.

Using the same sign for both expansions is incorrect. The first expression has a negative second term and the second has a positive second term, so the general terms must be written carefully before comparing coefficients.
Equating the powers incorrectly is a common error. The power of $x$ in the general term is $13-3r$ , not $13-r$ , because $1/bx^2$ contributes $x^{-2}$ along with the power from $ax$ .
Comparing the terms instead of only their coefficients is wrong. The question asks for equality of coefficients of specific powers of $x$ , so first identify the required $r$ values and then equate only the coefficient parts.

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