If the coefficient of x5 in the expansion of ax2+1bx13 is equal to the coefficient of x-5 in the expansion of ax-1bx213, then ab=

TS EAMCET · Maths · Binomial Theorem

If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{1}{b x})}^{13}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{13}$ , then $a b =$

A $1$
B $\frac{1}{6}$
C $\frac{7}{6}$
D $\frac{4}{2}$

Verified Solution

Answer & Solution

Correct Answer

(A) $1$

Step-by-step Solution

Detailed explanation

In the expansion of ax2+1bx13 Tr+1=Cr13ax213-r1bxr=Cr13a13-rbrx26-3r Here, 26-3r=5 ⇒26-5=3r ⇒3r=21⇒r=7 Coefficient of term containing x5 is C713a13-7b7=C713a6b7. In the expansion of ax-1bx213 Tr+1=Cr13ax13-r-1bx2r=-1r×Cr13a13-rbrx13-3r ⇒13-3r=-5…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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