WBJEE · Maths · Binomial Theorem
If the coefficient of \(x^{8}\) in \(\left(a x^{2}+\frac{1}{b x}\right)^{13}\) is equal to the coefficient of \(x^{-8}\) in \(\left(a x-\frac{1}{b x^{2}}\right)^{13},\) then \(a\) and \(b\) will satisfy the relation
- A \(a b+1=0\)
- B \(a b=1\)
- C \(a=1-b\)
- D \(a+b=-1\)
Answer & Solution
Correct Answer
(A) \(a b+1=0\)
Step-by-step Solution
Detailed explanation
The general term in \(\left(a x^{2}+\frac{1}{b x}\right)^{13}\) is \(T_{r+1}={ }^{13} C_{r}\left(a x^{2}\right)^{13-r}\left(\frac{1}{b x}\right)^{r}\) \[ ={ }^{13} C_{r} a^{13-r} \times b^{-r}(x)^{26-3 r} \] For coefficient of \(x^{8}\). put \(26-3 r=8\)…
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