JEE Mains · Maths · STD 11 - 7. binomial theoram
The term independent of \(x\) in the expansion of \(\left( {\frac{1}{{60}} - \frac{{{x^8}}}{{81}}} \right).{\left( {2{x^2} - \frac{3}{{{x^2}}}} \right)^6}\) is equal to
- A \(36\)
- B \(-36\)
- C \(-108\)
- D \(-72\)
Answer & Solution
Correct Answer
(B) \(-36\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{1}{60}-\frac{x^{8}}{81}\right)\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}\) Term independent of \(x\) will be \(\frac{1}{60} \times\) independent of \(x\) in \(\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}-\frac{1}{8} \times\) Termof \(x^{-8}\) in…
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