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Question : 78 of 160
Marks:
+1,
-0
Solution:
‌‌log[‌]=f(x)dx ‌| 2n×(2n−1)×(2n−2)×...×(n+1)×n! |
| n×n×n×...×n×n! |
=2×(2−‌)×(2−‌)×(2−‌)×...×(1+‌) =(1+‌)(1+‌)(1+‌)×...(1+‌) (‌)[log(1+‌)+log(1+‌)+...log(1+‌)] =(‌)‌log(1+‌) Lower
limit=(1+‌)=1 Upper limit
=(1+‌)=2=log(x)‌d‌x ⇒‌‌f(x)=log‌x
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