# Doubt from limit as a sum

Question 10,8 and 6 of mypat jee main test paper

\displaystyle logL = \frac 1n × \left ( log (n) + log (n-1) + . .............Log1 - n×(log(n) + log(k) ) \right )
\displaystyle logL = \frac 1n × \left ( log (1) + log (1- \frac 1n ) + . ............. + nlog(k) \right )
Now this is limit as a sum ..
Solve further

What about question 8 and 6

3- \int_{0}^{3} \sqrt\frac{1}{1+x} dx

r u sure 2nd question correct i think in denominator should be n^6 sigma should be abolished from their

How can you further expand and tell how you got this

No it is correct answer is 7/12

neglect post 5
the correct is \frac{(\int_0^1 x^2 dx )(\int_0^1 x^3)}{\int_0^1 x^6 dx}

What's the ans to q10?? 1/ke?

I think so upload your solution please it would be better

1 Like

Q 8 you can do using limits concept alone as the Summation of x^n is proportional to x^n+1/n+1 and since it is infinity/infinity form the answer is simply the coefficient of highest power in numerator upon coefficient of highest power in denominator

1 Like

Woah nice method

easiest approch for 1st
f(x)=\frac{n!}{n})^{1/n} = 1/e
by stirling approximation
ln{n!}=nln(n)-n
on taking log in f(x) you will get same result
so ans 1/ke

ans is 2 and 3?

Yes i couldn't understand option 2 can you further express it

concept is \color{blue} r/n=t
now put limit match to it to \int_{1}^{2}