For \(\mathrm{a} \in \mathbb{R},|\mathrm{a}|>1\), let \(\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{1+\sqrt[3]{2}+\ldots . .+\sqrt[3]{\mathrm{n}}}{\mathrm{n}^{7 / 3}\left(\frac{1}{(\mathrm{an}+1)^{2}}+\frac{1}{(\mathrm{an}+2)^{2}}+\ldots \ldots .+\frac{1}{(\mathrm{an}+\mathrm{n})^{2}}\right)}\right)=54\). Then the possible value(s) of a is/are :

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Limits XII

For

a \in R, ∣ a ∣ > 1

, let

lim_{n \to \infty} (\frac{1 + 3 2 + \dots .. + 3 n}{n ^{7/3} ( \frac{1}{( an + 1 ) ^{2}} + \frac{1}{( an + 2 ) ^{2}} + \dots\dots . + \frac{1}{( an + n ) ^{2}} )}) = 54

. Then the possible value(s) of a is/are :

Answer: (A), (B)

Rewrite the limit as

lim_{n \to \infty} (\frac{\frac{1}{n} \sum _{r = 1}^{n} ( \frac{r}{n} ) ^{1/3}}{\frac{1}{n} \sum _{r = 1}^{n} \frac{1}{( a + r / n ) ^{2}}}) = 54

. Using Riemann sums, this becomes

\frac{\int _{0}^{1} x ^{1/3} d x}{\int _{0}^{1} \frac{1}{( a + x ) ^{2}} d x} = 54

. Evaluating:

\int_{0}^{1} x^{1/3} d x = \frac{3}{4}

and

\int_{0}^{1} \frac{1}{( a + x ) ^{2}} d x = \frac{1}{a ( a + 1 )}

. Thus

\frac{3/4}{1/ ( a ( a + 1 ))} = 54

, giving

a (a + 1) = 72

, so

a^{2} + a - 72 = 0

. Solving:

a = - 9

a = 8

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options