JEE Advanced 2019 Paper 2Question 1
For $a\in\mathbb{R}$, $|a|>1$, let $$\lim_{n\to\infty}\left(\frac{1+\sqrt[3]{2}+\cdots+\sqrt[3]{n}}{n^{7/3}\left(\frac{1}{(an+1)^2}+\frac{1}{(an+2)^2}+\cdots+\frac{1}{(an+n)^2}\right)}\right)=54.$$ Then the possible value(s) of $a$ is/are
Limits & Continuity
Solution
Concept. Turn both the numerator and the bracket into Riemann sums to find their growth in $n$. Formulas used. (1) $\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}g\big(\frac kn\big)=\int_0^1g(x)\,dx$(2) $\int_0^1x^{1/3}dx=\frac34$(3) $\int_0^1\frac{dx}{(a+x)^2}=\frac1a-\frac1{a+1}=\frac{1}{a(a+1)}$ (valid as $|a|>1$)Step 1 - Numerator (1), (2). $$\sum_{k=1}^{n}k^{1/3}=n^{4/3}\cdot\frac1n\sum_{k=1}^{n}\Big(\frac kn\Big)^{1/3}\sim\frac34n^{4/3}$$ Step 2 - Bracket. $$\sum_{r=1}^{n}\frac{1}{(an+r)^2}=\frac1n\cdot\frac1n\sum_{r=1}^{n}\frac{1}{(a+r/n)^2}\sim\frac1n\cdot\frac{1}{a(a+1)}\quad\text{(3)}$$ Step 3 - Denominator. $n^{7/3}\cdot\frac{1}{n\,a(a+1)}=\frac{n^{4/3}}{a(a+1)}$. Step 4 - Limit. $\frac{\frac34n^{4/3}}{n^{4/3}/(a(a+1))}=\frac34a(a+1)$. Step 5 - Solve. $\frac34a(a+1)=54\Rightarrow a^2+a-72=0\Rightarrow(a+9)(a-8)=0$. $a=8$ or $a=-9$ (both have $|a|>1$). Answer: A,D