QJEEAdv19Ma5fcb: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma5fcb Solution: For , let lim n → ∞ ⁡ ( 1 + 2 3 + . . . n 3 n 7 / 3 ( 1 ( a n + …

Question

Verified Solution • 2025–26 Session moderate

For

a \in R,\,|a|\, &gt; 1

, let

lim_{n \to \infty} (\frac{1 + \sqrt[3]{2} + . . . \sqrt[3]{n}}{n^{7 / 3} (\frac{1}{{(a n + 1)}^{2}} + \frac{1}{{(a n + 2)}^{2}} + . . . + \frac{1}{{(a n + n)}^{2}})}) = 54

Accepted Answer

lim_{n \to \infty} [\frac{1 + \sqrt[3]{2} + \sqrt[3]{3} + . . . \sqrt[3]{n}}{n^{7 / 3} (\frac{1}{{(a n + 1)}^{2}} + \frac{1}{{(a n + 2)}^{2}} + . . . + \frac{1}{{(a n + n)}^{2}})}]

,

a \in R,\,|a|\, &gt; 1

= \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {({r^{1/3}})} } \over {{n^{7/3}}\sum\limits_{r = 1}^n {{1 \over {{{(an + r)}^2}}}} }}

= \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^{1/3}}{1 \over n}} } \over {\sum\limits_{r = 1}^n {{1 \over {{{(a + {r \over n})}^2}}}{1 \over n}} }}

= {{\int\limits_0^1 {{x^{1/3}}dx} } \over {\int\limits_0^1 {{{dx} \over {{{(a + x)}^2}}}} }} = 54

, (given)

\Rightarrow {{{3 \over 4}[{x^{4/3}}]_0^1} \over {\left[ { - {1 \over {x + a}}} \right]_0^1}} = 54

\Rightarrow {{{3 \over 4}} \over { - {1 \over {a + 1}} + {1 \over a}}} = 54

\Rightarrow {3 \over {4 \times 54}} = {1 \over {a(a + 1)}}

\Rightarrow {a^2} + a = 72

\Rightarrow {a^2} + 9a - 8a - 72 = 0

\Rightarrow a(a + 9) - 8(a + 9) = 0

\Rightarrow (a - 8)(a + 9) = 0

\Rightarrow a = 8

or

- 9

Hence, options (c) and (d) are correct.

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