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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Mathematics
Chapter
Limits

JEE Advanced 2019 · Paper 2 · Mathematics · Limits

For a ∈ R, |a| > 1, let _ n → ∞ ( 1 + 23 + + n3 n ^ 7 / 3 ( 1 (an +...

For a∈R,∣a∣>1,a \in R, \left|a\right| > 1, let lim⁡n→∞(1+23+…+n3n7/3(1(an+1)2+1(an+2)2+…+1(an+n)2))=54.\lim_{n \to \infty} \left(\dfrac{1 + 23 + \ldots + n3}{{n}^{7 / 3}\left(\dfrac{1}{{\left(an + 1\right)}^{2}} + \dfrac{1}{{\left(an + 2\right)}^{2}} + \ldots + \dfrac{1}{{\left(an + n\right)}^{2}}\right)}\right) = 54 . Then the possible value(s)ofaof a is/are:

Options

  1. A88
  2. B−9- 9
  3. C−6- 6
  4. D77

Correct answer

A. 88

Step-by-step solution

Let S=lim⁡n→∞(1+213+313+…+n13n73(1(an+1)2+…+1(an+n)2))S = \lim_{n \to \infty} \left(\dfrac{1 + {2}^{\dfrac{1}{3}} + {3}^{\dfrac{1}{3}} + \ldots + {n}^{\dfrac{1}{3}}}{{n}^{\dfrac{7}{3}}\left(\dfrac{1}{{\left(an + 1\right)}^{2}} + \ldots + \dfrac{1}{{\left(an + n\right)}^{2}}\right)}\right)
⇒S=lim⁡n→∞∑r=1nr13n73∑r=1n1( na +r)2⇒ S = \lim_{n \to \infty} \dfrac{\mathop{∑}_{r = 1}^{n}{r}^{\dfrac{1}{3}}}{{n}^{\dfrac{7}{3}}\mathop{∑}_{r = 1}^{n}\dfrac{1}{{\left(\text{ na } + r\right)}^{2}}}<br/><br/>S = \lim_{n \to \infty} \dfrac{\left(\mathop{∑}_{r = 1}^{n}{\left(\dfrac{r}{n}\right)}^{\dfrac{1}{3}}\dfrac{1}{n}\right) \times {n}^{\dfrac{1}{3}}}{{n}^{\dfrac{7}{3}}\mathop{∑}_{r = 1}^{n}\dfrac{1}{{\left(a + \dfrac{r}{n}\right)}^{2}}\dfrac{1}{{n}^{2}}}<br/><br/>S = \lim_{n \to \infty} \dfrac{\mathop{∑}_{r = 1}^{n}{\left(\dfrac{r}{n}\right)}^{\dfrac{1}{3}} . \dfrac{1}{n}}{\mathop{∑}_{r = 1}^{n}{\left(\dfrac{1}{a + \dfrac{r}{n}}\right)}^{2}\dfrac{1}{n}}<br/><br/>S = \dfrac{\mathop{∫}_{0}^{1}{x}^{\dfrac{1}{3}}dx}{\mathop{∫}_{0}^{1}\dfrac{1}{{\left(a + x\right)}^{2}} . dx}<br/><br/>S = \dfrac{{\left[\dfrac{{x}^{\dfrac{4}{3}}}{\dfrac{4}{3}}\right]}_{0}^{1}}{{\left[- \dfrac{1}{x + a}\right]}_{0}^{1}}<br/><br/>S = \dfrac{\dfrac{3}{4}}{\dfrac{1}{a} - \dfrac{1}{a + 1}} = 54<br/><br/>\dfrac{1}{a\left(a + 1\right)} = \dfrac{1}{72}<br/><br/>{a}^{2} + a - 72 = 0<br/><br/>a = 8orora = - 9
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