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 let Then the possible value(s) is/are:
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Let
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}}}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}}S = \dfrac{\mathop{∫}_{0}^{1}{x}^{\dfrac{1}{3}}dx}{\mathop{∫}_{0}^{1}\dfrac{1}{{\left(a + x\right)}^{2}} . dx}S = \dfrac{{\left[\dfrac{{x}^{\dfrac{4}{3}}}{\dfrac{4}{3}}\right]}_{0}^{1}}{{\left[- \dfrac{1}{x + a}\right]}_{0}^{1}}S = \dfrac{\dfrac{3}{4}}{\dfrac{1}{a} - \dfrac{1}{a + 1}} = 54\dfrac{1}{a\left(a + 1\right)} = \dfrac{1}{72}{a}^{2} + a - 72 = 0a = 8a = - 9
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}}}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}}S = \dfrac{\mathop{∫}_{0}^{1}{x}^{\dfrac{1}{3}}dx}{\mathop{∫}_{0}^{1}\dfrac{1}{{\left(a + x\right)}^{2}} . dx}S = \dfrac{{\left[\dfrac{{x}^{\dfrac{4}{3}}}{\dfrac{4}{3}}\right]}_{0}^{1}}{{\left[- \dfrac{1}{x + a}\right]}_{0}^{1}}S = \dfrac{\dfrac{3}{4}}{\dfrac{1}{a} - \dfrac{1}{a + 1}} = 54\dfrac{1}{a\left(a + 1\right)} = \dfrac{1}{72}{a}^{2} + a - 72 = 0a = 8a = - 9