A) 0
B) 1
C) 2018
D) 2019
Correct Answer: D
Solution :
| Let \[L=\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{2}}\int_{-1/n}^{1/n}{(2018\sin x+2019\cos x)|x|dx}\] |
| \[L=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2\int_{0}^{1/n}{2019\cos x|x|dx}}{\frac{1}{{{n}^{2}}}}\] |
| \[[\because 2018\sin x|x|\text{is}\,\text{odd}\,\text{function }\!\!]\!\!\text{ }\] |
| \[\Rightarrow \]\[L=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2\cdot 2019\cos \left( \frac{1}{n} \right)\left( \frac{1}{n} \right)\times \left( \frac{-1}{{{n}^{2}}} \right)}{\frac{-\,2}{{{n}^{3}}}}\] |
| [\[\because \]by Leibnitz rule] |
| \[\Rightarrow \]\[L=\underset{n\to \infty }{\mathop{\lim }}\,2019\cos \left( \frac{1}{n} \right)=2019\] |
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