Question

Let $a_{n}= \int\limits_{-\pi}^{\pi} \left|x-1\right| \cos\,nx dx$ for all natural numbers $n$ Then sequence $(a_{n})_{n \ge\,0}$ satisfies

• A. $\displaystyle\lim_{n\to\infty} a_{n} =\infty$
• B. $\displaystyle\lim_{n\to\infty} a_{n} =-\infty$
• C. $\displaystyle\lim_{n\to\infty} a_{n} $ exists and is positive
• D. $\displaystyle\lim_{n\to\infty} a_{n} =0$

Answer 1

Answer: (D)

We have,
$a_{n}=\int\limits_{-\pi}^{\pi}|x-1| \cos n x d x$
$a_{n}=\int\limits_{-\pi}^{1}-(x-1) \cos n x d x$
$-\int\limits_{1}^{\pi}(x-1) \cos n x d x$
$a_{n}=-\left[\frac{(x-1) \sin n x}{n}\right]_{-\pi}^{1}-\left[\frac{\cos n \pi}{n^{2}}\right]_{-\pi}^{1}$
$+\left[\frac{(x-1) \sin n x}{n}\right]_{1}^{\pi}-\left[\frac{\cos n \pi}{n^{2}}\right]_{1}^{\pi}$
$a_{n}=\frac{2 \pi \sin n \pi}{n}+\frac{2}{n^{2}} \cos n \pi-\frac{2}{n^{2}} \cos x \pi$
$a_{n}=\frac{2 \pi \sin n \pi}{n}$
$\displaystyle\lim _{x \rightarrow \infty} a_{n}=\displaystyle\lim _{n \rightarrow \infty} \frac{2 \pi \sin n \pi}{n}=0$