86717
\(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x, \quad\) where \([t] \quad\) denotes greatest integer less than or equal to \(t\), is equal to :
1 -3
2 -2
3 2
4 0
Explanation:
(D) : Given, \(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x\) Let, \(I=\int_{0}^{2} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x +\int_{2}^{4} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(+\int_{4}^{5} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(=\int_{0}^{2}[\cos \pi(\mathrm{x}-0) \mathrm{dx}]+\int_{2}^{4} \cos (\pi \mathrm{x}-\pi) \mathrm{dx}+\int_{4}^{5} \cos (\pi \mathrm{x}-2 \pi) \mathrm{dx}\) \(=\int_{0}^{2} \cos (\pi x) d x-\int_{2}^{4} \cos (\pi x) d x+\int_{4}^{5} \cos (\pi x) d x\) \(=\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{0}^{2}-\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{2}^{4}+\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{4}^{5}\) \(=0\)
JEE Main-2022-29.06.2022
Integral Calculus
86718
If \(I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x\). Then
1 \(\frac{\pi}{2}\lt \) I \(\lt \frac{3 \pi}{4}\)
2 \(\frac{\pi}{5}\lt \) I \(\lt \frac{5 \pi}{12}\)
86717
\(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x, \quad\) where \([t] \quad\) denotes greatest integer less than or equal to \(t\), is equal to :
1 -3
2 -2
3 2
4 0
Explanation:
(D) : Given, \(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x\) Let, \(I=\int_{0}^{2} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x +\int_{2}^{4} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(+\int_{4}^{5} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(=\int_{0}^{2}[\cos \pi(\mathrm{x}-0) \mathrm{dx}]+\int_{2}^{4} \cos (\pi \mathrm{x}-\pi) \mathrm{dx}+\int_{4}^{5} \cos (\pi \mathrm{x}-2 \pi) \mathrm{dx}\) \(=\int_{0}^{2} \cos (\pi x) d x-\int_{2}^{4} \cos (\pi x) d x+\int_{4}^{5} \cos (\pi x) d x\) \(=\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{0}^{2}-\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{2}^{4}+\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{4}^{5}\) \(=0\)
JEE Main-2022-29.06.2022
Integral Calculus
86718
If \(I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x\). Then
1 \(\frac{\pi}{2}\lt \) I \(\lt \frac{3 \pi}{4}\)
2 \(\frac{\pi}{5}\lt \) I \(\lt \frac{5 \pi}{12}\)
86717
\(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x, \quad\) where \([t] \quad\) denotes greatest integer less than or equal to \(t\), is equal to :
1 -3
2 -2
3 2
4 0
Explanation:
(D) : Given, \(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x\) Let, \(I=\int_{0}^{2} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x +\int_{2}^{4} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(+\int_{4}^{5} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(=\int_{0}^{2}[\cos \pi(\mathrm{x}-0) \mathrm{dx}]+\int_{2}^{4} \cos (\pi \mathrm{x}-\pi) \mathrm{dx}+\int_{4}^{5} \cos (\pi \mathrm{x}-2 \pi) \mathrm{dx}\) \(=\int_{0}^{2} \cos (\pi x) d x-\int_{2}^{4} \cos (\pi x) d x+\int_{4}^{5} \cos (\pi x) d x\) \(=\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{0}^{2}-\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{2}^{4}+\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{4}^{5}\) \(=0\)
JEE Main-2022-29.06.2022
Integral Calculus
86718
If \(I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x\). Then
1 \(\frac{\pi}{2}\lt \) I \(\lt \frac{3 \pi}{4}\)
2 \(\frac{\pi}{5}\lt \) I \(\lt \frac{5 \pi}{12}\)
86717
\(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x, \quad\) where \([t] \quad\) denotes greatest integer less than or equal to \(t\), is equal to :
1 -3
2 -2
3 2
4 0
Explanation:
(D) : Given, \(\int_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x\) Let, \(I=\int_{0}^{2} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x +\int_{2}^{4} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(+\int_{4}^{5} \cos \left[\pi\left(x-\left[\frac{x}{2}\right]\right)\right] d x\) \(=\int_{0}^{2}[\cos \pi(\mathrm{x}-0) \mathrm{dx}]+\int_{2}^{4} \cos (\pi \mathrm{x}-\pi) \mathrm{dx}+\int_{4}^{5} \cos (\pi \mathrm{x}-2 \pi) \mathrm{dx}\) \(=\int_{0}^{2} \cos (\pi x) d x-\int_{2}^{4} \cos (\pi x) d x+\int_{4}^{5} \cos (\pi x) d x\) \(=\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{0}^{2}-\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{2}^{4}+\left[\frac{\sin \pi \mathrm{x}}{\pi}\right]_{4}^{5}\) \(=0\)
JEE Main-2022-29.06.2022
Integral Calculus
86718
If \(I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x\). Then
1 \(\frac{\pi}{2}\lt \) I \(\lt \frac{3 \pi}{4}\)
2 \(\frac{\pi}{5}\lt \) I \(\lt \frac{5 \pi}{12}\)
