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  • Let <img alt="\mathrm{f(x)=[\cos x+\sin x], 0<x<2 \pi }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5B%5Ccos%20x&plus;%5Csin%20x%5D%2C%200%3Cx%3C2%20%5Cp

Let \mathrm{f(x)=[\cos x+\sin x], 0<x<2 \pi }, where \mathrm{ [x] } denotes the greatest integer less than or equal to \mathrm{x }. The number of points of discontinuity of \mathrm{f(x), } is _

Option: 1

4


Option: 2

7


Option: 3

5


Option: 4

4


Answers (1)

best_answer

We have, \mathrm{f(x)=[\cos x+\sin x]=\left[\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)\right] }
We know that [x] is discontinuous at integral points. Therefore, f(x) is discontinuous at points between 0 and \mathrm{2 \pi\, \, where\, \, \sqrt{2} \cos \left(x-\frac{\pi}{4}\right) } assumes integral values.
Clearly, such points are \mathrm{x=\frac{\pi}{2}, \frac{\pi}{2}+\frac{\pi}{4}, \pi, \frac{3 \pi}{2}, \frac{3 \pi}{2}+\frac{\pi}{4} }

Posted by

Divya Prakash Singh

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