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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

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

6


Option: 2

5


Option: 3

4


Option: 4

3


Answers (1)

best_answer

\begin{aligned} & \mathrm{\cos x+\sin x=\sqrt{2} \cos \left(x-\frac{\pi}{4}\right), \text { So, } f(x)=\left[\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)\right] }\text {. } \\ \end{aligned}

We know that [x] is discontinuous at integral values of x.

Now, \begin{aligned} & \mathrm{\sqrt{2} \cos \left(x-\frac{\pi}{4}\right) } \\ \end{aligned}  is an integer at \begin{aligned} & \mathrm{ x=\frac{\pi}{2}, \frac{\pi}{2}+\frac{\pi}{4}, \pi+\frac{\pi}{4}, \frac{3 \pi}{2}+\frac{\pi}{4}.} \\ \end{aligned}

Posted by

himanshu.meshram

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