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  • <img alt="\mathrm{\text { If } f(x)= \begin{cases}\sin \left(\frac{a-x}{2}\right) \tan \left[\frac{\pi x}{2 a}\right] & \text { for } x>a \\ \frac{\left[\cos \left(\frac{\pi x}{2 a}\right)\r

\mathrm{\text { If } f(x)= \begin{cases}\sin \left(\frac{a-x}{2}\right) \tan \left[\frac{\pi x}{2 a}\right] & \text { for } x>a \\ \frac{\left[\cos \left(\frac{\pi x}{2 a}\right)\right]}{a-x} & \text { for } x<a\end{cases}}
 

(where \mathrm{[x]} is the greatest integer function of \mathrm{x} ), and \mathrm{a>0}, then

Option: 1

\mathrm{f\left(a\right)<0}


Option: 2

f has a removable discontinuity at x=a


Option: 3

f has in incmovable discontinuity at x=a


Option: 4

f\left ( a \right )< 0


Answers (1)

best_answer

\mathrm{f\left(a^{-}\right)=\lim _{h \rightarrow 0} \frac{\left[\cos \left(\frac{\pi}{2 a}(a-h)\right)\right]}{h}=\lim _{h \rightarrow 0} \frac{\left[\sin \frac{\pi h}{2 a}\right]}{h}=0 ; }

\mathrm{f\left(a^{+}\right)=\lim _{h \rightarrow 0}\left(-\frac{\sin h}{2}\right) \tan \left[\frac{\pi}{2}+\frac{\pi h}{2}\right]=0 ; }

\mathrm{Hence \: f\left(a^{+}\right)=f\left(a^{-}\right)}.
Thus, \mathrm{f(x)} has removable discontinuity at \mathrm{x=a}.

Posted by

Irshad Anwar

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