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  • If <img alt="f(x)=\left\{\begin{array}{cl}\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]}{

If f(x)=\left\{\begin{array}{cl}\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{array}\right. where [x]
is the greatest integer function of x and a>0, then

 

 

Option: 1

f\left(a^{-}\right)<0
 


Option: 2

 f has a removable discontinuity at x=a
 


Option: 3

 f has an irremovable discontinuity at x=a
 


Option: 4

f\left(a^{+}\right)>0


Answers (1)

best_answer

 Let x=a-h
\mathrm{\therefore \quad f\left(a^{-}\right)=\lim _{h \rightarrow 0} \frac{\left[\cos \left(\frac{\pi}{2 a}(a-h)\right)\right]}{h}=\frac{\left[\sin \frac{\pi h}{2 a}\right]}{h}=0 }
Again, let x=a+h
\mathrm{\therefore \quad f\left(a^{+}\right)=\lim _{h \rightarrow 0}-\sin \left(\frac{h}{2}\right) \tan \left[\frac{\pi}{2}+\frac{\pi h}{2 a}\right]=0 }
Hence \mathrm{f\left(a^{+}\right)=f\left(a^{-}\right) }

Posted by

Gautam harsolia

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