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  • Let <img alt="f(x)=\left\{\begin{array}{cc}-2 \sin x, & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, & \text { if }-\frac{\pi}{2}<x<\frac{\pi}{2} . \\ \cos x, & \text { if } x

Let f(x)=\left\{\begin{array}{cc}-2 \sin x, & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, & \text { if }-\frac{\pi}{2}<x<\frac{\pi}{2} . \\ \cos x, & \text { if } x \geq \frac{\pi}{2}\end{array}\right.Then

 

 

Option: 1

 f is discontinuous for all A and B
 


Option: 2

 f is continuous for all A=-1 and B=1
 


Option: 3

 f is continuous for all A=1 and B=-1
 


Option: 4

f is continuous for all real values of A, B.


Answers (1)

Only points of discontinuity may be \frac{\pi}{2} or -\frac{\pi}{2} because sine and cosine functions are always continuous.
For continuity at \mathrm{x=-\frac{\pi}{2},-A+B=2( R.H.L. = L.H.L. ) }
For continuity at \mathrm{x=\frac{\pi}{2}, A+B=0 \quad( L.H.L. = R.H.L. ) }
\mathrm{\therefore \quad A=-1, B=1 }
So, for A=-1, B=1, f(x) is always continuous.

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Kshitij

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