If f(x)=-2sinx if x≤-π2Asinx+B if -π2

If is a continuous function then

1. A

2. B

3. C

4. D

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Solution:

$f\left(-\frac{\pi }{2}\right)=-2\mathrm{sin}\left(-\frac{\pi }{2}\right)=2$

$\underset{x\to \infty }{\mathrm{lim}}f\left(x\right)=\underset{x\to \infty }{\mathrm{lim}}\left(A\mathrm{sin}x+B\right)=-A+B$

so

Also $f\left(\frac{\pi }{2}\right)=\mathrm{cos}\frac{\pi }{2}=0$

$\underset{x\to \frac{\pi }{2}-}{\mathrm{lim}}f\left(x\right)=\underset{x\to \pi /2-}{\mathrm{lim}}\left(A\mathrm{sin}x+B\right)=A+B$

Hence A + B = 0            (ii)

Solving (i) and (ii), we get .

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