If f(x)=-2sinx if x≤-π2Asinx+B if -π2<x<π2cosx if x≥π2is a continuous function then

If f(x)=-2sinx if x≤-π2Asinx+B if -π2

$f (- \frac{π}{2}) = - 2 \sin (- \frac{π}{2}) = 2$

$\lim_{x \to \infty} f (x) = \lim_{x \to \infty} (A \sin x + B) = - A + B$

so $2 = - A + B \dots (i)$

Also $f (\frac{π}{2}) = \cos \frac{π}{2} = 0$

$\lim_{x \to \frac{π}{2} -} f (x) = \lim_{x \to π / 2 -} (A \sin x + B) = A + B$

Hence A + B = 0 (ii)

Solving (i) and (ii), we get $A = - 1, B = 1$ .