Find `alpha and beta,` so the function f (x) defined by
`f(x) =-2 sin x, "for" -pi le x le-pi/2`
`=alpha sinx+beta"for" -pi/2lt x lt pi/2`
`=cosx , "for" pi/2 le x le pi,`
is continuous on `[-pi, pi].`

Explanation

Given, f is continuous on `[-pi,pi]`
`:.f` is continuous at `x=-(pi)/(2) ` and `x=(pi)/(2)`
Continuity at `x=-(pi)/(2)`:
`f(x)=-2 sin x`, for `-pi le x le - (pi)/(2)`
`:.f(-(pi)/(2))=-2* sin (-(pi)/(2))`
`= 2 sin. (pi)/(2)=2xx(1)=2`
`f(x)=alpha sin x + beta` , for `-(pi)/(2) lt x lt (pi)/(2)` ... (1)
`underset ( x rarr (-pi//2)^(+))f(x)=underset(x rarr (-pi//2)^(+))lim (alpha sin x + beta)`
`= alpha sin (-(pi)/(2))+beta`
`=-alpha sin . (pi)/(2) + beta`
`=-alpha* (1) + beta = - alpha + beta `
f is continuous at `x= - (pi)/(2)`
`:.underset(x rarr (-pi//2)^(+))lim f(x)=f(-(pi)/(2))`
`:. - alpha + beta = 2` ...(2)
Continuity at `x=(pi)/(2)`
`underset(x rarr (pi//2)^(-))lim f(x) = underset(x rarr (pi//2)^(-)) limalpha sin x + beta ` [ From (1) ]
`= alpha* sin.(pi)/(2) + beta`
`= alpha * (1) + beta = alpha + beta`
`f(x)= cos x`, for `(pi)/(2) le x le pi`
`:.f(pi/2)= cos.(pi)/(2)=0`
f is continuous at `x=(pi)/(2)`
`underset( x rarr (pi//2)^(-)) lim f(x) = f(pi/2)`
`:.alpha+beta=0` ...(3)
Adding equation (2) and (3),
`2 beta = 2`
`:.beta=1`
From equation (3),
`alpha + 1 = 0 `
`:.alpha=-1`
Hence, `alpha=1 and beta = 1`.