InterviewSolution
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InterviewSolution
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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].` |
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Answer» 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`. |
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