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Question

Let f:RR and g:RR be defined as
f(x)={x+a,x<0|x1|,x0 and g(x)={x+1,x<0(x1)2+b,x0,
where a,b are non-negative real numbers. If (gof)(x) is continuous for all xR, then a+b is equal to

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Solution

g[f(x)]={f(x)+1,f(x)<0(f(x)1)2+b,f(x)0
g[f(x)]=⎪ ⎪ ⎪⎪ ⎪ ⎪x+a+1,x+a<0 and x<0|x1|+1,|x1|<0 and x0(x+a1)2+b,x+a0 and x<0(|x1|1)2+b,|x1|0 and x0
g[f(x)]=⎪ ⎪ ⎪⎪ ⎪ ⎪x+a+1,x(,a) and x(,0)|x1|+1,xϕ(x+a1)2+b,x[a,) and x(,0)(|x1|1)2+b,xR and x[0,)
g[f(x)]=x+a+1,x(,a)(x+a1)2+b,x[a,0)(|x1|1)2+b,x[0,)
Given g(f(x)) is continuous for all xR.
So, at x=a and at x=0,
1=b+1 and (a1)2+b=b
b=0 and a=1
a+b=1

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