5
Question
Let f:R→R and g∶R→R be defined as f(x)={x+a, x<0|x−1|, x≥0 and g(x)={x+1, x<0(x−1)2+b, x≥0. Where a, b are non-negative real numbers. If (g∘f)(x) is continuous for all x∈R, then a+b is equal to
Let f:R→R and g∶R→R be defined as f(x)={x+a, x<0|x−1|, x≥0 and g(x)={x+1, x<0(x−1)2+b, x≥0. Where a, b are non-negative real numbers. If (g∘f)(x) is continuous for all x∈R, 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
=⎧⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪⎩x+a+1 x+a<0 and x<0|x−1|+1 |x−1|<0 and x≥0(x+a−1)2+b x+a≥0 and x<0(|x−1|−1)2+b |x−1|≥0 and x≥0
=⎧⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪⎩x+a+1, x∈(−∞,−a) and x∈(−∞,0)|x−1|+1,x∈ϕ(x+a−1)2+b,x∈[−a,∞) and x∈(−∞,0)(|x−1|+1)2+b,x∈R and x∈[0,∞)
g[f(x)]=⎧⎪
⎪⎨⎪
⎪⎩x+a+1 x∈(−∞,−a)(x+a−1)2+b x∈[−a,0)(|x−1|−1)2+b x∈[0,∞)
Given that g(f(x)) is continuous
at x=−a & at x=0
⇒1=b+1 & (a−1)2+b=b
⇒b=0 & a=1
∴a+b=1
=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩x+a+1 x+a<0 and x<0|x−1|+1 |x−1|<0 and x≥0(x+a−1)2+b x+a≥0 and x<0(|x−1|−1)2+b |x−1|≥0 and x≥0
=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩x+a+1, x∈(−∞,−a) and x∈(−∞,0)|x−1|+1,x∈ϕ(x+a−1)2+b,x∈[−a,∞) and x∈(−∞,0)(|x−1|+1)2+b,x∈R and x∈[0,∞)
g[f(x)]=⎧⎪ ⎪⎨⎪ ⎪⎩x+a+1 x∈(−∞,−a)(x+a−1)2+b x∈[−a,0)(|x−1|−1)2+b x∈[0,∞)
Given that g(f(x)) is continuous
at x=−a & at x=0
⇒1=b+1 & (a−1)2+b=b
⇒b=0 & a=1
∴a+b=1
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