Let f: R→R and g: R → R be defined as f(x) = {x+a, if x<0; |x-1|, if x≥0} and g(x) = {x+1, if x<0; (x-1)²+b, if x≥0}, where a, b are non-negative real numbers. If (gof)(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, if x<0; |x-1|, if x≥0} and g(x) = {x+1, if x<0; (x-1)²+b, if x≥0}, where a, b are non-negative real numbers. If (gof)(x) is continuous for all x ∈ R, then a + b is equal to..........
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1 Answer
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f (x) = {x+a, if x<0; |x-1|, if x≥0}
g (x) = {x+1, if x<0; (x-1)²+b, if x≥0}
g (f (x) must be continuous. The potential points of discontinuity are where the definitions of f (x) and g (f (x) change. This is at x=0 and where f (x)=0.
f (x)=0 when x=-a (if a>0) or when x=1.
Continuity at x = 0:
lim (x→0? ) g (f (x) = lim (x→0? ) g (x+a). Since a could be anything, let's analyze f (0? )=a. So, lim is g (a).
lim (x→0? ) g (f (x) = g (f (0? ) = g (|0-1|) = g (1). Since 1≥0, g (1) = (1-1)²+b = b.
g (f (0) = g (|0-1|) = g (1) = b.
So we need g (a) = b.
Case 1: a < 0. g (a) = a+1. So a+1=b.
Case 2: a ≥ 0. g (a) = (a-1)²+b. So (a-1)²+b=b => (a-1)²=0 => a=1.
Now consider continui...more
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