Question

# If f(x)=loge(1−x) and g(x)=[x, then determine each of the following functions : (i) f+g (ii) fg (iii) fg (iv) gf Also, find (f+g)(−1),(fg)(0),(fg)(12),(gf)(12)

Solution

## We have f(x)=loge(1−x) and g(x)=[x] f(x)=loge(1−x) is defined, if 1−x>0 ⇒1>x⇒x<1⇒xϵ(−∞,1) g(x) = [x] is defined for all xϵR ∴ Domain (g) = R ∴Domain(f)∩RDomain(g)=(−∞,1)∩R=(−∞,1) (i) f+g:(−∞,1)→R defined by (f+g)(x)=f(x)+g(x)=loge(1−x)+[x] (ii) fg:(−∞,1)→R defined by (fg)(x)=f(x)×g(x) =loge(1−x)×[x]=[x]loge(1−x) (iii) g(x) = [x] ∴ [x] = 0 ⇒xϵ(0,1) So, domain (fg)=domain(f)∩domain(g)−{x:g(x)=0}=(−∞,0) ∴fg:(−∞,0)→ defined by (fg)(x)=loge(1−x)[x] (iv) We have, f(x)=loge(1−x) ⇒1f(x)=1loge(1−x) is defined, and loge(1−x)≠0 ⇒1−x>0 and 1−x≠0 ⇒x<1 and x≠0 ⇒xϵ(−∞,0)∪(0,1) ∴ Domain (gf)=(−∞,0)∪(0,1) gf:(−∞,0)∪(0,1)→R defined by (gf)(x)=[x]loge(1−x) Now, (f+g)(−1)=f(−1)+g(−1)=loge(1−(−1))+[−1]=loge2−1⇒(f+g)(−1)=loge2−1 (v) fg(0)=loge(1−0)×[0] (vi) (fg)(12)= does not exist (vii) (gf)(12)=(12)loge(1−12)=0 MathematicsRD SharmaStandard XI

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