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Question

# Which of the following is/are true?

A
If f:RR is a function satisfying f(xf(y))=f(f(y))+xf(y)+f(x)1 x,yR then f(x) can be 1x22.
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B
If f(x)=ax+bcxa, xac, a,b,cR{0}, then f(f(x))=x
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C
If f be a real valued integrable function satisfying f(x)+f(x+4)=f(x+2)+f(x+6) and g(x)=x+8xf(t)dt, xR, then g(x) can be a constant function.
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D
Domain of the function 1f(x) is [0,), where f3(x)=1x33xf(x), xR and f(1)1
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Solution

## The correct options are A If f:R→R is a function satisfying f(x−f(y))=f(f(y))+xf(y)+f(x)−1 ∀ x,y∈R then f(x) can be 1−x22. B If f(x)=ax+bcx−a, x≠ac, a,b,c∈R−{0}, then f(f(x))=x C If f be a real valued integrable function satisfying f(x)+f(x+4)=f(x+2)+f(x+6) and g(x)=x+8∫xf(t)dt,∀ x∈R, then g(x) can be a constant function. D Domain of the function √1−f(x) is [0,∞), where f3(x)=1−x3−3xf(x), x∈R and f(−1)≠−1→ For f(x−f(y))=f(f(y))+xf(y)+f(x)−1 ∀ x,y∈R Put x=0, f(y)=0⇒f(0)=1 Put x=a, f(y)=a⇒f(a)=1−a22 → For f(x)=ax+bcx−a, f(f(x))=x → For f(x)+f(x+4)=f(x+2)+f(x+6) x→x+2 f(x)=f(x+8) ⇒f(x) is periodic with period 8 x+8∫xf(t) dt=8∫0f(t) dt=g(x) ∴g(x) is a constant function. → Let a=f(x),b=x,c=−1 then a3+b3+c3=3abc ⇒(a+b+c)×12[(a−b)2+(b−c)2+(c−a)2]=0 f(−1)≠−1⇒a+b+c=0⇒f(x)=1−x

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