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Let f:RR and g:RR be functions satisfying f(x+y)=f(x)+f(y)+f(x)f(y) and f(x)=xg(x) for all x,yR. If limx0g(x)=1, then which of the following statements is/are TRUE?

A
f is differentiable at every xR
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B
If g(0)=1, then g is differentiable at every xR
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C
The derivative f(1) is equal to 1
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D
The derivative f(0) is equal to 1
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Solution

The correct option is D The derivative f(0) is equal to 1
f(x)=limh0f(x+h)f(x)h
=limh0f(x)+f(h)+f(x)f(h)f(x)h
=limh0f(h)(f(x)+1)h
=limh0hg(h)(f(x)+1)h
=f(x)+1 (limx0g(x)=1)
f(x)=f(x)+1

f(x)1+f(x)=1
On integrating both sides, we get f(x)1+f(x)dx=dx
ln(f(x)+1)=x+C (1)

f(x+h)=f(x)+f(h)+f(x)f(h) for all x,hR
limh0f(x+h)=limh0 [f(x)+f(h)+f(x)f(h)]
limh0f(x+h)=f(x)+limh0 [hg(h)+f(x)hg(h)]
limh0f(x+h)=f(x)+0=f(x)
Similarly, limh0f(xh)=f(x)
f is continuous for all xR
f(0)=limh0f(h)=0

Now, putting x=0 in equation (1)
ln(1+f(0))=C
C=0
1+f(x)=ex
f(x)=ex1
f(x)=ex
f(1)=e and f(0)=1

g(x)=f(x)x=ex1x,x01,x=0

We have to check differentiability at x=0
g(0+)=limh 0⎜ ⎜ ⎜eh1h1h⎟ ⎟ ⎟
=limh 0(eh1hh2)=12
g(0)=limh 0⎜ ⎜ ⎜ ⎜eh1h1h⎟ ⎟ ⎟ ⎟
=limh 0(eh1+hh2)=12
If g(0)=1, then g is differentiable at x=0 and hence differentiable at every xR

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