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# Let f:R→R and g:R→R be functions satisfying f(x+y)=f(x)+f(y)+f(x)f(y) and f(x)=xg(x) for all x,y∈R. If limx→0g(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 1f′(x)=limh→0f(x+h)−f(x)h =limh→0f(x)+f(h)+f(x)f(h)−f(x)h =limh→0f(h)(f(x)+1)h =limh→0hg(h)(f(x)+1)h =f(x)+1 (∵limx→0g(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,h∈R ⇒limh→0f(x+h)=limh→0 [f(x)+f(h)+f(x)f(h)] ⇒limh→0f(x+h)=f(x)+limh→0 [hg(h)+f(x)hg(h)] ⇒limh→0f(x+h)=f(x)+0=f(x) Similarly, limh→0f(x−h)=f(x) ∴f is continuous for all x∈R ⇒f(0)=limh→0f(h)=0 Now, putting x=0 in equation (1) ln(1+f(0))=C ⇒C=0 ∴1+f(x)=ex f(x)=ex−1 f′(x)=ex f′(1)=e and f′(0)=1 g(x)=f(x)x=⎧⎨⎩ex−1x,x≠01,x=0 We have to check differentiability at x=0 g′(0+)=limh→ 0⎛⎜ ⎜ ⎜⎝eh−1h−1h⎞⎟ ⎟ ⎟⎠ =limh→ 0(eh−1−hh2)=12 g′(0−)=limh→ 0⎛⎜ ⎜ ⎜ ⎜⎝e−h−1−h−1−h⎞⎟ ⎟ ⎟ ⎟⎠ =limh→ 0(e−h−1+hh2)=12 If g(0)=1, then g is differentiable at x=0 and hence differentiable at every x∈R  Suggest Corrections  0      Similar questions  Explore more