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If f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪2(1286x)17(4x+16)142,x0k ,x=0 is continuous at x=0 and g is a function with the property that g(x+y)=g(x)+g(y)+xy and limh0g(h)h=28f(0)

A
g is a differentiable function
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B
g(x)=3x+x22+1
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C
g(x)=3x+x22
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D
g(x)=3+x
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Solution

The correct option is C g(x)=3x+x22
limx0f(x)=limx02(1286x)17(4x+16)142 (00)
Applying L'Hospital Rule,
=limx0017(6)(1286x)6/714×4(4x+16)3/40
=6×(27)6/77×(24)3/4
=328

limh0g(h)h=28f(0)=3
Now we know that,
g(x)=limh0g(x+h)g(x)h
g(x)=limh0g(x)+g(h)+xhg(x)h
g(x)=limh0g(h)h+xg(x)=3+x
Integrating both the sides, we get
g(x)=3x+x22+c
Putting x=x,y=0 in g(x+y) we get g(0)=0
g(x)=3x+x22

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