    Question

# If f(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2−(128−6x)17(4x+16)14−2,x≠0k ,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 limh→0g(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 limx→0f(x)=limx→02−(128−6x)17(4x+16)14−2 (00) Applying L'Hospital Rule, =limx→00−17(−6)(128−6x)−6/714×4(4x+16)−3/4−0 =6×(27)−6/77×(24)−3/4 =328 limh→0g(h)h=28f(0)=3 Now we know that, g′(x)=limh→0g(x+h)−g(x)h ⇒g′(x)=limh→0g(x)+g(h)+xh−g(x)h ⇒g′(x)=limh→0g(h)h+x⇒g′(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  Suggest Corrections  0      Similar questions  Explore more