`f(x)=x^(2)+xg'(1)+g''(2)and g(x)=f(1)x^(2)+xf'(x)+f'(x).`
The domain of the function `sqrt((f(x))/(g(x)))` is

Options

A. `(-oo,1]cup(2,3]`
B. `(-2,0]cup(1,oo)`
C. `(-oo,0]cup(2//3,3]`
D. none of these

Explanation

Here put `g'(1) =a, g''(2)=b`
Then,
`f(x)=x^(2)+ax+b,`
`f(1)=1+a+b or f'(x)=2x+a,f''(x)=2.` Therefore,
`g(x)=(1+a+b)x^(2)+(2x+a)x+2`
`=x^(3)(3+a+b)+ax+2`
`"or "g'(x)=2x(3+a+b)+a and g''(x)=2(3+a+b).`
`"Hence, "g'(1)=2(3+a+b)+a" (2)"`
`g''(2)=2(3+a+b)" (3)"`
From (1), (2), and (3), we have
`a=2(3+a+b)+a and b=2(3+a+b)`
`"or "3+a+b=0 and b+2a+6=0`
`"Hence, "b=0 and a=-3. So f(x)=x^(2)-3x and g(x)=-3x+2.`
`sqrt((f(x))/(g(x)))=sqrt((x^(2)-3x)/(-3x+2))" is defined if "(x^(2)-3x)/(-3x+2)ge0`
`"or "(x(-3))/((x-2//3))le0`
`rArr" "x in (-oo,0]cup(2//3,3]`