Consider two differentiable functions `f(x),g(x)` satisfying
`6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c and 2 int(g(x)dx)/(f(x))=x^(2)+c, " where " f(x) gt 0 AA x in R.`
`int (g(x)-f(x))dx` is equal to

Options

A. `(x^(4))/(4)-(x^(2))/(2)+x+c`
B. `(x^(4))/(4)+(x^(2))/(2)-(x^(3))/(3)+x+c`
C. `(x^(4))/(4)-(x^(3))/(3)+(x^(2))/(2)-x+c`
D. `(x^(4))/(4)+(x^(3))/(3)+c`

Answer

B

Explanation

We have, `6intf(x)g(x)dx=x^(6)+3x^(4)+3x^(2)+c`
Differentiating both sides w.r.t.x, we get
`6f(x)g(x)=6x^(5)+12x^(3)+6x`
` :. f(x)g(x)=x^(5)+2x^(3)+x " ...(1)" `
`2 int (g(x)dx)/(f(x))=x^(2)+c`
Differentiating both sides w.r.t.x, we get
`(g(x))/(f(x))=x " ...(2)" `
Multiplying (1) and (2), we get
`(g(x))^(2)=x^(6)+2x^(4)+x^(2)=(x^(3)+x)^(2)`
` :. g(x)=x^(3)+x`
` :. f(x)=x^(2)+1 " " ("as"f(x) gt 0).`
Now ` int (g(x)-f(x))dx=int (x^(3)+x-x^(2)-1)dx`
`=(x^(4))/(4)+(x^(2))/(2)-(x^(3))/(3)-x+c`
`underset (x to 0)(lim)(log(f(x)))/(g(x))=underset(x to 0)(lim)(log(1+x^(2)))/(x+x^(3))`
`=underset(x to 0)(lim)(log(1+x^(2)))/(x^(2))*(x^(2))/(x+x^(3))`
`=underset(x to 0)(lim)(x)/(1+x^(2))=0`