Question

If `d/(dx) f(x) = 4x^3 - 3/x^4`, such that `f(2) = 0`, then `f(x)` is

Options

• `x^4 + 1/x^3 - 129/8`

• `x^3 + 1/x^4 + 129/8`

• `x^4 + 1/x^3 + 129/8`

• `x^3 + 1/x^4 - 129/8`

Answer 1

`x^4 + 1/x^3 - 129/8`

Explanation:

`d/(dx) f(x) = 4x^3 - 3/x^4`

⇒ `f(x) = int(4x^3 - 3/x^4) dx`

= `4int  x^3 dx - 3 int x^-4 dx`

= `4 * x^4/4 - 3 * x^(-4 + 1)/(-4 + 1) + c`

= `x^4 - (3x^(-3))/(-3) + c`

= `x^4 + 1/x^3 + c`  .......(i)

But `f(2)` = 0

⇒ `x^4 + 1/x^3 + c` = 0 or `16 + 1/8 + c` = 0

`c = - 129/8`

Putting `c = - 129/8` = in (i)

∴ `f(x) = x^4 + 1/x^3 - 129/8`