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Mathematics Question on integral

If ddxf(x)=4x3−3x4\frac{d}{dx}f(x) = 4x^3-\frac{3}{x^4} such that f(2)=0f(2)=0, then f(x)f(x) is

A

x4+1x3−1298x^4+\frac{1}{x^3}-\frac{129}{8}

B

x3+1x4+1298x^3+\frac{1}{x^4}+\frac{129}{8}

C

x4+1x3+1298x^4+\frac{1}{x^3}+\frac{129}{8}

D

x3+1x4−1298x^3+\frac{1}{x^4}-\frac{129}{8}

Answer

x4+1x3−1298x^4+\frac{1}{x^3}-\frac{129}{8}

Explanation

Solution

It is given that,

ddxf(x)=4x3−3x4\frac{d}{dx}f(x) = 4x^3-\frac{3}{x^4}

∴ Anti derivative of 4x3−3x4 4x^3-\frac{3}{x^4} = f(X)

∴ f(x) = ∫4x3−3x4dx\int 4x^3-\frac{3}{x^4}dx

f(x) = 4∫x3dx−3∫(x−4)dx4\int x^3dx-3\int (x-4)dx

f(x) = 4(x44)−3x−3−3+C4\bigg(\frac{x^4}{4}\bigg)-3\frac{x^{-3}}{-3}+C

∴ f(x) = x4 + 1x3\frac{1}{x^3} + C

Also,
f(2) = 0

∴ f(2) = (2)4+1(2)3(2)^4+\frac{1}{(2)^3}= 0

⇒ 16+18\frac{1}{8}+C = 0

⇒ C = -(16+18\frac{1}{8})

⇒ C = -−1298-\frac{129}{8}

∴ f(x) = x4 + 1x3−1298\frac{1}{x^3}-\frac{129}{8}

Hence, the correct Answer is A.