Question

यदि ${\displaystyle \frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}}$ जिसमें f(2) = 0 तो f(x) है:

Options

• ${\displaystyle x^4+\frac{1}{x^3}-\frac{129}{8}}$

• ${\displaystyle x^3+\frac{1}{x^4}+\frac{129}{8}}$

• ${\displaystyle x^4+\frac{1}{x^3}+\frac{129}{8}}$

• ${\displaystyle x^3+\frac{1}{x^4}-\frac{129}{8}}$

Answer 1

${\displaystyle x^4+\frac{1}{x^3}-\frac{129}{8}}$

स्पष्टीकरण:

${\displaystyle \frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}}$

= f (x) ${\displaystyle =\int (4x^3-\frac{3}{x^4})dx}$

${\displaystyle =\frac{4}{4}{x}^4-\frac{3}{-3}\frac{.1}{x^3}+C}$

${\displaystyle =x^4+\frac{1}{x^3}}$ + C

परन्तु f(2) = 0

${\displaystyle {\left(2\right)}^4+\frac{1}{{\left(2\right)}^3}+C=0}$

${\displaystyle =16+\frac{1}{8}+C=0}$

⇒ ${\displaystyle C=-\frac{129}{8}}$

⇒ ${\displaystyle f\left(x\right)=x^4+\frac{1}{x^3}-\frac{129}{8}}$