Question

guys plez answer this question......​

Answer 1

$\huge {\boxed {\mathbb {QUESTION}}}$

$If\:\frac{d}{dx} f(x) =4{x}{3}-\frac{3}{{x}^{4}}\:such\:that\:f(2)=0.\:Then\:f(x)\:is$

$(A) \:{x}^{4}+\frac{1}{{x}^{3}}-\frac{129}{8}$

$(B) \:{x}^{3}+\frac{1}{{x}^{4}}+\frac{129}{8}$

$(C) \:{x}^{4}+\frac{1}{{x}^{3}}+\frac{129}{8}$

$(D) \:{x}^{3}+\frac{1}{{x}^{4}}-\frac{129}{8}$

$\huge {\boxed {\mathbb {ANSWER}}}$

${\boxed {\mathbb {GIVEN}}}$

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

${\boxed {\mathbb {SOLUTION}}}$

$Integrate\: the\: both\: sides$

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

$\implies \int \frac{d}{dx} f(x)=4 \int {x}^{3} dx-3\int \frac{1}{{x}^{4}} dx$

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

$\implies f(x) =4\frac{{x}^{3+1}}{3+1} - 3\frac{{x}^{-4+1}}{-4+1} +C$

$(because\: \int {x}^{n} dx=\frac{{x}^{n+1}}{n+1}+C)$

$\implies f(x) =4\frac{{x}^{4}}{4}-3\frac{{x}^{-3}}{-3} +C$

$\implies f(x)={x}^{4}+{x}^{-3}+C$

$\implies {\boxed {\boxed {f(x)={x}^{4}+\frac{1}{{x}^{3}}+C}}} ---(1)$

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$\huge {f(2)=0}$

$x=2$

$Substitute \:the \:value \:in\:(1)$

$\implies f(2)={(2)}^{4}+\frac{1}{{2}^{3}} +C$

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

$(because \:f(2)=0)$

$\implies 0=\frac{128+1}{8} +C$

$\implies 0=\frac{129}{8} +C$

$\implies {\boxed {\boxed {C=\frac{-129}{8}}}}$

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$\huge {C=\frac{-129}{8}}$

$Substitute \:the \:value \:in\:(1)$

$\implies f(x) ={x}^{4}+\frac{1}{{x}^{3}}+C$

$\implies{\boxed {\boxed {\boxed {f(x) ={x}^{4}+\frac{1}{{x}^{3}}+\frac{-129}{8}}}}}$

$\therefore Option\: (A) \:is\:correct$

$\huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$Substitutional \:Method$

$Solving \: one\: of \:the\: equations\: for\: one\: of\\ the\: variables\: and\: then\: plugging\: this\: back\\ into\: the \:other\: equation\: "substituting"\: for\: the\\ chosen\: variable \:and\: solving\: for\: the \:other.\\ Then\: you\: back\:solve\: for\: the\: first \:variable.\\Is \:called\: substitution\: method$

${\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5070}}}}}}}}}}}}}}}$

Answer 2

Answer:

a) is answer dear friend