Question

$\Large {\sf{\pmb{\underline{\underline{Challenge...}}}}}$

$\sf \circ\ {If\ \dfrac{d}{dx}f(x)\ =\ 4x^3\ -\ \dfrac{3}{x^4}\ such\ that\ f(2)\ =\ 0,\ then\ f(x)\ is:-}$

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

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

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Answer 1

Answer :

( A )

SolutioN :

$\tt \: If\ \dfrac{d}{dx}f(x)\ =\ 4x^3\ -\ \dfrac{3}{x^4}$

Then,

→ Multiple both sides by dx.

$\tt :\implies d \: f(x)\ = \bigg(4x^3 - \dfrac{3}{x^4} \bigg) dx$

Taking both side integration, We get.

$\tt :\implies \int d \: f(x)\ = \int \bigg(4x^3 - \dfrac{3}{x^4} \bigg) dx$

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

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

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

Now, Let find the value of C if f ( 2 ) is 0.

( Condition given )

$\tt :\implies f(2)\ = 0.$

$\tt :\implies f(2)\ = (2)^4 + {(2)}^{ - 3} + C$

$\tt :\implies 0 = (2)^4 + {(2)}^{ - 3} + C$

$\tt :\implies 0 = 16 + \dfrac{1}{8} + C$

$\tt :\implies C = - \dfrac{129}{8} .$

Now, Compare f ( x ) We get.

$\tt :\implies f(x)\ = x^4 + {x}^{ - 3} - \dfrac{129}{8}$

$\tt :\implies f(x)\ = x^4 + \dfrac{1} {{x}^{3}} - \dfrac{129}{8}$

★ Correct option A )

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

Answer 2

Solution

Given :-

• $\tt{\:if\:\dfrac{d}{dx}f(x)\:=\:4x^3-\dfrac{3}{x^4}\:Such\:that\:f(2)\:=\:0\:\:then\:Value\:of\:f(x)\:is}$

Find :-

• Value of f(x)

Explanation

We Have,

➠ $\tt{\dfrac{d}{dx}f(x)\:=\:4x^3-\dfrac{3}{x^4}}$_____(1)

Integrate equ(1)

$\implies\tt{\displaystyle\int\dfrac{d}{dx}f(x)\:dx\:=\:\int\big(4x^3-\dfrac{3}{x^4}\big)dx}$

We Know,

$\star\tt{\red{\displaystyle\int\:x^n\:dx\:=\:\dfrac{x^{n+1}}{n+1}+C}}$

$\implies\tt{\displaystyle\:f(x)\:=\:4\dfrac{x^{3+1}}{3+1}-3\dfrac{x^{-4+1}}{-4+1}+C}$

$\implies\tt{\:f(x)\:=\:x^4+\dfrac{1}{x^3}+C}$_________(2)

We Have,

• f(2) = 0

So, Keep x = 2 , in equ(2)

$\implies\tt{\:0\:=\:2^4+\dfrac{1}{2^3}+C}$

$\implies\tt{\:0\:=\:16+\dfrac{1}{8}+C}$

$\implies\tt{\:0\:=\:\dfrac{(128+1)}{8}+C}$

$\implies\tt{\:0\:=\:\dfrac{129}{8}+C}$

$\implies\tt{\:C\:=\:\dfrac{-129}{8}}$_______(3)

Keep Value of X in equ(2)

$\implies\tt{\red{\:f(x)\:=\:x^4-\dfrac{1}{x^3}-\dfrac{129}{8}}}$

Hence

• Your Answer will be option Number (A).

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