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Let $f:\left[ {1,\infty } \right) \to \left[ {2,\infty } \right)$ be a differentiable function such that $f\left( 1 \right) = 2$.
If $6\int\limits_1^x {f(t)dt} = 3xf(x) - {x^3}$ for all $x \geqslant 1$, then the value of $f\left( 2 \right)$ is:
A. $0$
B. $1$
C. $6$
D. $3$

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Last updated date: 25th Jun 2024
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Answer
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Hint: Here given that the function has the domain from 1 to $\infty $, whereas the function has the co-domain 2 to $\infty $. To solve this problem we have to have an idea on a few concepts about differentiation, integrations, which includes definite integrals and indefinite integrals, and also about how to solve a linear differential equation.
The general solution of a differential equation \[\dfrac{{dy}}{{dx}} + P(x).y = Q(x)\], is given by:
$ \Rightarrow y.I(x) = \int {I(x).Q(x)dx + c} $
Where $I(x) = {e^{\int {P(x)dx} }}$, is called the integral factor.

Complete answer:
Given that $f\left( 1 \right) = 2$, and also
$ \Rightarrow 6\int\limits_1^x {f(t)dt} = 3xf(x) - {x^3}$
Differentiating the above equation on both sides with respect to $x$, as given below:
$ \Rightarrow 6f(x) = 3x\dfrac{d}{{dx}}\left( {f(x)} \right) + 3f(x)\dfrac{d}{{dx}}\left( x \right) - \dfrac{d}{{dx}}\left( {{x^3}} \right)$
$ \Rightarrow 6f(x) = 3xf'(x) + 3f(x)(1) - 3{x^2}$
As the differentiation of $x$ is 1, as given below:
$ \Rightarrow 6f(x) = 3xf'(x) + 3f(x) - 3{x^2}$
Arranging the like and unlike terms together, as given below:
$ \Rightarrow xf'(x) - f(x) = {x^2}$
Dividing the above equation by 3, and rearranging the terms so that it appears as a linear differential equation, as given below:
Now let $f(x) = y$ and hence $f'(x) = \dfrac{{dy}}{{dx}}$, substituting these in the above expression, as given below:
\[ \Rightarrow x\dfrac{{dy}}{{dx}} - y = {x^2}\]
Dividing the above equation by x, as given below:
\[ \Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{y}{x} = \dfrac{{{x^2}}}{x}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{y}{x} = x\]
Now solving the above linear differential equation, which is in the form of \[\dfrac{{dy}}{{dx}} + P(x).y = Q(x)\]
Where the general solution of the above expression would be, as given below:
$ \Rightarrow y.I(x) = \int {I(x).Q(x)dx + c} $
Here $I(x) = {e^{\int {P(x)dx} }}$
Now solving the obtained linear differential equation \[\dfrac{{dy}}{{dx}} - \dfrac{y}{x} = x\], as given below:
\[ \Rightarrow \dfrac{{dy}}{{dx}} + \left( { - \dfrac{1}{x}} \right)y = x\]
Here $P(x) = - \dfrac{1}{x}$ and $Q(x) = x$
Calculating $I(x) = {e^{\int {P(x)dx} }}$
\[ \Rightarrow I(x) = {e^{ - \int {\dfrac{1}{x}dx} }}\]
\[ \Rightarrow I(x) = {e^{ - {{\log }_e}x}} = {e^{{{\log }_e}{x^{ - 1}}}}\]
\[ \Rightarrow I(x) = {e^{{{\log }_e}\dfrac{1}{x}}}\]
$\therefore I(x) = \dfrac{1}{x}$
Now solving the general equation of the differential equation, as given below:
$ \Rightarrow y.I(x) = \int {I(x).Q(x)dx + c} $
$ \Rightarrow y.\left( {\dfrac{1}{x}} \right) = \int {\left( {\dfrac{1}{x}} \right).xdx + c} $
$ \Rightarrow \dfrac{y}{x} = \int {dx + c} $
$ \Rightarrow \dfrac{y}{x} = x + c$
Now multiplying the above equation with $x$ on both sides, as given below:
$ \Rightarrow y = {x^2} + cx$
$\because y = f(x)$
$\therefore f(x) = {x^2} + cx$
Given that $f\left( 1 \right) = 2$, now substituting this in the above equation to get the value of c, the constant of integration, as given below:
Substituting the value of $x$ as 1, as $f\left( 1 \right) = 2$, as given below:
$ \Rightarrow f(1) = {\left( 1 \right)^2} + c\left( 1 \right)$
$ \Rightarrow 2 = 1 + c$
$ \Rightarrow c = 1$
Substituting the value of c, the constant of integration in the $f(x)$ expression, as given below:
$\therefore f(x) = {x^2} + x$
Now we have to find the value of $f\left( 2 \right)$, by substituting the value of $x = 2$, as given below:
$ \Rightarrow f(2) = {\left( 2 \right)^2} + 2$
$ \Rightarrow f(2) = 4 + 2$
$\therefore f(2) = 6$

The value of $f(2)$ is 6.

Note:
Here while solving this problem, there are a few basic formulas which are applied here from differentiation such as the chain rule which is the differentiation of the function which is a product of two functions, which is given by $\dfrac{d}{{dx}}\left( {{f_1}(x).{f_2}(x)} \right) = {f_1}(x)f_2'(x) + f_1'(x){f_2}(x) $ One more point to note here is that an important basic formula from logarithms which is also applied here \[{e^{{{\log }_e}a}} = a\].