Question

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

Answer 1

Hint: Here given that the function has the domain from 1 to $\mathrm{\infty }$, whereas the function has the co-domain 2 to $\mathrm{\infty }$. 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 $${\displaystyle \frac{dy}{dx}}+P(x).y=Q(x)$$, 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
$\Rightarrow 6\underset{1}{\overset{x}{\int }}f(t)dt=3xf(x)-x^3$
Differentiating the above equation on both sides with respect to $x$, as given below:
$\Rightarrow 6f(x)=3x{\displaystyle \frac{d}{dx}}\left(f(x)\right)+3f(x){\displaystyle \frac{d}{dx}}\left(x\right)-{\displaystyle \frac{d}{dx}}\left(x^3\right)$
$\Rightarrow 6f(x)=3x{f}^{\prime }(x)+3f(x)(1)-3x^2$
As the differentiation of $x$ is 1, as given below:
$\Rightarrow 6f(x)=3x{f}^{\prime }(x)+3f(x)-3x^2$
Arranging the like and unlike terms together, as given below:
$\Rightarrow x{f}^{\prime }(x)-f(x)=x^2$
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 $f^{\prime }(x)={\displaystyle \frac{dy}{dx}}$, substituting these in the above expression, as given below:
$$\Rightarrow x{\displaystyle \frac{dy}{dx}}-y=x^2$$
Dividing the above equation by x, as given below:
$$\Rightarrow {\displaystyle \frac{dy}{dx}}-{\displaystyle \frac{y}{x}}={\displaystyle \frac{x^2}{x}}$$
$$\Rightarrow {\displaystyle \frac{dy}{dx}}-{\displaystyle \frac{y}{x}}=x$$
Now solving the above linear differential equation, which is in the form of $${\displaystyle \frac{dy}{dx}}+P(x).y=Q(x)$$
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 $${\displaystyle \frac{dy}{dx}}-{\displaystyle \frac{y}{x}}=x$$, as given below:
$$\Rightarrow {\displaystyle \frac{dy}{dx}}+\left(-{\displaystyle \frac{1}{x}}\right)y=x$$
Here $P(x)=-{\displaystyle \frac{1}{x}}$ and $Q(x)=x$
Calculating $I(x)=e^{\int P(x)dx}$
$$\Rightarrow I(x)=e^{-\int {\displaystyle \frac{1}{x}}dx}$$
$$\Rightarrow I(x)=e^{-{\log }_{e}x}=e^{{\log }_e{x}^{-1}}$$
$$\Rightarrow I(x)=e^{{\log }_e{\displaystyle \frac{1}{x}}}$$
$\therefore I(x)={\displaystyle \frac{1}{x}}$
Now solving the general equation of the differential equation, as given below:
$\Rightarrow y.I(x)=\int I(x).Q(x)dx+c$
$\Rightarrow y.\left({\displaystyle \frac{1}{x}}\right)=\int \left({\displaystyle \frac{1}{x}}\right).xdx+c$
$\Rightarrow {\displaystyle \frac{y}{x}}=\int dx+c$
$\Rightarrow {\displaystyle \frac{y}{x}}=x+c$
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 ${\displaystyle \frac{d}{dx}}\left(f_1(x).f_2(x)\right)=f_1(x)f_2^{\prime }(x)+f_1^{\prime }(x)f_2(x)$ One more point to note here is that an important basic formula from logarithms which is also applied here $$e^{{\log }_{e}a}=a$$.