Question

Let $f:[1,\infty )\to [2,\infty )$ be a differentiable function such that $f\left(1\right)=2$.

If $6{\int }_1^{x}f\left(t\right)dt=3xf\left(x\right)-x^3$ for all $x\ge 1$, then the value of $f\left(2\right)$ is :

• A. $0$
• B. $1$
• C. $6$
• D. $3$

Answer 1

The correct option is C $6$

$6{\int }_1^{x}f\left(t\right)dt=3xf\left(x\right)-x^3$
Differentiating w.r.t. x, we get (Use Newton Leibnitz theorem for differentiating a definite integral)
$\Rightarrow 6f\left(x\right)=3x{f}^{{}^{\prime }}\left(x\right)+3f\left(x\right)-3x^2$
$\Rightarrow x{f}^{{}^{\prime }}\left(x\right)=f\left(x\right)+x^2$
or, $\frac{dy}{dx}-\frac{y}{x}=x$
This is a first order linear differential equation with Integrating factor $e^{-\int \frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$
Its general solution is $\frac{y}{x}=\int x\times \frac{1}{x}dx+C$
$\Rightarrow \frac{y}{x}=x+C$
Since $f\left(1\right)=2$
$\Rightarrow C=1$
So, $y=x^2+x$
$\Rightarrow f\left(2\right)=6$