Question

Let $f:\left[1,\infty \right)\to \left[2,\infty \right)$ be a differentiable function such that $f\left(1\right)=2.$ if $6\underset{1}{\overset{x}{\int }}f\left(t\right)dt=3xf\left(x\right)-x^3$ for value of $f\left(2\right)$ is

• A. $3$
• B. $4$
• C. $5$
• D. None of these

Answer 1

Answer: (D)

None of these

$6{\int }_1^{x}f\left(t\right)dt=3xf\left(x\right)-x^3$

Difference both sides.

$6f\left(x\right)=3f\left(x\right)-3x^2+3x{f}^{\prime }\left(x\right)$

$f\left(x\right)=x{f}^{\prime }\left(x\right)-x^2$

$y+x^2=x\frac{dy}{dx}$

$\frac{y}{x}+x=\frac{dy}{dx}$

$y=vx\frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x=v+x\frac{dv}{dx}$

$\int dv=\int dx$

$v=x+c$

$y=x^2+cx=f\left(x\right)$

$f\left(1\right)=2$

$1+c=2$

$c=1$

$f\left(2\right)=(2{)}^2+2=6$

$D$ is correct.