MCQ
Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int_1^x f(t) d t=3 x f(x)-x^3$ for all $x \geq 1$, then the value of $f(2)$ is
  • $6$
  • B
    $3$
  • C
    $0$
  • D
    $1$

Answer

Correct option: A.
$6$
a
$6 \int_1^x f(t) d t=3 x f(x)-x^3$

Differentiating w.r.t. $x$, we get (Use Newton Leibnitz theorem for differentiating a definite integral)

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

$\Rightarrow x f^{\prime}(x)=f(x)+x^2 $

$\text { or, } \frac{d y}{d x}-\frac{y}{x}=x$

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 $\frac{y}{x}=\int x \times \frac{1}{x} d x+C$

$\Rightarrow \frac{y}{x}=x+C $

$\text { Since } f(1)=2 $

$\Rightarrow C=1$

So $y=x^2+x $

$\Rightarrow f(2)=6$

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