Question

Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f (1)=1$. If the $y$-intercept of the tangent at any point $P ( x , y )$ on the curve $y = f ( x )$ is equal to the cube of the abscissa of $P$, then the value of $f(-3)$ is equal to _____

Answer 1

$y - y _{1}= m \left( x - x _{1}\right)$
Put $x=0$, to get $y$ intercept
$y_{1}-m x_{1}=x_{1}^{3} $
$y_{1}-x_{1} \frac{d y}{d x}=x_{1}^{3} $
$x \frac{d y}{d x}-y=-x^{3}$
$\frac{d y}{d x}-\frac{y}{x}=-x^{2}$
$e^{\int \frac{1}{x} dx}=e^{-\ln x}=\frac{1}{x} $
$y \times \frac{1}{x}=\int-x^{2} \times \frac{1}{x} d x $
$\frac{y}{x}=-\int x d x $
$\Rightarrow \frac{y}{x}=-\frac{x^{2}}{2}+c $
$\Rightarrow f(x)=-\frac{x^{3}}{2}+\frac{3}{2} x$
$ \therefore f(-3)=9$