Question

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

Answer 1

The equation of the tangent at $(x,y)$ to the given curve $y=f\left(x\right)$ is
$Y-y=\frac{dy}{dx}(X-x)$
$Y-$intercept $=y-x\frac{dy}{dx}$
According to the question,
$x^3=y-x\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}-\frac{y}{x}=-x^2$
which is linear is $x$
I.F. $=e^{\int \frac{-1}{x}dx}$
$=\frac{1}{x}$
Required solution is
$y\frac{1}{x}=\int -x^2\frac{1}{x}dx$
$\frac{y}{x}=\frac{-x^2}{2}+C$
$y=\frac{-x^3}{2}+Cx$
at $x=1,y=1$
$\therefore 1=\frac{-1}{2}+c$
$\Rightarrow C=\frac{3}{2}$
Now, $f(-3)=\frac{27}{2}+\frac{3}{2}(-3)$
$=\frac{27-9}{2}=9$