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

9

The equation of tangent at $P(x, y)$ to $y=f(x)$ is

$Y-y=f'(x)(X-x)$

This cuts y-axis at $\left(0, y-x f'(x)\right)$

It is given that

$y-x f'(x)=x^3 \Rightarrow \frac{d f(x)}{d x}+\left(-\frac{1}{x}\right) f(x)=-x^2$

This is a linear differential equation with integrating factor

$e^{\int-\frac{1}{x} d x}=\frac{1}{x}$

So, its solution is given by

$f(x) \times \frac{1}{x}=\int-x^2 \times \frac{1}{x} d x+C$

or, $f(x)=-\frac{x^3}{2}+C x$            .....(i)

$\Rightarrow f(1)=-\frac{1}{2}+C \Rightarrow 1=-\frac{1}{2}+C \Rightarrow C=\frac{3}{2}$

∴     $f(x)=-\frac{x^3}{2}+\frac{3}{2} x$            [Putting $c-\frac{3}{2}$ in (i)]

$\Rightarrow f(-3)=\frac{27}{2}-\frac{9}{2}=9$