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?

• A. $-3$
• B. $3$
• C. $9$
• D. $-9$

Answer 1

The correct option is C $9$

$P\equiv (x,y)$

Let slope by $\frac{dy}{dx}$

So, equation of tangent will be

$(X-m)\frac{dy}{dx}=(Y-y)$

for Y- intercept , $x=0$

So, $Y=y-x\frac{dy}{dx}$

Now according to question ,

$x^3=y-x\frac{dy}{dx}\frac{dy}{dx}-\frac{y}{x}=x^2\frac{xdy-ydx}{x^2}=-xdxd\left(\frac{y}{x}\right)=-xdx$

On integrating , we get

$\frac{y}{x}=\frac{-x^2}{2}+C$

Now, $y\left(1\right)=1$

So, we get $\frac{y}{x}=\frac{-x^2}{2}+\frac{3}{2}$

So, $y(-3)=f(-3)=9$