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

• A. 9
• B. -9
• C. 0
• D. 4

Answer 1

Answer: (A)

$Y-y=m(X-x) $
$ Y $-intercept$ (X=0) $
$ Y=y-m x$
Given that $y-m x=x^3 \Rightarrow x \frac{d y}{d x}-y=-x^3$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x^2$
Intergrating factor (I.F.) $= e ^{-\int \frac{1}{x} d x}=\frac{1}{x}$
solution $y \cdot \frac{1}{x}=\int \frac{1}{x} \cdot\left(-x^2\right) d x $
$ \Rightarrow f(x)=y=-\frac{x^3}{2}+c x$
Given $f(1)=1 \Rightarrow c=\frac{3}{2}$
$\therefore f(x)=-\frac{x^3}{2}+\frac{3 x}{2} \Rightarrow f(-3)=9$