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