JEE Advanced 2010 Paper 1Question 1
Let f be a real-valued differentiable function on $\mathbf R$ (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
Differential Equation
Answer$9$
Solution
Concept. The $y$-intercept of the tangent at $(x,y)$ is $y-xy'$. Formulas used. (1) $\left(\dfrac yx\right)'=\dfrac{xy'-y}{x^2}$Step 1 - Equation. $y-xy'=x^3$, i.e. $xy'-y=-x^3$. Step 2 - Integrate. By (1), $\left(\dfrac yx\right)'=-x$, so $y=-\dfrac{x^3}{2}+cx$. Step 3 - Use $f(1)=1$. $-\frac12+c=1$, so $c=\frac32$. Step 4 - Value. $f(-3)=\dfrac{27}{2}-\dfrac92=9$. Answer: 9