Let `f` be a real-valued differentiable function on `R` (the set of all real numbers) such that `f(1)=1.` If the `y-in t e r c e p t` 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________

Options

A. 3
B. 6
C. 9
D. 0

Answer

C

Explanation

The equation of tangent at `P(x,y)" to "y=f(x)` is
`Y-y=f'(x)(X-x)`
This cute y-axis at `(0,y-xf'(x))`
It is given that
`y-xf'(x)=x^(3)rArr (df(x))/(dx)+(-(1)/(x))f(x)=-x^(2)`
This is a linear differential equation with integrating factor
`e^(int-(1)/(x)dx)=(1)/(x)`
So, its solution is given by
`f(x)xx(1)/(x)=int-x^(2)xx(1)/(x)dx+C`
or, `f(x)=-(x^(3))/(2)+Cx" ...(i)"`
`rArr" "f(1)=(1)/(2)+CrArr 1=-(1)/(2)+CrArrC=(3)/(2)`
`therefore" "f(x)=-(x^(3))/(2)+(3)/(2)x" "["Putting "C-(3)/(2)" in (i)"]`
`rArr" "f(-3)=(27)/(2)-(9)/(2)=9`