A beautiful problem involving the concept of relation-mapping from IIT JAM 2014.
Let $f:(0,\infty)\to \mathbb R$ be a differentiable function such that $f'(x^2)=1-x^3$ for all $x>0$ and $f(1)=0$. Then $f(4)$ equals
Relation/Mapping
Differentiation
Integration
Answer: (A) $ \frac{-47}{5}$
The above problem can be done in many ways we will try to solve this by the simplest method.
Now, as the function is given as $f'(x^2)=1-x^3$
So first try to change this $x^3$ into $x^2$. Try this. It's very easy !!!
To change $x^3$ into $x^2$ we can easily do
$f'(x^2)=1-(x^2)^{\frac32}$
Now we have to find the value of $f(4)$ so we have to change the second degree term, i.e., $x^2$ into some linear form. Can you cook this up ???
Let us assume $x^2=y$
i.e., $f'(y)=1-y^{\frac32}$
Now you know from previous knowledge that integration is also known as anti-derivative. So $f'(y)$ can be changed into $f(y)$ by integrating it with respect to $y$. Try to do this integration and we are half way done !!!
On integrating both side w.r.t $y$ we get :
$f(y)=y-\frac25y^{\frac52}+c$, (where $c$ is a integrating constant.)
Now we find the value to $c$
We know $f(1)=0$
$\Rightarrow c=-\frac35$
i.e., $f(y)=y-\frac25y^{\frac52}-\frac35$
Can you find the answer now ?
Now simply, putting $y=4$
we get $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$
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