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Explore the Back-StoryA 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

- $\frac{-47}{5}$
- $\frac{-47}{10}$
- $\frac{-16}{5}$
- $\frac{-8}{5}$

**Relation/Mapping**

**Differentiation**

**Integration**

Answer: $\textbf{(A)} \frac{-47}{5}$.

Question 33 (IIT JAM 2014)

Real Analysis (Willy)

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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