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
But try the problem first...
Answer: $\textbf{(A)} \frac{-47}{5}$.
Question 33 (IIT JAM 2014)
Real Analysis (Willy)
First hint
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 !!!
Second Hint
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 ???
Third Hint
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 !!!
Fourth Hint
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 ?
Final Step
Now simply, putting $y=4$
we get $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$
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
But try the problem first...
Answer: $\textbf{(A)} \frac{-47}{5}$.
Question 33 (IIT JAM 2014)
Real Analysis (Willy)
First hint
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 !!!
Second Hint
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 ???
Third Hint
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 !!!
Fourth Hint
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 ?
Final Step
Now simply, putting $y=4$
we get $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$
