What is the NO-SHORTCUT approach for learning great Mathematics?

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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

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

**Relation/Mapping**

**Differentiation**

**Integration**

But try the problem first...

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

Source

Suggested Reading

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 $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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