How Cheenta works to ensure student success?
Explore the Back-Story

Relation Mapping (IIT JAM 2014)

Question 33 - Relation-Mapping (IIT JAM 2014)


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}

Key Concepts


Relation/Mapping

Differentiation

Integration

Check the Answer


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

Question 33 (IIT JAM 2014)

Real Analysis (Willy)

Try with Hints


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}

Subscribe to Cheenta at Youtube


Question 33 - Relation-Mapping (IIT JAM 2014)


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}

Key Concepts


Relation/Mapping

Differentiation

Integration

Check the Answer


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

Question 33 (IIT JAM 2014)

Real Analysis (Willy)

Try with Hints


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}

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
magic-wandrockethighlight