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: (A) $ \frac{-47}{5}$
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 $f(4)=4-\frac25(4)^{\frac52}-\frac35 \\=4-\frac{64}{5}-\frac35 \\= \frac{20-67}{5} \\= -\frac{47}{5}$
The range($ST$) =$S_p\{c_1\}=$ Nullspace($P$).
range($P$)=$S_p\{c_2\}=$ Nullspace($ST$)
& rank($T$) $=1=$ rank ($S$)
But, Nullity of $T=1 \neq 2=$ Nullity of $S$.
And dim($U$) = dim($W$) = 2
So, the option $(A)$ and $(B)$ are incorrect.
[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]Consider $U= \mathbb{R}^3,V= \mathbb{R}^4, W= \mathbb{R}^5 $
range ($ST$) = $S_p\{c_1,c_2\}=$ Nullspace($P$)
Nullspace ($ST$)=$S_p\{c_3\}$=Range($P$)
rank($T$)=$2$=rank($S$)
and $P$ is not zero map
So, option $(D)$ is not correct.
Hence option $(C)$ is the one left which has to be true, Now lets prove that.
[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]We'll prove that $P$ is non zero.
Suppose $P=0$
$U\longrightarrow V\longrightarrow W$
dim($U$) = $4$, dim($V$) =$3$
Now,
range ($P$) $=0=$ Nullspace ($ST$)
$\Rightarrow ST $ is injective
$\Rightarrow T $ is injective
$\Rightarrow $ dim ($V\geq 4$)
Which is a contradiction.
Hence $P=0$
And we are done[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" custom_margin="20px||20px||false|false" global_module="50750"][et_pb_fullwidth_header title="College Mathematics Program" button_one_text="Learn more" button_one_url="https://cheenta.com/collegeprogram/" header_image_url="https://cheenta.com/wp-content/uploads/2018/03/College-1.png" _builder_version="4.2.2" background_color="#12876f" custom_button_one="on" button_one_text_color="#12876f" button_one_bg_color="#ffffff"]