TIFR Mathematics 2025 | Part A | Question: 9

Question

Consider the linear map $T: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ defined by $T(p(x))=$ $\dfrac{d}{d x}(x p(x))$. Which of the following statements is correct?

• A. $T$ is injective but not surjective.
• B. $T$ is surjective but not injective.
• C. $T$ is bijective.
• D. $T$ is neither injective nor surjective.

Comments

Let, P(x)=a(x)^2 + b(x) + c..
T(xp(x))=3a(x)^2 + 2b(x) + c..

So, for a=0,b=0,c=0, ker(T(x(p(x)))={0}..  So, it is injective...

And, for every image, there will be an x to hold true, since the constants 3a, 2b, c can take any values...(simply being in Real to real mapping).
Hence, it is bijective

Answer 1

Remember:
    Degree Drops ($x^n \mapsto x^{n-1}$) = Not Injective
    Degree Rises ($x^n \mapsto x^{n+1}$) = Not Surjective
    Degree stays the same ($x^n \mapsto x^n$) = Bijective