Let \(V=\{a_0+a_1x+a_2x^2 : a_0, a_1, a_2 \in \mathbb{R}\}\) be a vector space and \(T:V\rightarrow V\) is a linear transformation such that \(T(f)=f+\dfrac{df}{dx}\), the the value of \((T^3-3T^2+3T)^{2025}(x)\) is-
(a) \(x\qquad \qquad\)
(b) \(x+1\qquad \qquad\)
(c) \(2025! x\qquad \qquad\)
(d) \(2025! x+1\qquad \qquad\)
Explanation: Let basis of \(V\) is \(\beta=\{1,x,x^2\}\). Then matrix of \(T\) is given by \begin{pmatrix}1 & 1 & 0\\ 0& 1& 2\\0 &0 & 1\end{pmatrix} Thus eigenvalues of \(T\) are 1, 1, 1. Now let \(B=(T^3-3T^2+3T)^{2025}\), then 1, 1, 1 are also eigenvalues of \(B\). Thus \(Bx=\lambda x\) implies \(Bx=x\), i.e. \((T^3-3T^2+3T)^{2025}(x)=x\). Hence option (a) is correct.