Question

Let $P_{12}\left(x\right)$ be the real vector space of polynomials in the variable 𝑥 with real
coefficients and having degree at most 12, together with the zero polynomial.
Define

$V=\{f\in P_{12}\left(x\right):f\left(-x\right)=f\left(x\right)\text{for all}x\in ℝ\text{and}f\left(2024\right)=0\}\mathrm{.}$

Then, the dimension of V is _____________

Answer: $\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_}{6}$

Solution :-

Given that $P_{12}\left(x\right)=\{a_0+a_{1}x+a_2{x}^2+\cdots +a_{12}{x}^{12}:a_0,a_1,\cdots ,a_{12}\in ℝ\}$

We know that basis for $P_{12}\left(x\right)=\{1,x,x^2,\cdots ,x^{12}\}\mathrm{.}$

Let $W=\{f\in P_{12}\left(x\right):f\left(-x\right)=f\left(x\right)\text{for all}x\in ℝ\}\mathrm{.}$ We know that W is subspace of $P_{12}\left(x\right)$ and basis for W is set $\{1,x^2,x^4,x^6,x^8,x^{10},x^{12}\}\mathrm{.}$ So, dimension of W is 7.

Now, $V=\{f\in P_{12}\left(x\right):f\left(-x\right)=f\left(x\right)\text{for all}x\in ℝ\text{and}f\left(2024\right)=0\}$ is a subspace of W and one linearly independent constraint ${\prime }^{}f\left(2024\right)=0^{\prime }$ is given on V.

So, the correct answer is $\to$ $\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_}{6}$