Solution.

A subset S is a subspace of V if $\alpha x+\beta y\in S$ for all x, $y\in S$ and $\alpha ,\beta \in R$

A) Not a subspace

$p(x)={x}^{2}+5,\text{}g(x)={x}^{2}-5$

$p(1)=6,\text{}p(0)=5$ and $p(1)>p(0)$

$q(1)=-4,\text{}q(0)=-5$ and $q(1)>q(0)$

$\Rightarrow p(x),q(x)\in S$

Consider $R(x)=p(x)-2q(x)$

$R(1)=p(1)-2q(1)=6+8=24$

$R(0)=p(0)-2q(0)=5+10=25$

Here $R(1)<R(0)$

$\Rightarrow R(x)\ne \in S$

$\Rightarrow S$ is not a subspace

B) Not a subspace

$(-1,-1,0)\in S$ as $(-1)-(-1)6+0=-1+6=5$

Consider $-1(-1,-1,0)=(1,1,0)$

$1-6+0=-5\ne 5$

$\Rightarrow (1,1,0)\ne \in S$

$\Rightarrow S$ is not subspace

C) Subspace

Let $x,y\in S$ then $Ax=0$ and $Ay=0$

$A(\alpha x+\beta y)=\alpha Ax+\beta Ay$

$=\alpha (0)+\beta (0)$

$=0$

$\Rightarrow \alpha x+\beta y\in S$

$\Rightarrow S$ is a subspace

D) Subspace

Let $x,y\in S$ then ${x}^{\u2033}-4{x}^{\prime}+3x=0$

and ${y}^{\u2033}-4{y}^{\prime}+3y=0$

Consider $\alpha x+\beta y$

$(\alpha x+\beta y{)}^{\u2033}-4(\alpha x+\beta y{)}^{\prime}+3(\alpha x+\beta y)$

$=\alpha {x}^{\u2033}+\beta {y}^{\u2033}-4\alpha {x}^{\prime}-4\beta {y}^{\prime}+3\alpha x+3\beta y$

$=\alpha ({x}^{\u2033}-4{x}^{\prime}+3x)+\beta ({y}^{\u2033}-4{y}^{\prime}+3y)$

$=\alpha (0)+\beta (0)$

$=0$

$\Rightarrow \alpha x+\beta y\in S$

$\Rightarrow S$ is a subspace

E) Not a subspace

Let $f,g\in S$ then $f(a)=5$ and $g(a)=5$

$\alpha f(a)+\beta g(a)=5\alpha +5\beta $

Let $\alpha =1$ and $\beta =-1$

$\Rightarrow \alpha f(a)+\beta g(a)=5-5=0$

$\Rightarrow \alpha f(a)+Bg(a)\ne \in S$

$\Rightarrow S$ is not a subspace

F) Subspace

Let $p,g\ne \in S$ then $p(0)=0$ and $g(0)=0$

Consider $\alpha p+\beta g$