Let f and g be twice differentiable functions on R such that $f\text{'}\text{'}\left(x\right)=g\text{'}\text{'}\left(x\right)+6x,f\text{'}\left(1\right)=4g\text{'}\left(1\right)-3=9,f\left(2\right)=3g\left(2\right)=12$ then which of the following is TRUE ?

$f^{\mathbf{/}\mathbf{/}}\left(x\right)=g^{\mathbf{/}\mathbf{/}}\left(x\right)+6x$ 
$f^{\mathbf{/}}\left(x\right)-g^{\mathbf{/}}\left(x\right)=3x^2+C$ 
$f^{\mathbf{/}}\left(1\right)=9,g^{\mathbf{/}}\left(1\right)=3,f\left(2\right)=12,g\left(2\right)=4$ 
At x = 1, $f^{/}\left(1\right)-g^{/}\left(1\right)=3+C$  
$\Rightarrow 9-3=3+C\Rightarrow C=3$ 
$f^{/}\left(x\right)-g^{/}\left(x\right)=3x^2+3$ 
$f\left(x\right)=g\left(x\right)+x^3+3x+D$ 
$f\left(2\right)=12\Rightarrow f\left(2\right)=g\left(2\right)+8+16+D$ 
$g\left(2\right)=4$  12 = 4 + 8 + 6 + D
 D = –6
$f\left(x\right)=g\left(x\right)+x^3+3x-6$ 
$h\left(x\right)=f\left(x\right)-g\left(x\right)=x^3+3x-6$ 
h(2) = 8 + 6 – 6 = 8
h(–2) = –8 – 6 – 6 = –20
For –1 < x < 2, h(x) = f(x) – g(x) = $x^3+3x-6$  
$h^{/}\left(x\right)=3x^2+3\uparrow forx\in R$ 
$h^{/}(-1)=f^{/}(-1)-g^{/}(-1)\Rightarrow |3x^2+3|<6$ 
$\Rightarrow x^2<1\Rightarrow -1<x<1$ 
For $x\in (-1,1)\Rightarrow |f^{/}\left(x\right)-g^{/}\left(x\right)|<6$  
D) $f\left(1\right)-g\left(x\right)=0\Rightarrow x^3+3x-6=0$  
$h\left(1\right)=-ve,h\left(\frac{3}{2}\right)=+ve$ 
So, there exists $x_0\in (1,\frac{3}{2})$ such that $f\left(x_0\right)=g\left(x_0\right)$