Consider the two statements:
Statement I : y = sin kt satisfy the differential equation $y\text{'}\text{'}+9y=0.$  
Statement II : $y=e^{kt}$ satisfy the differential equation $y\text{'}\text{'}+y\text{'}-6y=0.$  
The value of k for which both the statements are correct is 

Statement I: y = sin kt, y’ = k cos kt; $y\text{'}\text{'}=-k^2\sin kt$  
$\therefore -k^2\sin kt+9\sin kt=0$ 
$\sin kt[9-k^2]=0\Rightarrow k=0,k=3,k=-3$ 
Statement II: $y=e^{kt},y\text{'}=k{e}^{kt};y\text{'}\text{'}=k^2{e}^{kt}$ 
$\therefore k^2{e}^{kt}+k{e}^{kt}-6e^{kt}=0$  
$e^{kt}(k^2+k-6)=0$  
$(k+3)(k-2)=0$  
 K = -3 or 2
Common value is k = -3
Hence, (a) is the correct answer.