Then A is:
A) invertible only if t \[t=\frac{\pi }{2}\]
B) not invertible for any\[t\in R\]
C) invertible for all \[t\in R\]
D) invertible only if \[t\in \pi \]
Correct Answer: C
Solution :
\[=\,{{e}^{-1}}\left[ 5{{\cos }^{2}}t+5{{\sin }^{2}}t \right]\forall t\in R\] \[=5{{e}^{-t}}\ne 0\,\forall \,t\in R.\]
You need to login to perform this action.
You will be redirected in
3 sec
