A) 1
B) 2
C) 4
D) None of these
Correct Answer: D
Solution :
\[f(0-)=\underset{x\to 0-}{\mathop{\lim }}\,\,k(2x-{{x}^{2}})=0\]; \[f(0+)=\underset{x\to 0+}{\mathop{\lim }}\,\,\cos x=1\] \[\therefore \,\,\,f(0)=\cos x=1\] Hence no value of k can make \[f(0-)=1.\]You need to login to perform this action.
You will be redirected in
3 sec