A) \[k>-1\]
B) \[k>-1\]
C) \[k>1\]
D) None of these
Correct Answer: A
Solution :
Given, \[f(x)=kx-\cos x\] and \[x\in R\] \[f'(x)=k+\sin x\] Since, the function \[f(x)\] is monotonically increase for all \[x\in R\]. \[\therefore \] \[f'(x)>0,\,\,\,\forall \,\,\,\,x\in R\] \[\Rightarrow \] \[f'\left( \frac{\pi }{2} \right)>0\] \[\Rightarrow \] \[k+\sin \frac{\pi }{2}>0\] \[\Rightarrow \] \[k+1>0\] \[\Rightarrow \] \[k>-1\]You need to login to perform this action.
You will be redirected in
3 sec