A) 4
B) 7
C) 9
D) 13
Correct Answer: D
$ \therefore $ $ x=-3 $ For maximum value, we take $ f’(x) $ $ f’(x)=-,2 $ Since, value of $ f’(x) $ is negative. So, $ f,(x) $ is maximum at $ x=-3. $ Putting $ x=-3 $ in $ f,(x), $ we get $ f,(-3)=4-(6)(-3)-{{(-3)}^{2}} $ $ =4+18-9=13 $ The maximum value of $ 4-6x-x^{2} $ is 13.