Quantitative Aptitude Ques 1897

Question: For any real number x, the maximum value of $ 4-6x-x^{2} $ is

Options:

A) 4

B) 7

C) 9

D) 13

Show Answer

Answer:

Correct Answer: D

Solution:

  • Let the given equation be represented as $ f,(x)=4-6x-x^{2} $ Now, differentiating above function w.r.t. x, we get $ f’(x)=-6-2x $ For value of x put $ f’(x)=0 $ $ -6-2x=0 $

$ \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.