Quantitative Aptitude Ques 26
Question: For any real number x, the maximum value of $ 4-6x-x^{2} $
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 in $ x=-,3 $ in $ f(x), $ we get $ f(-,3)=4-(6)(-,3)-{{(-,3)}^{2}} $ $ =4+18-9=13 $ So, the maximum value of $ =4-6x-x^{2} $ is 13.
BETA
