Applications Of Derivatives Question 182
Question: The minimum value of $ 2x^{2}+x-1 $ is
[EAMCET 2003]
Options:
A) $ -\frac{1}{4} $
B) $ \frac{3}{2} $
C) $ \frac{-9}{8} $
D) $ \frac{9}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=2x^{2}+x-1 $
therefore $ f’(x)=4x+1\Rightarrow f’(x)=0\Rightarrow x=-\frac{1}{4} $
$ {f}’’,(x)=4=+ve $
$ \therefore {{[f(-1/4)]}_{\min }}=\frac{2}{16}-\frac{1}{4}-1=\frac{-9}{8} $ .
BETA
