Applications Of Derivatives Question 325
Question: The real number which most exceeds its cube is
[MP PET 2000]
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{\sqrt{3}} $
C) $ \frac{1}{\sqrt{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let Number = x, then cube = $ x^{3} $
Now $ f(x)=x-x^{3} $ (Maximum)
therefore $ f’(x)=1-3x^{2} $ Put $ {f}’(x)=0 $
therefore $ 1-3x^{2}=0 $
therefore $ x=\pm \frac{1}{\sqrt{3}} $
Because $ {f}’’(x)=-6x=-ve. $ when $ x=+\frac{1}{\sqrt{3}} $ .
BETA
