Applications Of Derivatives Question 207
Question: The sum of two non-zero numbers is 4. The minimum value of the sum of their reciprocals is
[Kurukshetra CEE 1998]
Options:
A) $ \frac{3}{4} $
B) $ \frac{6}{5} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ x+y=4 $ or $ y=4-x $
$ \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy} $ or $ f(x)=\frac{4}{xy} $
$ =\frac{4}{x(4-x)} $ ; $ f(x)=\frac{4}{4x-x^{2}} $ Now $ f’(x)=\frac{-4}{{{(4x-x^{2})}^{2}}}.(4-2x) $ Put $ f’(x)=0 $ , then $ 4-2x=0 $
$ \therefore $ $ x=2 $ and $ y=2 $ ;
$ \therefore \min ( \frac{1}{x}+\frac{1}{y} )=\frac{1}{2}+\frac{1}{2}=1 $ .
BETA
