Equations And Inequalities Question 46
Question: If $ R\ge r>0 $ and $ d>0, $ then $ 0<\frac{d^{2}+R^{2}-r^{2}}{2dR}\le 1 $
Options:
A) Is satisfied if $ | d-R |\le r $
B) Is satisfied if $ | d-R |\le 2r $
C) Is satisfied if $ | d-R |\ge r $
D) Is not satisfied at all
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Given $ R\ge r>0 $ and $ d>0 $
$ \Rightarrow 0<\frac{d^{2}+R^{2}-r^{2}}{2dR}\le 1 $
$ \Rightarrow 0<(d+R-r)(d+R+r)\le 2dR; $ which is true iff $ (d^{2}+R^{2}-r^{2})\le 2dR, $ which is true iff $ d^{2}+R^{2}-2dR\le r^{2} $
$ \Rightarrow {{(d-R)}^{2}}\le r^{2} $ $ | d-R |\le r, $ which is also $ -r\le (d-R)\le r $ $ | d-R |\le r, $ which is also $ -r\le (d-R)\le r $
BETA
