Applications Of Derivatives Question 387
Question: If $ P=(1,,1) $ , $ Q=(3,,2) $ and R is a point on x-axis then the value of $ PR+RQ $ will be minimum at
[AMU 2005]
Options:
A) $ ( \frac{5}{3},,0 ) $
B) $ ( \frac{1}{3},,0 ) $
C) (3, 0)
D) (1, 0)
Show Answer
Answer:
Correct Answer: A
Solution:
Let co-ordinate of R (x, 0) Given $ P(1,,1) $ and $ Q,(3,,2) $
$ PR+RQ=\sqrt{{{(x-1)}^{2}}+{{(0-1)}^{2}}}+\sqrt{{{(x-3)}^{2}}+{{(0-2)}^{2}}} $ = $ \sqrt{x^{2}-2x+2}+\sqrt{x^{2}-6x+13} $ For minimum value of PR + RQ, $ \frac{d}{dx}(PR+RQ)=0 $
therefore $ \frac{d}{dx}(\sqrt{x^{2}-2x+2})+\frac{d}{dx}(\sqrt{x^{2}-6x+13})=0 $
therefore $ \frac{(x-1)}{\sqrt{x^{2}-2x+2}}=-\frac{(x-3)}{\sqrt{x^{2}-6x+13}} $
Squaring both sides, $ \frac{{{(x-1)}^{2}}}{(x^{2}-2x+2)}=\frac{{{(x-3)}^{2}}}{x^{2}-6x+13} $
therefore $ 3x^{2}-2x-5=0 $
therefore $ (3x-5),(x+1)=0 $ , $ x=\frac{5}{3},,-1 $ . Also $ 1<x<3 $ .
$ \therefore $ $ R=(5/3,,0) $ .
BETA
