A) 3
B) 1
C) 2
D) None
Correct Answer: A
Solution :
Let\[{{x}^{\ln \frac{y}{z}}}+{{y}^{\ln \frac{z}{x}}}+{{z}^{\ln \frac{x}{y}}}\]and\[{{A}_{1}}={{x}^{\ln \frac{y}{z}}},{{A}_{2}}={{y}^{\ln \frac{z}{x}}}\]and\[{{A}_{3}}={{z}^{\ln \frac{x}{y}}}\] Also \[\ln {{A}_{1}}+\ln {{A}_{2}}+\ln {{A}_{3}}\] \[=\left( \ln \frac{y}{z} \right)\ln x+\left( \ln \frac{z}{x} \right)\ln y+\left( \ln \frac{x}{y} \right)\ln z\] \[=(lny-lnz)lnx+(lnz-lnx)lny+(lnx-lny)lnz\] \[\ln {{A}_{1}}+\ln {{A}_{2}}+\ln {{A}_{3}}=0\]or\[{{A}_{1}}{{A}_{2}}{{A}_{3}}={{e}^{0}}=1\] Using\[A.M\ge G.M\] \[\frac{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}}{3}\ge {{({{A}_{1}}{{A}_{2}}{{A}_{3}})}^{1/3}}=1\] or\[{{A}_{1}}+{{A}_{2}}+{{A}_{3}}=A\ge 3\] so that minimum value of A is 3.
You need to login to perform this action.
You will be redirected in
3 sec
