JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 13)
Let m be the minimum possible value of $${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}})$$, where $${y_1},{y_2},{y_3}$$ are real numbers for which $${{y_1} + {y_2} + {y_3}}$$ = 9. Let M be the maximum possible value of $$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$$, where $${x_1},{x_2},{x_3}$$ are positive real numbers for which $${{x_1} + {x_2} + {x_3}}$$ = 9. Then the value of $${\log _2}({m^3}) + {\log _3}({M^2})$$ is ...........
Answer
8
Explanation
For real numbers y1, y2, y3, the quantities $${{3^{{y_1}}}}$$, $${{3^{{y_2}}}}$$ and $${{3^{{y_3}}}}$$ are positive real numbers, so according to the AM-GM inequality, we have
$${{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}} \over 3} \ge {({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$$
$$ \Rightarrow {3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \ge 3{({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$$
On applying logarithm with base '3', we get
$${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}) \ge \left[ {1 + {1 \over 3}({y_1} + {y_2} + {y_3}} \right.)]$$
= 1 + 3
= 4
{$$ \because $$ $${{y_1} + {y_2} + {y_3}}$$ = 9}
$$ \therefore $$ m = 4
Now, for positive real numbers x1, x2 and x3 according to AM-GM inequality, we have
$${{{x_1} + {x_2} + {x_3}} \over 3} \ge {({x_1}{x_2}{x_3})^{{1 \over 3}}}$$
On applying logarithm with base '3', we get
$${\log _3}\left( {{{{x_1} + {x_2} + {x_3}} \over 3}} \right) \ge {1 \over 3}$$$$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$$
$$ \Rightarrow $$ $$1 \ge {1 \over 3}\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$$
{$$ \because $$ x1 + x2 + x3 = 9}
$$ \therefore $$ M = 3
Now, $${\log _2}({m^3}) + {\log _3}({M^2})$$
$$ = 3lo{g_2}(4) + 2lo{g_2}(3)$$ = 6 + 2 = 8
$${{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}} \over 3} \ge {({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$$
$$ \Rightarrow {3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \ge 3{({3^{{y_1}}}\,.\,{3^{{y_2}}}\,.\,{3^{{y_3}}})^{{1 \over 3}}}$$
On applying logarithm with base '3', we get
$${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}) \ge \left[ {1 + {1 \over 3}({y_1} + {y_2} + {y_3}} \right.)]$$
= 1 + 3
= 4
{$$ \because $$ $${{y_1} + {y_2} + {y_3}}$$ = 9}
$$ \therefore $$ m = 4
Now, for positive real numbers x1, x2 and x3 according to AM-GM inequality, we have
$${{{x_1} + {x_2} + {x_3}} \over 3} \ge {({x_1}{x_2}{x_3})^{{1 \over 3}}}$$
On applying logarithm with base '3', we get
$${\log _3}\left( {{{{x_1} + {x_2} + {x_3}} \over 3}} \right) \ge {1 \over 3}$$$$({\log _3}{x_1} + {\log _3}{x_2} + {\log _3}{x_3})$$
$$ \Rightarrow $$ $$1 \ge {1 \over 3}\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$$
{$$ \because $$ x1 + x2 + x3 = 9}
$$ \therefore $$ M = 3
Now, $${\log _2}({m^3}) + {\log _3}({M^2})$$
$$ = 3lo{g_2}(4) + 2lo{g_2}(3)$$ = 6 + 2 = 8
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