JEE Main201911 Jan 2019Evening ShiftMathematicsApplication of DerivativesActual

Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression x^ m y^ n (1+x^ 2 ~m ) (1+y^ 2 n ) is :

Options

A1
B1 2
C1 4
Dm+n 6 m n

Correct answer

C. 1 4

Step-by-step solution

A= x^ m y^ n (1+x^ 2 m ) (1+y^ 2 n ) = 1 (x^ -m +x^ m ) (y^ -n +y^ n ) x^ m +y^ -m 2 (x^ n+1 , x^ -m )^ 1 2 x^ n+ +x^ -n 2 In the same way, y^ n +y^ n 2 Then , (x^ n +x^ -n ) (y^ -n +y^ n ) 4 1 (x^ "1 +x^ -m ) (y^ -n +y^ n ) 1 4

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