JEE MAINS 2019 Question

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

(A) $\frac{1}{2}$

(B) $\frac{1}{4}$

(C) $\frac{m+n}{6mn}$

(D) 1

Answer: B

Solution:

Using A.M $\ge$ G.M inequality, we get

$\frac{1+x^{2m}+y^{2n}+x^{2m}.y^{2n}}{4}$ $\ge$ $(x^{2m}.y^{2n}.x^{2m}.y^{2n})^{\frac {1}{4}}$

$\therefore$ $\frac{(1+x^{2m})(1+y^{2n})}{4}$ $\ge$ $x^{m}.y^{n}$

$\Longrightarrow$ $\frac{x^{m}.y^{n}}{(1+x^{2m})(1+y^{2n})} \le \frac{1}{4}$

Therefore, maximum value is $\frac{1}{4}$ .

Thanks for reading the post and visiting the website.

Some More Questions to Read:

Some Important Concept to Read:

Let x, y be positive real numbers and m, n positive integers..