CAT 2022 Slot 3 QA Question 1: If c = $\dfrac{16 x}{y} + \dfrac{49 y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

Question

If c = $\dfrac{16 x}{y} + \dfrac{49 y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

Accepted Answer

Let $t = \dfrac{x}{y}$. Since $x$ and $y$ are non-zero, $t$ is also non-zero.
$c = 16t + \dfrac{49}{t}$
If $t$ is positive, AM-GM gives the minimum value.
$16t + \dfrac{49}{t} \geq 2 \sqrt{16t 	imes \dfrac{49}{t}} = 56$
So, for $t > 0$, $c \geq 56$.
If $t$ is negative, write $t = -u$ where $u > 0$.

If c = 16xy+49yx\dfrac{16 x}{y} + \dfrac{49 y}{x}y16x​+x49y​ for some non-zero real numbers x and y, then c cannot take the value

If c = $\dfrac{16 x}{y} + \dfrac{49 y}{x}$ for some non-zero real numbers x and y, then c cannot take the value