cat 2022 Question 15

Question

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

Accepted Answer

Let $\dfrac{x}{y}$ be $t$

Therefore, $c = 16t + \dfrac{49}{t}$

Applying AM ≥ GM

$\frac{16t + \frac{49}{t}}{2} \ge (16t imes \frac{49}{t})^{\frac12}$

$16t + \dfrac{49}{t} \ge 56$

When $t$ is positive then $c$ is greater than equal to $56$.

When $t$ is negative then $c$ is less than equal to $-56$.

Therefore $c \in (-\infty, -56] \cup [56, \infty]$

Question 15.