Please submit your concern

Question:

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

Explanation:

Given, c = $\frac{16x}{y}$ + $\frac{49y}{x}$ 

Let $\frac{\mathrm{x}}{\mathrm{y}}$ = a

⇒ c = 16a + 49/a

Now, for two positive numbers, AM ≥ GM
If a > 0
⇒ $\frac{16\mathrm{a}+{\displaystyle \frac{49}{\mathrm{a}}}}{2}$ ≥ $\sqrt{16\mathrm{a}\times \frac{49}{\mathrm{a}}}$

⇒ $\frac{\mathrm{c}}{2}$ ≥ 28

⇒ c ≥ 56

∴ option (c) is rejected.

Now, for two negative numbers, AM ≤ - GM
If a < 0
⇒ $\frac{16\mathrm{a}+{\displaystyle \frac{49}{\mathrm{a}}}}{2}$ ≤ - $\sqrt{16\mathrm{a}\times \frac{49}{\mathrm{a}}}$

⇒ $\frac{\mathrm{c}}{2}$ ≤ - 28

⇒ c ≤ - 56

∴ option (a) and (b) are rejected.

∴ c cannot be equal to -50

Hence, option (d).