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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

1. $-60$
2. $-50$
3. $60$
4. $-70$

Let z=(x/y). Now, c = 16z + 49/z  => 16z$^{2}$ – cz + 49 = 0. Since x & y are non zero real number. So, z is also a non zero real number. So, the equation 16z$^{2}$ – cz + 49 = 0 satisfies for some real number z, if b$^{2}$ – 4ac >= 0  => c$^{2}$ – 56$^{2}$ >= 0  => |c| >= 56.

Therefore the absolute value of c is always >= 56. So, value of c can’t be -50.

Ans is (B).
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