Question:

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

Updated On: Sep 30, 2024
  • -70
  • 60
  • -50
  • -60
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The Correct Option is B

Solution and Explanation

The given expression is c=16xy+49yxc = \frac{16x}{y} + \frac{49y}{x}, where xx and yy are non-zero real numbers. We need to determine which of the given values 70-70, 6060, 50-50, and 60-60 cannot be taken by cc. First, let's simplify the expression: c=16x2+49y2xyc = \frac{16x^2 + 49y^2}{xy} Now, we can rewrite the expression as: c=16x2xy+49y2xy=16(xy)+49(yx)=16k+49kc = \frac{16x^2}{xy} + \frac{49y^2}{xy} = 16\left(\frac{x}{y}\right) + 49\left(\frac{y}{x}\right) = 16k + \frac{49}{k} Where k=xyk = \frac{x}{y}. Since xx and yy are real and non-zero, kk is also real. We can now create a quadratic equation in terms of kk: 16k2ck+49=016k^2 - ck + 49 = 0 For kk to be real, the discriminant of this quadratic equation must be greater than or equal to 00: c2416490c^2 - 4 \cdot 16 \cdot 49 \geq 0 c231360c^2 - 3136 \geq 0 Solving for cc, we get: c56|c| \geq 56 This means that the absolute value of cc must be greater than or equal to 5656. Therefore, cc cannot take the value 50-50, as 50=50<56|-50| = 50 < 56. So, out of the given options, cc cannot take the value 50-50.
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