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UPSC NDA I 2024 Mathematics Paper

Question : 28 of 120

Marks: +1, -0

For

x \geq y>1

\log_x \left(\frac{x}{y}\right) + \log_y \left(\frac{y}{x}\right) = k

k

Solution:

We can simplify the given expression using the change of base formula and properties of logarithms:

\log_x \left(\frac{x}{y}\right) + \log_y \left(\frac{y}{x}\right) = \frac{\log\left(\frac{x}{y}\right)}{\log x} + \frac{\log\left(\frac{y}{x}\right)}{\log y}

= \frac{\log x - \log y}{\log x} + \frac{\log y - \log x}{\log y}

= 1 - \frac{\log y}{\log x} + \frac{\log y}{\log y} - \frac{\log x}{\log y}

= 1 + \frac{\log y}{\log y} - \left( \frac{\log y}{\log x} + \frac{\log x}{\log y} \right)

= 2 - \left( \frac{\log y}{\log x} + \frac{\log x}{\log y} \right)

Now, let's analyze the expression

\frac{\log y}{\log x} + \frac{\log x}{\log y}

. Since

x \geq y>1

, we know that

\log x \geq \log y > 0

. Therefore, both

\frac{\log y}{\log x}

and

\frac{\log x}{\log y}

are positive. By AM-GM,

\frac{\log y}{\log x} + \frac{\log x}{\log y} \geq 2 \sqrt{ \frac{\log y}{\log x} \cdot \frac{\log x}{\log y} } \geq 2 \text{. }

This means

2 - \left( \frac{\log y}{\log x} + \frac{\log x}{\log y} \right) \leq 0

. In other words,

k \leq 0

. Therefore,

k

can never be equal to 1 .

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