If x = _{c}(ab), y = _{a}(bc), z = _{b}(ca), then which of the following is correct?

Correct Answer is : (1+x)−1+(1+y)−1+(1+z)−1=1(1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} = 1(1+x)−1+(1+y)−1+(1+z)−1=1

(1 + x)^{-2} + (1 + y)^{-2} + (1 + z)^{-2} = 1

Correct Answer is : (1+x)−1+(1+y)−1+(1+z)−1=1(1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} = 1(1+x)−1+(1+y)−1+(1+z)−1=1

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UPSC NDA I and II 2020 Math Solved Paper

Question : 37 of 120

Marks: +1, -0

x = \log_{c}(ab)

y = \log_{a}(bc)

z = \log_{b}(ca)

, then which of the following is correct?

Solution:

Concept:

Use logarithm base change and product rule to simplify expressions.

Explanation:

Given

x = \log_{c}(ab)

y = \log_{a}(bc)

z = \log_{b}(ca)

Using the base change formula:

x = \frac{\log(ab)}{\log c}

y = \frac{\log(bc)}{\log a}

z = \frac{\log(ca)}{\log b}

(logarithms to any common base).

Now compute

1+x

1+y

1+z

1+x = 1 + \frac{\log(ab)}{\log c} = \frac{\log c + \log(ab)}{\log c} = \frac{\log(abc)}{\log c}

Similarly,

1+y = \frac{\log(abc)}{\log a}

1+z = \frac{\log(abc)}{\log b}

Then

(1+x)^{-1} + (1+y)^{-1} + (1+z)^{-1} = \frac{\log c}{\log(abc)} + \frac{\log a}{\log(abc)} + \frac{\log b}{\log(abc)}

Combine the fractions:

\frac{\log a + \log b + \log c}{\log(abc)} = \frac{\log(abc)}{\log(abc)} = 1

Thus option C is correct. Options A, B, and D do not hold (simple verification shows they are false).

Answer:

Option C:

(1+x)^{-1} + (1+y)^{-1} + (1+z)^{-1} = 1

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