NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

If $x = \log_{a} bc$ , $y = \log_{b} ca$ and $z = \log_{c} ab$ , then $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$ is equal to:

Choose the correct answer:

Explanation

1. Simplify the denominator terms:

We know that $1 = \log_{a} a$ .

1 + x = \log_{a} a + \log_{a} bc

Using the property $\log_m n + \log_m p = \log_m (np)$ :

1 + x = \log_{a} (abc)

Similarly:

1 + y = \log_{b} b + \log_{b} ca = \log_{b} (abc)

1 + z = \log_{c} c + \log_{c} ab = \log_{c} (abc)

2. Simplify the reciprocals:

Using the base-change property $\frac{1}{\log_{m} n} = \log_{n} m$ :

\frac{1}{1+x} = \frac{1}{\log_{a} abc} = \log_{abc} a

\frac{1}{1+y} = \frac{1}{\log_{b} abc} = \log_{abc} b

\frac{1}{1+z} = \frac{1}{\log_{c} abc} = \log_{abc} c

3. Sum the terms:

\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \log_{abc} a + \log_{abc} b + \log_{abc} c

\text{Sum} = \log_{abc} (a \cdot b \cdot c)

\text{Sum} = \log_{abc} (abc)

\text{Sum} = 1

Conclusion:

The expression simplifies to $1$ .

Correct Option: (c)

Choose the correct answer:

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