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CAT Exam 2021 Slot 1 Solved Paper

Question : 60 of 66

Marks: +1, -0

5 - \log_{10} \sqrt{1+x} + 4\log_{10} \sqrt{1-x} = \log_{10} \frac{1}{\sqrt{1-x^2}}

, then 100x equals

Solution:

5 - \log_{10} \sqrt{1+x} + 4\log_{10} \sqrt{1-x} = \log_{10} \frac{1}{\sqrt{1-x^2}}

we can re-write the equition as :

5 - \log_{10} \sqrt{1+x} \ 4 \log_{10} \sqrt{1-x} = \log_{10} \left( \sqrt{1+x} \times \sqrt{1-x} \right)^{-1}

5 - \log_{10} \sqrt{1+x} + 4\log_{10} \sqrt{1-x} = (-1) \log_{10}(\sqrt{1+x}) + (-1) \log_{10}(\sqrt{1-x})

5 = -\log_{10} \sqrt{1+x} + \log_{10} \sqrt{1+x} - \log_{10} \sqrt{1-x} - 4\log_{10} \sqrt{1-x}

5 = -5 \log_{10} \sqrt{1-x}

\sqrt{1-x} = \frac{1}{100}

squaring both sides :

\left( \sqrt{1-x} \right)^2 = \frac{1}{100}

∴

x = 1 - \frac{1}{100} = \frac{99}{100}

hence,

100x = 100 \times \frac{99}{100} = 99

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