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Principle Of Mathematical Induction

Question
CBSEENMA11015450
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Let straight p space equals space limit as straight x rightwards arrow 0 to the power of plus of left parenthesis 1 plus tan squared square root of straight x right parenthesis to the power of 1 divided by 2 straight x end exponent comma space then log p is equal to 

  • 2

  • 1

  • 1/2

  • 1/4

Solution

C.

1/2

Given,
straight p space equals limit as straight x rightwards arrow 0 to the power of plus of space left parenthesis 1 plus tan squared square root of straight x right parenthesis to the power of fraction numerator 1 over denominator 2 straight x end fraction end exponent space left square bracket 1 to the power of infinity space form right square bracket equals space straight e to the power of straight x rightwards arrow stack 0 to the power of plus with lim on top space fraction numerator tan squared square root of straight x over denominator 2 straight x end fraction end exponent space equals space straight e to the power of 1 half space straight x rightwards arrow stack 0 to the power of plus with lim on top space fraction numerator tan squared square root of straight x over denominator 2 straight x end fraction space space end exponent space equals space straight e to the power of 1 divided by 2 end exponent log space straight p equals space log space straight e to the power of 1 divided by 2 end exponent space equals space 1 divided by 2

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