CBSE Class 12-science Answered

Please help to find k given in the problem

Asked by rajuharan | 03 Jul, 2016, 10:14: AM

Expert Answer

$limit as straight x rightwards arrow 0 to the power of plus of straight f left parenthesis straight x right parenthesis space equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator square root of 1 plus kx end root space minus space square root of 1 minus kx end root over denominator straight x end fraction equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator open square brackets square root of 1 plus kx end root space minus space square root of 1 minus kx end root close square brackets open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets over denominator straight x open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator open square brackets 1 plus kx minus 1 plus kx close square brackets over denominator straight x open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator 2 kx over denominator straight x open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator 2 straight k over denominator open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction equals fraction numerator 2 straight k over denominator open square brackets square root of 1 plus 0 end root space plus space square root of 1 minus 0 end root close square brackets end fraction equals straight k space limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis equals space limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis fraction numerator 2 straight x plus 1 over denominator straight x minus 2 end fraction equals fraction numerator 0 plus 1 over denominator 0 minus 2 end fraction equals negative 1 half Given space that space straight f left parenthesis straight x right parenthesis space is space continuous space at space space straight x space equals space 0 therefore limit as straight x rightwards arrow 0 to the power of plus of straight f left parenthesis straight x right parenthesis space space equals space limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis space rightwards double arrow straight k space equals space minus 1 half$

Answered by Vijaykumar Wani | 04 Jul, 2016, 10:50: AM

CBSE Class 12-science Answered

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CBSE Class 12-science Answered

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