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CBSE Class 12-science Answered

the 8th one
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Asked by rajeshbangia69 | 28 May, 2020, 08:13: PM
answered-by-expert Expert Answer
Question: If the function f(x) satisfies the relation f(x+y)=y|x-1|/(x-1) f(x) + f(y) with f(1)=2, then lim x tends to 1 f'(x) is
Solution:
Given colon space straight f open parentheses straight x plus straight y close parentheses equals straight y fraction numerator open vertical bar straight x minus 1 close vertical bar over denominator open parentheses straight x minus 1 close parentheses end fraction straight f open parentheses straight x close parentheses plus straight f open parentheses straight y close parentheses rightwards double arrow fraction numerator straight f open parentheses straight x plus straight y close parentheses minus straight f open parentheses straight y close parentheses over denominator straight y end fraction equals fraction numerator open vertical bar straight x minus 1 close vertical bar over denominator open parentheses straight x minus 1 close parentheses end fraction straight f open parentheses straight x close parentheses rightwards double arrow limit as straight y rightwards arrow 0 of fraction numerator straight f open parentheses straight x plus straight y close parentheses minus straight f open parentheses straight y close parentheses over denominator straight y end fraction equals fraction numerator open vertical bar straight x minus 1 close vertical bar over denominator open parentheses straight x minus 1 close parentheses end fraction straight f open parentheses straight x close parentheses rightwards double arrow straight f apostrophe open parentheses straight x close parentheses equals fraction numerator open vertical bar straight x minus 1 close vertical bar over denominator open parentheses straight x minus 1 close parentheses end fraction straight f open parentheses straight x close parentheses limit as straight x rightwards arrow 1 to the power of plus of straight f apostrophe open parentheses straight x close parentheses equals limit as straight x rightwards arrow 1 of fraction numerator straight x minus 1 over denominator open parentheses straight x minus 1 close parentheses end fraction straight f open parentheses straight x close parentheses equals straight f open parentheses 1 close parentheses equals 2 limit as straight x rightwards arrow 1 to the power of minus of straight f apostrophe open parentheses straight x close parentheses equals limit as straight x rightwards arrow 1 of fraction numerator negative open parentheses straight x minus 1 close parentheses over denominator open parentheses straight x minus 1 close parentheses end fraction straight f open parentheses straight x close parentheses equals negative straight f open parentheses 1 close parentheses equals negative 2 Since comma space RHL not equal to LHL limit as straight x rightwards arrow 1 of straight f apostrophe open parentheses straight x close parentheses space does space not space exist space Hence comma space the space answer space is space none space of space these.
Answered by Renu Varma | 29 May, 2020, 10:26: AM

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