Let k>0 and λ=limx→0 k1−4k2−x2x2k2−x2 be finite. Then the value of λk, is

Let k>0 and λ=limx0k14k2x2x2k2x2 be finite. Then the value of λk, is

  1. A

    λ=8, k=12

  2. B

    λ=8, k=14

  3. C

    λ=4, k=12

  4. D

    6λ=4, k=14

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    Solution:

    It is given that

           λ=limx0k14k2x2x2k2x2 exists

     limx0k14k2x2=014k=0k=14 λ=limx01116x2x2116x2

     λ=limx01116x2x21+116x2116x2 λ=16limx011+116x2116x2=8

    Hence, λ=8 and k=14

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