Let f(x)=∫1x 2−t2dt, Then the real roots of the equation x2−f′(x)=0 are

Let f(x)=1x2t2dt, Then the real roots of the equation x2f(x)=0 are

  1. A

    ±1

  2. B

    ±12

  3. C

    ±12

  4. D

    0 and 1

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    Solution:

    We have, 

    f(x)=1x2t2dtf(x)=2x2x2f(x)=0x22x2=0x4=2x2x4+x22=0x2+2x21=0x=±1        x2+20

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