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Question

If f(x)=x1dt2+t4, then
  1. f(2)<13
  2. f(2)>13
  3. f(2)<1
  4. f(2)=13


Solution

The correct option is A f(2)<13
f(x)=12+x4
By LMVT f(C)=f(2)f(1)21 for some cϵ(1,2)
f(2)=12+c4 as f(1)=01<c<23<2+c4<18f(2)<13

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