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Let f(x)= cases x+1 & , 0 x 1 2 x^2-6 x+6 & , 1 x 2 cases And g(t)= _ t-1 ^t f(x) d x, t [1,2] . Which of the following is true statement

Options

  1. Af(x) is continuous and differentiable x [0,2]
  2. Bg ( t ) is maximum at t = 3 2
  3. Cg ( t ) is minimum at t =2
  4. Dall the above

Correct answer

C. g ( t ) is minimum at t =2

Step-by-step solution

f(x) is non differential at x=1 (Point A in figure) But f(x) is cont. x [0,2] g^ (t)=f(t)-f(t-1), t [1,2]= (2 t^2-6 t+6 )-((t-1)+1)=2 t^2-7 t+6=(2 t-3)(t-2) Maximum when t= 3 2 and minimum when t=2 .

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