Let f : R → R be a continuous and differentiable function such that f 2 = 6 and f ' 2 = 1 48 . If…

Let f:R→R be a continuous and differentiable function such that f2=6 and f'2=148. If ∫6f(x)4t3dt=x-2gx, then limx→2gx is equal to
  1. 24
  2. 18
  3. 12
  4. 36

Solution

limx→2gx=limx→2∫6fx4t3dtx-2      00 form.

By using L'Hospital rule
=limx→24.f3(x)⋅f'x1
=4f32f'2=4×6×6×6×148=18.

Asked in: JEE Main 2019 (12 Apr Shift 1)

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