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Manipal MET2015MathematicsApplication of Derivatives

The equation of the tangent to the curve y=(2 x-1) e^ 2(1-x) at the point of its maximum, is

Options

  1. Ay-1=0
  2. Bx-1=0
  3. Cx+y-1=0
  4. Dx-y+1=0

Correct answer

A. y-1=0

Step-by-step solution

We have, y=(2 x-1) e^ 2(1-x) On differentiating both sides w.r.t. x , we get aligned d y d x & =2 e^ 2(1-x) -2(2 x-1) e^ 2(1-x) d y d x & =2 e^ 2(1-x) (2-2 x)=4 e^ 2(1-x) (1-x) aligned At points of maximum, we must have d y d x =0 x=1 Now, d^2 y d x^2 =-8 e^ 2(1-x) (1-x)-4 e^ 2(1-x) [ d^2 y d x^2 ]_ x=1 =-4 0 So, y is maximum at x=1 . Clearly, y=1 for x=1 . Thus, the point of maximum is (1,1) . The equation is the tangent at (1,1) is y-1=0(x-1) y=1

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