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MHT CET20228 Aug 2022Evening ShiftMathematicsApplication of DerivativesActual

The tangent to the curve y=x^3+a x-b at the point (1,-5) is perpendicular to the line y-x+4=0 , then which one of the following points lies on the curve?

Options

  1. A(2,-2)
  2. B(-2,2)
  3. C(-2,1)
  4. D(2,-1)

Correct answer

A. (2,-2)

Step-by-step solution

y=x^3+a x-b slope of tangent = d y ~d x =3 x^2+a slope of the line y-x+4=0 is 1 aligned & d y ~d x _ a t(1-5) =-1 & 3 1^2+a=-1 & a=-4 aligned also (1,-5) lies on the curve y=x^3+a x-b aligned & -5=1^3+a 1-b=1+(-4) 1-b & b=2 aligned Hence, the curve is y=x^3-4 x-2 which is satisfied by (2,-2)

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