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Manipal MET2014MathematicsApplication of Derivatives

The triangle formed by the tangent to the curve f(x)=x^2+b x-b at the point (1,1) and the coordinate axes lies in the first quadrant. If its area is 2 , then the value of b is

Options

  1. A-1
  2. B3
  3. C-3
  4. D1

Correct answer

C. -3

Step-by-step solution

Given curve is y=f(x)=x^2+b x-b On differentiating w.r.t.x, we get d y d x =2 x+b The equation of the tagent at (1,1) is array rlrl & & y-1= ( d y d x )_ (1,1) (x-1) & y-1=(b+2)(x-1) & (2+b) x-y=1+b & x (1+b) /(2+b) - y (1+b) =1 array So, O A= 1+b 2+b and O B=-(1+b) Now, area of A O B array rrrl & = 1 2 (1+b) (2+b) [-(1+b)]=2 (given) & 4(2+b)+(1+b)^2=0 & 8+4 b+1+b^2+2 b=0 & b^2+6 b+9=0 &(b+3)^2=0 b=-3 array

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