Area under the curveHard

Question

The triangle formed by the tangent to the curve f(x) = x2 + bx -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

A.- 1
B.3
C.-3
D.1

Solution

  
Let y = f (x) = x2 + bx -b
The equation of the tangent at P (1, 1)
to the curve y = x2 + bx - b is
y + 1 = 2x.1 + b(x + 1) - 2b ⇒ y = (2 + b)x - (1+ b)
Its meet the coordinate axes at
      and xB = -(1 + b)
∴     Area of ᐃ OAB = OA × AB = -     (given)
⇒       (1 + b)2 + 4(2 + b) = 0
⇒       b2 + 6b + 9 = 0
⇒       (b + 3)2 = 0 ⇒ b = - 3 and x = b is (b + 1) sin (3b + 4). Then, f(x) is

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