The triangle formed by the tangent to the curve f (x) = x2 + bx - b at the point (1, 1) and the coordinate axes lie in the first quadrant. If its area is 2, then the value of b is

1) -1

2) 3

3) -3

4) 1

Solution: (3) -3

y = f (x) = x2 + bx – b

m = dy / dx = 2x + b = (2 + b)

The equation of the tangent at (1, 1)

y – y1 = m (x – x1)

y – 1 = m (x – x1)

y – 1 = (b + 2) (x – 1)

y – 1 = (b + 2) x – (b + 2)

y – (b + 2) x = – (b + 2) + 1

y – (b + 2) x = – b – 2 + 1

y – (b + 2) x + (b + 1) = 0

Area of triangle = (1 / 2) bh

2 = (1 / 2) *(b + 1) / (b + 2) – [b + 1]

4 = – [b + 1]2 / [b + 2]

4b + 8 = – b2 – 2b – 1

b2 + 6b + 9 = 0

(b + 3)2 = 0

b = – 3

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