Question

A triangle is formed by the tangents to the curve f(x) = x² + bx – b at the point (1, 1) and the coordinate axes, lies in the first Quadrant. If the area is 2, then value of b is :

(a) – 1

(b) 3

(c) – 3

(d) 1

Answer 1

The correct option (c) – 3

Explanation:

f(x) = x² + bx – b

f'(x) = 2x + b

f'(1) = 2 + b

equation of tangent at (1, 1) will be y – 1 = (2 + b)(x – 1)

put y = 0 ⇒ – 1 = (2 + b)(x – 1)

∴ (x – 1) = [(– 1)/(2 + b)]

∴ x = [(1 + b)/(2 + b)]

∴ Tangent cut x-axis at [{(1 + b)/(2 + b)}, 0]

putting x = 0,  ⇒ y – 1 = (2 + b)(0 – 1)

∴ y = – 1 – b

Triangle is in 1^st quadrant ⇒ – 1 – b ≥ 0

∴ Tangent cut y axis at b(0, – 1 – b)

area = (1/2) × [(1 + b)/(2 + b)](– 1 – b) = 2

∴ (1 + b)(– 1 – b) = 4(2 + b)

∴ b² + 6b + 9 = 0 ⇒ (b + 3)² = 0 ⇒ b = – 3.