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VITEEE 2014 Paper

Section: Mathematics

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Question : 102 of 120

Marks: +1, -0

The triangle formed by the tangent to the curve f (x) =

x^2

+ 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

- 1
3
- 3
1

Solution:

Given curve is y = f (x) =

x^2

+ bx - b

⇒ f ' (x) = 2x + b

The equation of tangent at point (1, 1) is

y - 1 =

\left(\frac{dy}{dx}\right)_{(1,1)}

(x - 1)

⇒ y - 1 = (b + 2) (x - 1)

⇒ (2 + b) x - y = 1 + b

⇒

\frac{x}{\frac{1+b}{2+b}}

-

\frac{y}{1+b}

= 1

So, OA =

\frac{1+b}{2+b}

and OB = -(1 +b)

Now, area of Δ AOB =

\frac{1}{2}

×

\frac{(1+b)[-(1+b)]}{2+b}

= 2

⇒ 4 (2 + b) +

(1+b)^2

= 0

⇒

b^2

+ 6b + 9 = 0

⇒

(b+3)^2

⇒ b = - 3

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