Question

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 (a) `-1` (b) `3` (c) `-3` (d) `1`
A. -1
B. 3
C. -3
D. 1

Answer 1

Correct Answer - C
Let ` y=f(x)=x^(2)+bx-b`
The equation of the tangent at P(1,1)
to the curve ` 2y=2x^(2)+2bx-2b `is
` y+1=2x*1+b(x+1)-2b `
` rArr y=(2+b)x-(1+b) `
Its meet the coordinate axes at
`x_(A)=(1+b)/(2+b) " and " y_(B)=-(1+b) `
` therefore " Area of " triangle OAB=(1)/(2)OA xx OB`
` =-(1)/(2)xx((1+b)^(2))/((2+b))=2 " "["given "]`
` rArr (1+b)^(2)+4(2+b)=0 rArr b^(2)+6b+9=0 `
` rArr (b+3)^(2)=0 rArr b=-3`