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A tangent drawn to the curve f(x)=x²+bx-b at the point (1,1) cuts the coordinate axes so that the triangle enclosed by the tangent and the axes lies entirely in the first quadrant. If the area of this triangle is 2, what is the value of b?
- −1
- 3
- −3
- 1
Correct answer: −3
Solution
The point (1,1) lies on y=x^2+bx-b for any b, and slope is 2+b. The tangent meets the axes at x=1-1/(2+b) and y=-1-b; area=1/2|xy|=2 with both intercepts positive requires b=-3, giving intercepts (2,0) and (0,2) and area 2.
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