The triangle formed by the tangent to the curve f(x)=x2+bx−b at the point (1,1) and the coordinates axes, lies in the first quadrant. If its area is 2 then the value of b is
A
-1
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B
3
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C
-3
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D
1
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Solution
The correct option is B -3 f′(x)=2x+b ⇒m=2+b So equation of tangent at (1,1) is (y−1)=(b+2)(x−1) intercept on the axes are (b+1)/(b+2) and −(b+1) ⇒∣−12(b+1)(b+2)(b+1)∣=2 '⇒∣(b+1)2b+2∣=4 ⇒b=−3 is only solution