The triangle formed by the tangent to the curve f(x)=x2+bx−b at the point (1, 1) and the co-ordinate axes, lies in the first quadrant. If its area is 2, then the value of 'b' is
A
-3
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B
-2
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C
-1
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D
0
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Solution
The correct option is A -3
Given,
f(x)=x2+bx−b
f′(x)=2x+b
f′(x)(1,1)=2(1)+b=2+b
Equation of tangent at (1,1)
y−1=(2+b)(x−1)
⇒x1+b2+b+y−(1+b)=1
Therefore area of triangle =∣∣∣12×(1+b2+b)×−(1+b)∣∣∣=2sq.units