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Question

The triangle formed by the tangent to the curve f(x)=x2+bxb 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+bxb

f(x)=2x+b

f(x)(1,1)=2(1)+b=2+b

Equation of tangent at (1,1)

y1=(2+b)(x1)

x1+b2+b+y(1+b)=1

Therefore area of triangle =12×(1+b2+b)×(1+b)=2sq.units

(1+b)22+b=4

1+b2+2b=84b

(b+3)2=0

b=3

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