JEE Main & Advanced Mathematics Definite Integration Question Bank Self Evaluation Test - Application of Integrals

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

Correct Answer: C

Solution :

[c] \[f(x)={{x}^{2}}+bx-b;f'(x)=2x+b\]

\[\Rightarrow f'(1)=b+2\]

Equation of tangent: \[y-1=(b+2)(x-1)\]

Putting \[x=0\Rightarrow y=1-b-2=-b-1>0\]

\[\Rightarrow b<-1\]

Putting \[y=0\Rightarrow x-1=-\frac{1}{b+2}\Rightarrow x=\frac{-1}{b+2}+1\]

\[=\frac{b+1}{b+2}>0\Rightarrow b<-2\] or \[b>-1\]

Combining, the two conditions \[=b<-2\]

Now, \[\frac{1}{2}\left| -b-1 \right|\left| \frac{b+1}{b+2} \right|=2;{{(b+1)}^{2}}=4\left| b+2 \right|\]

\[=-4b-8\]

\[\Rightarrow {{(b+3)}^{2}}=0\Rightarrow b=-3\] follows the condition \[b<-2\]